Matsumura Commutative Algebra

Daniel Murfet

October 5 2006

These notes closely follow Matsumurarsquos book [Mat80] on commutative algebra Proofs arethe ones given there sometimes with slightly more detail Our focus is on the results needed inalgebraic geometry so some topics in the book do not occur here or are not treated in their fulldepth In particular material the reader can find in the more elementary [AM69] is often omittedReferences on dimension theory are usually to Robert Ashrsquos webnotes since the author prefers thisapproach to that of [AM69]

Contents

1 General Rings 1

2 Flatness 621 Faithful Flatness 1122 Going-up and Going-down 1323 Constructible Sets 13

3 Associated Primes 15

4 Dimension 1541 Homomorphism and Dimension 17

5 Depth 1851 Cohen-Macaulay Rings 27

6 Normal and Regular Rings 3261 Classical Theory 3262 Homological Theory 3963 Koszul Complexes 4464 Unique Factorisation 52

1 General Rings

Throughout these notes all rings are commutative and unless otherwise specified all modules areleft modules A local ring A is a commutative ring with a single maximal ideal (we do not requireA to be noetherian)

Lemma 1 (Nakayama) Let A be a ring M a finitely generated A-module and I an ideal of ASuppose that IM = M Then there exists an element a isin A of the form a = 1+x x isin I such thataM = 0 If moreover I is contained in the Jacobson radical then M = 0

Corollary 2 Let A be a ring M an A-module N and N prime submodules of M and I an ideal ofA Suppose that M = N + IN prime and that either (a) I is nilpotent or (b) I is contained in theJacobson radical and N prime is finitely generated Then M = N

1

Proof In case (a) we have MN = I(MN) = I2(MN) = middot middot middot = 0 In (b) apply NakayamarsquosLemma to MN

In particular let (Am k) be a local ring and M an A-module Suppose that either m isnilpotent or M is finitely generated Then a subset G of M generates M iff its image inMmM = M otimesA k generates M otimesA k as a k-vector space In fact if N is submodule generated byG and if the image of G generates M otimesA k then M = N +mM whence M = N by the CorollarySince M otimesA k is a finitely generated vector space over the field k it has a finite basis and if wetake an arbitrary preimage of each element this collection generates M A set of elements whichbecomes a basis in MmM (and therefore generates M) is called a minimal basis If M is a finitelygenerated free A-module then it is clear that

rankAM = rankk(MmM)

In fact of rankAM = n ge 1 and x1 xn is a basis of M then x1 + mM xn + mM isa basis of the k-module MmM Or equivalently the xi otimes 1 are a basis of the k-module M otimes k

Let A be a ring and α Z minusrarr A The kernel is (n) for some integer n ge 0 which we call thecharacteristic of A The characteristic of a field is either 0 or a prime number and if A is localthe characteristic ch(A) is either 0 or a power of a prime number (m is a primary ideal and thecontraction of primary ideals are primary and 0 and (pn) are the only primary ideals in Z)

Lemma 3 Let A be an integral domain with quotient field K all localisations of A can be viewedas subrings of K and in this sense A =

⋂mAm where the intersection is over all maximal ideals

Proof Given x isin K we put D = a isin A | ax isin A we call D the ideal of denominators of x Theelement x is in A iff D = A and x isin Ap iff D p Therefore if x isin A there exists a maximalideal m such that D sube p and x isin Am for this m

Lemma 4 Let A be a ring and S sube T multiplicatively closed subsets Then

(a) There is a canonical isomorphism of Sminus1A-algebras Tminus1A sim= Tminus1(Sminus1A) defined by at 7rarr(a1)(t1)

(b) If M is an A-module then there is a canonical isomorphism of Sminus1A-modules Tminus1M sim=Tminus1(Sminus1M) defined by mt 7rarr (m1)(t1)

Proof (a) Just using the universal property of localisation we can see Tminus1A sim= Tminus1(Sminus1A) asSminus1A-algebras via the map at 7rarr (a1)(t1) (b) is also easily checked

Lemma 5 Let A be an integral domain with quotient field K and B a subring of K containingA If Q is the quotient field of B then there is a canonical isomorphism of B-algebras K sim= Q

If φ A minusrarr B is a ring isomorphism and S sube A is multiplicatively closed (denote also by S theimage in B) then there is an isomorphism of rings Sminus1A sim= Sminus1B making the following diagramcommute

Sminus1A +3 Sminus1B

Aφ

+3

OO

B

OO

Lemma 6 Let A be a ring S sube A a multiplicatively closed subset and p a prime ideal withp cap S = empty Let B = Sminus1A Then there is a canonical ring isomorphism BpB

sim= Ap

Proof A minusrarr B minusrarr BpB sends elements of A not in p to units so we have an induced ringmorphism Ap minusrarr BpB defined by as 7rarr (a1)(s1) and it is easy to check this is an isomorphism

2

Let ψ A minusrarr B be a morphism of rings and I an ideal of A The extended ideal IB consistsof sums

sumψ(ai)bi with ai isin I bi isin B Consider the exact sequence of A-modules

0 minusrarr I minusrarr A minusrarr AI minusrarr 0

Tensoring with B gives an exact sequence of B-modules

I otimesA B minusrarr AotimesA B minusrarr (AI)otimesA B minusrarr 0

The image of I otimesA B in B sim= A otimesA B is simply IB So there is an isomorphism of B-modulesBIB sim= (AI)otimesA B defined by b+ IB 7rarr 1otimes b In fact this is an isomorphism of rings as wellOf course for any two A-algebras EF twisting gives a ring isomorphism E otimesA F sim= F otimesA E

Lemma 7 Let φ A minusrarr B be a morphism of rings S a multiplicatively closed subset of A andset T = φ(S) Then for any B-module M there is a canonical isomorphism of Sminus1A-modulesnatural in M

α Sminus1M minusrarr Tminus1M

α(ms) = mφ(s)

In particular there is a canonical isomorphism of Sminus1A-algebras Sminus1B sim= Tminus1B

Proof One checks easily that α is a well-defined isomorphism of Sminus1A-modules In the caseM = B the Sminus1A-module Sminus1B becomes a ring in the obvious way and α preserves this ringstructure

In particular let S be a multiplicatively closed subset of a ring A let I be an ideal of A andlet T denote the image of S in AI Then there is a canonical isomorphism of rings

Tminus1(AI) sim= AI otimesA Sminus1A sim= Sminus1AI(Sminus1A)

(a+ I)(s+ I) 7rarr as+ I(Sminus1A)

Definition 1 A ring A is catenary if for each pair of prime ideals q sub p the height of the primeideal pq in Aq is finite and is equal to the length of any maximal chain of prime ideals between pand q Clearly the catenary property is stable under isomorphism and any quotient of a catenaryring is catenary If S sube A is a multiplicatively closed subset and A is catenary then so is Sminus1A

Lemma 8 Let A be a ring Then the following are equivalent

(i) A is catenary

(ii) Ap is catenary for every prime ideal p

(iii) Am is catenary for every maximal ideal m

Proof The implications (i) rArr (ii) rArr (iii) are obvious (iii) rArr (i) If q sub p are primes find amaximal ideal m containing p and pass to the catenary ring Am to see that the required propertyis satisfied for q p

Lemma 9 Let A be a noetherian ring Then A is catenary if for every pair of prime ideals q sub pwe have ht(pq) = htpminus htq

Proof Since A is noetherian all involved heights are finite Suppose A satisfies the condition andlet q sub p be prime ideals Obviously ht(pq) is finite and there is at least one maximal chainbetween p and q with length ht(pq) Let

q = q0 sub q1 sub middot middot middot sub qn = p

be a maximal chain of length n Then by assumption 1 = ht(qiqiminus1) = htqi minus htqiminus1 for1 le i le n Hence htp = htq + n so n = ht(pq) as required

3

Definition 2 A ring A is universally catenary if A is noetherian and every finitely generatedA-algebra is catenary Equivalently a noetherian ring A is universally catenary if A is catenaryand A[x1 xn] is catenary for n ge 1

Lemma 10 Let A be a ring and S sube A a multiplicatively closed subset Then there is a canonicalring isomorphism Sminus1(A[x]) sim= (Sminus1A)[x] In particular if q is a prime ideal of A[x] and p = qcapAthen A[x]q sim= Ap[x]qAp[x]

Proof The ring morphism A minusrarr Sminus1A induced A[x] minusrarr (Sminus1A)[x] which sends elements ofS sube A[x] to units So there is an induced ring morphism ϕ Sminus1(A[x]) minusrarr (Sminus1A)[x] defined by

ϕ

(a0 + a1x+ middot middot middot+ anx

n

s

)=a0

s+a1

sx+ middot middot middot+ an

sxn

This is easily checked to be an isomorphism In the second claim there is an isomorphismAp[x] sim= A[x]p where the second ring denotes (Aminusp)minus1(A[x]) and qAp[x] denotes the prime idealof Ap[x] corresponding to qA[x]p Using the isomorphism ϕ it is clear that qAp[x] cap Ap = pApUsing Lemma 6 there is clearly an isomorphism of rings A[x]q sim= Ap[x]qAp[x]

If R is a ring and M an R-module then let Z(M) denote the set of zero-divisors in M Thatis all elements r isin R with rm = 0 for some nonzero m isinM

Lemma 11 Let R be a nonzero reduced noetherian ring Then Z(R) =⋃i pi with the union

being taken over all minimal prime ideals pi

Proof Since R is reduced⋂i pi = 0 If ab = 0 with b 6= 0 then b isin pj for some j and therefore

a isin pj sube⋃i pi The reverse inclusion follows from the fact that no minimal prime can contain a

regular element (since otherwise by Krullrsquos PID Theorem it would have height ge 1)

Lemma 12 Let R be a nonzero reduced noetherian ring Assume that every element of R iseither a unit or a zero-divisor Then dim(R) = 0

Proof Let p1 pn be the minimal primes of R Then by Lemma 11 Z(R) = p1 cup middot middot middot cup pnLet p be a prime ideal Since p is proper p sube Z(R) and therefore p sube pi for some i Since pi isminimal p = pi so the pi are the only primes in R Since these all have height zero it is clearthat dim(R) = 0

Lemma 13 Let R be a reduced ring p a minimal prime ideal of R Then Rp is a field

Proof If p = 0 this is trivial so assume p 6= 0 Since pRp is the only prime ideal in Rp it is alsothe nilradical So if x isin p then txn = 0 for some t isin p and n gt 0 But this implies that tx isnilpotent and therefore zero since R is reduced Therefore pRp = 0 and Rp is a field

Let A1 An be rings Let A be the product ring A =prodni=1Ai Ideals of A are in bijection

with sequences I1 In with Ii an ideal of Ai This sequence corresponds to

I1 times middot middot middot times In

This bijection identifies the prime ideals of A with sequences I1 In in which every Ii = Aiexcept for a single Ij which is a prime ideal of Aj So the primes look like

A1 times middot middot middot times pi times middot middot middot timesAn

for some i and some prime ideal pi of Ai Given i and a prime ideal pi of Ai let p be the primeideal A1 times middot middot middot times pi times middot middot middotAn Then the projection of rings A minusrarr Ai gives rise to a ring morphism

Ap minusrarr (Ai)pi

(a1 ai an)(b1 bi bn) 7rarr aibi

4

It is easy to check that this is an isomorphism An orthogonal set of idempotents in a ring A is aset e1 er with 1 = e1 + middot middot middot+ er e2i = ei and eiej = 0 for i 6= j If A =

prodni=1Ai is a product

of rings then the elements e1 = (1 0 0) en = (0 0 1) are clearly such a setConversely if e1 er is an orthogonal set of idempotents in a ring A then the ideal eiA

becomes a ring with identity ei The map

A minusrarr e1Atimes middot middot middot times erA

a 7rarr (e1a era)

is a ring isomorphism

Proposition 14 Any nonzero artinian ring A is a finite direct product of local artinian rings

Proof See [Eis95] Corollary 216 This shows that there is a finite list of maximal ideals m1 mn

(allowing repeats) and a ring isomorphism A minusrarrprodni=1Ami

defined by a 7rarr (a1 a1)

Proposition 15 Let ϕ A minusrarr B be a surjective morphism of rings M an A-module andp isin SpecB There is a canonical morphism of Bp-modules natural in M

κ HomA(BM)p minusrarr HomAϕminus1p(BpMϕminus1p)

κ(us)(bt) = u(b)ϕminus1(st)

If A is noetherian this is an isomorphism

Proof Let a morphism of A-modules u B minusrarrM s t isin B p and b isin B be given Choose k isin Awith ϕ(k) = st We claim the fraction u(b)k isin Mϕminus1p doesnrsquot depend on the choice of k If wehave ϕ(l) = st also then

ku(b) = u(kb) = u(ϕ(k)b) = u(ϕ(l)b) = u(lb) = lu(b)

so u(b)l = u(b)k as claimed Throughout the proof given x isin B we write ϕminus1(x) for anarbitrary element in the inverse image of x One checks the result does not depend on this choiceWe can now define a morphism of Bp-modules κ(us)(bt) = u(b)ϕminus1(st) which one checks iswell-defined and natural in M

Now assume that A is noetherian In showing that κ is an isomorphism we may as well assumeϕ is the canonical projection A minusrarr Aa for some ideal a In that case the prime ideal p is qa forsome prime q of A containing a and if we set S = A q and T = ϕ(S) we have by Lemma 7 anisomorphism

HomA(BM)p = Tminus1HomA(AaM)sim= Sminus1HomA(AaM)sim= HomSminus1A(Sminus1(Aa) Sminus1M)sim= HomSminus1A(Tminus1(Aa) Sminus1M)= HomAϕminus1p

(BpMϕminus1p)

We have use the fact that A is noetherian to see that Aa is finitely presented so we have thesecond isomorphism in the above sequence One checks easily that this isomorphism agrees withκ completing the proof

Remark 1 The right adjoint HomA(Bminus) to the restriction of scalars functor exists for anymorphism of rings ϕ A minusrarr B but as we have just seen this functor is not local unless thering morphism is surjective This explains why the right adjoint f to the direct image functor inalgebraic geometry essentially only exists for closed immersions

5

2 Flatness

Definition 3 Let A be a ring and M an A-module We say M is flat if the functor minus otimesA M AMod minusrarr AMod is exact (equivalently M otimesA minus is exact) Equivalently M is flat if wheneverwe have an injective morphism of modules N minusrarr N prime the morphism N otimesA M minusrarr N prime otimesA M isinjective This property is stable under isomorphism

We say M is faithfully flat if a morphism N minusrarr N prime is injective if and only if N otimesA M minusrarrN prime otimesAM is injective This property is also stable under isomorphism An A-algebra A minusrarr B isflat if B is a flat A-module and we say A minusrarr B is a flat morphism

Example 1 Nonzero free modules are faithfully flat

Lemma 16 We have the following fundamental properties of flatness

bull Transitivity If φ A minusrarr B is a flat morphism of rings and N a flat B-module then N isalso flat over A

bull Change of Base If φ A minusrarr B is a morphism of rings and M is a flat A-module thenM otimesA B is a flat B-module

bull Localisation If A is a ring and S a multiplicatively closed subset then Sminus1A is flat over A

Proof The second and third claims are done in our Atiyah amp Macdonald notes To prove the firstclaim let M minusrarr M prime be a monomorphism of A-modules and consider the following commutativediagram of abelian groups

M otimesA N

M prime otimesA N

M otimesA (B otimesB N)

M prime otimesA (B otimesB N)

(M otimesA B)otimesB N (M prime otimesA B)otimesB N

Since B is a flat A-module and N is a flat B-module the bottom row is injective hence so is thetop row

Lemma 17 Let φ A minusrarr B be a morphism of rings and N a B-module which is flat over A IfS is a multiplicatively closed subset of B then Sminus1N is flat over A In particular any localisationof a flat A-module is flat

Proof If M minusrarrM prime is a monomorphism of A-modules then we have a commutative diagram

M otimesA Sminus1N

M prime otimesA Sminus1N

(M otimesA N)otimesB Sminus1B (M prime otimesA N)otimesB Sminus1B

The bottom row is clearly injective and hence so is the top row which shows that Sminus1N is flatover A

Lemma 18 Let A be a ring and MN flat A-modules Then M otimesA N is also flat over A

Lemma 19 Let φ A minusrarr B be a flat morphism of rings and S a multiplicatively closed subsetof A Then T = φ(S) is a multiplicatively closed subset of B and Tminus1B is flat over Sminus1A

Proof This follows from Lemma 7 and stability of flatness under base change

6

Lemma 20 Let A minusrarr B be a morphism of rings Then the functor minusotimesAB AMod minusrarr BModpreserves projectives

Proof The functor minusotimesA B is left adjoint to the restriction of scalars functor This latter functoris clearly exact so since any functor with an exact right adjoint must preserve projectives P otimesABis a projective B-module for any projective A-module P

Lemma 21 Let A minusrarr B be a flat morphism of rings If I is an injective B-module then it isalso an injective A-module

Proof The restriction of scalars functor has an exact left adjoint minus otimesA B AMod minusrarr BModand therefore preserves injectives

Lemma 22 Let φ A minusrarr B be a flat morphism of rings and let MN be A-modules Then thereis an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (MotimesABNotimesAB) If A is noetherianand M finitely generated over A there is an isomorphism of B-modules ExtiA(MN) otimesA B sim=ExtiB(M otimesA BN otimesA B)

Proof Let X middot middot middot minusrarr X1 minusrarr X0 minusrarr M minusrarr 0 be a projective resolution of the A-module M Since B is flat the sequence

X otimesA B middot middot middot minusrarr X1 otimesA B minusrarr X0 otimesA B minusrarrM otimesA B minusrarr 0

is a projective resolution of M otimesA B The chain complex of B-modules (X otimesA B)otimesB (B otimesA N)is isomorphic to (X otimesA N)otimesA B The exact functor minusotimesA B commutes with taking homology sothere is an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (M otimesABN otimesAB) as required

If A is noetherian andM finitely generated we can assume that theXi are finite free A-modulesThen ExtiA(MN) is the i-cohomology module of the sequence

0 minusrarr Hom(X0 N) minusrarr Hom(X1 N) minusrarr Hom(X2 N) minusrarr middot middot middot

Since tensoring with B is exact ExtiA(MN) otimesA B is isomorphic as a B-module to the i-thcohomology of the following sequence

0 minusrarr Hom(X0 N)otimesA B minusrarr Hom(X1 N)otimesA B minusrarr middot middot middot

After a bit of work we see that this cochain complex is isomorphic to HomB(X otimesA BN otimesA B)and the i-th cohomology of this complex is ExtiB(M otimesA BN otimesA B) as required

In particular for a ring A and prime ideal p sube A we have isomorphisms of Ap-modules for i ge 0

TorAp

i (Mp Np) sim= TorAi (MN)p

ExtiAp(Mp Np) sim= ExtiA(MN)p

the latter being valid for A noetherian and M finitely generated

Lemma 23 Let A be a ring and M an A-module Then the following are equivalent

(i) M is a flat A-module

(ii) Mp is a flat Ap-module for each prime ideal p

(iii) Mm is a flat Am-module for each maximal ideal m

Proof See [AM69] or any book on commutative algebra

Proposition 24 Let (Am k) be a local ring and M an A-module Suppose that either m isnilpotent or M is finitely generated over A Then M is free hArr M is projective hArr M is flat

7

Proof It suffices to show that if M is flat then it is free We prove that any minimal basis of Mis a basis of M If MmM = 0 then M = 0 and M is trivially free Otherwise it suffices to showthat if x1 xn isinM are elements whose images in MmM = M otimesA k are linearly independentover k then they are linearly independent over A We use induction on n For n = 1 let ax = 0Then there exist y1 yr isin M and b1 br isin A such that abi = 0 for all i and x =

sumbiyi

Since x+ mM 6= 0 not all bi are in m Suppose b1 isin m Then b1 is a unit in A and ab1 = 0 hencea = 0

Suppose n gt 1 andsumni=1 aixi = 0 Then there exist y1 yr isin M and bij isin A(1 le j le r)

such that xi =sumj bijyj and

sumi aibij = 0 Since xn isin mM we have bnj isin m for at least one j

Since a1b1j + middot middot middot+ anbnj = 0 and bnj is a unit we have

an =nminus1sumi=1

ciai ci = minusbijbnj

Then

0 =nsumi=1

aixi = a1(x1 + c1xn) + middot middot middot+ anminus1(xnminus1 + cnminus1xn)

Since the residues of x1+c1xn xnminus1+cnminus1xn are linearly independent over k by the inductivehypothesis we get a1 = middot middot middot = anminus1 = 0 and an =

sumciai = 0

Corollary 25 Let A be a ring and M a finitely generated A-module Then the following areequivalent

(i) M is a flat A-module

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof This is immediate from the previous two results

Proposition 26 Let A be a ring and M a finitely presented A-module Then M is flat if andonly if it is projective

Proof See Stenstrom Chapter 1 Corollary 115

Corollary 27 Let A be a noetherian ring M a finitely generated A-module Then the followingconditions are equivalent

(i) M is projective

(ii) M is flat

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof Since A is noetherian M is finitely presented so (i) hArr (ii) is an immediate consequenceof Proposition 26 The rest of the proof follows from Corollary 25

Lemma 28 Let A minusrarr B be a flat morphism of rings and let I J be ideals of A Then (IcapJ)B =IB cap JB and (I J)B = (IB JB) if J is finitely generated

Proof Consider the exact sequence of A-modules

I cap J minusrarr A minusrarr AI oplusAJ

Tensoring with B we get an exact sequence

(I cap J)otimesA B = (I cap J)B minusrarr B minusrarr BIB oplusBJB

8

This means (I cap J)B = IB cap JB For the second claim suppose firstly that J is a principal idealaA and use the exact sequence

(I aA) i Af AI

where i is the injection and f(x) = ax+I Tensoring withB we get the formula (I a)B = (IB a)In the general case if J = (a1 an) we have (I J) =

⋂i(I ai) so that

(I J)B =⋂

(I ai)B =⋂

(IB ai) = (IB JB)

Example 2 Let A = k[x y] be a polynomial ring over a field k and put B = A(x) sim= k[y] ThenB is not flat over A since y isin A is regular but is not regular on B Let I = (x+ y) and J = (y)Then I cap J = (xy + y2) and IB = JB = yB (I cap J)B = y2B 6= IB cap JB

Example 3 Let k be a field put A = k[x y] and let K be the quotient field of A Let B be thesubring k[x yx] of K (ie the k-subalgebra generated by x and z = yx) Then A sub B sub K LetI = xA J = yA Then I cap J = xyA and (I cap J)B = x2zB IB cap JB = xzB so B is not flat overA The map SpecB minusrarr SpecA corresponding to A minusrarr B is the projection to the (x y)-planeof the surface F xz = y in (x y z)-space Note F contains the whole z-axis so it does not lookldquoflatrdquo over the (x y)-plane

Proposition 29 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB there is a canonical isomorphism of Bp-modules natural in both variables

κ MpcapA otimesApcapANp minusrarr (M otimesA N)p

κ(msotimes nt) = (motimes n)ϕ(s)t

Proof Fix p isin SpecB and q = pcapA There is a canonical ring morphism Aq minusrarr Bp and we makeNp into an Aq-module using this morphism One checks that the following map is well-definedand Aq-bilinear

ε Mq timesNp minusrarr (M otimesA N)p

ε(ms nt) = (motimes n)ϕ(s)t

We show that in fact this is a tensor product of Aq-modules Let Z be an abelian group andψ Mq timesNp minusrarr Z an Aq-bilinear map

Mq timesNp

ε

ψ Z

(M otimesA N)p

φ

(1)

We have to define a morphism of abelian groups φ unique making this diagram commute Fors isin p we define an A-bilinear morphism φprimes M times N minusrarr Z by φprimes(mn) = ψ(m1 ns) Thisinduces a morphism of abelian groups

φprimeprimes M otimesA N minusrarr Z

φprimeprimes (motimes b) = ψ(m1 bs)

We make some observations about these morphisms

bull Suppose ws = wprimesprime in (M otimesA N)p with say w =sumimi otimes ni w

prime =sumim

primei otimes nprimei and t isin p

such that tsprimew = tswprime That issumimi otimes tsprimeni =

sumim

primei otimes tsnprimei Applying φprimeprimetssprime to both sides

of this equality gives φprimeprimes (w) = φprimeprimesprime(wprime)

9

bull For wswprimesprime isin (M otimesA N)p we have φprimeprimes (w) + φprimeprimesprime(wprime) = φprimeprimessprime(s

primew + swprime)

It follows that φ(ws) = φprimeprimes (w) gives a well-defined morphism of abelian groups φ (MotimesAN)p minusrarrZ which is clearly unique making (1) commute By uniqueness of the tensor product there is aninduced isomorphism of abelian groups κ Mq otimesAq Np minusrarr (M otimesA N)p with κ(ms otimes nt) =(m otimes n)ϕ(s)t One checks that this is a morphism of Bp-modules The inverse is defined by(motimes n)t 7rarr m1otimes nt Naturality in both variables is easily checked

Corollary 30 Let ϕ A minusrarr B be a morphism of rings M an A-module and p isin SpecB Thenthere is a canonical isomorphism of Bp-modules MpcapA otimesApcapA

Bp minusrarr (M otimesA B)p natural in M

We will not actually use the next result in these notes so the reader not familiar with homo-logical δ-functors can safely skip it Alternatively one can provide a proof by following the onegiven in Matsumura (the proof we give is more elegant provided you know about δ-functors)

Proposition 31 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB and i ge 0 there is a canonical isomorphism of Bp-modules natural inM

κi TorAi (NM)p minusrarr TorApcapA

i (NpMpcapA)

Proof Fix p isin SpecB and a B-module N and set q = p capA Then N is a B-A-bimodule and Np

is a Bp-Aq-bimodule so by (TORSection 51) the abelian group TorAi (NM) acquires a canonicalB-module structure and Tor

Aq

i (NpMq) acquires a canonical Bp-module structure for any A-module M and i ge 0 Using (TORLemma 14) and (DFDefinition 23) we have two homologicalδ-functors between AMod and BpMod

TorAi (Nminus)pige0 TorAq

i (Np (minus)q)ige0

For i gt 0 these functors all vanish on free A-modules so by (DFTheorem 74) both δ-functors areuniversal For i = 0 we have the canonical natural equivalence of Proposition 29

κ0 TorA0 (Nminus)psim= (N otimesA minus)p

sim= Np otimesAq (minus)qsim= Tor

Aq

0 (Np (minus)q)

By universality this lifts to a canonical isomorphism of homological δ-functors κ In particularfor each i ge 0 we have a canonical natural equivalence κi TorAi (Nminus)p minusrarr Tor

Aq

i (Np (minus)q) asrequired

We know from Lemma 23 that flatness is a local property We are now ready to show thatrelative flatness (ie flatness with respect to a morphism of rings) is also local This is particularlyimportant in algebraic geometry The reader who skipped Proposition 31 will also have to skipthe implication (iii) rArr (i) in the next result but this will not affect their ability to read the restof these notes

Corollary 32 Let A minusrarr B be a morphism of rings and N a B-module Then the followingconditions are equivalent

(i) N is flat over A

(ii) Np is flat over ApcapA for all prime ideals p of B

(iii) Nm is flat over AmcapA for all maximal ideals m of B

Proof (i) rArr (ii) If N is flat over A then NpcapA is flat over ApcapA for any prime p of B By anargument similar to the one given in Lemma 19 we see that Np is isomorphic as a BpcapA-moduleto a localisation of NpcapA Applying Lemma 17 to the ring morphism ApcapA minusrarr BpcapA we see thatNp is flat over ApcapA as required (ii) rArr (iii) is trivial (iii) rArr (i) For every A-module M andmaximal ideal m of B we have by Proposition 31

TorA1 (NM)msim= TorAmcapA

1 (NmMmcapA) = 0

since by assumption Nm is flat over AmcapA Therefore TorA1 (NM) = 0 for every A-module M which implies that N is flat over A and completes the proof

10

Proof In case (a) we have MN = I(MN) = I2(MN) = middot middot middot = 0 In (b) apply NakayamarsquosLemma to MN

In particular let (Am k) be a local ring and M an A-module Suppose that either m isnilpotent or M is finitely generated Then a subset G of M generates M iff its image inMmM = M otimesA k generates M otimesA k as a k-vector space In fact if N is submodule generated byG and if the image of G generates M otimesA k then M = N +mM whence M = N by the CorollarySince M otimesA k is a finitely generated vector space over the field k it has a finite basis and if wetake an arbitrary preimage of each element this collection generates M A set of elements whichbecomes a basis in MmM (and therefore generates M) is called a minimal basis If M is a finitelygenerated free A-module then it is clear that

rankAM = rankk(MmM)

In fact of rankAM = n ge 1 and x1 xn is a basis of M then x1 + mM xn + mM isa basis of the k-module MmM Or equivalently the xi otimes 1 are a basis of the k-module M otimes k

Let A be a ring and α Z minusrarr A The kernel is (n) for some integer n ge 0 which we call thecharacteristic of A The characteristic of a field is either 0 or a prime number and if A is localthe characteristic ch(A) is either 0 or a power of a prime number (m is a primary ideal and thecontraction of primary ideals are primary and 0 and (pn) are the only primary ideals in Z)

Lemma 3 Let A be an integral domain with quotient field K all localisations of A can be viewedas subrings of K and in this sense A =

⋂mAm where the intersection is over all maximal ideals

Proof Given x isin K we put D = a isin A | ax isin A we call D the ideal of denominators of x Theelement x is in A iff D = A and x isin Ap iff D p Therefore if x isin A there exists a maximalideal m such that D sube p and x isin Am for this m

Lemma 4 Let A be a ring and S sube T multiplicatively closed subsets Then

(a) There is a canonical isomorphism of Sminus1A-algebras Tminus1A sim= Tminus1(Sminus1A) defined by at 7rarr(a1)(t1)

(b) If M is an A-module then there is a canonical isomorphism of Sminus1A-modules Tminus1M sim=Tminus1(Sminus1M) defined by mt 7rarr (m1)(t1)

Proof (a) Just using the universal property of localisation we can see Tminus1A sim= Tminus1(Sminus1A) asSminus1A-algebras via the map at 7rarr (a1)(t1) (b) is also easily checked

Lemma 5 Let A be an integral domain with quotient field K and B a subring of K containingA If Q is the quotient field of B then there is a canonical isomorphism of B-algebras K sim= Q

If φ A minusrarr B is a ring isomorphism and S sube A is multiplicatively closed (denote also by S theimage in B) then there is an isomorphism of rings Sminus1A sim= Sminus1B making the following diagramcommute

Sminus1A +3 Sminus1B

Aφ

+3

OO

B

OO

Lemma 6 Let A be a ring S sube A a multiplicatively closed subset and p a prime ideal withp cap S = empty Let B = Sminus1A Then there is a canonical ring isomorphism BpB

sim= Ap

Proof A minusrarr B minusrarr BpB sends elements of A not in p to units so we have an induced ringmorphism Ap minusrarr BpB defined by as 7rarr (a1)(s1) and it is easy to check this is an isomorphism

2

Let ψ A minusrarr B be a morphism of rings and I an ideal of A The extended ideal IB consistsof sums

sumψ(ai)bi with ai isin I bi isin B Consider the exact sequence of A-modules

0 minusrarr I minusrarr A minusrarr AI minusrarr 0

Tensoring with B gives an exact sequence of B-modules

I otimesA B minusrarr AotimesA B minusrarr (AI)otimesA B minusrarr 0

The image of I otimesA B in B sim= A otimesA B is simply IB So there is an isomorphism of B-modulesBIB sim= (AI)otimesA B defined by b+ IB 7rarr 1otimes b In fact this is an isomorphism of rings as wellOf course for any two A-algebras EF twisting gives a ring isomorphism E otimesA F sim= F otimesA E

Lemma 7 Let φ A minusrarr B be a morphism of rings S a multiplicatively closed subset of A andset T = φ(S) Then for any B-module M there is a canonical isomorphism of Sminus1A-modulesnatural in M

α Sminus1M minusrarr Tminus1M

α(ms) = mφ(s)

In particular there is a canonical isomorphism of Sminus1A-algebras Sminus1B sim= Tminus1B

Proof One checks easily that α is a well-defined isomorphism of Sminus1A-modules In the caseM = B the Sminus1A-module Sminus1B becomes a ring in the obvious way and α preserves this ringstructure

In particular let S be a multiplicatively closed subset of a ring A let I be an ideal of A andlet T denote the image of S in AI Then there is a canonical isomorphism of rings

Tminus1(AI) sim= AI otimesA Sminus1A sim= Sminus1AI(Sminus1A)

(a+ I)(s+ I) 7rarr as+ I(Sminus1A)

Definition 1 A ring A is catenary if for each pair of prime ideals q sub p the height of the primeideal pq in Aq is finite and is equal to the length of any maximal chain of prime ideals between pand q Clearly the catenary property is stable under isomorphism and any quotient of a catenaryring is catenary If S sube A is a multiplicatively closed subset and A is catenary then so is Sminus1A

Lemma 8 Let A be a ring Then the following are equivalent

(i) A is catenary

(ii) Ap is catenary for every prime ideal p

(iii) Am is catenary for every maximal ideal m

Proof The implications (i) rArr (ii) rArr (iii) are obvious (iii) rArr (i) If q sub p are primes find amaximal ideal m containing p and pass to the catenary ring Am to see that the required propertyis satisfied for q p

Lemma 9 Let A be a noetherian ring Then A is catenary if for every pair of prime ideals q sub pwe have ht(pq) = htpminus htq

Proof Since A is noetherian all involved heights are finite Suppose A satisfies the condition andlet q sub p be prime ideals Obviously ht(pq) is finite and there is at least one maximal chainbetween p and q with length ht(pq) Let

q = q0 sub q1 sub middot middot middot sub qn = p

be a maximal chain of length n Then by assumption 1 = ht(qiqiminus1) = htqi minus htqiminus1 for1 le i le n Hence htp = htq + n so n = ht(pq) as required

3

Definition 2 A ring A is universally catenary if A is noetherian and every finitely generatedA-algebra is catenary Equivalently a noetherian ring A is universally catenary if A is catenaryand A[x1 xn] is catenary for n ge 1

Lemma 10 Let A be a ring and S sube A a multiplicatively closed subset Then there is a canonicalring isomorphism Sminus1(A[x]) sim= (Sminus1A)[x] In particular if q is a prime ideal of A[x] and p = qcapAthen A[x]q sim= Ap[x]qAp[x]

Proof The ring morphism A minusrarr Sminus1A induced A[x] minusrarr (Sminus1A)[x] which sends elements ofS sube A[x] to units So there is an induced ring morphism ϕ Sminus1(A[x]) minusrarr (Sminus1A)[x] defined by

ϕ

(a0 + a1x+ middot middot middot+ anx

n

s

)=a0

s+a1

sx+ middot middot middot+ an

sxn

This is easily checked to be an isomorphism In the second claim there is an isomorphismAp[x] sim= A[x]p where the second ring denotes (Aminusp)minus1(A[x]) and qAp[x] denotes the prime idealof Ap[x] corresponding to qA[x]p Using the isomorphism ϕ it is clear that qAp[x] cap Ap = pApUsing Lemma 6 there is clearly an isomorphism of rings A[x]q sim= Ap[x]qAp[x]

If R is a ring and M an R-module then let Z(M) denote the set of zero-divisors in M Thatis all elements r isin R with rm = 0 for some nonzero m isinM

Lemma 11 Let R be a nonzero reduced noetherian ring Then Z(R) =⋃i pi with the union

being taken over all minimal prime ideals pi

Proof Since R is reduced⋂i pi = 0 If ab = 0 with b 6= 0 then b isin pj for some j and therefore

a isin pj sube⋃i pi The reverse inclusion follows from the fact that no minimal prime can contain a

regular element (since otherwise by Krullrsquos PID Theorem it would have height ge 1)

Lemma 12 Let R be a nonzero reduced noetherian ring Assume that every element of R iseither a unit or a zero-divisor Then dim(R) = 0

Proof Let p1 pn be the minimal primes of R Then by Lemma 11 Z(R) = p1 cup middot middot middot cup pnLet p be a prime ideal Since p is proper p sube Z(R) and therefore p sube pi for some i Since pi isminimal p = pi so the pi are the only primes in R Since these all have height zero it is clearthat dim(R) = 0

Lemma 13 Let R be a reduced ring p a minimal prime ideal of R Then Rp is a field

Proof If p = 0 this is trivial so assume p 6= 0 Since pRp is the only prime ideal in Rp it is alsothe nilradical So if x isin p then txn = 0 for some t isin p and n gt 0 But this implies that tx isnilpotent and therefore zero since R is reduced Therefore pRp = 0 and Rp is a field

Let A1 An be rings Let A be the product ring A =prodni=1Ai Ideals of A are in bijection

with sequences I1 In with Ii an ideal of Ai This sequence corresponds to

I1 times middot middot middot times In

This bijection identifies the prime ideals of A with sequences I1 In in which every Ii = Aiexcept for a single Ij which is a prime ideal of Aj So the primes look like

A1 times middot middot middot times pi times middot middot middot timesAn

for some i and some prime ideal pi of Ai Given i and a prime ideal pi of Ai let p be the primeideal A1 times middot middot middot times pi times middot middot middotAn Then the projection of rings A minusrarr Ai gives rise to a ring morphism

Ap minusrarr (Ai)pi

(a1 ai an)(b1 bi bn) 7rarr aibi

4

It is easy to check that this is an isomorphism An orthogonal set of idempotents in a ring A is aset e1 er with 1 = e1 + middot middot middot+ er e2i = ei and eiej = 0 for i 6= j If A =

prodni=1Ai is a product

of rings then the elements e1 = (1 0 0) en = (0 0 1) are clearly such a setConversely if e1 er is an orthogonal set of idempotents in a ring A then the ideal eiA

becomes a ring with identity ei The map

A minusrarr e1Atimes middot middot middot times erA

a 7rarr (e1a era)

is a ring isomorphism

Proposition 14 Any nonzero artinian ring A is a finite direct product of local artinian rings

Proof See [Eis95] Corollary 216 This shows that there is a finite list of maximal ideals m1 mn

(allowing repeats) and a ring isomorphism A minusrarrprodni=1Ami

defined by a 7rarr (a1 a1)

Proposition 15 Let ϕ A minusrarr B be a surjective morphism of rings M an A-module andp isin SpecB There is a canonical morphism of Bp-modules natural in M

κ HomA(BM)p minusrarr HomAϕminus1p(BpMϕminus1p)

κ(us)(bt) = u(b)ϕminus1(st)

If A is noetherian this is an isomorphism

Proof Let a morphism of A-modules u B minusrarrM s t isin B p and b isin B be given Choose k isin Awith ϕ(k) = st We claim the fraction u(b)k isin Mϕminus1p doesnrsquot depend on the choice of k If wehave ϕ(l) = st also then

ku(b) = u(kb) = u(ϕ(k)b) = u(ϕ(l)b) = u(lb) = lu(b)

so u(b)l = u(b)k as claimed Throughout the proof given x isin B we write ϕminus1(x) for anarbitrary element in the inverse image of x One checks the result does not depend on this choiceWe can now define a morphism of Bp-modules κ(us)(bt) = u(b)ϕminus1(st) which one checks iswell-defined and natural in M

Now assume that A is noetherian In showing that κ is an isomorphism we may as well assumeϕ is the canonical projection A minusrarr Aa for some ideal a In that case the prime ideal p is qa forsome prime q of A containing a and if we set S = A q and T = ϕ(S) we have by Lemma 7 anisomorphism

HomA(BM)p = Tminus1HomA(AaM)sim= Sminus1HomA(AaM)sim= HomSminus1A(Sminus1(Aa) Sminus1M)sim= HomSminus1A(Tminus1(Aa) Sminus1M)= HomAϕminus1p

(BpMϕminus1p)

We have use the fact that A is noetherian to see that Aa is finitely presented so we have thesecond isomorphism in the above sequence One checks easily that this isomorphism agrees withκ completing the proof

Remark 1 The right adjoint HomA(Bminus) to the restriction of scalars functor exists for anymorphism of rings ϕ A minusrarr B but as we have just seen this functor is not local unless thering morphism is surjective This explains why the right adjoint f to the direct image functor inalgebraic geometry essentially only exists for closed immersions

5

2 Flatness

Definition 3 Let A be a ring and M an A-module We say M is flat if the functor minus otimesA M AMod minusrarr AMod is exact (equivalently M otimesA minus is exact) Equivalently M is flat if wheneverwe have an injective morphism of modules N minusrarr N prime the morphism N otimesA M minusrarr N prime otimesA M isinjective This property is stable under isomorphism

We say M is faithfully flat if a morphism N minusrarr N prime is injective if and only if N otimesA M minusrarrN prime otimesAM is injective This property is also stable under isomorphism An A-algebra A minusrarr B isflat if B is a flat A-module and we say A minusrarr B is a flat morphism

Example 1 Nonzero free modules are faithfully flat

Lemma 16 We have the following fundamental properties of flatness

bull Transitivity If φ A minusrarr B is a flat morphism of rings and N a flat B-module then N isalso flat over A

bull Change of Base If φ A minusrarr B is a morphism of rings and M is a flat A-module thenM otimesA B is a flat B-module

bull Localisation If A is a ring and S a multiplicatively closed subset then Sminus1A is flat over A

Proof The second and third claims are done in our Atiyah amp Macdonald notes To prove the firstclaim let M minusrarr M prime be a monomorphism of A-modules and consider the following commutativediagram of abelian groups

M otimesA N

M prime otimesA N

M otimesA (B otimesB N)

M prime otimesA (B otimesB N)

(M otimesA B)otimesB N (M prime otimesA B)otimesB N

Since B is a flat A-module and N is a flat B-module the bottom row is injective hence so is thetop row

Lemma 17 Let φ A minusrarr B be a morphism of rings and N a B-module which is flat over A IfS is a multiplicatively closed subset of B then Sminus1N is flat over A In particular any localisationof a flat A-module is flat

Proof If M minusrarrM prime is a monomorphism of A-modules then we have a commutative diagram

M otimesA Sminus1N

M prime otimesA Sminus1N

(M otimesA N)otimesB Sminus1B (M prime otimesA N)otimesB Sminus1B

The bottom row is clearly injective and hence so is the top row which shows that Sminus1N is flatover A

Lemma 18 Let A be a ring and MN flat A-modules Then M otimesA N is also flat over A

Lemma 19 Let φ A minusrarr B be a flat morphism of rings and S a multiplicatively closed subsetof A Then T = φ(S) is a multiplicatively closed subset of B and Tminus1B is flat over Sminus1A

Proof This follows from Lemma 7 and stability of flatness under base change

6

Lemma 20 Let A minusrarr B be a morphism of rings Then the functor minusotimesAB AMod minusrarr BModpreserves projectives

Proof The functor minusotimesA B is left adjoint to the restriction of scalars functor This latter functoris clearly exact so since any functor with an exact right adjoint must preserve projectives P otimesABis a projective B-module for any projective A-module P

Lemma 21 Let A minusrarr B be a flat morphism of rings If I is an injective B-module then it isalso an injective A-module

Proof The restriction of scalars functor has an exact left adjoint minus otimesA B AMod minusrarr BModand therefore preserves injectives

Lemma 22 Let φ A minusrarr B be a flat morphism of rings and let MN be A-modules Then thereis an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (MotimesABNotimesAB) If A is noetherianand M finitely generated over A there is an isomorphism of B-modules ExtiA(MN) otimesA B sim=ExtiB(M otimesA BN otimesA B)

Proof Let X middot middot middot minusrarr X1 minusrarr X0 minusrarr M minusrarr 0 be a projective resolution of the A-module M Since B is flat the sequence

X otimesA B middot middot middot minusrarr X1 otimesA B minusrarr X0 otimesA B minusrarrM otimesA B minusrarr 0

is a projective resolution of M otimesA B The chain complex of B-modules (X otimesA B)otimesB (B otimesA N)is isomorphic to (X otimesA N)otimesA B The exact functor minusotimesA B commutes with taking homology sothere is an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (M otimesABN otimesAB) as required

If A is noetherian andM finitely generated we can assume that theXi are finite free A-modulesThen ExtiA(MN) is the i-cohomology module of the sequence

0 minusrarr Hom(X0 N) minusrarr Hom(X1 N) minusrarr Hom(X2 N) minusrarr middot middot middot

Since tensoring with B is exact ExtiA(MN) otimesA B is isomorphic as a B-module to the i-thcohomology of the following sequence

0 minusrarr Hom(X0 N)otimesA B minusrarr Hom(X1 N)otimesA B minusrarr middot middot middot

After a bit of work we see that this cochain complex is isomorphic to HomB(X otimesA BN otimesA B)and the i-th cohomology of this complex is ExtiB(M otimesA BN otimesA B) as required

In particular for a ring A and prime ideal p sube A we have isomorphisms of Ap-modules for i ge 0

TorAp

i (Mp Np) sim= TorAi (MN)p

ExtiAp(Mp Np) sim= ExtiA(MN)p

the latter being valid for A noetherian and M finitely generated

Lemma 23 Let A be a ring and M an A-module Then the following are equivalent

(i) M is a flat A-module

(ii) Mp is a flat Ap-module for each prime ideal p

(iii) Mm is a flat Am-module for each maximal ideal m

Proof See [AM69] or any book on commutative algebra

Proposition 24 Let (Am k) be a local ring and M an A-module Suppose that either m isnilpotent or M is finitely generated over A Then M is free hArr M is projective hArr M is flat

7

Proof It suffices to show that if M is flat then it is free We prove that any minimal basis of Mis a basis of M If MmM = 0 then M = 0 and M is trivially free Otherwise it suffices to showthat if x1 xn isinM are elements whose images in MmM = M otimesA k are linearly independentover k then they are linearly independent over A We use induction on n For n = 1 let ax = 0Then there exist y1 yr isin M and b1 br isin A such that abi = 0 for all i and x =

sumbiyi

Since x+ mM 6= 0 not all bi are in m Suppose b1 isin m Then b1 is a unit in A and ab1 = 0 hencea = 0

Suppose n gt 1 andsumni=1 aixi = 0 Then there exist y1 yr isin M and bij isin A(1 le j le r)

such that xi =sumj bijyj and

sumi aibij = 0 Since xn isin mM we have bnj isin m for at least one j

Since a1b1j + middot middot middot+ anbnj = 0 and bnj is a unit we have

an =nminus1sumi=1

ciai ci = minusbijbnj

Then

0 =nsumi=1

aixi = a1(x1 + c1xn) + middot middot middot+ anminus1(xnminus1 + cnminus1xn)

Since the residues of x1+c1xn xnminus1+cnminus1xn are linearly independent over k by the inductivehypothesis we get a1 = middot middot middot = anminus1 = 0 and an =

sumciai = 0

Corollary 25 Let A be a ring and M a finitely generated A-module Then the following areequivalent

(i) M is a flat A-module

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof This is immediate from the previous two results

Proposition 26 Let A be a ring and M a finitely presented A-module Then M is flat if andonly if it is projective

Proof See Stenstrom Chapter 1 Corollary 115

Corollary 27 Let A be a noetherian ring M a finitely generated A-module Then the followingconditions are equivalent

(i) M is projective

(ii) M is flat

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof Since A is noetherian M is finitely presented so (i) hArr (ii) is an immediate consequenceof Proposition 26 The rest of the proof follows from Corollary 25

Lemma 28 Let A minusrarr B be a flat morphism of rings and let I J be ideals of A Then (IcapJ)B =IB cap JB and (I J)B = (IB JB) if J is finitely generated

Proof Consider the exact sequence of A-modules

I cap J minusrarr A minusrarr AI oplusAJ

Tensoring with B we get an exact sequence

(I cap J)otimesA B = (I cap J)B minusrarr B minusrarr BIB oplusBJB

8

This means (I cap J)B = IB cap JB For the second claim suppose firstly that J is a principal idealaA and use the exact sequence

(I aA) i Af AI

where i is the injection and f(x) = ax+I Tensoring withB we get the formula (I a)B = (IB a)In the general case if J = (a1 an) we have (I J) =

⋂i(I ai) so that

(I J)B =⋂

(I ai)B =⋂

(IB ai) = (IB JB)

Example 2 Let A = k[x y] be a polynomial ring over a field k and put B = A(x) sim= k[y] ThenB is not flat over A since y isin A is regular but is not regular on B Let I = (x+ y) and J = (y)Then I cap J = (xy + y2) and IB = JB = yB (I cap J)B = y2B 6= IB cap JB

Example 3 Let k be a field put A = k[x y] and let K be the quotient field of A Let B be thesubring k[x yx] of K (ie the k-subalgebra generated by x and z = yx) Then A sub B sub K LetI = xA J = yA Then I cap J = xyA and (I cap J)B = x2zB IB cap JB = xzB so B is not flat overA The map SpecB minusrarr SpecA corresponding to A minusrarr B is the projection to the (x y)-planeof the surface F xz = y in (x y z)-space Note F contains the whole z-axis so it does not lookldquoflatrdquo over the (x y)-plane

Proposition 29 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB there is a canonical isomorphism of Bp-modules natural in both variables

κ MpcapA otimesApcapANp minusrarr (M otimesA N)p

κ(msotimes nt) = (motimes n)ϕ(s)t

Proof Fix p isin SpecB and q = pcapA There is a canonical ring morphism Aq minusrarr Bp and we makeNp into an Aq-module using this morphism One checks that the following map is well-definedand Aq-bilinear

ε Mq timesNp minusrarr (M otimesA N)p

ε(ms nt) = (motimes n)ϕ(s)t

We show that in fact this is a tensor product of Aq-modules Let Z be an abelian group andψ Mq timesNp minusrarr Z an Aq-bilinear map

Mq timesNp

ε

ψ Z

(M otimesA N)p

φ

(1)

We have to define a morphism of abelian groups φ unique making this diagram commute Fors isin p we define an A-bilinear morphism φprimes M times N minusrarr Z by φprimes(mn) = ψ(m1 ns) Thisinduces a morphism of abelian groups

φprimeprimes M otimesA N minusrarr Z

φprimeprimes (motimes b) = ψ(m1 bs)

We make some observations about these morphisms

bull Suppose ws = wprimesprime in (M otimesA N)p with say w =sumimi otimes ni w

prime =sumim

primei otimes nprimei and t isin p

such that tsprimew = tswprime That issumimi otimes tsprimeni =

sumim

primei otimes tsnprimei Applying φprimeprimetssprime to both sides

of this equality gives φprimeprimes (w) = φprimeprimesprime(wprime)

9

bull For wswprimesprime isin (M otimesA N)p we have φprimeprimes (w) + φprimeprimesprime(wprime) = φprimeprimessprime(s

primew + swprime)

It follows that φ(ws) = φprimeprimes (w) gives a well-defined morphism of abelian groups φ (MotimesAN)p minusrarrZ which is clearly unique making (1) commute By uniqueness of the tensor product there is aninduced isomorphism of abelian groups κ Mq otimesAq Np minusrarr (M otimesA N)p with κ(ms otimes nt) =(m otimes n)ϕ(s)t One checks that this is a morphism of Bp-modules The inverse is defined by(motimes n)t 7rarr m1otimes nt Naturality in both variables is easily checked

Corollary 30 Let ϕ A minusrarr B be a morphism of rings M an A-module and p isin SpecB Thenthere is a canonical isomorphism of Bp-modules MpcapA otimesApcapA

Bp minusrarr (M otimesA B)p natural in M

We will not actually use the next result in these notes so the reader not familiar with homo-logical δ-functors can safely skip it Alternatively one can provide a proof by following the onegiven in Matsumura (the proof we give is more elegant provided you know about δ-functors)

Proposition 31 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB and i ge 0 there is a canonical isomorphism of Bp-modules natural inM

κi TorAi (NM)p minusrarr TorApcapA

i (NpMpcapA)

Proof Fix p isin SpecB and a B-module N and set q = p capA Then N is a B-A-bimodule and Np

is a Bp-Aq-bimodule so by (TORSection 51) the abelian group TorAi (NM) acquires a canonicalB-module structure and Tor

Aq

i (NpMq) acquires a canonical Bp-module structure for any A-module M and i ge 0 Using (TORLemma 14) and (DFDefinition 23) we have two homologicalδ-functors between AMod and BpMod

TorAi (Nminus)pige0 TorAq

i (Np (minus)q)ige0

For i gt 0 these functors all vanish on free A-modules so by (DFTheorem 74) both δ-functors areuniversal For i = 0 we have the canonical natural equivalence of Proposition 29

κ0 TorA0 (Nminus)psim= (N otimesA minus)p

sim= Np otimesAq (minus)qsim= Tor

Aq

0 (Np (minus)q)

By universality this lifts to a canonical isomorphism of homological δ-functors κ In particularfor each i ge 0 we have a canonical natural equivalence κi TorAi (Nminus)p minusrarr Tor

Aq

i (Np (minus)q) asrequired

We know from Lemma 23 that flatness is a local property We are now ready to show thatrelative flatness (ie flatness with respect to a morphism of rings) is also local This is particularlyimportant in algebraic geometry The reader who skipped Proposition 31 will also have to skipthe implication (iii) rArr (i) in the next result but this will not affect their ability to read the restof these notes

Corollary 32 Let A minusrarr B be a morphism of rings and N a B-module Then the followingconditions are equivalent

(i) N is flat over A

(ii) Np is flat over ApcapA for all prime ideals p of B

(iii) Nm is flat over AmcapA for all maximal ideals m of B

Proof (i) rArr (ii) If N is flat over A then NpcapA is flat over ApcapA for any prime p of B By anargument similar to the one given in Lemma 19 we see that Np is isomorphic as a BpcapA-moduleto a localisation of NpcapA Applying Lemma 17 to the ring morphism ApcapA minusrarr BpcapA we see thatNp is flat over ApcapA as required (ii) rArr (iii) is trivial (iii) rArr (i) For every A-module M andmaximal ideal m of B we have by Proposition 31

TorA1 (NM)msim= TorAmcapA

1 (NmMmcapA) = 0

since by assumption Nm is flat over AmcapA Therefore TorA1 (NM) = 0 for every A-module M which implies that N is flat over A and completes the proof

10

Let ψ A minusrarr B be a morphism of rings and I an ideal of A The extended ideal IB consistsof sums

sumψ(ai)bi with ai isin I bi isin B Consider the exact sequence of A-modules

0 minusrarr I minusrarr A minusrarr AI minusrarr 0

Tensoring with B gives an exact sequence of B-modules

I otimesA B minusrarr AotimesA B minusrarr (AI)otimesA B minusrarr 0

The image of I otimesA B in B sim= A otimesA B is simply IB So there is an isomorphism of B-modulesBIB sim= (AI)otimesA B defined by b+ IB 7rarr 1otimes b In fact this is an isomorphism of rings as wellOf course for any two A-algebras EF twisting gives a ring isomorphism E otimesA F sim= F otimesA E

Lemma 7 Let φ A minusrarr B be a morphism of rings S a multiplicatively closed subset of A andset T = φ(S) Then for any B-module M there is a canonical isomorphism of Sminus1A-modulesnatural in M

α Sminus1M minusrarr Tminus1M

α(ms) = mφ(s)

In particular there is a canonical isomorphism of Sminus1A-algebras Sminus1B sim= Tminus1B

Proof One checks easily that α is a well-defined isomorphism of Sminus1A-modules In the caseM = B the Sminus1A-module Sminus1B becomes a ring in the obvious way and α preserves this ringstructure

In particular let S be a multiplicatively closed subset of a ring A let I be an ideal of A andlet T denote the image of S in AI Then there is a canonical isomorphism of rings

Tminus1(AI) sim= AI otimesA Sminus1A sim= Sminus1AI(Sminus1A)

(a+ I)(s+ I) 7rarr as+ I(Sminus1A)

Definition 1 A ring A is catenary if for each pair of prime ideals q sub p the height of the primeideal pq in Aq is finite and is equal to the length of any maximal chain of prime ideals between pand q Clearly the catenary property is stable under isomorphism and any quotient of a catenaryring is catenary If S sube A is a multiplicatively closed subset and A is catenary then so is Sminus1A

Lemma 8 Let A be a ring Then the following are equivalent

(i) A is catenary

(ii) Ap is catenary for every prime ideal p

(iii) Am is catenary for every maximal ideal m

Proof The implications (i) rArr (ii) rArr (iii) are obvious (iii) rArr (i) If q sub p are primes find amaximal ideal m containing p and pass to the catenary ring Am to see that the required propertyis satisfied for q p

Lemma 9 Let A be a noetherian ring Then A is catenary if for every pair of prime ideals q sub pwe have ht(pq) = htpminus htq

Proof Since A is noetherian all involved heights are finite Suppose A satisfies the condition andlet q sub p be prime ideals Obviously ht(pq) is finite and there is at least one maximal chainbetween p and q with length ht(pq) Let

q = q0 sub q1 sub middot middot middot sub qn = p

be a maximal chain of length n Then by assumption 1 = ht(qiqiminus1) = htqi minus htqiminus1 for1 le i le n Hence htp = htq + n so n = ht(pq) as required

3

Definition 2 A ring A is universally catenary if A is noetherian and every finitely generatedA-algebra is catenary Equivalently a noetherian ring A is universally catenary if A is catenaryand A[x1 xn] is catenary for n ge 1

Lemma 10 Let A be a ring and S sube A a multiplicatively closed subset Then there is a canonicalring isomorphism Sminus1(A[x]) sim= (Sminus1A)[x] In particular if q is a prime ideal of A[x] and p = qcapAthen A[x]q sim= Ap[x]qAp[x]

Proof The ring morphism A minusrarr Sminus1A induced A[x] minusrarr (Sminus1A)[x] which sends elements ofS sube A[x] to units So there is an induced ring morphism ϕ Sminus1(A[x]) minusrarr (Sminus1A)[x] defined by

ϕ

(a0 + a1x+ middot middot middot+ anx

n

s

)=a0

s+a1

sx+ middot middot middot+ an

sxn

This is easily checked to be an isomorphism In the second claim there is an isomorphismAp[x] sim= A[x]p where the second ring denotes (Aminusp)minus1(A[x]) and qAp[x] denotes the prime idealof Ap[x] corresponding to qA[x]p Using the isomorphism ϕ it is clear that qAp[x] cap Ap = pApUsing Lemma 6 there is clearly an isomorphism of rings A[x]q sim= Ap[x]qAp[x]

If R is a ring and M an R-module then let Z(M) denote the set of zero-divisors in M Thatis all elements r isin R with rm = 0 for some nonzero m isinM

Lemma 11 Let R be a nonzero reduced noetherian ring Then Z(R) =⋃i pi with the union

being taken over all minimal prime ideals pi

Proof Since R is reduced⋂i pi = 0 If ab = 0 with b 6= 0 then b isin pj for some j and therefore

a isin pj sube⋃i pi The reverse inclusion follows from the fact that no minimal prime can contain a

regular element (since otherwise by Krullrsquos PID Theorem it would have height ge 1)

Lemma 12 Let R be a nonzero reduced noetherian ring Assume that every element of R iseither a unit or a zero-divisor Then dim(R) = 0

Proof Let p1 pn be the minimal primes of R Then by Lemma 11 Z(R) = p1 cup middot middot middot cup pnLet p be a prime ideal Since p is proper p sube Z(R) and therefore p sube pi for some i Since pi isminimal p = pi so the pi are the only primes in R Since these all have height zero it is clearthat dim(R) = 0

Lemma 13 Let R be a reduced ring p a minimal prime ideal of R Then Rp is a field

Proof If p = 0 this is trivial so assume p 6= 0 Since pRp is the only prime ideal in Rp it is alsothe nilradical So if x isin p then txn = 0 for some t isin p and n gt 0 But this implies that tx isnilpotent and therefore zero since R is reduced Therefore pRp = 0 and Rp is a field

Let A1 An be rings Let A be the product ring A =prodni=1Ai Ideals of A are in bijection

with sequences I1 In with Ii an ideal of Ai This sequence corresponds to

I1 times middot middot middot times In

This bijection identifies the prime ideals of A with sequences I1 In in which every Ii = Aiexcept for a single Ij which is a prime ideal of Aj So the primes look like

A1 times middot middot middot times pi times middot middot middot timesAn

for some i and some prime ideal pi of Ai Given i and a prime ideal pi of Ai let p be the primeideal A1 times middot middot middot times pi times middot middot middotAn Then the projection of rings A minusrarr Ai gives rise to a ring morphism

Ap minusrarr (Ai)pi

(a1 ai an)(b1 bi bn) 7rarr aibi

4

It is easy to check that this is an isomorphism An orthogonal set of idempotents in a ring A is aset e1 er with 1 = e1 + middot middot middot+ er e2i = ei and eiej = 0 for i 6= j If A =

prodni=1Ai is a product

of rings then the elements e1 = (1 0 0) en = (0 0 1) are clearly such a setConversely if e1 er is an orthogonal set of idempotents in a ring A then the ideal eiA

becomes a ring with identity ei The map

A minusrarr e1Atimes middot middot middot times erA

a 7rarr (e1a era)

is a ring isomorphism

Proposition 14 Any nonzero artinian ring A is a finite direct product of local artinian rings

Proof See [Eis95] Corollary 216 This shows that there is a finite list of maximal ideals m1 mn

(allowing repeats) and a ring isomorphism A minusrarrprodni=1Ami

defined by a 7rarr (a1 a1)

Proposition 15 Let ϕ A minusrarr B be a surjective morphism of rings M an A-module andp isin SpecB There is a canonical morphism of Bp-modules natural in M

κ HomA(BM)p minusrarr HomAϕminus1p(BpMϕminus1p)

κ(us)(bt) = u(b)ϕminus1(st)

If A is noetherian this is an isomorphism

Proof Let a morphism of A-modules u B minusrarrM s t isin B p and b isin B be given Choose k isin Awith ϕ(k) = st We claim the fraction u(b)k isin Mϕminus1p doesnrsquot depend on the choice of k If wehave ϕ(l) = st also then

ku(b) = u(kb) = u(ϕ(k)b) = u(ϕ(l)b) = u(lb) = lu(b)

so u(b)l = u(b)k as claimed Throughout the proof given x isin B we write ϕminus1(x) for anarbitrary element in the inverse image of x One checks the result does not depend on this choiceWe can now define a morphism of Bp-modules κ(us)(bt) = u(b)ϕminus1(st) which one checks iswell-defined and natural in M

Now assume that A is noetherian In showing that κ is an isomorphism we may as well assumeϕ is the canonical projection A minusrarr Aa for some ideal a In that case the prime ideal p is qa forsome prime q of A containing a and if we set S = A q and T = ϕ(S) we have by Lemma 7 anisomorphism

HomA(BM)p = Tminus1HomA(AaM)sim= Sminus1HomA(AaM)sim= HomSminus1A(Sminus1(Aa) Sminus1M)sim= HomSminus1A(Tminus1(Aa) Sminus1M)= HomAϕminus1p

(BpMϕminus1p)

We have use the fact that A is noetherian to see that Aa is finitely presented so we have thesecond isomorphism in the above sequence One checks easily that this isomorphism agrees withκ completing the proof

Remark 1 The right adjoint HomA(Bminus) to the restriction of scalars functor exists for anymorphism of rings ϕ A minusrarr B but as we have just seen this functor is not local unless thering morphism is surjective This explains why the right adjoint f to the direct image functor inalgebraic geometry essentially only exists for closed immersions

5

2 Flatness

Definition 3 Let A be a ring and M an A-module We say M is flat if the functor minus otimesA M AMod minusrarr AMod is exact (equivalently M otimesA minus is exact) Equivalently M is flat if wheneverwe have an injective morphism of modules N minusrarr N prime the morphism N otimesA M minusrarr N prime otimesA M isinjective This property is stable under isomorphism

We say M is faithfully flat if a morphism N minusrarr N prime is injective if and only if N otimesA M minusrarrN prime otimesAM is injective This property is also stable under isomorphism An A-algebra A minusrarr B isflat if B is a flat A-module and we say A minusrarr B is a flat morphism

Example 1 Nonzero free modules are faithfully flat

Lemma 16 We have the following fundamental properties of flatness

bull Transitivity If φ A minusrarr B is a flat morphism of rings and N a flat B-module then N isalso flat over A

bull Change of Base If φ A minusrarr B is a morphism of rings and M is a flat A-module thenM otimesA B is a flat B-module

bull Localisation If A is a ring and S a multiplicatively closed subset then Sminus1A is flat over A

Proof The second and third claims are done in our Atiyah amp Macdonald notes To prove the firstclaim let M minusrarr M prime be a monomorphism of A-modules and consider the following commutativediagram of abelian groups

M otimesA N

M prime otimesA N

M otimesA (B otimesB N)

M prime otimesA (B otimesB N)

(M otimesA B)otimesB N (M prime otimesA B)otimesB N

Since B is a flat A-module and N is a flat B-module the bottom row is injective hence so is thetop row

Lemma 17 Let φ A minusrarr B be a morphism of rings and N a B-module which is flat over A IfS is a multiplicatively closed subset of B then Sminus1N is flat over A In particular any localisationof a flat A-module is flat

Proof If M minusrarrM prime is a monomorphism of A-modules then we have a commutative diagram

M otimesA Sminus1N

M prime otimesA Sminus1N

(M otimesA N)otimesB Sminus1B (M prime otimesA N)otimesB Sminus1B

The bottom row is clearly injective and hence so is the top row which shows that Sminus1N is flatover A

Lemma 18 Let A be a ring and MN flat A-modules Then M otimesA N is also flat over A

Lemma 19 Let φ A minusrarr B be a flat morphism of rings and S a multiplicatively closed subsetof A Then T = φ(S) is a multiplicatively closed subset of B and Tminus1B is flat over Sminus1A

Proof This follows from Lemma 7 and stability of flatness under base change

6

Lemma 20 Let A minusrarr B be a morphism of rings Then the functor minusotimesAB AMod minusrarr BModpreserves projectives

Proof The functor minusotimesA B is left adjoint to the restriction of scalars functor This latter functoris clearly exact so since any functor with an exact right adjoint must preserve projectives P otimesABis a projective B-module for any projective A-module P

Lemma 21 Let A minusrarr B be a flat morphism of rings If I is an injective B-module then it isalso an injective A-module

Proof The restriction of scalars functor has an exact left adjoint minus otimesA B AMod minusrarr BModand therefore preserves injectives

Lemma 22 Let φ A minusrarr B be a flat morphism of rings and let MN be A-modules Then thereis an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (MotimesABNotimesAB) If A is noetherianand M finitely generated over A there is an isomorphism of B-modules ExtiA(MN) otimesA B sim=ExtiB(M otimesA BN otimesA B)

Proof Let X middot middot middot minusrarr X1 minusrarr X0 minusrarr M minusrarr 0 be a projective resolution of the A-module M Since B is flat the sequence

X otimesA B middot middot middot minusrarr X1 otimesA B minusrarr X0 otimesA B minusrarrM otimesA B minusrarr 0

is a projective resolution of M otimesA B The chain complex of B-modules (X otimesA B)otimesB (B otimesA N)is isomorphic to (X otimesA N)otimesA B The exact functor minusotimesA B commutes with taking homology sothere is an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (M otimesABN otimesAB) as required

If A is noetherian andM finitely generated we can assume that theXi are finite free A-modulesThen ExtiA(MN) is the i-cohomology module of the sequence

0 minusrarr Hom(X0 N) minusrarr Hom(X1 N) minusrarr Hom(X2 N) minusrarr middot middot middot

Since tensoring with B is exact ExtiA(MN) otimesA B is isomorphic as a B-module to the i-thcohomology of the following sequence

0 minusrarr Hom(X0 N)otimesA B minusrarr Hom(X1 N)otimesA B minusrarr middot middot middot

After a bit of work we see that this cochain complex is isomorphic to HomB(X otimesA BN otimesA B)and the i-th cohomology of this complex is ExtiB(M otimesA BN otimesA B) as required

In particular for a ring A and prime ideal p sube A we have isomorphisms of Ap-modules for i ge 0

TorAp

i (Mp Np) sim= TorAi (MN)p

ExtiAp(Mp Np) sim= ExtiA(MN)p

the latter being valid for A noetherian and M finitely generated

Lemma 23 Let A be a ring and M an A-module Then the following are equivalent

(i) M is a flat A-module

(ii) Mp is a flat Ap-module for each prime ideal p

(iii) Mm is a flat Am-module for each maximal ideal m

Proof See [AM69] or any book on commutative algebra

Proposition 24 Let (Am k) be a local ring and M an A-module Suppose that either m isnilpotent or M is finitely generated over A Then M is free hArr M is projective hArr M is flat

7

Proof It suffices to show that if M is flat then it is free We prove that any minimal basis of Mis a basis of M If MmM = 0 then M = 0 and M is trivially free Otherwise it suffices to showthat if x1 xn isinM are elements whose images in MmM = M otimesA k are linearly independentover k then they are linearly independent over A We use induction on n For n = 1 let ax = 0Then there exist y1 yr isin M and b1 br isin A such that abi = 0 for all i and x =

sumbiyi

Since x+ mM 6= 0 not all bi are in m Suppose b1 isin m Then b1 is a unit in A and ab1 = 0 hencea = 0

Suppose n gt 1 andsumni=1 aixi = 0 Then there exist y1 yr isin M and bij isin A(1 le j le r)

such that xi =sumj bijyj and

sumi aibij = 0 Since xn isin mM we have bnj isin m for at least one j

Since a1b1j + middot middot middot+ anbnj = 0 and bnj is a unit we have

an =nminus1sumi=1

ciai ci = minusbijbnj

Then

0 =nsumi=1

aixi = a1(x1 + c1xn) + middot middot middot+ anminus1(xnminus1 + cnminus1xn)

Since the residues of x1+c1xn xnminus1+cnminus1xn are linearly independent over k by the inductivehypothesis we get a1 = middot middot middot = anminus1 = 0 and an =

sumciai = 0

Corollary 25 Let A be a ring and M a finitely generated A-module Then the following areequivalent

(i) M is a flat A-module

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof This is immediate from the previous two results

Proposition 26 Let A be a ring and M a finitely presented A-module Then M is flat if andonly if it is projective

Proof See Stenstrom Chapter 1 Corollary 115

Corollary 27 Let A be a noetherian ring M a finitely generated A-module Then the followingconditions are equivalent

(i) M is projective

(ii) M is flat

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof Since A is noetherian M is finitely presented so (i) hArr (ii) is an immediate consequenceof Proposition 26 The rest of the proof follows from Corollary 25

Lemma 28 Let A minusrarr B be a flat morphism of rings and let I J be ideals of A Then (IcapJ)B =IB cap JB and (I J)B = (IB JB) if J is finitely generated

Proof Consider the exact sequence of A-modules

I cap J minusrarr A minusrarr AI oplusAJ

Tensoring with B we get an exact sequence

(I cap J)otimesA B = (I cap J)B minusrarr B minusrarr BIB oplusBJB

8

This means (I cap J)B = IB cap JB For the second claim suppose firstly that J is a principal idealaA and use the exact sequence

(I aA) i Af AI

where i is the injection and f(x) = ax+I Tensoring withB we get the formula (I a)B = (IB a)In the general case if J = (a1 an) we have (I J) =

⋂i(I ai) so that

(I J)B =⋂

(I ai)B =⋂

(IB ai) = (IB JB)

Example 2 Let A = k[x y] be a polynomial ring over a field k and put B = A(x) sim= k[y] ThenB is not flat over A since y isin A is regular but is not regular on B Let I = (x+ y) and J = (y)Then I cap J = (xy + y2) and IB = JB = yB (I cap J)B = y2B 6= IB cap JB

Example 3 Let k be a field put A = k[x y] and let K be the quotient field of A Let B be thesubring k[x yx] of K (ie the k-subalgebra generated by x and z = yx) Then A sub B sub K LetI = xA J = yA Then I cap J = xyA and (I cap J)B = x2zB IB cap JB = xzB so B is not flat overA The map SpecB minusrarr SpecA corresponding to A minusrarr B is the projection to the (x y)-planeof the surface F xz = y in (x y z)-space Note F contains the whole z-axis so it does not lookldquoflatrdquo over the (x y)-plane

Proposition 29 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB there is a canonical isomorphism of Bp-modules natural in both variables

κ MpcapA otimesApcapANp minusrarr (M otimesA N)p

κ(msotimes nt) = (motimes n)ϕ(s)t

Proof Fix p isin SpecB and q = pcapA There is a canonical ring morphism Aq minusrarr Bp and we makeNp into an Aq-module using this morphism One checks that the following map is well-definedand Aq-bilinear

ε Mq timesNp minusrarr (M otimesA N)p

ε(ms nt) = (motimes n)ϕ(s)t

We show that in fact this is a tensor product of Aq-modules Let Z be an abelian group andψ Mq timesNp minusrarr Z an Aq-bilinear map

Mq timesNp

ε

ψ Z

(M otimesA N)p

φ

(1)

We have to define a morphism of abelian groups φ unique making this diagram commute Fors isin p we define an A-bilinear morphism φprimes M times N minusrarr Z by φprimes(mn) = ψ(m1 ns) Thisinduces a morphism of abelian groups

φprimeprimes M otimesA N minusrarr Z

φprimeprimes (motimes b) = ψ(m1 bs)

We make some observations about these morphisms

bull Suppose ws = wprimesprime in (M otimesA N)p with say w =sumimi otimes ni w

prime =sumim

primei otimes nprimei and t isin p

such that tsprimew = tswprime That issumimi otimes tsprimeni =

sumim

primei otimes tsnprimei Applying φprimeprimetssprime to both sides

of this equality gives φprimeprimes (w) = φprimeprimesprime(wprime)

9

bull For wswprimesprime isin (M otimesA N)p we have φprimeprimes (w) + φprimeprimesprime(wprime) = φprimeprimessprime(s

primew + swprime)

It follows that φ(ws) = φprimeprimes (w) gives a well-defined morphism of abelian groups φ (MotimesAN)p minusrarrZ which is clearly unique making (1) commute By uniqueness of the tensor product there is aninduced isomorphism of abelian groups κ Mq otimesAq Np minusrarr (M otimesA N)p with κ(ms otimes nt) =(m otimes n)ϕ(s)t One checks that this is a morphism of Bp-modules The inverse is defined by(motimes n)t 7rarr m1otimes nt Naturality in both variables is easily checked

Corollary 30 Let ϕ A minusrarr B be a morphism of rings M an A-module and p isin SpecB Thenthere is a canonical isomorphism of Bp-modules MpcapA otimesApcapA

Bp minusrarr (M otimesA B)p natural in M

We will not actually use the next result in these notes so the reader not familiar with homo-logical δ-functors can safely skip it Alternatively one can provide a proof by following the onegiven in Matsumura (the proof we give is more elegant provided you know about δ-functors)

Proposition 31 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB and i ge 0 there is a canonical isomorphism of Bp-modules natural inM

κi TorAi (NM)p minusrarr TorApcapA

i (NpMpcapA)

Proof Fix p isin SpecB and a B-module N and set q = p capA Then N is a B-A-bimodule and Np

is a Bp-Aq-bimodule so by (TORSection 51) the abelian group TorAi (NM) acquires a canonicalB-module structure and Tor

Aq

i (NpMq) acquires a canonical Bp-module structure for any A-module M and i ge 0 Using (TORLemma 14) and (DFDefinition 23) we have two homologicalδ-functors between AMod and BpMod

TorAi (Nminus)pige0 TorAq

i (Np (minus)q)ige0

For i gt 0 these functors all vanish on free A-modules so by (DFTheorem 74) both δ-functors areuniversal For i = 0 we have the canonical natural equivalence of Proposition 29

κ0 TorA0 (Nminus)psim= (N otimesA minus)p

sim= Np otimesAq (minus)qsim= Tor

Aq

0 (Np (minus)q)

By universality this lifts to a canonical isomorphism of homological δ-functors κ In particularfor each i ge 0 we have a canonical natural equivalence κi TorAi (Nminus)p minusrarr Tor

Aq

i (Np (minus)q) asrequired

We know from Lemma 23 that flatness is a local property We are now ready to show thatrelative flatness (ie flatness with respect to a morphism of rings) is also local This is particularlyimportant in algebraic geometry The reader who skipped Proposition 31 will also have to skipthe implication (iii) rArr (i) in the next result but this will not affect their ability to read the restof these notes

Corollary 32 Let A minusrarr B be a morphism of rings and N a B-module Then the followingconditions are equivalent

(i) N is flat over A

(ii) Np is flat over ApcapA for all prime ideals p of B

(iii) Nm is flat over AmcapA for all maximal ideals m of B

Proof (i) rArr (ii) If N is flat over A then NpcapA is flat over ApcapA for any prime p of B By anargument similar to the one given in Lemma 19 we see that Np is isomorphic as a BpcapA-moduleto a localisation of NpcapA Applying Lemma 17 to the ring morphism ApcapA minusrarr BpcapA we see thatNp is flat over ApcapA as required (ii) rArr (iii) is trivial (iii) rArr (i) For every A-module M andmaximal ideal m of B we have by Proposition 31

TorA1 (NM)msim= TorAmcapA

1 (NmMmcapA) = 0

since by assumption Nm is flat over AmcapA Therefore TorA1 (NM) = 0 for every A-module M which implies that N is flat over A and completes the proof

10

Definition 2 A ring A is universally catenary if A is noetherian and every finitely generatedA-algebra is catenary Equivalently a noetherian ring A is universally catenary if A is catenaryand A[x1 xn] is catenary for n ge 1

Lemma 10 Let A be a ring and S sube A a multiplicatively closed subset Then there is a canonicalring isomorphism Sminus1(A[x]) sim= (Sminus1A)[x] In particular if q is a prime ideal of A[x] and p = qcapAthen A[x]q sim= Ap[x]qAp[x]

Proof The ring morphism A minusrarr Sminus1A induced A[x] minusrarr (Sminus1A)[x] which sends elements ofS sube A[x] to units So there is an induced ring morphism ϕ Sminus1(A[x]) minusrarr (Sminus1A)[x] defined by

ϕ

(a0 + a1x+ middot middot middot+ anx

n

s

)=a0

s+a1

sx+ middot middot middot+ an

sxn

This is easily checked to be an isomorphism In the second claim there is an isomorphismAp[x] sim= A[x]p where the second ring denotes (Aminusp)minus1(A[x]) and qAp[x] denotes the prime idealof Ap[x] corresponding to qA[x]p Using the isomorphism ϕ it is clear that qAp[x] cap Ap = pApUsing Lemma 6 there is clearly an isomorphism of rings A[x]q sim= Ap[x]qAp[x]

If R is a ring and M an R-module then let Z(M) denote the set of zero-divisors in M Thatis all elements r isin R with rm = 0 for some nonzero m isinM

Lemma 11 Let R be a nonzero reduced noetherian ring Then Z(R) =⋃i pi with the union

being taken over all minimal prime ideals pi

Proof Since R is reduced⋂i pi = 0 If ab = 0 with b 6= 0 then b isin pj for some j and therefore

a isin pj sube⋃i pi The reverse inclusion follows from the fact that no minimal prime can contain a

regular element (since otherwise by Krullrsquos PID Theorem it would have height ge 1)

Lemma 12 Let R be a nonzero reduced noetherian ring Assume that every element of R iseither a unit or a zero-divisor Then dim(R) = 0

Proof Let p1 pn be the minimal primes of R Then by Lemma 11 Z(R) = p1 cup middot middot middot cup pnLet p be a prime ideal Since p is proper p sube Z(R) and therefore p sube pi for some i Since pi isminimal p = pi so the pi are the only primes in R Since these all have height zero it is clearthat dim(R) = 0

Lemma 13 Let R be a reduced ring p a minimal prime ideal of R Then Rp is a field

Proof If p = 0 this is trivial so assume p 6= 0 Since pRp is the only prime ideal in Rp it is alsothe nilradical So if x isin p then txn = 0 for some t isin p and n gt 0 But this implies that tx isnilpotent and therefore zero since R is reduced Therefore pRp = 0 and Rp is a field

Let A1 An be rings Let A be the product ring A =prodni=1Ai Ideals of A are in bijection

with sequences I1 In with Ii an ideal of Ai This sequence corresponds to

I1 times middot middot middot times In

This bijection identifies the prime ideals of A with sequences I1 In in which every Ii = Aiexcept for a single Ij which is a prime ideal of Aj So the primes look like

A1 times middot middot middot times pi times middot middot middot timesAn

for some i and some prime ideal pi of Ai Given i and a prime ideal pi of Ai let p be the primeideal A1 times middot middot middot times pi times middot middot middotAn Then the projection of rings A minusrarr Ai gives rise to a ring morphism

Ap minusrarr (Ai)pi

(a1 ai an)(b1 bi bn) 7rarr aibi

4

It is easy to check that this is an isomorphism An orthogonal set of idempotents in a ring A is aset e1 er with 1 = e1 + middot middot middot+ er e2i = ei and eiej = 0 for i 6= j If A =

prodni=1Ai is a product

of rings then the elements e1 = (1 0 0) en = (0 0 1) are clearly such a setConversely if e1 er is an orthogonal set of idempotents in a ring A then the ideal eiA

becomes a ring with identity ei The map

A minusrarr e1Atimes middot middot middot times erA

a 7rarr (e1a era)

is a ring isomorphism

Proposition 14 Any nonzero artinian ring A is a finite direct product of local artinian rings

Proof See [Eis95] Corollary 216 This shows that there is a finite list of maximal ideals m1 mn

(allowing repeats) and a ring isomorphism A minusrarrprodni=1Ami

defined by a 7rarr (a1 a1)

Proposition 15 Let ϕ A minusrarr B be a surjective morphism of rings M an A-module andp isin SpecB There is a canonical morphism of Bp-modules natural in M

κ HomA(BM)p minusrarr HomAϕminus1p(BpMϕminus1p)

κ(us)(bt) = u(b)ϕminus1(st)

If A is noetherian this is an isomorphism

Proof Let a morphism of A-modules u B minusrarrM s t isin B p and b isin B be given Choose k isin Awith ϕ(k) = st We claim the fraction u(b)k isin Mϕminus1p doesnrsquot depend on the choice of k If wehave ϕ(l) = st also then

ku(b) = u(kb) = u(ϕ(k)b) = u(ϕ(l)b) = u(lb) = lu(b)

so u(b)l = u(b)k as claimed Throughout the proof given x isin B we write ϕminus1(x) for anarbitrary element in the inverse image of x One checks the result does not depend on this choiceWe can now define a morphism of Bp-modules κ(us)(bt) = u(b)ϕminus1(st) which one checks iswell-defined and natural in M

Now assume that A is noetherian In showing that κ is an isomorphism we may as well assumeϕ is the canonical projection A minusrarr Aa for some ideal a In that case the prime ideal p is qa forsome prime q of A containing a and if we set S = A q and T = ϕ(S) we have by Lemma 7 anisomorphism

HomA(BM)p = Tminus1HomA(AaM)sim= Sminus1HomA(AaM)sim= HomSminus1A(Sminus1(Aa) Sminus1M)sim= HomSminus1A(Tminus1(Aa) Sminus1M)= HomAϕminus1p

(BpMϕminus1p)

We have use the fact that A is noetherian to see that Aa is finitely presented so we have thesecond isomorphism in the above sequence One checks easily that this isomorphism agrees withκ completing the proof

Remark 1 The right adjoint HomA(Bminus) to the restriction of scalars functor exists for anymorphism of rings ϕ A minusrarr B but as we have just seen this functor is not local unless thering morphism is surjective This explains why the right adjoint f to the direct image functor inalgebraic geometry essentially only exists for closed immersions

5

2 Flatness

Definition 3 Let A be a ring and M an A-module We say M is flat if the functor minus otimesA M AMod minusrarr AMod is exact (equivalently M otimesA minus is exact) Equivalently M is flat if wheneverwe have an injective morphism of modules N minusrarr N prime the morphism N otimesA M minusrarr N prime otimesA M isinjective This property is stable under isomorphism

We say M is faithfully flat if a morphism N minusrarr N prime is injective if and only if N otimesA M minusrarrN prime otimesAM is injective This property is also stable under isomorphism An A-algebra A minusrarr B isflat if B is a flat A-module and we say A minusrarr B is a flat morphism

Example 1 Nonzero free modules are faithfully flat

Lemma 16 We have the following fundamental properties of flatness

bull Transitivity If φ A minusrarr B is a flat morphism of rings and N a flat B-module then N isalso flat over A

bull Change of Base If φ A minusrarr B is a morphism of rings and M is a flat A-module thenM otimesA B is a flat B-module

bull Localisation If A is a ring and S a multiplicatively closed subset then Sminus1A is flat over A

Proof The second and third claims are done in our Atiyah amp Macdonald notes To prove the firstclaim let M minusrarr M prime be a monomorphism of A-modules and consider the following commutativediagram of abelian groups

M otimesA N

M prime otimesA N

M otimesA (B otimesB N)

M prime otimesA (B otimesB N)

(M otimesA B)otimesB N (M prime otimesA B)otimesB N

Since B is a flat A-module and N is a flat B-module the bottom row is injective hence so is thetop row

Lemma 17 Let φ A minusrarr B be a morphism of rings and N a B-module which is flat over A IfS is a multiplicatively closed subset of B then Sminus1N is flat over A In particular any localisationof a flat A-module is flat

Proof If M minusrarrM prime is a monomorphism of A-modules then we have a commutative diagram

M otimesA Sminus1N

M prime otimesA Sminus1N

(M otimesA N)otimesB Sminus1B (M prime otimesA N)otimesB Sminus1B

The bottom row is clearly injective and hence so is the top row which shows that Sminus1N is flatover A

Lemma 18 Let A be a ring and MN flat A-modules Then M otimesA N is also flat over A

Lemma 19 Let φ A minusrarr B be a flat morphism of rings and S a multiplicatively closed subsetof A Then T = φ(S) is a multiplicatively closed subset of B and Tminus1B is flat over Sminus1A

Proof This follows from Lemma 7 and stability of flatness under base change

6

Lemma 20 Let A minusrarr B be a morphism of rings Then the functor minusotimesAB AMod minusrarr BModpreserves projectives

Proof The functor minusotimesA B is left adjoint to the restriction of scalars functor This latter functoris clearly exact so since any functor with an exact right adjoint must preserve projectives P otimesABis a projective B-module for any projective A-module P

Lemma 21 Let A minusrarr B be a flat morphism of rings If I is an injective B-module then it isalso an injective A-module

Proof The restriction of scalars functor has an exact left adjoint minus otimesA B AMod minusrarr BModand therefore preserves injectives

Lemma 22 Let φ A minusrarr B be a flat morphism of rings and let MN be A-modules Then thereis an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (MotimesABNotimesAB) If A is noetherianand M finitely generated over A there is an isomorphism of B-modules ExtiA(MN) otimesA B sim=ExtiB(M otimesA BN otimesA B)

Proof Let X middot middot middot minusrarr X1 minusrarr X0 minusrarr M minusrarr 0 be a projective resolution of the A-module M Since B is flat the sequence

X otimesA B middot middot middot minusrarr X1 otimesA B minusrarr X0 otimesA B minusrarrM otimesA B minusrarr 0

is a projective resolution of M otimesA B The chain complex of B-modules (X otimesA B)otimesB (B otimesA N)is isomorphic to (X otimesA N)otimesA B The exact functor minusotimesA B commutes with taking homology sothere is an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (M otimesABN otimesAB) as required

If A is noetherian andM finitely generated we can assume that theXi are finite free A-modulesThen ExtiA(MN) is the i-cohomology module of the sequence

0 minusrarr Hom(X0 N) minusrarr Hom(X1 N) minusrarr Hom(X2 N) minusrarr middot middot middot

Since tensoring with B is exact ExtiA(MN) otimesA B is isomorphic as a B-module to the i-thcohomology of the following sequence

0 minusrarr Hom(X0 N)otimesA B minusrarr Hom(X1 N)otimesA B minusrarr middot middot middot

After a bit of work we see that this cochain complex is isomorphic to HomB(X otimesA BN otimesA B)and the i-th cohomology of this complex is ExtiB(M otimesA BN otimesA B) as required

In particular for a ring A and prime ideal p sube A we have isomorphisms of Ap-modules for i ge 0

TorAp

i (Mp Np) sim= TorAi (MN)p

ExtiAp(Mp Np) sim= ExtiA(MN)p

the latter being valid for A noetherian and M finitely generated

Lemma 23 Let A be a ring and M an A-module Then the following are equivalent

(i) M is a flat A-module

(ii) Mp is a flat Ap-module for each prime ideal p

(iii) Mm is a flat Am-module for each maximal ideal m

Proof See [AM69] or any book on commutative algebra

Proposition 24 Let (Am k) be a local ring and M an A-module Suppose that either m isnilpotent or M is finitely generated over A Then M is free hArr M is projective hArr M is flat

7

Proof It suffices to show that if M is flat then it is free We prove that any minimal basis of Mis a basis of M If MmM = 0 then M = 0 and M is trivially free Otherwise it suffices to showthat if x1 xn isinM are elements whose images in MmM = M otimesA k are linearly independentover k then they are linearly independent over A We use induction on n For n = 1 let ax = 0Then there exist y1 yr isin M and b1 br isin A such that abi = 0 for all i and x =

sumbiyi

Since x+ mM 6= 0 not all bi are in m Suppose b1 isin m Then b1 is a unit in A and ab1 = 0 hencea = 0

Suppose n gt 1 andsumni=1 aixi = 0 Then there exist y1 yr isin M and bij isin A(1 le j le r)

such that xi =sumj bijyj and

sumi aibij = 0 Since xn isin mM we have bnj isin m for at least one j

Since a1b1j + middot middot middot+ anbnj = 0 and bnj is a unit we have

an =nminus1sumi=1

ciai ci = minusbijbnj

Then

0 =nsumi=1

aixi = a1(x1 + c1xn) + middot middot middot+ anminus1(xnminus1 + cnminus1xn)

Since the residues of x1+c1xn xnminus1+cnminus1xn are linearly independent over k by the inductivehypothesis we get a1 = middot middot middot = anminus1 = 0 and an =

sumciai = 0

Corollary 25 Let A be a ring and M a finitely generated A-module Then the following areequivalent

(i) M is a flat A-module

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof This is immediate from the previous two results

Proposition 26 Let A be a ring and M a finitely presented A-module Then M is flat if andonly if it is projective

Proof See Stenstrom Chapter 1 Corollary 115

Corollary 27 Let A be a noetherian ring M a finitely generated A-module Then the followingconditions are equivalent

(i) M is projective

(ii) M is flat

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof Since A is noetherian M is finitely presented so (i) hArr (ii) is an immediate consequenceof Proposition 26 The rest of the proof follows from Corollary 25

Lemma 28 Let A minusrarr B be a flat morphism of rings and let I J be ideals of A Then (IcapJ)B =IB cap JB and (I J)B = (IB JB) if J is finitely generated

Proof Consider the exact sequence of A-modules

I cap J minusrarr A minusrarr AI oplusAJ

Tensoring with B we get an exact sequence

(I cap J)otimesA B = (I cap J)B minusrarr B minusrarr BIB oplusBJB

8

This means (I cap J)B = IB cap JB For the second claim suppose firstly that J is a principal idealaA and use the exact sequence

(I aA) i Af AI

where i is the injection and f(x) = ax+I Tensoring withB we get the formula (I a)B = (IB a)In the general case if J = (a1 an) we have (I J) =

⋂i(I ai) so that

(I J)B =⋂

(I ai)B =⋂

(IB ai) = (IB JB)

Example 2 Let A = k[x y] be a polynomial ring over a field k and put B = A(x) sim= k[y] ThenB is not flat over A since y isin A is regular but is not regular on B Let I = (x+ y) and J = (y)Then I cap J = (xy + y2) and IB = JB = yB (I cap J)B = y2B 6= IB cap JB

Example 3 Let k be a field put A = k[x y] and let K be the quotient field of A Let B be thesubring k[x yx] of K (ie the k-subalgebra generated by x and z = yx) Then A sub B sub K LetI = xA J = yA Then I cap J = xyA and (I cap J)B = x2zB IB cap JB = xzB so B is not flat overA The map SpecB minusrarr SpecA corresponding to A minusrarr B is the projection to the (x y)-planeof the surface F xz = y in (x y z)-space Note F contains the whole z-axis so it does not lookldquoflatrdquo over the (x y)-plane

Proposition 29 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB there is a canonical isomorphism of Bp-modules natural in both variables

κ MpcapA otimesApcapANp minusrarr (M otimesA N)p

κ(msotimes nt) = (motimes n)ϕ(s)t

Proof Fix p isin SpecB and q = pcapA There is a canonical ring morphism Aq minusrarr Bp and we makeNp into an Aq-module using this morphism One checks that the following map is well-definedand Aq-bilinear

ε Mq timesNp minusrarr (M otimesA N)p

ε(ms nt) = (motimes n)ϕ(s)t

We show that in fact this is a tensor product of Aq-modules Let Z be an abelian group andψ Mq timesNp minusrarr Z an Aq-bilinear map

Mq timesNp

ε

ψ Z

(M otimesA N)p

φ

(1)

We have to define a morphism of abelian groups φ unique making this diagram commute Fors isin p we define an A-bilinear morphism φprimes M times N minusrarr Z by φprimes(mn) = ψ(m1 ns) Thisinduces a morphism of abelian groups

φprimeprimes M otimesA N minusrarr Z

φprimeprimes (motimes b) = ψ(m1 bs)

We make some observations about these morphisms

bull Suppose ws = wprimesprime in (M otimesA N)p with say w =sumimi otimes ni w

prime =sumim

primei otimes nprimei and t isin p

such that tsprimew = tswprime That issumimi otimes tsprimeni =

sumim

primei otimes tsnprimei Applying φprimeprimetssprime to both sides

of this equality gives φprimeprimes (w) = φprimeprimesprime(wprime)

9

bull For wswprimesprime isin (M otimesA N)p we have φprimeprimes (w) + φprimeprimesprime(wprime) = φprimeprimessprime(s

primew + swprime)

It follows that φ(ws) = φprimeprimes (w) gives a well-defined morphism of abelian groups φ (MotimesAN)p minusrarrZ which is clearly unique making (1) commute By uniqueness of the tensor product there is aninduced isomorphism of abelian groups κ Mq otimesAq Np minusrarr (M otimesA N)p with κ(ms otimes nt) =(m otimes n)ϕ(s)t One checks that this is a morphism of Bp-modules The inverse is defined by(motimes n)t 7rarr m1otimes nt Naturality in both variables is easily checked

Corollary 30 Let ϕ A minusrarr B be a morphism of rings M an A-module and p isin SpecB Thenthere is a canonical isomorphism of Bp-modules MpcapA otimesApcapA

Bp minusrarr (M otimesA B)p natural in M

We will not actually use the next result in these notes so the reader not familiar with homo-logical δ-functors can safely skip it Alternatively one can provide a proof by following the onegiven in Matsumura (the proof we give is more elegant provided you know about δ-functors)

Proposition 31 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB and i ge 0 there is a canonical isomorphism of Bp-modules natural inM

κi TorAi (NM)p minusrarr TorApcapA

i (NpMpcapA)

Proof Fix p isin SpecB and a B-module N and set q = p capA Then N is a B-A-bimodule and Np

is a Bp-Aq-bimodule so by (TORSection 51) the abelian group TorAi (NM) acquires a canonicalB-module structure and Tor

Aq

i (NpMq) acquires a canonical Bp-module structure for any A-module M and i ge 0 Using (TORLemma 14) and (DFDefinition 23) we have two homologicalδ-functors between AMod and BpMod

TorAi (Nminus)pige0 TorAq

i (Np (minus)q)ige0

For i gt 0 these functors all vanish on free A-modules so by (DFTheorem 74) both δ-functors areuniversal For i = 0 we have the canonical natural equivalence of Proposition 29

κ0 TorA0 (Nminus)psim= (N otimesA minus)p

sim= Np otimesAq (minus)qsim= Tor

Aq

0 (Np (minus)q)

By universality this lifts to a canonical isomorphism of homological δ-functors κ In particularfor each i ge 0 we have a canonical natural equivalence κi TorAi (Nminus)p minusrarr Tor

Aq

i (Np (minus)q) asrequired

We know from Lemma 23 that flatness is a local property We are now ready to show thatrelative flatness (ie flatness with respect to a morphism of rings) is also local This is particularlyimportant in algebraic geometry The reader who skipped Proposition 31 will also have to skipthe implication (iii) rArr (i) in the next result but this will not affect their ability to read the restof these notes

Corollary 32 Let A minusrarr B be a morphism of rings and N a B-module Then the followingconditions are equivalent

(i) N is flat over A

(ii) Np is flat over ApcapA for all prime ideals p of B

(iii) Nm is flat over AmcapA for all maximal ideals m of B

Proof (i) rArr (ii) If N is flat over A then NpcapA is flat over ApcapA for any prime p of B By anargument similar to the one given in Lemma 19 we see that Np is isomorphic as a BpcapA-moduleto a localisation of NpcapA Applying Lemma 17 to the ring morphism ApcapA minusrarr BpcapA we see thatNp is flat over ApcapA as required (ii) rArr (iii) is trivial (iii) rArr (i) For every A-module M andmaximal ideal m of B we have by Proposition 31

TorA1 (NM)msim= TorAmcapA

1 (NmMmcapA) = 0

since by assumption Nm is flat over AmcapA Therefore TorA1 (NM) = 0 for every A-module M which implies that N is flat over A and completes the proof

10

It is easy to check that this is an isomorphism An orthogonal set of idempotents in a ring A is aset e1 er with 1 = e1 + middot middot middot+ er e2i = ei and eiej = 0 for i 6= j If A =

prodni=1Ai is a product

of rings then the elements e1 = (1 0 0) en = (0 0 1) are clearly such a setConversely if e1 er is an orthogonal set of idempotents in a ring A then the ideal eiA

becomes a ring with identity ei The map

A minusrarr e1Atimes middot middot middot times erA

a 7rarr (e1a era)

is a ring isomorphism

Proposition 14 Any nonzero artinian ring A is a finite direct product of local artinian rings

Proof See [Eis95] Corollary 216 This shows that there is a finite list of maximal ideals m1 mn

(allowing repeats) and a ring isomorphism A minusrarrprodni=1Ami

defined by a 7rarr (a1 a1)

Proposition 15 Let ϕ A minusrarr B be a surjective morphism of rings M an A-module andp isin SpecB There is a canonical morphism of Bp-modules natural in M

κ HomA(BM)p minusrarr HomAϕminus1p(BpMϕminus1p)

κ(us)(bt) = u(b)ϕminus1(st)

If A is noetherian this is an isomorphism

Proof Let a morphism of A-modules u B minusrarrM s t isin B p and b isin B be given Choose k isin Awith ϕ(k) = st We claim the fraction u(b)k isin Mϕminus1p doesnrsquot depend on the choice of k If wehave ϕ(l) = st also then

ku(b) = u(kb) = u(ϕ(k)b) = u(ϕ(l)b) = u(lb) = lu(b)

so u(b)l = u(b)k as claimed Throughout the proof given x isin B we write ϕminus1(x) for anarbitrary element in the inverse image of x One checks the result does not depend on this choiceWe can now define a morphism of Bp-modules κ(us)(bt) = u(b)ϕminus1(st) which one checks iswell-defined and natural in M

Now assume that A is noetherian In showing that κ is an isomorphism we may as well assumeϕ is the canonical projection A minusrarr Aa for some ideal a In that case the prime ideal p is qa forsome prime q of A containing a and if we set S = A q and T = ϕ(S) we have by Lemma 7 anisomorphism

HomA(BM)p = Tminus1HomA(AaM)sim= Sminus1HomA(AaM)sim= HomSminus1A(Sminus1(Aa) Sminus1M)sim= HomSminus1A(Tminus1(Aa) Sminus1M)= HomAϕminus1p

(BpMϕminus1p)

We have use the fact that A is noetherian to see that Aa is finitely presented so we have thesecond isomorphism in the above sequence One checks easily that this isomorphism agrees withκ completing the proof

Remark 1 The right adjoint HomA(Bminus) to the restriction of scalars functor exists for anymorphism of rings ϕ A minusrarr B but as we have just seen this functor is not local unless thering morphism is surjective This explains why the right adjoint f to the direct image functor inalgebraic geometry essentially only exists for closed immersions

5

2 Flatness

Definition 3 Let A be a ring and M an A-module We say M is flat if the functor minus otimesA M AMod minusrarr AMod is exact (equivalently M otimesA minus is exact) Equivalently M is flat if wheneverwe have an injective morphism of modules N minusrarr N prime the morphism N otimesA M minusrarr N prime otimesA M isinjective This property is stable under isomorphism

We say M is faithfully flat if a morphism N minusrarr N prime is injective if and only if N otimesA M minusrarrN prime otimesAM is injective This property is also stable under isomorphism An A-algebra A minusrarr B isflat if B is a flat A-module and we say A minusrarr B is a flat morphism

Example 1 Nonzero free modules are faithfully flat

Lemma 16 We have the following fundamental properties of flatness

bull Transitivity If φ A minusrarr B is a flat morphism of rings and N a flat B-module then N isalso flat over A

bull Change of Base If φ A minusrarr B is a morphism of rings and M is a flat A-module thenM otimesA B is a flat B-module

bull Localisation If A is a ring and S a multiplicatively closed subset then Sminus1A is flat over A

Proof The second and third claims are done in our Atiyah amp Macdonald notes To prove the firstclaim let M minusrarr M prime be a monomorphism of A-modules and consider the following commutativediagram of abelian groups

M otimesA N

M prime otimesA N

M otimesA (B otimesB N)

M prime otimesA (B otimesB N)

(M otimesA B)otimesB N (M prime otimesA B)otimesB N

Since B is a flat A-module and N is a flat B-module the bottom row is injective hence so is thetop row

Lemma 17 Let φ A minusrarr B be a morphism of rings and N a B-module which is flat over A IfS is a multiplicatively closed subset of B then Sminus1N is flat over A In particular any localisationof a flat A-module is flat

Proof If M minusrarrM prime is a monomorphism of A-modules then we have a commutative diagram

M otimesA Sminus1N

M prime otimesA Sminus1N

(M otimesA N)otimesB Sminus1B (M prime otimesA N)otimesB Sminus1B

The bottom row is clearly injective and hence so is the top row which shows that Sminus1N is flatover A

Lemma 18 Let A be a ring and MN flat A-modules Then M otimesA N is also flat over A

Lemma 19 Let φ A minusrarr B be a flat morphism of rings and S a multiplicatively closed subsetof A Then T = φ(S) is a multiplicatively closed subset of B and Tminus1B is flat over Sminus1A

Proof This follows from Lemma 7 and stability of flatness under base change

6

Lemma 20 Let A minusrarr B be a morphism of rings Then the functor minusotimesAB AMod minusrarr BModpreserves projectives

Proof The functor minusotimesA B is left adjoint to the restriction of scalars functor This latter functoris clearly exact so since any functor with an exact right adjoint must preserve projectives P otimesABis a projective B-module for any projective A-module P

Lemma 21 Let A minusrarr B be a flat morphism of rings If I is an injective B-module then it isalso an injective A-module

Proof The restriction of scalars functor has an exact left adjoint minus otimesA B AMod minusrarr BModand therefore preserves injectives

Lemma 22 Let φ A minusrarr B be a flat morphism of rings and let MN be A-modules Then thereis an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (MotimesABNotimesAB) If A is noetherianand M finitely generated over A there is an isomorphism of B-modules ExtiA(MN) otimesA B sim=ExtiB(M otimesA BN otimesA B)

Proof Let X middot middot middot minusrarr X1 minusrarr X0 minusrarr M minusrarr 0 be a projective resolution of the A-module M Since B is flat the sequence

X otimesA B middot middot middot minusrarr X1 otimesA B minusrarr X0 otimesA B minusrarrM otimesA B minusrarr 0

is a projective resolution of M otimesA B The chain complex of B-modules (X otimesA B)otimesB (B otimesA N)is isomorphic to (X otimesA N)otimesA B The exact functor minusotimesA B commutes with taking homology sothere is an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (M otimesABN otimesAB) as required

If A is noetherian andM finitely generated we can assume that theXi are finite free A-modulesThen ExtiA(MN) is the i-cohomology module of the sequence

0 minusrarr Hom(X0 N) minusrarr Hom(X1 N) minusrarr Hom(X2 N) minusrarr middot middot middot

Since tensoring with B is exact ExtiA(MN) otimesA B is isomorphic as a B-module to the i-thcohomology of the following sequence

0 minusrarr Hom(X0 N)otimesA B minusrarr Hom(X1 N)otimesA B minusrarr middot middot middot

After a bit of work we see that this cochain complex is isomorphic to HomB(X otimesA BN otimesA B)and the i-th cohomology of this complex is ExtiB(M otimesA BN otimesA B) as required

In particular for a ring A and prime ideal p sube A we have isomorphisms of Ap-modules for i ge 0

TorAp

i (Mp Np) sim= TorAi (MN)p

ExtiAp(Mp Np) sim= ExtiA(MN)p

the latter being valid for A noetherian and M finitely generated

Lemma 23 Let A be a ring and M an A-module Then the following are equivalent

(i) M is a flat A-module

(ii) Mp is a flat Ap-module for each prime ideal p

(iii) Mm is a flat Am-module for each maximal ideal m

Proof See [AM69] or any book on commutative algebra

Proposition 24 Let (Am k) be a local ring and M an A-module Suppose that either m isnilpotent or M is finitely generated over A Then M is free hArr M is projective hArr M is flat

7

Proof It suffices to show that if M is flat then it is free We prove that any minimal basis of Mis a basis of M If MmM = 0 then M = 0 and M is trivially free Otherwise it suffices to showthat if x1 xn isinM are elements whose images in MmM = M otimesA k are linearly independentover k then they are linearly independent over A We use induction on n For n = 1 let ax = 0Then there exist y1 yr isin M and b1 br isin A such that abi = 0 for all i and x =

sumbiyi

Since x+ mM 6= 0 not all bi are in m Suppose b1 isin m Then b1 is a unit in A and ab1 = 0 hencea = 0

Suppose n gt 1 andsumni=1 aixi = 0 Then there exist y1 yr isin M and bij isin A(1 le j le r)

such that xi =sumj bijyj and

sumi aibij = 0 Since xn isin mM we have bnj isin m for at least one j

Since a1b1j + middot middot middot+ anbnj = 0 and bnj is a unit we have

an =nminus1sumi=1

ciai ci = minusbijbnj

Then

0 =nsumi=1

aixi = a1(x1 + c1xn) + middot middot middot+ anminus1(xnminus1 + cnminus1xn)

Since the residues of x1+c1xn xnminus1+cnminus1xn are linearly independent over k by the inductivehypothesis we get a1 = middot middot middot = anminus1 = 0 and an =

sumciai = 0

Corollary 25 Let A be a ring and M a finitely generated A-module Then the following areequivalent

(i) M is a flat A-module

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof This is immediate from the previous two results

Proposition 26 Let A be a ring and M a finitely presented A-module Then M is flat if andonly if it is projective

Proof See Stenstrom Chapter 1 Corollary 115

Corollary 27 Let A be a noetherian ring M a finitely generated A-module Then the followingconditions are equivalent

(i) M is projective

(ii) M is flat

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof Since A is noetherian M is finitely presented so (i) hArr (ii) is an immediate consequenceof Proposition 26 The rest of the proof follows from Corollary 25

Lemma 28 Let A minusrarr B be a flat morphism of rings and let I J be ideals of A Then (IcapJ)B =IB cap JB and (I J)B = (IB JB) if J is finitely generated

Proof Consider the exact sequence of A-modules

I cap J minusrarr A minusrarr AI oplusAJ

Tensoring with B we get an exact sequence

(I cap J)otimesA B = (I cap J)B minusrarr B minusrarr BIB oplusBJB

8

This means (I cap J)B = IB cap JB For the second claim suppose firstly that J is a principal idealaA and use the exact sequence

(I aA) i Af AI

where i is the injection and f(x) = ax+I Tensoring withB we get the formula (I a)B = (IB a)In the general case if J = (a1 an) we have (I J) =

⋂i(I ai) so that

(I J)B =⋂

(I ai)B =⋂

(IB ai) = (IB JB)

Example 2 Let A = k[x y] be a polynomial ring over a field k and put B = A(x) sim= k[y] ThenB is not flat over A since y isin A is regular but is not regular on B Let I = (x+ y) and J = (y)Then I cap J = (xy + y2) and IB = JB = yB (I cap J)B = y2B 6= IB cap JB

Example 3 Let k be a field put A = k[x y] and let K be the quotient field of A Let B be thesubring k[x yx] of K (ie the k-subalgebra generated by x and z = yx) Then A sub B sub K LetI = xA J = yA Then I cap J = xyA and (I cap J)B = x2zB IB cap JB = xzB so B is not flat overA The map SpecB minusrarr SpecA corresponding to A minusrarr B is the projection to the (x y)-planeof the surface F xz = y in (x y z)-space Note F contains the whole z-axis so it does not lookldquoflatrdquo over the (x y)-plane

Proposition 29 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB there is a canonical isomorphism of Bp-modules natural in both variables

κ MpcapA otimesApcapANp minusrarr (M otimesA N)p

κ(msotimes nt) = (motimes n)ϕ(s)t

Proof Fix p isin SpecB and q = pcapA There is a canonical ring morphism Aq minusrarr Bp and we makeNp into an Aq-module using this morphism One checks that the following map is well-definedand Aq-bilinear

ε Mq timesNp minusrarr (M otimesA N)p

ε(ms nt) = (motimes n)ϕ(s)t

We show that in fact this is a tensor product of Aq-modules Let Z be an abelian group andψ Mq timesNp minusrarr Z an Aq-bilinear map

Mq timesNp

ε

ψ Z

(M otimesA N)p

φ

(1)

We have to define a morphism of abelian groups φ unique making this diagram commute Fors isin p we define an A-bilinear morphism φprimes M times N minusrarr Z by φprimes(mn) = ψ(m1 ns) Thisinduces a morphism of abelian groups

φprimeprimes M otimesA N minusrarr Z

φprimeprimes (motimes b) = ψ(m1 bs)

We make some observations about these morphisms

bull Suppose ws = wprimesprime in (M otimesA N)p with say w =sumimi otimes ni w

prime =sumim

primei otimes nprimei and t isin p

such that tsprimew = tswprime That issumimi otimes tsprimeni =

sumim

primei otimes tsnprimei Applying φprimeprimetssprime to both sides

of this equality gives φprimeprimes (w) = φprimeprimesprime(wprime)

9

bull For wswprimesprime isin (M otimesA N)p we have φprimeprimes (w) + φprimeprimesprime(wprime) = φprimeprimessprime(s

primew + swprime)

It follows that φ(ws) = φprimeprimes (w) gives a well-defined morphism of abelian groups φ (MotimesAN)p minusrarrZ which is clearly unique making (1) commute By uniqueness of the tensor product there is aninduced isomorphism of abelian groups κ Mq otimesAq Np minusrarr (M otimesA N)p with κ(ms otimes nt) =(m otimes n)ϕ(s)t One checks that this is a morphism of Bp-modules The inverse is defined by(motimes n)t 7rarr m1otimes nt Naturality in both variables is easily checked

Corollary 30 Let ϕ A minusrarr B be a morphism of rings M an A-module and p isin SpecB Thenthere is a canonical isomorphism of Bp-modules MpcapA otimesApcapA

Bp minusrarr (M otimesA B)p natural in M

We will not actually use the next result in these notes so the reader not familiar with homo-logical δ-functors can safely skip it Alternatively one can provide a proof by following the onegiven in Matsumura (the proof we give is more elegant provided you know about δ-functors)

Proposition 31 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB and i ge 0 there is a canonical isomorphism of Bp-modules natural inM

κi TorAi (NM)p minusrarr TorApcapA

i (NpMpcapA)

Proof Fix p isin SpecB and a B-module N and set q = p capA Then N is a B-A-bimodule and Np

is a Bp-Aq-bimodule so by (TORSection 51) the abelian group TorAi (NM) acquires a canonicalB-module structure and Tor

Aq

i (NpMq) acquires a canonical Bp-module structure for any A-module M and i ge 0 Using (TORLemma 14) and (DFDefinition 23) we have two homologicalδ-functors between AMod and BpMod

TorAi (Nminus)pige0 TorAq

i (Np (minus)q)ige0

For i gt 0 these functors all vanish on free A-modules so by (DFTheorem 74) both δ-functors areuniversal For i = 0 we have the canonical natural equivalence of Proposition 29

κ0 TorA0 (Nminus)psim= (N otimesA minus)p

sim= Np otimesAq (minus)qsim= Tor

Aq

0 (Np (minus)q)

By universality this lifts to a canonical isomorphism of homological δ-functors κ In particularfor each i ge 0 we have a canonical natural equivalence κi TorAi (Nminus)p minusrarr Tor

Aq

i (Np (minus)q) asrequired

We know from Lemma 23 that flatness is a local property We are now ready to show thatrelative flatness (ie flatness with respect to a morphism of rings) is also local This is particularlyimportant in algebraic geometry The reader who skipped Proposition 31 will also have to skipthe implication (iii) rArr (i) in the next result but this will not affect their ability to read the restof these notes

Corollary 32 Let A minusrarr B be a morphism of rings and N a B-module Then the followingconditions are equivalent

(i) N is flat over A

(ii) Np is flat over ApcapA for all prime ideals p of B

(iii) Nm is flat over AmcapA for all maximal ideals m of B

Proof (i) rArr (ii) If N is flat over A then NpcapA is flat over ApcapA for any prime p of B By anargument similar to the one given in Lemma 19 we see that Np is isomorphic as a BpcapA-moduleto a localisation of NpcapA Applying Lemma 17 to the ring morphism ApcapA minusrarr BpcapA we see thatNp is flat over ApcapA as required (ii) rArr (iii) is trivial (iii) rArr (i) For every A-module M andmaximal ideal m of B we have by Proposition 31

TorA1 (NM)msim= TorAmcapA

1 (NmMmcapA) = 0

since by assumption Nm is flat over AmcapA Therefore TorA1 (NM) = 0 for every A-module M which implies that N is flat over A and completes the proof

10

2 Flatness

Definition 3 Let A be a ring and M an A-module We say M is flat if the functor minus otimesA M AMod minusrarr AMod is exact (equivalently M otimesA minus is exact) Equivalently M is flat if wheneverwe have an injective morphism of modules N minusrarr N prime the morphism N otimesA M minusrarr N prime otimesA M isinjective This property is stable under isomorphism

We say M is faithfully flat if a morphism N minusrarr N prime is injective if and only if N otimesA M minusrarrN prime otimesAM is injective This property is also stable under isomorphism An A-algebra A minusrarr B isflat if B is a flat A-module and we say A minusrarr B is a flat morphism

Example 1 Nonzero free modules are faithfully flat

Lemma 16 We have the following fundamental properties of flatness

bull Transitivity If φ A minusrarr B is a flat morphism of rings and N a flat B-module then N isalso flat over A

bull Change of Base If φ A minusrarr B is a morphism of rings and M is a flat A-module thenM otimesA B is a flat B-module

bull Localisation If A is a ring and S a multiplicatively closed subset then Sminus1A is flat over A

Proof The second and third claims are done in our Atiyah amp Macdonald notes To prove the firstclaim let M minusrarr M prime be a monomorphism of A-modules and consider the following commutativediagram of abelian groups

M otimesA N

M prime otimesA N

M otimesA (B otimesB N)

M prime otimesA (B otimesB N)

(M otimesA B)otimesB N (M prime otimesA B)otimesB N

Since B is a flat A-module and N is a flat B-module the bottom row is injective hence so is thetop row

Lemma 17 Let φ A minusrarr B be a morphism of rings and N a B-module which is flat over A IfS is a multiplicatively closed subset of B then Sminus1N is flat over A In particular any localisationof a flat A-module is flat

Proof If M minusrarrM prime is a monomorphism of A-modules then we have a commutative diagram

M otimesA Sminus1N

M prime otimesA Sminus1N

(M otimesA N)otimesB Sminus1B (M prime otimesA N)otimesB Sminus1B

The bottom row is clearly injective and hence so is the top row which shows that Sminus1N is flatover A

Lemma 18 Let A be a ring and MN flat A-modules Then M otimesA N is also flat over A

Lemma 19 Let φ A minusrarr B be a flat morphism of rings and S a multiplicatively closed subsetof A Then T = φ(S) is a multiplicatively closed subset of B and Tminus1B is flat over Sminus1A

Proof This follows from Lemma 7 and stability of flatness under base change

6

Lemma 20 Let A minusrarr B be a morphism of rings Then the functor minusotimesAB AMod minusrarr BModpreserves projectives

Proof The functor minusotimesA B is left adjoint to the restriction of scalars functor This latter functoris clearly exact so since any functor with an exact right adjoint must preserve projectives P otimesABis a projective B-module for any projective A-module P

Lemma 21 Let A minusrarr B be a flat morphism of rings If I is an injective B-module then it isalso an injective A-module

Proof The restriction of scalars functor has an exact left adjoint minus otimesA B AMod minusrarr BModand therefore preserves injectives

Lemma 22 Let φ A minusrarr B be a flat morphism of rings and let MN be A-modules Then thereis an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (MotimesABNotimesAB) If A is noetherianand M finitely generated over A there is an isomorphism of B-modules ExtiA(MN) otimesA B sim=ExtiB(M otimesA BN otimesA B)

Proof Let X middot middot middot minusrarr X1 minusrarr X0 minusrarr M minusrarr 0 be a projective resolution of the A-module M Since B is flat the sequence

X otimesA B middot middot middot minusrarr X1 otimesA B minusrarr X0 otimesA B minusrarrM otimesA B minusrarr 0

is a projective resolution of M otimesA B The chain complex of B-modules (X otimesA B)otimesB (B otimesA N)is isomorphic to (X otimesA N)otimesA B The exact functor minusotimesA B commutes with taking homology sothere is an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (M otimesABN otimesAB) as required

If A is noetherian andM finitely generated we can assume that theXi are finite free A-modulesThen ExtiA(MN) is the i-cohomology module of the sequence

0 minusrarr Hom(X0 N) minusrarr Hom(X1 N) minusrarr Hom(X2 N) minusrarr middot middot middot

Since tensoring with B is exact ExtiA(MN) otimesA B is isomorphic as a B-module to the i-thcohomology of the following sequence

0 minusrarr Hom(X0 N)otimesA B minusrarr Hom(X1 N)otimesA B minusrarr middot middot middot

After a bit of work we see that this cochain complex is isomorphic to HomB(X otimesA BN otimesA B)and the i-th cohomology of this complex is ExtiB(M otimesA BN otimesA B) as required

In particular for a ring A and prime ideal p sube A we have isomorphisms of Ap-modules for i ge 0

TorAp

i (Mp Np) sim= TorAi (MN)p

ExtiAp(Mp Np) sim= ExtiA(MN)p

the latter being valid for A noetherian and M finitely generated

Lemma 23 Let A be a ring and M an A-module Then the following are equivalent

(i) M is a flat A-module

(ii) Mp is a flat Ap-module for each prime ideal p

(iii) Mm is a flat Am-module for each maximal ideal m

Proof See [AM69] or any book on commutative algebra

Proposition 24 Let (Am k) be a local ring and M an A-module Suppose that either m isnilpotent or M is finitely generated over A Then M is free hArr M is projective hArr M is flat

7

Proof It suffices to show that if M is flat then it is free We prove that any minimal basis of Mis a basis of M If MmM = 0 then M = 0 and M is trivially free Otherwise it suffices to showthat if x1 xn isinM are elements whose images in MmM = M otimesA k are linearly independentover k then they are linearly independent over A We use induction on n For n = 1 let ax = 0Then there exist y1 yr isin M and b1 br isin A such that abi = 0 for all i and x =

sumbiyi

Since x+ mM 6= 0 not all bi are in m Suppose b1 isin m Then b1 is a unit in A and ab1 = 0 hencea = 0

Suppose n gt 1 andsumni=1 aixi = 0 Then there exist y1 yr isin M and bij isin A(1 le j le r)

such that xi =sumj bijyj and

sumi aibij = 0 Since xn isin mM we have bnj isin m for at least one j

Since a1b1j + middot middot middot+ anbnj = 0 and bnj is a unit we have

an =nminus1sumi=1

ciai ci = minusbijbnj

Then

0 =nsumi=1

aixi = a1(x1 + c1xn) + middot middot middot+ anminus1(xnminus1 + cnminus1xn)

Since the residues of x1+c1xn xnminus1+cnminus1xn are linearly independent over k by the inductivehypothesis we get a1 = middot middot middot = anminus1 = 0 and an =

sumciai = 0

Corollary 25 Let A be a ring and M a finitely generated A-module Then the following areequivalent

(i) M is a flat A-module

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof This is immediate from the previous two results

Proposition 26 Let A be a ring and M a finitely presented A-module Then M is flat if andonly if it is projective

Proof See Stenstrom Chapter 1 Corollary 115

Corollary 27 Let A be a noetherian ring M a finitely generated A-module Then the followingconditions are equivalent

(i) M is projective

(ii) M is flat

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof Since A is noetherian M is finitely presented so (i) hArr (ii) is an immediate consequenceof Proposition 26 The rest of the proof follows from Corollary 25

Lemma 28 Let A minusrarr B be a flat morphism of rings and let I J be ideals of A Then (IcapJ)B =IB cap JB and (I J)B = (IB JB) if J is finitely generated

Proof Consider the exact sequence of A-modules

I cap J minusrarr A minusrarr AI oplusAJ

Tensoring with B we get an exact sequence

(I cap J)otimesA B = (I cap J)B minusrarr B minusrarr BIB oplusBJB

8

This means (I cap J)B = IB cap JB For the second claim suppose firstly that J is a principal idealaA and use the exact sequence

(I aA) i Af AI

where i is the injection and f(x) = ax+I Tensoring withB we get the formula (I a)B = (IB a)In the general case if J = (a1 an) we have (I J) =

⋂i(I ai) so that

(I J)B =⋂

(I ai)B =⋂

(IB ai) = (IB JB)

Example 2 Let A = k[x y] be a polynomial ring over a field k and put B = A(x) sim= k[y] ThenB is not flat over A since y isin A is regular but is not regular on B Let I = (x+ y) and J = (y)Then I cap J = (xy + y2) and IB = JB = yB (I cap J)B = y2B 6= IB cap JB

Example 3 Let k be a field put A = k[x y] and let K be the quotient field of A Let B be thesubring k[x yx] of K (ie the k-subalgebra generated by x and z = yx) Then A sub B sub K LetI = xA J = yA Then I cap J = xyA and (I cap J)B = x2zB IB cap JB = xzB so B is not flat overA The map SpecB minusrarr SpecA corresponding to A minusrarr B is the projection to the (x y)-planeof the surface F xz = y in (x y z)-space Note F contains the whole z-axis so it does not lookldquoflatrdquo over the (x y)-plane

Proposition 29 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB there is a canonical isomorphism of Bp-modules natural in both variables

κ MpcapA otimesApcapANp minusrarr (M otimesA N)p

κ(msotimes nt) = (motimes n)ϕ(s)t

Proof Fix p isin SpecB and q = pcapA There is a canonical ring morphism Aq minusrarr Bp and we makeNp into an Aq-module using this morphism One checks that the following map is well-definedand Aq-bilinear

ε Mq timesNp minusrarr (M otimesA N)p

ε(ms nt) = (motimes n)ϕ(s)t

We show that in fact this is a tensor product of Aq-modules Let Z be an abelian group andψ Mq timesNp minusrarr Z an Aq-bilinear map

Mq timesNp

ε

ψ Z

(M otimesA N)p

φ

(1)

We have to define a morphism of abelian groups φ unique making this diagram commute Fors isin p we define an A-bilinear morphism φprimes M times N minusrarr Z by φprimes(mn) = ψ(m1 ns) Thisinduces a morphism of abelian groups

φprimeprimes M otimesA N minusrarr Z

φprimeprimes (motimes b) = ψ(m1 bs)

We make some observations about these morphisms

bull Suppose ws = wprimesprime in (M otimesA N)p with say w =sumimi otimes ni w

prime =sumim

primei otimes nprimei and t isin p

such that tsprimew = tswprime That issumimi otimes tsprimeni =

sumim

primei otimes tsnprimei Applying φprimeprimetssprime to both sides

of this equality gives φprimeprimes (w) = φprimeprimesprime(wprime)

9

bull For wswprimesprime isin (M otimesA N)p we have φprimeprimes (w) + φprimeprimesprime(wprime) = φprimeprimessprime(s

primew + swprime)

It follows that φ(ws) = φprimeprimes (w) gives a well-defined morphism of abelian groups φ (MotimesAN)p minusrarrZ which is clearly unique making (1) commute By uniqueness of the tensor product there is aninduced isomorphism of abelian groups κ Mq otimesAq Np minusrarr (M otimesA N)p with κ(ms otimes nt) =(m otimes n)ϕ(s)t One checks that this is a morphism of Bp-modules The inverse is defined by(motimes n)t 7rarr m1otimes nt Naturality in both variables is easily checked

Corollary 30 Let ϕ A minusrarr B be a morphism of rings M an A-module and p isin SpecB Thenthere is a canonical isomorphism of Bp-modules MpcapA otimesApcapA

Bp minusrarr (M otimesA B)p natural in M

We will not actually use the next result in these notes so the reader not familiar with homo-logical δ-functors can safely skip it Alternatively one can provide a proof by following the onegiven in Matsumura (the proof we give is more elegant provided you know about δ-functors)

Proposition 31 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB and i ge 0 there is a canonical isomorphism of Bp-modules natural inM

κi TorAi (NM)p minusrarr TorApcapA

i (NpMpcapA)

Proof Fix p isin SpecB and a B-module N and set q = p capA Then N is a B-A-bimodule and Np

is a Bp-Aq-bimodule so by (TORSection 51) the abelian group TorAi (NM) acquires a canonicalB-module structure and Tor

Aq

i (NpMq) acquires a canonical Bp-module structure for any A-module M and i ge 0 Using (TORLemma 14) and (DFDefinition 23) we have two homologicalδ-functors between AMod and BpMod

TorAi (Nminus)pige0 TorAq

i (Np (minus)q)ige0

For i gt 0 these functors all vanish on free A-modules so by (DFTheorem 74) both δ-functors areuniversal For i = 0 we have the canonical natural equivalence of Proposition 29

κ0 TorA0 (Nminus)psim= (N otimesA minus)p

sim= Np otimesAq (minus)qsim= Tor

Aq

0 (Np (minus)q)

By universality this lifts to a canonical isomorphism of homological δ-functors κ In particularfor each i ge 0 we have a canonical natural equivalence κi TorAi (Nminus)p minusrarr Tor

Aq

i (Np (minus)q) asrequired

We know from Lemma 23 that flatness is a local property We are now ready to show thatrelative flatness (ie flatness with respect to a morphism of rings) is also local This is particularlyimportant in algebraic geometry The reader who skipped Proposition 31 will also have to skipthe implication (iii) rArr (i) in the next result but this will not affect their ability to read the restof these notes

Corollary 32 Let A minusrarr B be a morphism of rings and N a B-module Then the followingconditions are equivalent

(i) N is flat over A

(ii) Np is flat over ApcapA for all prime ideals p of B

(iii) Nm is flat over AmcapA for all maximal ideals m of B

Proof (i) rArr (ii) If N is flat over A then NpcapA is flat over ApcapA for any prime p of B By anargument similar to the one given in Lemma 19 we see that Np is isomorphic as a BpcapA-moduleto a localisation of NpcapA Applying Lemma 17 to the ring morphism ApcapA minusrarr BpcapA we see thatNp is flat over ApcapA as required (ii) rArr (iii) is trivial (iii) rArr (i) For every A-module M andmaximal ideal m of B we have by Proposition 31

TorA1 (NM)msim= TorAmcapA

1 (NmMmcapA) = 0

since by assumption Nm is flat over AmcapA Therefore TorA1 (NM) = 0 for every A-module M which implies that N is flat over A and completes the proof

10

Lemma 20 Let A minusrarr B be a morphism of rings Then the functor minusotimesAB AMod minusrarr BModpreserves projectives

Proof The functor minusotimesA B is left adjoint to the restriction of scalars functor This latter functoris clearly exact so since any functor with an exact right adjoint must preserve projectives P otimesABis a projective B-module for any projective A-module P

Lemma 21 Let A minusrarr B be a flat morphism of rings If I is an injective B-module then it isalso an injective A-module

Proof The restriction of scalars functor has an exact left adjoint minus otimesA B AMod minusrarr BModand therefore preserves injectives

Lemma 22 Let φ A minusrarr B be a flat morphism of rings and let MN be A-modules Then thereis an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (MotimesABNotimesAB) If A is noetherianand M finitely generated over A there is an isomorphism of B-modules ExtiA(MN) otimesA B sim=ExtiB(M otimesA BN otimesA B)

Proof Let X middot middot middot minusrarr X1 minusrarr X0 minusrarr M minusrarr 0 be a projective resolution of the A-module M Since B is flat the sequence

X otimesA B middot middot middot minusrarr X1 otimesA B minusrarr X0 otimesA B minusrarrM otimesA B minusrarr 0

is a projective resolution of M otimesA B The chain complex of B-modules (X otimesA B)otimesB (B otimesA N)is isomorphic to (X otimesA N)otimesA B The exact functor minusotimesA B commutes with taking homology sothere is an isomorphism of B-modules TorAi (MN)otimesAB sim= TorBi (M otimesABN otimesAB) as required

If A is noetherian andM finitely generated we can assume that theXi are finite free A-modulesThen ExtiA(MN) is the i-cohomology module of the sequence

0 minusrarr Hom(X0 N) minusrarr Hom(X1 N) minusrarr Hom(X2 N) minusrarr middot middot middot

Since tensoring with B is exact ExtiA(MN) otimesA B is isomorphic as a B-module to the i-thcohomology of the following sequence

0 minusrarr Hom(X0 N)otimesA B minusrarr Hom(X1 N)otimesA B minusrarr middot middot middot

After a bit of work we see that this cochain complex is isomorphic to HomB(X otimesA BN otimesA B)and the i-th cohomology of this complex is ExtiB(M otimesA BN otimesA B) as required

In particular for a ring A and prime ideal p sube A we have isomorphisms of Ap-modules for i ge 0

TorAp

i (Mp Np) sim= TorAi (MN)p

ExtiAp(Mp Np) sim= ExtiA(MN)p

the latter being valid for A noetherian and M finitely generated

Lemma 23 Let A be a ring and M an A-module Then the following are equivalent

(i) M is a flat A-module

(ii) Mp is a flat Ap-module for each prime ideal p

(iii) Mm is a flat Am-module for each maximal ideal m

Proof See [AM69] or any book on commutative algebra

Proposition 24 Let (Am k) be a local ring and M an A-module Suppose that either m isnilpotent or M is finitely generated over A Then M is free hArr M is projective hArr M is flat

7

Proof It suffices to show that if M is flat then it is free We prove that any minimal basis of Mis a basis of M If MmM = 0 then M = 0 and M is trivially free Otherwise it suffices to showthat if x1 xn isinM are elements whose images in MmM = M otimesA k are linearly independentover k then they are linearly independent over A We use induction on n For n = 1 let ax = 0Then there exist y1 yr isin M and b1 br isin A such that abi = 0 for all i and x =

sumbiyi

Since x+ mM 6= 0 not all bi are in m Suppose b1 isin m Then b1 is a unit in A and ab1 = 0 hencea = 0

Suppose n gt 1 andsumni=1 aixi = 0 Then there exist y1 yr isin M and bij isin A(1 le j le r)

such that xi =sumj bijyj and

sumi aibij = 0 Since xn isin mM we have bnj isin m for at least one j

Since a1b1j + middot middot middot+ anbnj = 0 and bnj is a unit we have

an =nminus1sumi=1

ciai ci = minusbijbnj

Then

0 =nsumi=1

aixi = a1(x1 + c1xn) + middot middot middot+ anminus1(xnminus1 + cnminus1xn)

Since the residues of x1+c1xn xnminus1+cnminus1xn are linearly independent over k by the inductivehypothesis we get a1 = middot middot middot = anminus1 = 0 and an =

sumciai = 0

Corollary 25 Let A be a ring and M a finitely generated A-module Then the following areequivalent

(i) M is a flat A-module

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof This is immediate from the previous two results

Proposition 26 Let A be a ring and M a finitely presented A-module Then M is flat if andonly if it is projective

Proof See Stenstrom Chapter 1 Corollary 115

Corollary 27 Let A be a noetherian ring M a finitely generated A-module Then the followingconditions are equivalent

(i) M is projective

(ii) M is flat

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof Since A is noetherian M is finitely presented so (i) hArr (ii) is an immediate consequenceof Proposition 26 The rest of the proof follows from Corollary 25

Lemma 28 Let A minusrarr B be a flat morphism of rings and let I J be ideals of A Then (IcapJ)B =IB cap JB and (I J)B = (IB JB) if J is finitely generated

Proof Consider the exact sequence of A-modules

I cap J minusrarr A minusrarr AI oplusAJ

Tensoring with B we get an exact sequence

(I cap J)otimesA B = (I cap J)B minusrarr B minusrarr BIB oplusBJB

8

This means (I cap J)B = IB cap JB For the second claim suppose firstly that J is a principal idealaA and use the exact sequence

(I aA) i Af AI

where i is the injection and f(x) = ax+I Tensoring withB we get the formula (I a)B = (IB a)In the general case if J = (a1 an) we have (I J) =

⋂i(I ai) so that

(I J)B =⋂

(I ai)B =⋂

(IB ai) = (IB JB)

Example 2 Let A = k[x y] be a polynomial ring over a field k and put B = A(x) sim= k[y] ThenB is not flat over A since y isin A is regular but is not regular on B Let I = (x+ y) and J = (y)Then I cap J = (xy + y2) and IB = JB = yB (I cap J)B = y2B 6= IB cap JB

Example 3 Let k be a field put A = k[x y] and let K be the quotient field of A Let B be thesubring k[x yx] of K (ie the k-subalgebra generated by x and z = yx) Then A sub B sub K LetI = xA J = yA Then I cap J = xyA and (I cap J)B = x2zB IB cap JB = xzB so B is not flat overA The map SpecB minusrarr SpecA corresponding to A minusrarr B is the projection to the (x y)-planeof the surface F xz = y in (x y z)-space Note F contains the whole z-axis so it does not lookldquoflatrdquo over the (x y)-plane

Proposition 29 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB there is a canonical isomorphism of Bp-modules natural in both variables

κ MpcapA otimesApcapANp minusrarr (M otimesA N)p

κ(msotimes nt) = (motimes n)ϕ(s)t

Proof Fix p isin SpecB and q = pcapA There is a canonical ring morphism Aq minusrarr Bp and we makeNp into an Aq-module using this morphism One checks that the following map is well-definedand Aq-bilinear

ε Mq timesNp minusrarr (M otimesA N)p

ε(ms nt) = (motimes n)ϕ(s)t

We show that in fact this is a tensor product of Aq-modules Let Z be an abelian group andψ Mq timesNp minusrarr Z an Aq-bilinear map

Mq timesNp

ε

ψ Z

(M otimesA N)p

φ

(1)

We have to define a morphism of abelian groups φ unique making this diagram commute Fors isin p we define an A-bilinear morphism φprimes M times N minusrarr Z by φprimes(mn) = ψ(m1 ns) Thisinduces a morphism of abelian groups

φprimeprimes M otimesA N minusrarr Z

φprimeprimes (motimes b) = ψ(m1 bs)

We make some observations about these morphisms

bull Suppose ws = wprimesprime in (M otimesA N)p with say w =sumimi otimes ni w

prime =sumim

primei otimes nprimei and t isin p

such that tsprimew = tswprime That issumimi otimes tsprimeni =

sumim

primei otimes tsnprimei Applying φprimeprimetssprime to both sides

of this equality gives φprimeprimes (w) = φprimeprimesprime(wprime)

9

bull For wswprimesprime isin (M otimesA N)p we have φprimeprimes (w) + φprimeprimesprime(wprime) = φprimeprimessprime(s

primew + swprime)

It follows that φ(ws) = φprimeprimes (w) gives a well-defined morphism of abelian groups φ (MotimesAN)p minusrarrZ which is clearly unique making (1) commute By uniqueness of the tensor product there is aninduced isomorphism of abelian groups κ Mq otimesAq Np minusrarr (M otimesA N)p with κ(ms otimes nt) =(m otimes n)ϕ(s)t One checks that this is a morphism of Bp-modules The inverse is defined by(motimes n)t 7rarr m1otimes nt Naturality in both variables is easily checked

Corollary 30 Let ϕ A minusrarr B be a morphism of rings M an A-module and p isin SpecB Thenthere is a canonical isomorphism of Bp-modules MpcapA otimesApcapA

Bp minusrarr (M otimesA B)p natural in M

We will not actually use the next result in these notes so the reader not familiar with homo-logical δ-functors can safely skip it Alternatively one can provide a proof by following the onegiven in Matsumura (the proof we give is more elegant provided you know about δ-functors)

Proposition 31 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB and i ge 0 there is a canonical isomorphism of Bp-modules natural inM

κi TorAi (NM)p minusrarr TorApcapA

i (NpMpcapA)

Proof Fix p isin SpecB and a B-module N and set q = p capA Then N is a B-A-bimodule and Np

is a Bp-Aq-bimodule so by (TORSection 51) the abelian group TorAi (NM) acquires a canonicalB-module structure and Tor

Aq

i (NpMq) acquires a canonical Bp-module structure for any A-module M and i ge 0 Using (TORLemma 14) and (DFDefinition 23) we have two homologicalδ-functors between AMod and BpMod

TorAi (Nminus)pige0 TorAq

i (Np (minus)q)ige0

For i gt 0 these functors all vanish on free A-modules so by (DFTheorem 74) both δ-functors areuniversal For i = 0 we have the canonical natural equivalence of Proposition 29

κ0 TorA0 (Nminus)psim= (N otimesA minus)p

sim= Np otimesAq (minus)qsim= Tor

Aq

0 (Np (minus)q)

By universality this lifts to a canonical isomorphism of homological δ-functors κ In particularfor each i ge 0 we have a canonical natural equivalence κi TorAi (Nminus)p minusrarr Tor

Aq

i (Np (minus)q) asrequired

We know from Lemma 23 that flatness is a local property We are now ready to show thatrelative flatness (ie flatness with respect to a morphism of rings) is also local This is particularlyimportant in algebraic geometry The reader who skipped Proposition 31 will also have to skipthe implication (iii) rArr (i) in the next result but this will not affect their ability to read the restof these notes

Corollary 32 Let A minusrarr B be a morphism of rings and N a B-module Then the followingconditions are equivalent

(i) N is flat over A

(ii) Np is flat over ApcapA for all prime ideals p of B

(iii) Nm is flat over AmcapA for all maximal ideals m of B

Proof (i) rArr (ii) If N is flat over A then NpcapA is flat over ApcapA for any prime p of B By anargument similar to the one given in Lemma 19 we see that Np is isomorphic as a BpcapA-moduleto a localisation of NpcapA Applying Lemma 17 to the ring morphism ApcapA minusrarr BpcapA we see thatNp is flat over ApcapA as required (ii) rArr (iii) is trivial (iii) rArr (i) For every A-module M andmaximal ideal m of B we have by Proposition 31

TorA1 (NM)msim= TorAmcapA

1 (NmMmcapA) = 0

since by assumption Nm is flat over AmcapA Therefore TorA1 (NM) = 0 for every A-module M which implies that N is flat over A and completes the proof

10

Proof It suffices to show that if M is flat then it is free We prove that any minimal basis of Mis a basis of M If MmM = 0 then M = 0 and M is trivially free Otherwise it suffices to showthat if x1 xn isinM are elements whose images in MmM = M otimesA k are linearly independentover k then they are linearly independent over A We use induction on n For n = 1 let ax = 0Then there exist y1 yr isin M and b1 br isin A such that abi = 0 for all i and x =

sumbiyi

Since x+ mM 6= 0 not all bi are in m Suppose b1 isin m Then b1 is a unit in A and ab1 = 0 hencea = 0

Suppose n gt 1 andsumni=1 aixi = 0 Then there exist y1 yr isin M and bij isin A(1 le j le r)

such that xi =sumj bijyj and

sumi aibij = 0 Since xn isin mM we have bnj isin m for at least one j

Since a1b1j + middot middot middot+ anbnj = 0 and bnj is a unit we have

an =nminus1sumi=1

ciai ci = minusbijbnj

Then

0 =nsumi=1

aixi = a1(x1 + c1xn) + middot middot middot+ anminus1(xnminus1 + cnminus1xn)

Since the residues of x1+c1xn xnminus1+cnminus1xn are linearly independent over k by the inductivehypothesis we get a1 = middot middot middot = anminus1 = 0 and an =

sumciai = 0

Corollary 25 Let A be a ring and M a finitely generated A-module Then the following areequivalent

(i) M is a flat A-module

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof This is immediate from the previous two results

Proposition 26 Let A be a ring and M a finitely presented A-module Then M is flat if andonly if it is projective

Proof See Stenstrom Chapter 1 Corollary 115

Corollary 27 Let A be a noetherian ring M a finitely generated A-module Then the followingconditions are equivalent

(i) M is projective

(ii) M is flat

(ii) Mp is a free Ap-module for each prime ideal p

(iii) Mm is a free Am-module for each maximal ideal m

Proof Since A is noetherian M is finitely presented so (i) hArr (ii) is an immediate consequenceof Proposition 26 The rest of the proof follows from Corollary 25

Lemma 28 Let A minusrarr B be a flat morphism of rings and let I J be ideals of A Then (IcapJ)B =IB cap JB and (I J)B = (IB JB) if J is finitely generated

Proof Consider the exact sequence of A-modules

I cap J minusrarr A minusrarr AI oplusAJ

Tensoring with B we get an exact sequence

(I cap J)otimesA B = (I cap J)B minusrarr B minusrarr BIB oplusBJB

8

This means (I cap J)B = IB cap JB For the second claim suppose firstly that J is a principal idealaA and use the exact sequence

(I aA) i Af AI

where i is the injection and f(x) = ax+I Tensoring withB we get the formula (I a)B = (IB a)In the general case if J = (a1 an) we have (I J) =

⋂i(I ai) so that

(I J)B =⋂

(I ai)B =⋂

(IB ai) = (IB JB)

Example 2 Let A = k[x y] be a polynomial ring over a field k and put B = A(x) sim= k[y] ThenB is not flat over A since y isin A is regular but is not regular on B Let I = (x+ y) and J = (y)Then I cap J = (xy + y2) and IB = JB = yB (I cap J)B = y2B 6= IB cap JB

Example 3 Let k be a field put A = k[x y] and let K be the quotient field of A Let B be thesubring k[x yx] of K (ie the k-subalgebra generated by x and z = yx) Then A sub B sub K LetI = xA J = yA Then I cap J = xyA and (I cap J)B = x2zB IB cap JB = xzB so B is not flat overA The map SpecB minusrarr SpecA corresponding to A minusrarr B is the projection to the (x y)-planeof the surface F xz = y in (x y z)-space Note F contains the whole z-axis so it does not lookldquoflatrdquo over the (x y)-plane

Proposition 29 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB there is a canonical isomorphism of Bp-modules natural in both variables

κ MpcapA otimesApcapANp minusrarr (M otimesA N)p

κ(msotimes nt) = (motimes n)ϕ(s)t

Proof Fix p isin SpecB and q = pcapA There is a canonical ring morphism Aq minusrarr Bp and we makeNp into an Aq-module using this morphism One checks that the following map is well-definedand Aq-bilinear

ε Mq timesNp minusrarr (M otimesA N)p

ε(ms nt) = (motimes n)ϕ(s)t

We show that in fact this is a tensor product of Aq-modules Let Z be an abelian group andψ Mq timesNp minusrarr Z an Aq-bilinear map

Mq timesNp

ε

ψ Z

(M otimesA N)p

φ

(1)

We have to define a morphism of abelian groups φ unique making this diagram commute Fors isin p we define an A-bilinear morphism φprimes M times N minusrarr Z by φprimes(mn) = ψ(m1 ns) Thisinduces a morphism of abelian groups

φprimeprimes M otimesA N minusrarr Z

φprimeprimes (motimes b) = ψ(m1 bs)

We make some observations about these morphisms

bull Suppose ws = wprimesprime in (M otimesA N)p with say w =sumimi otimes ni w

prime =sumim

primei otimes nprimei and t isin p

such that tsprimew = tswprime That issumimi otimes tsprimeni =

sumim

primei otimes tsnprimei Applying φprimeprimetssprime to both sides

of this equality gives φprimeprimes (w) = φprimeprimesprime(wprime)

9

bull For wswprimesprime isin (M otimesA N)p we have φprimeprimes (w) + φprimeprimesprime(wprime) = φprimeprimessprime(s

primew + swprime)

It follows that φ(ws) = φprimeprimes (w) gives a well-defined morphism of abelian groups φ (MotimesAN)p minusrarrZ which is clearly unique making (1) commute By uniqueness of the tensor product there is aninduced isomorphism of abelian groups κ Mq otimesAq Np minusrarr (M otimesA N)p with κ(ms otimes nt) =(m otimes n)ϕ(s)t One checks that this is a morphism of Bp-modules The inverse is defined by(motimes n)t 7rarr m1otimes nt Naturality in both variables is easily checked

Corollary 30 Let ϕ A minusrarr B be a morphism of rings M an A-module and p isin SpecB Thenthere is a canonical isomorphism of Bp-modules MpcapA otimesApcapA

Bp minusrarr (M otimesA B)p natural in M

We will not actually use the next result in these notes so the reader not familiar with homo-logical δ-functors can safely skip it Alternatively one can provide a proof by following the onegiven in Matsumura (the proof we give is more elegant provided you know about δ-functors)

Proposition 31 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB and i ge 0 there is a canonical isomorphism of Bp-modules natural inM

κi TorAi (NM)p minusrarr TorApcapA

i (NpMpcapA)

Proof Fix p isin SpecB and a B-module N and set q = p capA Then N is a B-A-bimodule and Np

is a Bp-Aq-bimodule so by (TORSection 51) the abelian group TorAi (NM) acquires a canonicalB-module structure and Tor

Aq

i (NpMq) acquires a canonical Bp-module structure for any A-module M and i ge 0 Using (TORLemma 14) and (DFDefinition 23) we have two homologicalδ-functors between AMod and BpMod

TorAi (Nminus)pige0 TorAq

i (Np (minus)q)ige0

For i gt 0 these functors all vanish on free A-modules so by (DFTheorem 74) both δ-functors areuniversal For i = 0 we have the canonical natural equivalence of Proposition 29

κ0 TorA0 (Nminus)psim= (N otimesA minus)p

sim= Np otimesAq (minus)qsim= Tor

Aq

0 (Np (minus)q)

By universality this lifts to a canonical isomorphism of homological δ-functors κ In particularfor each i ge 0 we have a canonical natural equivalence κi TorAi (Nminus)p minusrarr Tor

Aq

i (Np (minus)q) asrequired

We know from Lemma 23 that flatness is a local property We are now ready to show thatrelative flatness (ie flatness with respect to a morphism of rings) is also local This is particularlyimportant in algebraic geometry The reader who skipped Proposition 31 will also have to skipthe implication (iii) rArr (i) in the next result but this will not affect their ability to read the restof these notes

Corollary 32 Let A minusrarr B be a morphism of rings and N a B-module Then the followingconditions are equivalent

(i) N is flat over A

(ii) Np is flat over ApcapA for all prime ideals p of B

(iii) Nm is flat over AmcapA for all maximal ideals m of B

Proof (i) rArr (ii) If N is flat over A then NpcapA is flat over ApcapA for any prime p of B By anargument similar to the one given in Lemma 19 we see that Np is isomorphic as a BpcapA-moduleto a localisation of NpcapA Applying Lemma 17 to the ring morphism ApcapA minusrarr BpcapA we see thatNp is flat over ApcapA as required (ii) rArr (iii) is trivial (iii) rArr (i) For every A-module M andmaximal ideal m of B we have by Proposition 31

TorA1 (NM)msim= TorAmcapA

1 (NmMmcapA) = 0

since by assumption Nm is flat over AmcapA Therefore TorA1 (NM) = 0 for every A-module M which implies that N is flat over A and completes the proof

10

This means (I cap J)B = IB cap JB For the second claim suppose firstly that J is a principal idealaA and use the exact sequence

(I aA) i Af AI

where i is the injection and f(x) = ax+I Tensoring withB we get the formula (I a)B = (IB a)In the general case if J = (a1 an) we have (I J) =

⋂i(I ai) so that

(I J)B =⋂

(I ai)B =⋂

(IB ai) = (IB JB)

Example 2 Let A = k[x y] be a polynomial ring over a field k and put B = A(x) sim= k[y] ThenB is not flat over A since y isin A is regular but is not regular on B Let I = (x+ y) and J = (y)Then I cap J = (xy + y2) and IB = JB = yB (I cap J)B = y2B 6= IB cap JB

Example 3 Let k be a field put A = k[x y] and let K be the quotient field of A Let B be thesubring k[x yx] of K (ie the k-subalgebra generated by x and z = yx) Then A sub B sub K LetI = xA J = yA Then I cap J = xyA and (I cap J)B = x2zB IB cap JB = xzB so B is not flat overA The map SpecB minusrarr SpecA corresponding to A minusrarr B is the projection to the (x y)-planeof the surface F xz = y in (x y z)-space Note F contains the whole z-axis so it does not lookldquoflatrdquo over the (x y)-plane

Proposition 29 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB there is a canonical isomorphism of Bp-modules natural in both variables

κ MpcapA otimesApcapANp minusrarr (M otimesA N)p

κ(msotimes nt) = (motimes n)ϕ(s)t

Proof Fix p isin SpecB and q = pcapA There is a canonical ring morphism Aq minusrarr Bp and we makeNp into an Aq-module using this morphism One checks that the following map is well-definedand Aq-bilinear

ε Mq timesNp minusrarr (M otimesA N)p

ε(ms nt) = (motimes n)ϕ(s)t

We show that in fact this is a tensor product of Aq-modules Let Z be an abelian group andψ Mq timesNp minusrarr Z an Aq-bilinear map

Mq timesNp

ε

ψ Z

(M otimesA N)p

φ

(1)

We have to define a morphism of abelian groups φ unique making this diagram commute Fors isin p we define an A-bilinear morphism φprimes M times N minusrarr Z by φprimes(mn) = ψ(m1 ns) Thisinduces a morphism of abelian groups

φprimeprimes M otimesA N minusrarr Z

φprimeprimes (motimes b) = ψ(m1 bs)

We make some observations about these morphisms

bull Suppose ws = wprimesprime in (M otimesA N)p with say w =sumimi otimes ni w

prime =sumim

primei otimes nprimei and t isin p

such that tsprimew = tswprime That issumimi otimes tsprimeni =

sumim

primei otimes tsnprimei Applying φprimeprimetssprime to both sides

of this equality gives φprimeprimes (w) = φprimeprimesprime(wprime)

9

bull For wswprimesprime isin (M otimesA N)p we have φprimeprimes (w) + φprimeprimesprime(wprime) = φprimeprimessprime(s

primew + swprime)

It follows that φ(ws) = φprimeprimes (w) gives a well-defined morphism of abelian groups φ (MotimesAN)p minusrarrZ which is clearly unique making (1) commute By uniqueness of the tensor product there is aninduced isomorphism of abelian groups κ Mq otimesAq Np minusrarr (M otimesA N)p with κ(ms otimes nt) =(m otimes n)ϕ(s)t One checks that this is a morphism of Bp-modules The inverse is defined by(motimes n)t 7rarr m1otimes nt Naturality in both variables is easily checked

Corollary 30 Let ϕ A minusrarr B be a morphism of rings M an A-module and p isin SpecB Thenthere is a canonical isomorphism of Bp-modules MpcapA otimesApcapA

Bp minusrarr (M otimesA B)p natural in M

We will not actually use the next result in these notes so the reader not familiar with homo-logical δ-functors can safely skip it Alternatively one can provide a proof by following the onegiven in Matsumura (the proof we give is more elegant provided you know about δ-functors)

Proposition 31 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB and i ge 0 there is a canonical isomorphism of Bp-modules natural inM

κi TorAi (NM)p minusrarr TorApcapA

i (NpMpcapA)

Proof Fix p isin SpecB and a B-module N and set q = p capA Then N is a B-A-bimodule and Np

is a Bp-Aq-bimodule so by (TORSection 51) the abelian group TorAi (NM) acquires a canonicalB-module structure and Tor

Aq

i (NpMq) acquires a canonical Bp-module structure for any A-module M and i ge 0 Using (TORLemma 14) and (DFDefinition 23) we have two homologicalδ-functors between AMod and BpMod

TorAi (Nminus)pige0 TorAq

i (Np (minus)q)ige0

For i gt 0 these functors all vanish on free A-modules so by (DFTheorem 74) both δ-functors areuniversal For i = 0 we have the canonical natural equivalence of Proposition 29

κ0 TorA0 (Nminus)psim= (N otimesA minus)p

sim= Np otimesAq (minus)qsim= Tor

Aq

0 (Np (minus)q)

By universality this lifts to a canonical isomorphism of homological δ-functors κ In particularfor each i ge 0 we have a canonical natural equivalence κi TorAi (Nminus)p minusrarr Tor

Aq

i (Np (minus)q) asrequired

We know from Lemma 23 that flatness is a local property We are now ready to show thatrelative flatness (ie flatness with respect to a morphism of rings) is also local This is particularlyimportant in algebraic geometry The reader who skipped Proposition 31 will also have to skipthe implication (iii) rArr (i) in the next result but this will not affect their ability to read the restof these notes

Corollary 32 Let A minusrarr B be a morphism of rings and N a B-module Then the followingconditions are equivalent

(i) N is flat over A

(ii) Np is flat over ApcapA for all prime ideals p of B

(iii) Nm is flat over AmcapA for all maximal ideals m of B

Proof (i) rArr (ii) If N is flat over A then NpcapA is flat over ApcapA for any prime p of B By anargument similar to the one given in Lemma 19 we see that Np is isomorphic as a BpcapA-moduleto a localisation of NpcapA Applying Lemma 17 to the ring morphism ApcapA minusrarr BpcapA we see thatNp is flat over ApcapA as required (ii) rArr (iii) is trivial (iii) rArr (i) For every A-module M andmaximal ideal m of B we have by Proposition 31

TorA1 (NM)msim= TorAmcapA

1 (NmMmcapA) = 0

since by assumption Nm is flat over AmcapA Therefore TorA1 (NM) = 0 for every A-module M which implies that N is flat over A and completes the proof

10

bull For wswprimesprime isin (M otimesA N)p we have φprimeprimes (w) + φprimeprimesprime(wprime) = φprimeprimessprime(s

primew + swprime)

It follows that φ(ws) = φprimeprimes (w) gives a well-defined morphism of abelian groups φ (MotimesAN)p minusrarrZ which is clearly unique making (1) commute By uniqueness of the tensor product there is aninduced isomorphism of abelian groups κ Mq otimesAq Np minusrarr (M otimesA N)p with κ(ms otimes nt) =(m otimes n)ϕ(s)t One checks that this is a morphism of Bp-modules The inverse is defined by(motimes n)t 7rarr m1otimes nt Naturality in both variables is easily checked

Corollary 30 Let ϕ A minusrarr B be a morphism of rings M an A-module and p isin SpecB Thenthere is a canonical isomorphism of Bp-modules MpcapA otimesApcapA

Bp minusrarr (M otimesA B)p natural in M

We will not actually use the next result in these notes so the reader not familiar with homo-logical δ-functors can safely skip it Alternatively one can provide a proof by following the onegiven in Matsumura (the proof we give is more elegant provided you know about δ-functors)

Proposition 31 Let ϕ A minusrarr B be a morphism of rings M an A-module and N a B-moduleThen for every p isin SpecB and i ge 0 there is a canonical isomorphism of Bp-modules natural inM

κi TorAi (NM)p minusrarr TorApcapA

i (NpMpcapA)

Proof Fix p isin SpecB and a B-module N and set q = p capA Then N is a B-A-bimodule and Np

is a Bp-Aq-bimodule so by (TORSection 51) the abelian group TorAi (NM) acquires a canonicalB-module structure and Tor

Aq

i (NpMq) acquires a canonical Bp-module structure for any A-module M and i ge 0 Using (TORLemma 14) and (DFDefinition 23) we have two homologicalδ-functors between AMod and BpMod

TorAi (Nminus)pige0 TorAq

i (Np (minus)q)ige0

For i gt 0 these functors all vanish on free A-modules so by (DFTheorem 74) both δ-functors areuniversal For i = 0 we have the canonical natural equivalence of Proposition 29

κ0 TorA0 (Nminus)psim= (N otimesA minus)p

sim= Np otimesAq (minus)qsim= Tor

Aq

0 (Np (minus)q)

By universality this lifts to a canonical isomorphism of homological δ-functors κ In particularfor each i ge 0 we have a canonical natural equivalence κi TorAi (Nminus)p minusrarr Tor

Aq

i (Np (minus)q) asrequired

We know from Lemma 23 that flatness is a local property We are now ready to show thatrelative flatness (ie flatness with respect to a morphism of rings) is also local This is particularlyimportant in algebraic geometry The reader who skipped Proposition 31 will also have to skipthe implication (iii) rArr (i) in the next result but this will not affect their ability to read the restof these notes

Corollary 32 Let A minusrarr B be a morphism of rings and N a B-module Then the followingconditions are equivalent

(i) N is flat over A

(ii) Np is flat over ApcapA for all prime ideals p of B

(iii) Nm is flat over AmcapA for all maximal ideals m of B

Proof (i) rArr (ii) If N is flat over A then NpcapA is flat over ApcapA for any prime p of B By anargument similar to the one given in Lemma 19 we see that Np is isomorphic as a BpcapA-moduleto a localisation of NpcapA Applying Lemma 17 to the ring morphism ApcapA minusrarr BpcapA we see thatNp is flat over ApcapA as required (ii) rArr (iii) is trivial (iii) rArr (i) For every A-module M andmaximal ideal m of B we have by Proposition 31

TorA1 (NM)msim= TorAmcapA

1 (NmMmcapA) = 0

since by assumption Nm is flat over AmcapA Therefore TorA1 (NM) = 0 for every A-module M which implies that N is flat over A and completes the proof

10

Lemma 33 Let A minusrarr B be a morphism of rings Then the following conditions are equivalent

(i) B is flat over A

(ii) Bp is flat over ApcapA for all prime ideals p of B

(iii) Bm is flat over AmcapA for all maximal ideals m of B

21 Faithful Flatness

Theorem 34 Let A be a ring and M an A-module The following conditions are equivalent

(i) M is faithfully flat over A

(ii) M is flat over A and for any nonzero A-module N we have N otimesAM 6= 0

(iii) M is flat over A and for any maximal ideal m of A we have mM 6= M

Proof (i) rArr (ii) Let N be an A-module and ϕ N minusrarr 0 the zero map Then if M is faithfullyflat and N otimesAM = 0 we have ϕotimesAM = 0 which means that ϕ is injective and therefore N = 0(ii) rArr (iii) Since Am 6= 0 we have (Am) otimesA M = MmM 6= 0 by hypothesis (iii) rArr (ii)Let N be a nonzero A-module and pick 0 6= x isin N Let ϕ A minusrarr N be 1 7rarr x If I = Kerϕthen there is an injective morphism of modules AI minusrarr N Let m be a maximal ideal containingI Then M sup mM supe IM so (AI) otimesA M = MIM 6= 0 Since M is flat the morphism(AI)otimesAM minusrarr N otimesAM is injective so N otimesAM 6= 0 (ii) rArr (i) Let ψ N minusrarr N prime be a morphismof modules with kernel K minusrarr N If N otimesAM minusrarr N prime otimesAM is injective then K otimesAM = 0 whichis only possible if K = 0

Corollary 35 Let A and B be local rings and ψ A minusrarr B a local morphism of rings Let M bea nonzero finitely generated B-module Then

M is flat over AlArrrArrM is faithfully flat over A

In particular B is flat over A if and only if it is faithfully flat over A

Proof Let m and n be the maximal ideals of A and B respectively Then mM sube nM since ψ islocal and nM 6= M by Nakayama so the assertion follows from the Theorem

Lemma 36 We have the following fundamental properties of flatness

bull Transitivity If φ A minusrarr B is a faithfully flat morphism of rings and N a faithfully flatB-module then N is a faithfully flat A-module

bull Change of Base If φ A minusrarr B is a morphism of rings and M is a faithfully flat A-modulethen M otimesA B is a faithfully flat B-module

bull Descent If φ A minusrarr B is a ring morphism and M is a faithfully flat B-module which isalso faithfully flat over A then B is faithfully flat over A

Proof The diagram in the proof of transitivity for flatness makes it clear that faithful flatness isalso transitive Similarly the flatness under base change proof in our Atiyah amp Macdonald notesshows that faithful flatness is also stable under base change The descent property is also easilychecked

Proposition 37 Let ψ A minusrarr B be a faithfully flat morphism of rings Then

(i) For any A-module N the map N minusrarr NotimesAB defined by x 7rarr xotimes1 is injective In particularψ is injective and A can be viewed as a subring of B

(ii) For any ideal I of A we have IB capA = I

(iii) The map Ψ Spec(B) minusrarr Spec(A) is surjective

11

(iv) If B is noetherian then so is A

Proof (i) Let 0 6= x isin N Then 0 6= Ax sube N and since B is flat we see that AxotimesAB is isomorphicto the submodule (xotimes 1)B of N otimesA B It follows from Theorem 34 that xotimes 1 6= 0

(ii) By change of base B otimesA (AI) = BIB is faithfully flat over AI Now the assertionfollows from (i) For (iii) let p isin Spec(A) The ring Bp = B otimesA Ap is faithfully flat over Ap soby (ii) pBp 6= Bp Take a maximal ideal m of Bp containing pBp Then m capAp supe pAp thereforem capAp = pAp since pAp is maximal Putting q = m capB we get

q capA = (m capB) capA = m capA = (m capAp) capA = pAp capA = p

as required (iv) Follows immediately from (ii)

Theorem 38 Let ψ A minusrarr B be a morphism of rings The following conditions are equivalent

(1) ψ is faithfully flat

(2) ψ is flat and Ψ Spec(B) minusrarr Spec(A) is surjective

(3) ψ is flat and for any maximal ideal m of A there is a maximal ideal n of B lying over m

Proof (1) rArr (2) was proved above (2) rArr (3) By assumption there exists q isin Spec(B) withqcapA = m If n is any maximal ideal of B containing q then ncapA = m as m is maximal (3) rArr (1)The existence of n implies mB 6= B so B is faithfully flat over A by Theorem 34

Definition 4 In algebraic geometry we say a morphism of schemes f X minusrarr Y is flat if thelocal morphisms OYf(x) minusrarr OXx are flat for all x isin X We say the morphism is faithfully flat ifit is flat and surjective

Lemma 39 Let A be a ring and B a faithfully flat A-algebra Let M be an A-module Then

(i) M is flat (resp faithfully flat) over A hArr M otimesA B is so over B

(ii) If A is local and M finitely generated over A we have M is A-free hArr M otimesA B is B-free

Proof (i) Let N minusrarr N prime be a morphism of A-modules Both claims follow from commutativity ofthe following diagram

(N otimesAM)otimesA B

(N prime otimesAM)otimesA B

N otimesA (M otimesA B)

N prime otimesA (M otimesA B)

N otimesA (B otimesB (M otimesA B))

N prime otimesA (B otimesB (M otimesA B))

(N otimesA B)otimesB (M otimesA B) (N prime otimesA B)otimesB (M otimesA B)

(ii) The functor minusotimesAB preserves coproducts so the implication (rArr) is trivial (lArr) follows from(i) because under the hypothesis freeness of M is equivalent to flatness as we saw in Proposition24

12

22 Going-up and Going-down

Definition 5 Let φ A minusrarr B be a morphism of rings We say that the going-up theorem holdsfor φ if the following condition is satisfied

(GU) For any p pprime isin Spec(A) such that p sub pprime and for any prime q isin Spec(B) lyingover p there exists qprime isin Spec(B) lying over pprime such that q sub qprime

Similarly we say that the going-down theorem holds for φ if the following condition is satisfied

(GD) For any p pprime isin Spec(A) such that p sub pprime and for any prime qprime isin Spec(B) lyingover pprime there exists q isin Spec(B) lying over p such that q sub qprime

Lemma 40 The condition (GD) is equivalent to the following condition (GDrsquo) For any p isinSpec(A) and any minimal prime overideal q of pB we have q capA = p

Proof (GD) rArr (GDrsquo) Clearly q cap A supe p If this inclusion is proper then by (GD) there exists aprime q1 of B with q1 sub q and q1 capA = p contradicting minimality of q (GDrsquo) rArr (GD) Supposeprimes p sub pprime of A are given and qprime cap A = pprime We can shrink qprime to a prime q minimal among allprime ideals containing pB and by assumption q capA = p which completes the proof

Let a be any proper radical ideal in a noetherian ring B Then a is the intersection of allits minimal primes p1 pn and the closed irreducible sets V (p1) sube V (a) are the irreduciblecomponents of the closed set V (a) in the noetherian space Spec(B)

Let φ A minusrarr B a morphism of rings put X = Spec(A) Y = Spec(B) and let Ψ Y minusrarr Xthe corresponding morphism of affine schemes and suppose B is noetherian Then (GDprime) can beformulated geometrically as follows let p isin X put X prime = V (p) sube X and let Y prime be an arbitraryirreducible component of Ψminus1(X prime) (which we assume is nonempty) Then Ψ maps Y prime genericallyonto X prime in the sense that the generic point of Y prime is mapped to the generic point p of X prime

Theorem 41 Let φ A minusrarr B be a flat morphism of rings Then the going-down theorem holdsfor φ

Proof Let pprime sub p be prime ideals of A and let q be a prime ideal of B lying over p Then Bq isflat over Ap by Lemma 33 hence faithfully flat since Ap minusrarr Bq is local Therefore Spec(Bq) minusrarrSpec(Ap) is surjective Let qprimeprime be a prime ideal of Bq lying over pprimeAp Then qprime = qprimeprimecapB is a primeideal of B lying over pprime and contained in q

Theorem 42 Let B be a ring and A a subring over which B is integral Then

(i) The canonical map Spec(B) minusrarr Spec(A) is surjective

(ii) If two prime ideals q sube qprime lie over the same prime ideal of A then they are equal

(iii) The going-up theorem holds for A sube B

(iv) If A is a local ring and m its maximal ideal then the prime ideals of B lying over m areprecisely the maximal ideals of B

(v) If A and B are integral domains and A is integrally closed then the going-down theoremholds for A sube B

Proof See [AM69] or [Mat80] Theorem 5

23 Constructible Sets

Definition 6 A topological space X is noetherian if the descending chain condition holds forthe closed sets in X The spectrum Spec(A) of a noetherian ring A is noetherian If a spaceis covered by a finite number of noetherian subspaces then it is noetherian Any subspace of anoetherian space is noetherian A noetherian space is quasi-compact In a noetherian space Xany nonempty closed set Z is uniquely decomposed into a finite number of irreducible closed setsZ = Z1 cup middot middot middot cupZn such that Zi Zj for i 6= j The Zi are called the irreducible components of Z

13

Lemma 43 (Noetherian Induction) Let X be a noetherian topological space and P a propertyof closed subsets of X Assume that for any closed subset Y of X if P holds for every properclosed subset of Y then P holds for Y (in particular P holds for the empty set) Then P holdsfor X

Proof Suppose that P does not hold for X and let Z be the set of all proper closed subsetsof X which do not satisfy P Then since X is noetherian Z has a minimal element Y SinceY is minimal every proper closed subset of Y must satisfy P and therefore Y satisfies Pcontradicting the fact that Y isin Z

Lemma 44 Let X be a noetherian topological space and P a property of general subsets of XAssume that for any subset Y of X if P holds for every proper subset Y prime of Y with Y prime sub Y thenP holds for Y (in particular P holds for the empty set) Then P holds for X

Proof Suppose that P does not hold for X and let Z be the set of all closures Q of proper subsetsQ of X with Q sub X and P not holding for Q Let Q be a minimal element of Z If Qprime is anyproper subset of Q with Qprime sub Q then Qprime must satisfy P otherwise Qprime would contradict minimalityof Q in Z But by assumption this implies that Q satisfies P which is a contradiction

Definition 7 Let X be a topological space and Z a subset of X We say Z is locally closed inX if it satisfies the following equivalent properties

(i) Every point z isin Z has an open neighborhood U in X such that U cap Z is closed in U

(ii) Z is the intersection of an open set in X and a closed set in X

(iii) Z is an open subset of its closure

Definition 8 Let X be a noetherian space We say a subset Z of X is a constructible set in Xif it is a finite union of locally closed sets in X so Z =

⋃mi=1(Ui capFi) with Ui open and Fi closed

The set F of all constructible subsets of X is the smallest collection of subsets of X containing allthe open sets which is closed with respect to the formation of finite intersections and complementsIt follows that all open and closed sets are constructible and F is also closed under finite unions

We say that a subset Z is pro-constructible (resp ind-constructible) if it is the intersection(resp union) of an arbitrary collection of constructible sets in X

Proposition 45 Let X be a noetherian space and Z a subset of X Then Z is constructible ifand only if the following condition is satisfied

(lowast) For each irreducible closed set X0 in X either X0capZ is not dense in X0 or X0capZcontains a nonempty open set of X0

Proof Assume that Z is constructible and Z cap X0 nonempty Then we can write X0 cap Z =⋃mi=1 Ui capFi for Ui open in X Fi closed and irreducible in X (by taking irreducible components)

and UicapFi nonempty for all i Then Ui cap Fi = Fi since Fi is irreducible therefore X0 cap Z =⋃i Fi

If X0 cap Z is dense in X0 we have X0 =⋃i Fi so that some Fi say F1 is equal to X0 Then

U1 capX0 = U1 cap F1 is a nonempty open subset of X0 contained in X0 cap ZNext we prove the converse We say that a subset T of X has the property P if whenever a

subset Z of T satisfies (lowast) it is constructible We need to show that X has the property P forwhich we use the form of noetherian induction given in Lemma 44 Suppose that Y is a subsetof X with P holding for every proper subset Y prime of Y with Y prime sub Y We need to show that Pholds for Y Let Z be a nonempty subset of Y satisfying (lowast) and let Z = F1 cup cup Fr be thedecomposition of Z into irreducible components Since Z = Z cap F1 cup middot middot middot cup Z cap Fr we have

F1 = F1 cap Z = F1 cap(Z cap F1 cup cup Z cap Fr

)= (F1 cap Z cap F1) cup middot middot middot cup (F1 cap Z cap Fr)

Since F1 is irreducible and not contained in any other Fi we must have F1 = Z cap F1 so F1 cap Zis dense in F1 whence by (lowast) there exists a proper closed subset F prime of F1 such that F1 F prime sube ZThen putting F lowast = F prime cupF2 cup middot middot middot cupFr we have Z = (F1 F prime)cup (Z capF lowast) The set F1 F prime is locally

14

closed in X so to complete the proof it suffices to show that Z cap F lowast is constructible in X SinceZ cap F lowast sube F lowast sub Z sube Y by the inductive hypothesis P holds for ZcapF lowast so it suffices to show thatZ cap F lowast satisfies (lowast) If X0 is irreducible and Z cap F lowast capX0 = X0 the closed set F lowast must containX0 and so Z cap F lowast capX0 = Z capX0 which contains a nonempty open subset of X0 since Z satisfies(lowast) and clearly Z capX0 is dense in X0

Lemma 46 Let φ A minusrarr B be a morphism of rings and f Spec(B) minusrarr Spec(A) the corre-sponding morphism of schemes Then f dominant if and only if Kerφ sube nil(A) In particular ifA is reduced the f dominant if and only if φ is injective

Proof Let X = Spec(A) and Y = Spec(B) The closure f(Y ) is the closed set V (I) defined bythe ideal I =

⋂pisinY φ

minus1p = φminus1⋂

pisinY p which is φminus1(nil(B)) Clearly Kerφ sube I Suppose thatf(Y ) is dense in X Then V (I) = X whence I = nil(A) and so Kerφ sube nil(A) Converselysuppose Kerφ sube nil(A) Then it is clear that I = φminus1(nil(B)) = nil(A) which means thatf(Y ) = V (I) = X

3 Associated Primes

This material can be found in [AM69] Chapter 11 webnotes of Robert Ash or in [Mat80] itselfThere is not much relevant to add here apart from a few small comments

Lemma 47 Let A be a ring and M an A-module Let a be an ideal in A that is maximal amongall annihilators of nonzero elements of M Then a is prime

Proof Say a = Ann(x) Given ab isin a we must show that a isin a or b isin a Assume a isin aThen ax 6= 0 We note that Ann(ax) supe a By hypothesis it cannot properly be larger HenceAnn(ax) = a Now b annihilates ax hence b isin a

Lemma 48 Let A be a noetherian ring and M an A-module If 0 6= a isin M then Ann(a) iscontained in an associated prime of M

Proposition 49 Let A be a noetherian ring and M a nonzero finitely generated A-module Amaximal ideal m is an associated prime of M if and only if no element of m is regular on M

Proof One implication is obvious If x isin m is not regular on M say x isin Ann(b) for some nonzerob then x is contained in an associated prime of M Thus m is contained in the finite union of theassociated primes of M and since m is maximal it must be one of them

Proposition 50 Let A be a nonzero noetherian ring I an ideal and M a nonzero finitelygenerated A-module If there exist elements x y isin I with x regular on A and y regular on M thenthere exists an element of I regular on both A and M

Proof Let p1 pn be the associated primes of A and q1 qm the associated primes of M By assumption I is not contained in any of these primes But if no element of I is regular on bothA and M then I is contained in the union p1 cup middot middot middot cup pn cup q1 cup middot middot middot cup qm and therefore containedin one of these primes which is a contradiction

4 Dimension

This is covered in [AM69] so we restrict ourselves here to mentioning some of the major pointsRecall that an ideal q sube R in a ring is primary if it is proper and if whenever xy isin q we haveeither x isin q or yn isin q for some n gt 0 Then the radical of q is a prime ideal p and we say q isa p-primary ideal If a is an ideal and b supe a is p-primary then in the ring Ra the ideal ba ispa-primary A minimal primary decomposition of an ideal b is an expression

b = q1 cap middot middot middot cap qn

15

where capj 6=iqj qi for all i and the primes pi = r(qi) are all distinct If a is an ideal contained inb then

ba = q1a cap middot middot middot cap qna

is a minimal primary decomposition of ba in AaLet A be a nonzero ring Recall that dimension of an A-module M is the Krull dimension of

the ring AAnn(M) and is defined for all modules M (minus1 if M = 0) The rank is defined for freeA-modules and is the common size of any basis (0 if M = 0) Throughout these notes dim(M)will denote the dimension not the rank

Definition 9 Let (Am k) be a noetherian local ring of dimension d An ideal of definition is anm-primary ideal Recall that the dimension of A is the size of the smallest collection of elementsof A which generates an m-primary ideal Recall that rankk(mm2) is equal to the size of thesmallest set of generators for m as an ideal so always d le rankk(mm2)

A system of parameters is a set of d elements generating an m-primary ideal If d = rankk(mm2)or equivalently there is a system of parameters generating m we say that A is a regular local ringand we call such a system of parameters a regular system of parameters

Proposition 51 Let (Am) be a noetherian local ring of dimension d ge 1 and let x1 xd bea system of parameters of A Then

dim(A(x1 xi)) = dminus i = dim(A)minus i

for each 1 le i le d

Proof Put A = A(x1 xi) If i = d then the zero ideal in A is an ideal of definition so clearlydim(A) = 0 If 1 le i lt d then dim(A) le dminus i since xi+1 xd generate an ideal of definition ofA Let dim(A) = p If p = 0 then (x1 xi) must be an ideal of definition contradicting i lt dSo p ge 1 and if y1 yp is a system of parameters of A then x1 xi y1 yp generate anideal of definition of A so that p+ i ge d That is p ge dminus i

Definition 10 Let A be a nonzero ring and I a proper ideal The height of I denoted htI isthe minimum of the heights of the prime ideals containing I

htI = infhtp | p supe I

This is a number in 0 1 2 infin Equivalently we can take the infimum over the heights ofprimes minimal over I Clearly ht0 = 0 and if I sube J are proper ideals then it is clear thathtI le htJ If I is a prime ideal then htI is the usual height of a prime ideal If A is a noetherianring then htI ltinfin for every proper ideal I since Ap is a local noetherian ring and htp = dim(Ap)

Lemma 52 Let A be a nonzero ring and I a proper ideal Then we have

htI = infhtIAm |m is a maximal ideal and I sube m

Lemma 53 Let A be a noetherian ring and suppose we have an exact sequence

0 minusrarrM prime minusrarrM minusrarrM primeprime minusrarr 0

in which M primeMM primeprime are nonzero and finitely generated Then dimM = maxdimM prime dimM primeprime

Proof We know that Supp(M) = Supp(M prime)cupSupp(M primeprime) and for all three modules the dimensionis the supremum of the coheights of prime ideals in the support So the result is straightforwardto check

16

41 Homomorphism and Dimension

Let φ A minusrarr B be a morphism of rings If p isin Spec(A) then put κ(p) = AppAp Let Bp denotethe ring Tminus1B where T = φ(Aminusp) There is an isomorphism of A-algebras Bp

sim= BotimesAAp Thereis a commutative diagram of rings

B

))RRRRRRRRRRRRRRRR Bp BppBp

B otimesA κ(p)

The vertical isomorphism is defined by bφ(s)+ pBp 7rarr botimes (1s+ pAp) We call Spec(BotimesA κ(p))the fibre over p of the map Φ Spec(B) minusrarr Spec(A) Since primes of BppBp clearly correspondto primes q of B with q cap A = p it is easy to see that the ring morphism B minusrarr B otimesA κ(p) givesrise to a continuous map Spec(B otimesA κ(p)) minusrarr SpecB which gives a homeomorphism betweenΦminus1(p) and Spec(B otimesA κ(p)) See [AM69] Chapter 3

Lemma 54 Let q be a prime ideal of B with q cap A = p and let P be the corresponding primeideal of B otimesA κ(p) Then there is an isomorphism of A-algebras

Bq otimesA κ(p) sim= (B otimesA κ(p))P

btotimes (as+ pAp) 7rarr (botimes (a1 + pAp))(totimes (s1 + pAp))

Proof It is not difficult to see that there is an isomorphism of rings Bqsim= (Bp)qBp defined by

bt 7rarr (b1)(t1) Consider the prime ideal qBppBp We know that there is a ring isomorphism

(B otimesA κ(p))Psim= (BppBp)qBppBp

sim= (Bp)qBpp(Bp)qBpsim= BqpBq

by the comments following Lemma 7 It is not hard to check there is a ring isomorphism BqpBqsim=

BqotimesA κ(p) defined by bt+ pBq 7rarr btotimes 1 (the inverse of btotimes (as+ pAp) is bφ(a)tφ(s)+ pBq)So by definition of P there is an isomorphism of rings (B otimesA κ(p))P

sim= Bq otimesA κ(p) and this isclearly an isomorphism of A-algebras

In particular if φ A minusrarr B is a ring morphism p isin Spec(A) and q isin Spec(B) such thatq capA = p then there is an isomorphism of rings (BpB)qpB

sim= BqpBq so we have ht(qpB) =dim(Bq otimesA κ(p))

Theorem 55 Let φ A minusrarr B be a morphism of noetherian rings Let q isin Spec(B) and putp = q capA Then

(1) htq le htp + ht(qpB) In other words dim(Bq) le dim(Ap) + dim(Bq otimesA κ(p))

(2) We have equality in (1) if the going-down theorem holds for φ (in particular if φ is flat)

(3) If Φ Spec(B) minusrarr Spec(A) is surjective and if the going-down theorem holds then we havedim(B) ge dim(A) and htI = ht(IB) for any proper ideal I of A

Proof (1) Replacing A by Ap and B by Bq we may suppose that (A p) and (B q) are noetherianlocal rings such that q cap A = p and we must show that dim(B) le dim(A) + dim(BpB) Let Ibe a p-primary ideal of A Then pn sube I for some n gt 0 so pnB sube IB sube pB Thus the ideals pBand IB have the same radical and so by definition dim(BpB) = dim(BIB) If dimA = 0 thenwe can take I = 0 and the result is trivial So assume dimA = r ge 1 and let I = (a1 ar) fora system of parameters a1 ar If dim(BIB) = 0 then IB is an q-primary ideal of B and sodim(B) le r as required Otherwise if dim(BIB) = s ge 1 let b1 + IB bs + IB be a systemof parameters of BIB Then b1 bs a1 ar generate an ideal of definition of B Hencedim(B) le r + s

(2) We use the same notation as above If ht(qpB) = s ge 0 then there exists a prime chainof length s q = q0 sup q1 sup middot middot middot sup qs such that qs supe pB As p = q cap A supe qi cap A supe p all the qi

17

lie over p If htp = r ge 0 then there exists a prime chain p = p0 sup p1 sup middot middot middot sup pr in A and bygoing-down there exists a prime chain qs = t0 sup t1 sup middot middot middot sup tr of B such that ti capA = pi Then

q = q0 sup middot middot middot sup qs sup t1 sup middot middot middot sup tr

is a prime chain of length r + s therefore htq ge r + s(3) (i) follows from (2) since dim(A) = suphtp | p isin Spec(A) (ii) Since Φ is surjective IB

is a proper ideal Let q be a minimal prime over IB such that htq = ht(IB) and put p = q capAThen ht(qpB) = 0 so by (2) we find that ht(IB) = htq = htp ge htI For the reverseinclusion let p be a minimal prime ideal over I such that htp = htI and take a prime q of Blying over p Replacing q if necessary we may assume that q is a minimal prime ideal over pBThen htI = htp = htq ge ht(IB)

Theorem 56 Let A be a nonzero subring of B and suppose that B is integral over A Then

(1) dim(A) = dim(B)

(2) Let q isin Spec(B) and set p = q capA Then we have cohtp = cohtq and htq le htp

(3) If the going-down theorem holds between A and B then for any ideal J of B with J capA 6= Awe have htJ = ht(J capA)

Proof (1) By Theorem 42 the going-up theorem holds for A sube B and Spec(B) minusrarr Spec(A) issurjective so we can lift any chain of prime ideals p0 sub p1 sub middot middot middot sub pn in A to a chain of primeideals q0 sub middot middot middot sub qn in B On the other hand if q sube qprime are prime ideals of B and q capA = qprime capAthen q = qprime so any chain of prime ideals in B restricts to a chain of the same length in A Hencedim(A) = dim(B)

(2) Since Bq is integral over Ap it is clear from (1) that cohtp = cohtq If q = q0 sup middot middot middot sup qnis a chain of prime ideals in B then intersecting with A gives a chain of length n descending fromp Hence htq le htp

(3) Given the going-down theorem it is clear that htq = htp in (2) Let J be a proper ideal ofB with J capA 6= A and let q be such that htJ = htq Then ht(J capA) le ht(qcapA) = htq = htJ On the other hand BJ is integral over BJ cap A so every prime ideal p of A containing J cap Acan be lifted to a prime ideal q of B containing J In particular we can lift a prime ideal p withht(J capA) = htp to see that htJ le htq = htp = ht(J capA) as required

5 Depth

Definition 11 Let A be a ring M an A-module and a1 ar a sequence of elements of AWe say a1 ar is an M -regular sequence (or simply M -sequence) if the following conditions aresatisfied

(1) For each 2 le i le r ai is regular on M(a1 aiminus1)M and a1 is regular on M

(2) M 6= (a1 an)M

If a1 ar is an M -regular sequence then so is a1 ai for any i le r When all ai belong to anideal I we say a1 ar is an M -regular sequence in I If moreover there is no b isin I such thata1 ar b is M -regular then a1 ar is said to be a maximal M -regular sequence in I Noticethat the notion of M -regular depends on the order of the elements in the sequence If MN areisomorphic A-modules then a sequence is regular on M iff it is regular on N

Lemma 57 A sequence a1 ar with r ge 2 is M -regular if and only if a1 is regular on Mand a2 ar is an Ma1M -regular sequence If the sequence a1 ar is a maximal M -regularsequence in I then a2 ar is a maximal Ma1M -regular sequence in I

18

Proof The key point is that for ideals a sube b there is a canonical isomorphism of A-modulesMbM sim= NbN where N = MaM If a1 ar is M -regular then a1 is regular on M a2 isregular on N = Ma1M and for 3 le i le r ai is regular on

M(a1 aiminus1)M sim= N(a2 aiminus1)N

Hence a2 ar is an N -regular sequence The converse follows from the same argument

More generally if a1 ar is an M -regular sequence and we set N = M(a1 ar)M andif b1 bs is an N -regular sequence then a1 ar b1 bs is an M -regular sequence

Lemma 58 If a1 ar is an A-regular sequence and M is a flat A-module then a1 ar isalso M -regular provided (a1 ar)M 6= M

Proof Left multiplication by a1 defines a monomorphism A minusrarr A since a1 is A-regular Tensoringwith M and using the fact that M is flat we see that left multiplication by a1 also gives amonomorphismM minusrarrM as required Similarly tensoring with the monomorphism a2 Aa1 minusrarrAa1 we get a monomorphism Ma1M minusrarrMa1M and so on

Lemma 59 Let A be a ring and M an A-module Given an integer n ge 1 a sequence a1 aris M -regular if and only if it is Mn-regular

Proof Suppose the sequence a1 ar is M -regular We prove it is Mn-regular by induction on rThe case r = 1 is trivial so assume r gt 1 By the inductive hypothesis the sequence a1 arminus1 isMn-regular Let L = (a1 arminus1)M Then (a1 arminus1)Mn = Ln and there is an isomorphismof A-modules MnLn sim= (ML)n So we need only show that ar is regular on (ML)n Since byassumption it is regular on ML this is not hard to check Clearly (a1 ar)Mn 6= Mn so thesequence a1 ar is Mn-regular as required The converse is similarly checked

Lemma 60 Let A be a nonzero ring M an A-module and a1 ar isin A If a1 ar isin Am isMm-regular for every maximal ideal m of A then the sequence a1 ar is M -regular

Proof This follows from the fact that given an A-module M an element a isin A is regular on M ifand only if its image in Am is regular on Mm for every maximal ideal m of A

Lemma 61 Suppose that a1 ar is M -regular and a1ξ1 + middot middot middot + arξr = 0 for ξi isin M Thenξi isin (a1 ar)M for all i

Proof By induction on r For r = 1 a1ξ1 = 0 implies that ξ1 = 0 Let r gt 1 Since ar is regularon M(a1 arminus1)M we have ξr =

sumrminus1i=1 aiηi so

sumrminus1i=1 ai(ξi + arηi) = 0 By the inductive

hypothesis for i lt r we have ξi + arηi isin (a1 arminus1)M so that ξi isin (a1 ar)M

Theorem 62 Let A be a ring and M an A-module and let a1 ar be an M -regular sequenceThen for every sequence n1 nr of integers gt 0 the sequence an1

1 anrr is M -regular

Proof Suppose we can prove the following statement

(lowast) Given an integer n gt 0 an A-module M and any M -regular sequence a1 arthe sequence an1 a2 ar is also M -regular

We prove the rest of the Theorem by induction on r For r = 1 this follows immediately from (lowast)Let r gt 1 and suppose a1 ar is M -regular Then by (lowast) an1

1 a2 ar is M -regular Hencea2 ar is Man1

1 M -regular By the inductive hypothesis an22 anr

r is Man11 M -regular and

therefore an11 anr

r is M -regular by Lemma 57So it only remains to prove (lowast) which we do by induction on n The case n = 1 is trivial so let

n gt 1 be given along with an A-moduleM and anM -regular sequence a1 ar By the inductivehypothesis anminus1

1 a2 ar is M -regular and clearly an1 is regular on M Since (an1 a2 ar) sube(anminus1

1 a2 ar) it is clear thatM 6= (an1 a2 ar)M Let i gt 1 and assume that an1 a2 aiminus1

is an M -regular sequence We need to show that ai is regular on M(an1 a2 aiminus1)M Suppose

19

that aiω = an1 ξ1 +a2ξ2 + middot middot middot+aiminus1ξiminus1 Then ω = anminus11 η1 +a2η2 + middot middot middot+aiminus1ηiminus1 by the inductive

hypothesis So

anminus11 (a1ξ1 minus aiη1) + a2(ξ2 minus aiη2) + middot middot middot+ aiminus1(ξiminus1 minus aiηiminus1) = 0

Hence a1ξ1minusaiη1 isin (anminus11 a2 aiminus1)M by Lemma 61 It follows that aiη1 isin (a1 a2 aiminus1)M

hence η1 isin (a1 aiminus1)M and so ω isin (an1 a2 aiminus1)M as required This proves (lowast) andtherefore completes the proof

Let A be a ring There is an isomorphism of A[x1 xn]-modules M otimesA A[x1 xn] sim=M [x1 xn] where the latter module consists of polynomials in x1 xn with coefficients in M(see our Polynomial Ring notes) For any f(x1 xn) isinM [x1 xn] and tuple (a1 an) isinAn we can define an element of M

f(a1 an) =sumα

aα11 middot middot middot aαn

n middot f(α)

For an element r isin R and h isinM [x1 xn]

(f + h)(a1 an) = f(a1 an) + h(a1 an)(r middot f)(a1 an) = r middot f(a1 an)

For an ideal I in R the R-submodule IM [x1 xn] consists of all polynomials whose coefficientsare in the R-submodule IM subeM

Let us review the definition of the associated graded ring and modules Let A be a ring and Ian ideal of A Then the abelian group

grI(A) = AI oplus II2 oplus I2I3 oplus middot middot middot

becomes a graded ring in a fairly obvious way For an A-module M we have the graded grI(A)-module

grI(M) = MIM oplus IMI2M oplus I2MI3M oplus middot middot middotIf A is noetherian and M is a finitely generated A-module then grI(A) is a noetherian ring andif grI(M) is a finitely generated grI(A)-module

Given elements a1 an isin A and I = (a1 an) we define a morphism of abelian groupsψ M [x1 xn] minusrarr grI(M) as follows if f is homogenous of degreem ge 0 define ψ(f) to be theimage of f(a1 an) in ImMIm+1M This defines a morphism of groups M [x1 xn]m minusrarrImMIm+1M and together these define the morphism of groups ψ Since ψ maps IM [x1 xn]to zero it induces a morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) and

Proposition 63 Let A be a ring and M an A-module Let a1 an isin A and set I =(a1 an) Then the following conditions are equivalent

(a) For every m gt 0 and for every homogenous polynomial f(x1 xn) isin M [x1 xn] ofdegree m such that f(a1 an) isin Im+1M we have f isin IM [x1 xn]

(b) If f(x1 xn) isin M [x1 xn] is homogenous and f(a1 an) = 0 then the coefficientsof f are in IM

(c) The morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) defined by mappinga homogenous polynomial f(x1 xn) of degree m to f(a1 an) isin ImMIm+1M is anisomorphism

Proof It is easy to see that (a) hArr (c) and (a) rArr (b) To show (b) rArr (a) let f isinM [x1 xn] bea homogenous polynomial of degree m gt 0 and suppose f(a1 an) isin Im+1M Any element ofIm+1M can be written as a sum of terms of the form aα1

1 middot middot middot aαnn middotm with

sumi αi = m+1 By shifting

one of the ai across we can write f(a1 an) = g(a1 an) for a homogenous polynomialg isinM [x1 xn] of degreem all of whose coefficients belong to IM Hence (fminusg)(a1 an) = 0so by (b) the coefficients of f minus g belong to IM and this implies implies that the coefficients of falso belong to IM as required

20

Definition 12 Let A be a ring and M an A-module A sequence a1 an isin A is M -quasiregularif it satisfies the equivalent conditions of the Proposition Obviously this concept does not dependon the order of the elements But a1 ai for i lt n need not be M -quasiregular

Recall that for an A-module M a submodule N sube M and x isin A the notation (N x) meansm isin M |xm isin N This is a submodule of M If A is a ring I an ideal and M an A-modulerecall that M is separated in the I-adic topology when

⋂n I

nM = 0

Theorem 64 Let A be a ring M a nonzero A-module a1 an isin A and I = (a1 an)Then

(i) If a1 an is M -quasiregular and x isin A is such that (IM x) = IM then (ImM x) =ImM for all m gt 0

(ii) If a1 an is M -regular then it is M -quasiregular

(iii) If MMa1MM(a1 a2)M M(a1 anminus1)M are separated in the I-adic topologythen the converse of (ii) is also true

Proof (i) By induction on m with the case m = 1 true by assumption Suppose m gt 1 andxξ isin ImM By the inductive hypothesis ξ isin Imminus1M Hence there exists a homogenous polynomialf isinM [x1 xn] of degree mminus 1 such that ξ = f(a1 an) Since xξ = xf(a1 an) isin ImMthe coefficients of f are in (IM x) = IM Therefore ξ = f(a1 an) isin ImM

(ii) By induction on n For n = 1 this is easy to check Let n gt 1 and suppose a1 an is M -regular Then by the induction hypothesis a1 anminus1 is M -quasiregular Let f isinM [x1 xn]be homogenous of degree m gt 0 such that f(a1 an) = 0 We prove that f isin IM [x1 xn]by induction on m (the case m = 0 being trivial) Write

f(x1 xn) = g(x1 xnminus1) + xnh(x1 xn)

Then g and h are homogenous of degrees m and mminus 1 respectively By (i) we have

h(a1 an) isin ((a1 anminus1)mM an) = (a1 anminus1)mM sube ImM

Since by assumption a1 an is regular on M so an is regular on M(a1 anminus1)M andhence ((a1 anminus1)M an) = (a1 anminus1)M So by the induction hypothesis on m we haveh isin IM [x1 xn] (by the argument of Proposition 63) Since h(a1 an) isin (a1 anminus1)mMthere exists H isin M [x1 xnminus1] which is homogenous of degree m such that h(a1 an) =H(a1 anminus1) Putting

G(x1 xnminus1) = g(x1 xnminus1) + anH(x1 xnminus1)

we have G(a1 anminus1) = 0 so by the inductive hypothesis on n we have G isin IM [x1 xn]hence g isin IM [x1 xn] and so f isin IM [x1 xn]

(iii) By induction on n ge 1 Assume that a1 an is M -quasiregular and the modulesMMa1M M(a1 anminus1)M are all separated in the I-adic topology If a1ξ = 0 thenξ isin IM hence ξ =

sumaiηi and

suma1aiηi = 0 hence ηi isin IM and so ξ isin I2M In this way we

see that ξ isin⋂t ItM = 0 Thus a1 is regular on M and this also takes care of the case n = 1

since M 6= IM by the separation condition So assume n gt 1 By Lemma 57 it suffices to showthat a2 an is an N -regular sequence where N = Ma1M Since there is an isomorphism ofA-modules for 2 le i le nminus 1

M(a1 ai)M sim= N(a2 ai)N

The modules NNa2N N(a2 anminus1)N are separated in the I-adic topology So by theinductive hypothesis it suffices to show that the sequence a2 an is N -quasiregular

It suffices to show that if f(x2 xn) isin M [x2 xn] is homogenous of degree m ge 1 withf(a2 an) isin a1M then the coefficients of f belong to IM Put f(a2 an) = a1ω We claim

21

that ω isin Imminus1M Let 0 le i le mminus1 be the largest integer with ω isin IiM Then ω = g(a1 an)for some homogenous polynomial of degree i and

f(a2 an) = a1g(a1 an) (2)

If i lt mminus1 then g isin IM [x1 xn] and so ω isin Ii+1M which is a contradiction Hence i = mminus1and so ω isin Imminus1M Again using (2) we see that f(x2 xn)minusx1g(x1 xn) isin IM [x1 xn]Since f does not involve x1 we have f isin IM [x1 xn] as required

The theorem shows that under the assumptions of (iii) any permutation of an M -regularsequence is M -regular

Corollary 65 Let A be a noetherian ring M a finitely generated A-module and let a1 anbe contained in the Jacobson radical of A Then a1 an is M -regular if and only if it is M -quasiregular In particular if a1 an is M -regular so is any permutation of the sequence

Proof From [AM69] we know that for any ideal I contained in the Jacobson radical the I-adictopology on any finitely generated A-module is separated

If A is a ring and M an A-module then any M -regular sequence a1 an isin A gives rise toa strictly increasing chain of submodules a1M (a1 a2)M (a1 an)M Hence the chain ofideals (a1) (a1 a2) (a1 an) must also be strictly increasing

Lemma 66 Let A be a noetherian ring and M an A-module Any M -regular sequence a1 anin an ideal I can be extended to a maximal M -regular sequence in I

Proof If a1 an is not maximal in I we can find an+1 isin I such that a1 an an+1 is anM -regular sequence Either this process terminates at a maximal M -regular sequence in I or itproduces a strictly ascending chain of ideals

(a1) sub (a1 a2) sub (a1 a2 a3) sub middot middot middot

Since A is noetherian we can exclude this latter possibility

Theorem 67 Let A be a noetherian ring M a finitely generated A-module and I an ideal of Awith IM 6= M Let n gt 0 be an integer Then the following are equivalent

(1) ExtiA(NM) = 0 for i lt n and every finitely generated A-module N with Supp(N) sube V (I)

(2) ExtiA(AIM) = 0 for i lt n

(3) There exists a finitely generated A-module N with Supp(N) = V (I) and ExtiA(NM) = 0for i lt n

(4) There exists an M -regular sequence a1 an of length n in I

Proof (1) rArr (2) rArr (3) is trivial With I fixed we show that (3) rArr (4) for every finitely generatedmodule M with IM 6= M by induction on n We have 0 = Ext0A(NM) sim= HomA(NM) SinceM is finitely generated and nonzero the set of associated primes of M is finite and nonempty Ifno elements of I are M -regular then I is contained in the union of these associated primes andhence I sube p for some p isin Ass(M) (see [AM69] for details) By definition there is a monomorphismof A-modules φ Ap minusrarrM There is an isomorphism of A-modules

(Ap)psim= ApotimesA Ap

sim= AppAp = k

It is not hard to check this is an isomorphism of Ap-modules as well Since φp is a monomorphismand k 6= 0 it follows that HomAp(kMp) 6= 0 Since p isin V (I) = Supp(N) we have Np 6= 0 andso the k-module NppNp 6= 0 is nonzero and therefore free so Homk(NppNp k) 6= 0 Sincek sim= (Ap)p as Ap-modules it follows that HomA(NAp)p

sim= HomAp(Np (Ap)p) 6= 0 Since Apis isomorphic to a submodule of M it follows that HomA(NM) 6= 0 which is a contradiction

22

therefore there exists an M -regular element a1 isin I which takes care of the case n = 1 If n gt 1then put M1 = Ma1M From the exact sequence

0 Ma1 M M1

0 (3)

we get the long exact sequence

middot middot middot minusrarr ExtiA(NM) minusrarr ExtiA(NM1) minusrarr Exti+1A (NM) minusrarr middot middot middot

which shows that ExtiA(NM1) = 0 for 0 le i lt n minus 1 By the inductive hypothesis on n thereexists an M1-regular sequence a2 an in I The sequence a1 an is then an M -regularsequence in I

(4) rArr (1) By induction on n with I fixed For n = 1 we have a1 isin I regular on M and so (3)gives an exact sequence of R-modules

0 HomA(NM)a1 HomA(NM)

Where a1 denotes left multiplication by a1 Since Supp(N) = V (Ann(N)) sube V (I) it fol-lows that I sube r(Ann(N)) and so ar1N = 0 for some r gt 0 It follows that ar1 annihilatesHomA(NM) as well but since the action of ar1 on HomA(NM) gives an injective map it followsthat HomA(NM) = 0 Now assume n gt 1 and put M1 = Ma1M Then a2 an is anM1-regular sequence so by the inductive hypothesis ExtiA(NM1) = 0 for i lt nminus 1 So the longexact sequence corresponding to (3) gives an exact sequence for 0 le i lt n

0 ExtiA(NM)a1 ExtiA(NM)

Here a1 denotes left multiplication by a1 which is equal to ExtiA(αM) = Exti(N β) where α βare the endomorphisms given by left multiplication by a1 on NM respectively Assume that ar1annihilates N with r gt 0 Then αr is the zero map so ExtiA(αM)r = 0 and so ar1 also annihilatesExtiA(NM) Since the a1 is regular on this module it follows that ExtiA(NM) = 0 for i lt nas required

Corollary 68 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA with IM 6= M If a1 an a maximal M -regular sequence in I then ExtiA(AIM) = 0 fori lt n and ExtnA(AIM) 6= 0

Proof We already know that ExtiA(AIM) = 0 for i lt n so with I fixed we prove by inductionon n that ExtnA(AIM) 6= 0 for any finitely generated module M with IM 6= M admitting amaximal M -regular sequence of length n For n = 1 we have a1 isin I regular on M and an exactsequence (3) where M1 = Ma1M Part of the corresponding long exact sequence is

Ext0A(AIM) minusrarr Ext0A(AIM1) minusrarr Ext1A(AIM)

We know from the Theorem 67 that Ext0A(AIM) = 0 so it suffices to show that we haveHomA(AIM1) 6= 0 But if HomA(AIM1) = 0 then it follows from the proof of (3) rArr (4)above that there would be b isin I regular on M1 so a1 b is an M -regular sequence This is acontradiction since the sequence a1 was maximal so we conclude that Ext1A(AIM) 6= 0

Now assume n gt 1 and let a1 an be a maximal M -regular sequence in I Then a2 anis a maximal M1-regular sequence in I so by the inductive hypothesis Extnminus1

A (AIM1) 6= 0 Sofrom the long exact sequence for (3) we conclude that ExtnA(AIM) 6= 0 also

It follows that under the conditions of the Corollary every maximal M -regular sequence in Ihas a common length and you can find this length by looking at the sequence of abelian groups

HomA(AIM) Ext1A(AIM) Ext2A(AIM) ExtnA(AIM)

If there are M -regular sequences in I then this sequence will start off with n minus 1 zero groupswhere n ge 1 is the common length of every maximal M -regular sequence The nth group will

23

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

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as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

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Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

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Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

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Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

(iv) If B is noetherian then so is A

Proof (i) Let 0 6= x isin N Then 0 6= Ax sube N and since B is flat we see that AxotimesAB is isomorphicto the submodule (xotimes 1)B of N otimesA B It follows from Theorem 34 that xotimes 1 6= 0

(ii) By change of base B otimesA (AI) = BIB is faithfully flat over AI Now the assertionfollows from (i) For (iii) let p isin Spec(A) The ring Bp = B otimesA Ap is faithfully flat over Ap soby (ii) pBp 6= Bp Take a maximal ideal m of Bp containing pBp Then m capAp supe pAp thereforem capAp = pAp since pAp is maximal Putting q = m capB we get

q capA = (m capB) capA = m capA = (m capAp) capA = pAp capA = p

as required (iv) Follows immediately from (ii)

Theorem 38 Let ψ A minusrarr B be a morphism of rings The following conditions are equivalent

(1) ψ is faithfully flat

(2) ψ is flat and Ψ Spec(B) minusrarr Spec(A) is surjective

(3) ψ is flat and for any maximal ideal m of A there is a maximal ideal n of B lying over m

Proof (1) rArr (2) was proved above (2) rArr (3) By assumption there exists q isin Spec(B) withqcapA = m If n is any maximal ideal of B containing q then ncapA = m as m is maximal (3) rArr (1)The existence of n implies mB 6= B so B is faithfully flat over A by Theorem 34

Definition 4 In algebraic geometry we say a morphism of schemes f X minusrarr Y is flat if thelocal morphisms OYf(x) minusrarr OXx are flat for all x isin X We say the morphism is faithfully flat ifit is flat and surjective

Lemma 39 Let A be a ring and B a faithfully flat A-algebra Let M be an A-module Then

(i) M is flat (resp faithfully flat) over A hArr M otimesA B is so over B

(ii) If A is local and M finitely generated over A we have M is A-free hArr M otimesA B is B-free

Proof (i) Let N minusrarr N prime be a morphism of A-modules Both claims follow from commutativity ofthe following diagram

(N otimesAM)otimesA B

(N prime otimesAM)otimesA B

N otimesA (M otimesA B)

N prime otimesA (M otimesA B)

N otimesA (B otimesB (M otimesA B))

N prime otimesA (B otimesB (M otimesA B))

(N otimesA B)otimesB (M otimesA B) (N prime otimesA B)otimesB (M otimesA B)

(ii) The functor minusotimesAB preserves coproducts so the implication (rArr) is trivial (lArr) follows from(i) because under the hypothesis freeness of M is equivalent to flatness as we saw in Proposition24

12

22 Going-up and Going-down

Definition 5 Let φ A minusrarr B be a morphism of rings We say that the going-up theorem holdsfor φ if the following condition is satisfied

(GU) For any p pprime isin Spec(A) such that p sub pprime and for any prime q isin Spec(B) lyingover p there exists qprime isin Spec(B) lying over pprime such that q sub qprime

Similarly we say that the going-down theorem holds for φ if the following condition is satisfied

(GD) For any p pprime isin Spec(A) such that p sub pprime and for any prime qprime isin Spec(B) lyingover pprime there exists q isin Spec(B) lying over p such that q sub qprime

Lemma 40 The condition (GD) is equivalent to the following condition (GDrsquo) For any p isinSpec(A) and any minimal prime overideal q of pB we have q capA = p

Proof (GD) rArr (GDrsquo) Clearly q cap A supe p If this inclusion is proper then by (GD) there exists aprime q1 of B with q1 sub q and q1 capA = p contradicting minimality of q (GDrsquo) rArr (GD) Supposeprimes p sub pprime of A are given and qprime cap A = pprime We can shrink qprime to a prime q minimal among allprime ideals containing pB and by assumption q capA = p which completes the proof

Let a be any proper radical ideal in a noetherian ring B Then a is the intersection of allits minimal primes p1 pn and the closed irreducible sets V (p1) sube V (a) are the irreduciblecomponents of the closed set V (a) in the noetherian space Spec(B)

Let φ A minusrarr B a morphism of rings put X = Spec(A) Y = Spec(B) and let Ψ Y minusrarr Xthe corresponding morphism of affine schemes and suppose B is noetherian Then (GDprime) can beformulated geometrically as follows let p isin X put X prime = V (p) sube X and let Y prime be an arbitraryirreducible component of Ψminus1(X prime) (which we assume is nonempty) Then Ψ maps Y prime genericallyonto X prime in the sense that the generic point of Y prime is mapped to the generic point p of X prime

Theorem 41 Let φ A minusrarr B be a flat morphism of rings Then the going-down theorem holdsfor φ

Proof Let pprime sub p be prime ideals of A and let q be a prime ideal of B lying over p Then Bq isflat over Ap by Lemma 33 hence faithfully flat since Ap minusrarr Bq is local Therefore Spec(Bq) minusrarrSpec(Ap) is surjective Let qprimeprime be a prime ideal of Bq lying over pprimeAp Then qprime = qprimeprimecapB is a primeideal of B lying over pprime and contained in q

Theorem 42 Let B be a ring and A a subring over which B is integral Then

(i) The canonical map Spec(B) minusrarr Spec(A) is surjective

(ii) If two prime ideals q sube qprime lie over the same prime ideal of A then they are equal

(iii) The going-up theorem holds for A sube B

(iv) If A is a local ring and m its maximal ideal then the prime ideals of B lying over m areprecisely the maximal ideals of B

(v) If A and B are integral domains and A is integrally closed then the going-down theoremholds for A sube B

Proof See [AM69] or [Mat80] Theorem 5

23 Constructible Sets

Definition 6 A topological space X is noetherian if the descending chain condition holds forthe closed sets in X The spectrum Spec(A) of a noetherian ring A is noetherian If a spaceis covered by a finite number of noetherian subspaces then it is noetherian Any subspace of anoetherian space is noetherian A noetherian space is quasi-compact In a noetherian space Xany nonempty closed set Z is uniquely decomposed into a finite number of irreducible closed setsZ = Z1 cup middot middot middot cupZn such that Zi Zj for i 6= j The Zi are called the irreducible components of Z

13

Lemma 43 (Noetherian Induction) Let X be a noetherian topological space and P a propertyof closed subsets of X Assume that for any closed subset Y of X if P holds for every properclosed subset of Y then P holds for Y (in particular P holds for the empty set) Then P holdsfor X

Proof Suppose that P does not hold for X and let Z be the set of all proper closed subsetsof X which do not satisfy P Then since X is noetherian Z has a minimal element Y SinceY is minimal every proper closed subset of Y must satisfy P and therefore Y satisfies Pcontradicting the fact that Y isin Z

Lemma 44 Let X be a noetherian topological space and P a property of general subsets of XAssume that for any subset Y of X if P holds for every proper subset Y prime of Y with Y prime sub Y thenP holds for Y (in particular P holds for the empty set) Then P holds for X

Proof Suppose that P does not hold for X and let Z be the set of all closures Q of proper subsetsQ of X with Q sub X and P not holding for Q Let Q be a minimal element of Z If Qprime is anyproper subset of Q with Qprime sub Q then Qprime must satisfy P otherwise Qprime would contradict minimalityof Q in Z But by assumption this implies that Q satisfies P which is a contradiction

Definition 7 Let X be a topological space and Z a subset of X We say Z is locally closed inX if it satisfies the following equivalent properties

(i) Every point z isin Z has an open neighborhood U in X such that U cap Z is closed in U

(ii) Z is the intersection of an open set in X and a closed set in X

(iii) Z is an open subset of its closure

Definition 8 Let X be a noetherian space We say a subset Z of X is a constructible set in Xif it is a finite union of locally closed sets in X so Z =

⋃mi=1(Ui capFi) with Ui open and Fi closed

The set F of all constructible subsets of X is the smallest collection of subsets of X containing allthe open sets which is closed with respect to the formation of finite intersections and complementsIt follows that all open and closed sets are constructible and F is also closed under finite unions

We say that a subset Z is pro-constructible (resp ind-constructible) if it is the intersection(resp union) of an arbitrary collection of constructible sets in X

Proposition 45 Let X be a noetherian space and Z a subset of X Then Z is constructible ifand only if the following condition is satisfied

(lowast) For each irreducible closed set X0 in X either X0capZ is not dense in X0 or X0capZcontains a nonempty open set of X0

Proof Assume that Z is constructible and Z cap X0 nonempty Then we can write X0 cap Z =⋃mi=1 Ui capFi for Ui open in X Fi closed and irreducible in X (by taking irreducible components)

and UicapFi nonempty for all i Then Ui cap Fi = Fi since Fi is irreducible therefore X0 cap Z =⋃i Fi

If X0 cap Z is dense in X0 we have X0 =⋃i Fi so that some Fi say F1 is equal to X0 Then

U1 capX0 = U1 cap F1 is a nonempty open subset of X0 contained in X0 cap ZNext we prove the converse We say that a subset T of X has the property P if whenever a

subset Z of T satisfies (lowast) it is constructible We need to show that X has the property P forwhich we use the form of noetherian induction given in Lemma 44 Suppose that Y is a subsetof X with P holding for every proper subset Y prime of Y with Y prime sub Y We need to show that Pholds for Y Let Z be a nonempty subset of Y satisfying (lowast) and let Z = F1 cup cup Fr be thedecomposition of Z into irreducible components Since Z = Z cap F1 cup middot middot middot cup Z cap Fr we have

F1 = F1 cap Z = F1 cap(Z cap F1 cup cup Z cap Fr

)= (F1 cap Z cap F1) cup middot middot middot cup (F1 cap Z cap Fr)

Since F1 is irreducible and not contained in any other Fi we must have F1 = Z cap F1 so F1 cap Zis dense in F1 whence by (lowast) there exists a proper closed subset F prime of F1 such that F1 F prime sube ZThen putting F lowast = F prime cupF2 cup middot middot middot cupFr we have Z = (F1 F prime)cup (Z capF lowast) The set F1 F prime is locally

14

closed in X so to complete the proof it suffices to show that Z cap F lowast is constructible in X SinceZ cap F lowast sube F lowast sub Z sube Y by the inductive hypothesis P holds for ZcapF lowast so it suffices to show thatZ cap F lowast satisfies (lowast) If X0 is irreducible and Z cap F lowast capX0 = X0 the closed set F lowast must containX0 and so Z cap F lowast capX0 = Z capX0 which contains a nonempty open subset of X0 since Z satisfies(lowast) and clearly Z capX0 is dense in X0

Lemma 46 Let φ A minusrarr B be a morphism of rings and f Spec(B) minusrarr Spec(A) the corre-sponding morphism of schemes Then f dominant if and only if Kerφ sube nil(A) In particular ifA is reduced the f dominant if and only if φ is injective

Proof Let X = Spec(A) and Y = Spec(B) The closure f(Y ) is the closed set V (I) defined bythe ideal I =

⋂pisinY φ

minus1p = φminus1⋂

pisinY p which is φminus1(nil(B)) Clearly Kerφ sube I Suppose thatf(Y ) is dense in X Then V (I) = X whence I = nil(A) and so Kerφ sube nil(A) Converselysuppose Kerφ sube nil(A) Then it is clear that I = φminus1(nil(B)) = nil(A) which means thatf(Y ) = V (I) = X

3 Associated Primes

This material can be found in [AM69] Chapter 11 webnotes of Robert Ash or in [Mat80] itselfThere is not much relevant to add here apart from a few small comments

Lemma 47 Let A be a ring and M an A-module Let a be an ideal in A that is maximal amongall annihilators of nonzero elements of M Then a is prime

Proof Say a = Ann(x) Given ab isin a we must show that a isin a or b isin a Assume a isin aThen ax 6= 0 We note that Ann(ax) supe a By hypothesis it cannot properly be larger HenceAnn(ax) = a Now b annihilates ax hence b isin a

Lemma 48 Let A be a noetherian ring and M an A-module If 0 6= a isin M then Ann(a) iscontained in an associated prime of M

Proposition 49 Let A be a noetherian ring and M a nonzero finitely generated A-module Amaximal ideal m is an associated prime of M if and only if no element of m is regular on M

Proof One implication is obvious If x isin m is not regular on M say x isin Ann(b) for some nonzerob then x is contained in an associated prime of M Thus m is contained in the finite union of theassociated primes of M and since m is maximal it must be one of them

Proposition 50 Let A be a nonzero noetherian ring I an ideal and M a nonzero finitelygenerated A-module If there exist elements x y isin I with x regular on A and y regular on M thenthere exists an element of I regular on both A and M

Proof Let p1 pn be the associated primes of A and q1 qm the associated primes of M By assumption I is not contained in any of these primes But if no element of I is regular on bothA and M then I is contained in the union p1 cup middot middot middot cup pn cup q1 cup middot middot middot cup qm and therefore containedin one of these primes which is a contradiction

4 Dimension

This is covered in [AM69] so we restrict ourselves here to mentioning some of the major pointsRecall that an ideal q sube R in a ring is primary if it is proper and if whenever xy isin q we haveeither x isin q or yn isin q for some n gt 0 Then the radical of q is a prime ideal p and we say q isa p-primary ideal If a is an ideal and b supe a is p-primary then in the ring Ra the ideal ba ispa-primary A minimal primary decomposition of an ideal b is an expression

b = q1 cap middot middot middot cap qn

15

where capj 6=iqj qi for all i and the primes pi = r(qi) are all distinct If a is an ideal contained inb then

ba = q1a cap middot middot middot cap qna

is a minimal primary decomposition of ba in AaLet A be a nonzero ring Recall that dimension of an A-module M is the Krull dimension of

the ring AAnn(M) and is defined for all modules M (minus1 if M = 0) The rank is defined for freeA-modules and is the common size of any basis (0 if M = 0) Throughout these notes dim(M)will denote the dimension not the rank

Definition 9 Let (Am k) be a noetherian local ring of dimension d An ideal of definition is anm-primary ideal Recall that the dimension of A is the size of the smallest collection of elementsof A which generates an m-primary ideal Recall that rankk(mm2) is equal to the size of thesmallest set of generators for m as an ideal so always d le rankk(mm2)

A system of parameters is a set of d elements generating an m-primary ideal If d = rankk(mm2)or equivalently there is a system of parameters generating m we say that A is a regular local ringand we call such a system of parameters a regular system of parameters

Proposition 51 Let (Am) be a noetherian local ring of dimension d ge 1 and let x1 xd bea system of parameters of A Then

dim(A(x1 xi)) = dminus i = dim(A)minus i

for each 1 le i le d

Proof Put A = A(x1 xi) If i = d then the zero ideal in A is an ideal of definition so clearlydim(A) = 0 If 1 le i lt d then dim(A) le dminus i since xi+1 xd generate an ideal of definition ofA Let dim(A) = p If p = 0 then (x1 xi) must be an ideal of definition contradicting i lt dSo p ge 1 and if y1 yp is a system of parameters of A then x1 xi y1 yp generate anideal of definition of A so that p+ i ge d That is p ge dminus i

Definition 10 Let A be a nonzero ring and I a proper ideal The height of I denoted htI isthe minimum of the heights of the prime ideals containing I

htI = infhtp | p supe I

This is a number in 0 1 2 infin Equivalently we can take the infimum over the heights ofprimes minimal over I Clearly ht0 = 0 and if I sube J are proper ideals then it is clear thathtI le htJ If I is a prime ideal then htI is the usual height of a prime ideal If A is a noetherianring then htI ltinfin for every proper ideal I since Ap is a local noetherian ring and htp = dim(Ap)

Lemma 52 Let A be a nonzero ring and I a proper ideal Then we have

htI = infhtIAm |m is a maximal ideal and I sube m

Lemma 53 Let A be a noetherian ring and suppose we have an exact sequence

0 minusrarrM prime minusrarrM minusrarrM primeprime minusrarr 0

in which M primeMM primeprime are nonzero and finitely generated Then dimM = maxdimM prime dimM primeprime

Proof We know that Supp(M) = Supp(M prime)cupSupp(M primeprime) and for all three modules the dimensionis the supremum of the coheights of prime ideals in the support So the result is straightforwardto check

16

41 Homomorphism and Dimension

Let φ A minusrarr B be a morphism of rings If p isin Spec(A) then put κ(p) = AppAp Let Bp denotethe ring Tminus1B where T = φ(Aminusp) There is an isomorphism of A-algebras Bp

sim= BotimesAAp Thereis a commutative diagram of rings

B

))RRRRRRRRRRRRRRRR Bp BppBp

B otimesA κ(p)

The vertical isomorphism is defined by bφ(s)+ pBp 7rarr botimes (1s+ pAp) We call Spec(BotimesA κ(p))the fibre over p of the map Φ Spec(B) minusrarr Spec(A) Since primes of BppBp clearly correspondto primes q of B with q cap A = p it is easy to see that the ring morphism B minusrarr B otimesA κ(p) givesrise to a continuous map Spec(B otimesA κ(p)) minusrarr SpecB which gives a homeomorphism betweenΦminus1(p) and Spec(B otimesA κ(p)) See [AM69] Chapter 3

Lemma 54 Let q be a prime ideal of B with q cap A = p and let P be the corresponding primeideal of B otimesA κ(p) Then there is an isomorphism of A-algebras

Bq otimesA κ(p) sim= (B otimesA κ(p))P

btotimes (as+ pAp) 7rarr (botimes (a1 + pAp))(totimes (s1 + pAp))

Proof It is not difficult to see that there is an isomorphism of rings Bqsim= (Bp)qBp defined by

bt 7rarr (b1)(t1) Consider the prime ideal qBppBp We know that there is a ring isomorphism

(B otimesA κ(p))Psim= (BppBp)qBppBp

sim= (Bp)qBpp(Bp)qBpsim= BqpBq

by the comments following Lemma 7 It is not hard to check there is a ring isomorphism BqpBqsim=

BqotimesA κ(p) defined by bt+ pBq 7rarr btotimes 1 (the inverse of btotimes (as+ pAp) is bφ(a)tφ(s)+ pBq)So by definition of P there is an isomorphism of rings (B otimesA κ(p))P

sim= Bq otimesA κ(p) and this isclearly an isomorphism of A-algebras

In particular if φ A minusrarr B is a ring morphism p isin Spec(A) and q isin Spec(B) such thatq capA = p then there is an isomorphism of rings (BpB)qpB

sim= BqpBq so we have ht(qpB) =dim(Bq otimesA κ(p))

Theorem 55 Let φ A minusrarr B be a morphism of noetherian rings Let q isin Spec(B) and putp = q capA Then

(1) htq le htp + ht(qpB) In other words dim(Bq) le dim(Ap) + dim(Bq otimesA κ(p))

(2) We have equality in (1) if the going-down theorem holds for φ (in particular if φ is flat)

(3) If Φ Spec(B) minusrarr Spec(A) is surjective and if the going-down theorem holds then we havedim(B) ge dim(A) and htI = ht(IB) for any proper ideal I of A

Proof (1) Replacing A by Ap and B by Bq we may suppose that (A p) and (B q) are noetherianlocal rings such that q cap A = p and we must show that dim(B) le dim(A) + dim(BpB) Let Ibe a p-primary ideal of A Then pn sube I for some n gt 0 so pnB sube IB sube pB Thus the ideals pBand IB have the same radical and so by definition dim(BpB) = dim(BIB) If dimA = 0 thenwe can take I = 0 and the result is trivial So assume dimA = r ge 1 and let I = (a1 ar) fora system of parameters a1 ar If dim(BIB) = 0 then IB is an q-primary ideal of B and sodim(B) le r as required Otherwise if dim(BIB) = s ge 1 let b1 + IB bs + IB be a systemof parameters of BIB Then b1 bs a1 ar generate an ideal of definition of B Hencedim(B) le r + s

(2) We use the same notation as above If ht(qpB) = s ge 0 then there exists a prime chainof length s q = q0 sup q1 sup middot middot middot sup qs such that qs supe pB As p = q cap A supe qi cap A supe p all the qi

17

lie over p If htp = r ge 0 then there exists a prime chain p = p0 sup p1 sup middot middot middot sup pr in A and bygoing-down there exists a prime chain qs = t0 sup t1 sup middot middot middot sup tr of B such that ti capA = pi Then

q = q0 sup middot middot middot sup qs sup t1 sup middot middot middot sup tr

is a prime chain of length r + s therefore htq ge r + s(3) (i) follows from (2) since dim(A) = suphtp | p isin Spec(A) (ii) Since Φ is surjective IB

is a proper ideal Let q be a minimal prime over IB such that htq = ht(IB) and put p = q capAThen ht(qpB) = 0 so by (2) we find that ht(IB) = htq = htp ge htI For the reverseinclusion let p be a minimal prime ideal over I such that htp = htI and take a prime q of Blying over p Replacing q if necessary we may assume that q is a minimal prime ideal over pBThen htI = htp = htq ge ht(IB)

Theorem 56 Let A be a nonzero subring of B and suppose that B is integral over A Then

(1) dim(A) = dim(B)

(2) Let q isin Spec(B) and set p = q capA Then we have cohtp = cohtq and htq le htp

(3) If the going-down theorem holds between A and B then for any ideal J of B with J capA 6= Awe have htJ = ht(J capA)

Proof (1) By Theorem 42 the going-up theorem holds for A sube B and Spec(B) minusrarr Spec(A) issurjective so we can lift any chain of prime ideals p0 sub p1 sub middot middot middot sub pn in A to a chain of primeideals q0 sub middot middot middot sub qn in B On the other hand if q sube qprime are prime ideals of B and q capA = qprime capAthen q = qprime so any chain of prime ideals in B restricts to a chain of the same length in A Hencedim(A) = dim(B)

(2) Since Bq is integral over Ap it is clear from (1) that cohtp = cohtq If q = q0 sup middot middot middot sup qnis a chain of prime ideals in B then intersecting with A gives a chain of length n descending fromp Hence htq le htp

(3) Given the going-down theorem it is clear that htq = htp in (2) Let J be a proper ideal ofB with J capA 6= A and let q be such that htJ = htq Then ht(J capA) le ht(qcapA) = htq = htJ On the other hand BJ is integral over BJ cap A so every prime ideal p of A containing J cap Acan be lifted to a prime ideal q of B containing J In particular we can lift a prime ideal p withht(J capA) = htp to see that htJ le htq = htp = ht(J capA) as required

5 Depth

Definition 11 Let A be a ring M an A-module and a1 ar a sequence of elements of AWe say a1 ar is an M -regular sequence (or simply M -sequence) if the following conditions aresatisfied

(1) For each 2 le i le r ai is regular on M(a1 aiminus1)M and a1 is regular on M

(2) M 6= (a1 an)M

If a1 ar is an M -regular sequence then so is a1 ai for any i le r When all ai belong to anideal I we say a1 ar is an M -regular sequence in I If moreover there is no b isin I such thata1 ar b is M -regular then a1 ar is said to be a maximal M -regular sequence in I Noticethat the notion of M -regular depends on the order of the elements in the sequence If MN areisomorphic A-modules then a sequence is regular on M iff it is regular on N

Lemma 57 A sequence a1 ar with r ge 2 is M -regular if and only if a1 is regular on Mand a2 ar is an Ma1M -regular sequence If the sequence a1 ar is a maximal M -regularsequence in I then a2 ar is a maximal Ma1M -regular sequence in I

18

Proof The key point is that for ideals a sube b there is a canonical isomorphism of A-modulesMbM sim= NbN where N = MaM If a1 ar is M -regular then a1 is regular on M a2 isregular on N = Ma1M and for 3 le i le r ai is regular on

M(a1 aiminus1)M sim= N(a2 aiminus1)N

Hence a2 ar is an N -regular sequence The converse follows from the same argument

More generally if a1 ar is an M -regular sequence and we set N = M(a1 ar)M andif b1 bs is an N -regular sequence then a1 ar b1 bs is an M -regular sequence

Lemma 58 If a1 ar is an A-regular sequence and M is a flat A-module then a1 ar isalso M -regular provided (a1 ar)M 6= M

Proof Left multiplication by a1 defines a monomorphism A minusrarr A since a1 is A-regular Tensoringwith M and using the fact that M is flat we see that left multiplication by a1 also gives amonomorphismM minusrarrM as required Similarly tensoring with the monomorphism a2 Aa1 minusrarrAa1 we get a monomorphism Ma1M minusrarrMa1M and so on

Lemma 59 Let A be a ring and M an A-module Given an integer n ge 1 a sequence a1 aris M -regular if and only if it is Mn-regular

Proof Suppose the sequence a1 ar is M -regular We prove it is Mn-regular by induction on rThe case r = 1 is trivial so assume r gt 1 By the inductive hypothesis the sequence a1 arminus1 isMn-regular Let L = (a1 arminus1)M Then (a1 arminus1)Mn = Ln and there is an isomorphismof A-modules MnLn sim= (ML)n So we need only show that ar is regular on (ML)n Since byassumption it is regular on ML this is not hard to check Clearly (a1 ar)Mn 6= Mn so thesequence a1 ar is Mn-regular as required The converse is similarly checked

Lemma 60 Let A be a nonzero ring M an A-module and a1 ar isin A If a1 ar isin Am isMm-regular for every maximal ideal m of A then the sequence a1 ar is M -regular

Proof This follows from the fact that given an A-module M an element a isin A is regular on M ifand only if its image in Am is regular on Mm for every maximal ideal m of A

Lemma 61 Suppose that a1 ar is M -regular and a1ξ1 + middot middot middot + arξr = 0 for ξi isin M Thenξi isin (a1 ar)M for all i

Proof By induction on r For r = 1 a1ξ1 = 0 implies that ξ1 = 0 Let r gt 1 Since ar is regularon M(a1 arminus1)M we have ξr =

sumrminus1i=1 aiηi so

sumrminus1i=1 ai(ξi + arηi) = 0 By the inductive

hypothesis for i lt r we have ξi + arηi isin (a1 arminus1)M so that ξi isin (a1 ar)M

Theorem 62 Let A be a ring and M an A-module and let a1 ar be an M -regular sequenceThen for every sequence n1 nr of integers gt 0 the sequence an1

1 anrr is M -regular

Proof Suppose we can prove the following statement

(lowast) Given an integer n gt 0 an A-module M and any M -regular sequence a1 arthe sequence an1 a2 ar is also M -regular

We prove the rest of the Theorem by induction on r For r = 1 this follows immediately from (lowast)Let r gt 1 and suppose a1 ar is M -regular Then by (lowast) an1

1 a2 ar is M -regular Hencea2 ar is Man1

1 M -regular By the inductive hypothesis an22 anr

r is Man11 M -regular and

therefore an11 anr

r is M -regular by Lemma 57So it only remains to prove (lowast) which we do by induction on n The case n = 1 is trivial so let

n gt 1 be given along with an A-moduleM and anM -regular sequence a1 ar By the inductivehypothesis anminus1

1 a2 ar is M -regular and clearly an1 is regular on M Since (an1 a2 ar) sube(anminus1

1 a2 ar) it is clear thatM 6= (an1 a2 ar)M Let i gt 1 and assume that an1 a2 aiminus1

is an M -regular sequence We need to show that ai is regular on M(an1 a2 aiminus1)M Suppose

19

that aiω = an1 ξ1 +a2ξ2 + middot middot middot+aiminus1ξiminus1 Then ω = anminus11 η1 +a2η2 + middot middot middot+aiminus1ηiminus1 by the inductive

hypothesis So

anminus11 (a1ξ1 minus aiη1) + a2(ξ2 minus aiη2) + middot middot middot+ aiminus1(ξiminus1 minus aiηiminus1) = 0

Hence a1ξ1minusaiη1 isin (anminus11 a2 aiminus1)M by Lemma 61 It follows that aiη1 isin (a1 a2 aiminus1)M

hence η1 isin (a1 aiminus1)M and so ω isin (an1 a2 aiminus1)M as required This proves (lowast) andtherefore completes the proof

Let A be a ring There is an isomorphism of A[x1 xn]-modules M otimesA A[x1 xn] sim=M [x1 xn] where the latter module consists of polynomials in x1 xn with coefficients in M(see our Polynomial Ring notes) For any f(x1 xn) isinM [x1 xn] and tuple (a1 an) isinAn we can define an element of M

f(a1 an) =sumα

aα11 middot middot middot aαn

n middot f(α)

For an element r isin R and h isinM [x1 xn]

(f + h)(a1 an) = f(a1 an) + h(a1 an)(r middot f)(a1 an) = r middot f(a1 an)

For an ideal I in R the R-submodule IM [x1 xn] consists of all polynomials whose coefficientsare in the R-submodule IM subeM

Let us review the definition of the associated graded ring and modules Let A be a ring and Ian ideal of A Then the abelian group

grI(A) = AI oplus II2 oplus I2I3 oplus middot middot middot

becomes a graded ring in a fairly obvious way For an A-module M we have the graded grI(A)-module

grI(M) = MIM oplus IMI2M oplus I2MI3M oplus middot middot middotIf A is noetherian and M is a finitely generated A-module then grI(A) is a noetherian ring andif grI(M) is a finitely generated grI(A)-module

Given elements a1 an isin A and I = (a1 an) we define a morphism of abelian groupsψ M [x1 xn] minusrarr grI(M) as follows if f is homogenous of degreem ge 0 define ψ(f) to be theimage of f(a1 an) in ImMIm+1M This defines a morphism of groups M [x1 xn]m minusrarrImMIm+1M and together these define the morphism of groups ψ Since ψ maps IM [x1 xn]to zero it induces a morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) and

Proposition 63 Let A be a ring and M an A-module Let a1 an isin A and set I =(a1 an) Then the following conditions are equivalent

(a) For every m gt 0 and for every homogenous polynomial f(x1 xn) isin M [x1 xn] ofdegree m such that f(a1 an) isin Im+1M we have f isin IM [x1 xn]

(b) If f(x1 xn) isin M [x1 xn] is homogenous and f(a1 an) = 0 then the coefficientsof f are in IM

(c) The morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) defined by mappinga homogenous polynomial f(x1 xn) of degree m to f(a1 an) isin ImMIm+1M is anisomorphism

Proof It is easy to see that (a) hArr (c) and (a) rArr (b) To show (b) rArr (a) let f isinM [x1 xn] bea homogenous polynomial of degree m gt 0 and suppose f(a1 an) isin Im+1M Any element ofIm+1M can be written as a sum of terms of the form aα1

1 middot middot middot aαnn middotm with

sumi αi = m+1 By shifting

one of the ai across we can write f(a1 an) = g(a1 an) for a homogenous polynomialg isinM [x1 xn] of degreem all of whose coefficients belong to IM Hence (fminusg)(a1 an) = 0so by (b) the coefficients of f minus g belong to IM and this implies implies that the coefficients of falso belong to IM as required

20

Definition 12 Let A be a ring and M an A-module A sequence a1 an isin A is M -quasiregularif it satisfies the equivalent conditions of the Proposition Obviously this concept does not dependon the order of the elements But a1 ai for i lt n need not be M -quasiregular

Recall that for an A-module M a submodule N sube M and x isin A the notation (N x) meansm isin M |xm isin N This is a submodule of M If A is a ring I an ideal and M an A-modulerecall that M is separated in the I-adic topology when

⋂n I

nM = 0

Theorem 64 Let A be a ring M a nonzero A-module a1 an isin A and I = (a1 an)Then

(i) If a1 an is M -quasiregular and x isin A is such that (IM x) = IM then (ImM x) =ImM for all m gt 0

(ii) If a1 an is M -regular then it is M -quasiregular

(iii) If MMa1MM(a1 a2)M M(a1 anminus1)M are separated in the I-adic topologythen the converse of (ii) is also true

Proof (i) By induction on m with the case m = 1 true by assumption Suppose m gt 1 andxξ isin ImM By the inductive hypothesis ξ isin Imminus1M Hence there exists a homogenous polynomialf isinM [x1 xn] of degree mminus 1 such that ξ = f(a1 an) Since xξ = xf(a1 an) isin ImMthe coefficients of f are in (IM x) = IM Therefore ξ = f(a1 an) isin ImM

(ii) By induction on n For n = 1 this is easy to check Let n gt 1 and suppose a1 an is M -regular Then by the induction hypothesis a1 anminus1 is M -quasiregular Let f isinM [x1 xn]be homogenous of degree m gt 0 such that f(a1 an) = 0 We prove that f isin IM [x1 xn]by induction on m (the case m = 0 being trivial) Write

f(x1 xn) = g(x1 xnminus1) + xnh(x1 xn)

Then g and h are homogenous of degrees m and mminus 1 respectively By (i) we have

h(a1 an) isin ((a1 anminus1)mM an) = (a1 anminus1)mM sube ImM

Since by assumption a1 an is regular on M so an is regular on M(a1 anminus1)M andhence ((a1 anminus1)M an) = (a1 anminus1)M So by the induction hypothesis on m we haveh isin IM [x1 xn] (by the argument of Proposition 63) Since h(a1 an) isin (a1 anminus1)mMthere exists H isin M [x1 xnminus1] which is homogenous of degree m such that h(a1 an) =H(a1 anminus1) Putting

G(x1 xnminus1) = g(x1 xnminus1) + anH(x1 xnminus1)

we have G(a1 anminus1) = 0 so by the inductive hypothesis on n we have G isin IM [x1 xn]hence g isin IM [x1 xn] and so f isin IM [x1 xn]

(iii) By induction on n ge 1 Assume that a1 an is M -quasiregular and the modulesMMa1M M(a1 anminus1)M are all separated in the I-adic topology If a1ξ = 0 thenξ isin IM hence ξ =

sumaiηi and

suma1aiηi = 0 hence ηi isin IM and so ξ isin I2M In this way we

see that ξ isin⋂t ItM = 0 Thus a1 is regular on M and this also takes care of the case n = 1

since M 6= IM by the separation condition So assume n gt 1 By Lemma 57 it suffices to showthat a2 an is an N -regular sequence where N = Ma1M Since there is an isomorphism ofA-modules for 2 le i le nminus 1

M(a1 ai)M sim= N(a2 ai)N

The modules NNa2N N(a2 anminus1)N are separated in the I-adic topology So by theinductive hypothesis it suffices to show that the sequence a2 an is N -quasiregular

It suffices to show that if f(x2 xn) isin M [x2 xn] is homogenous of degree m ge 1 withf(a2 an) isin a1M then the coefficients of f belong to IM Put f(a2 an) = a1ω We claim

21

that ω isin Imminus1M Let 0 le i le mminus1 be the largest integer with ω isin IiM Then ω = g(a1 an)for some homogenous polynomial of degree i and

f(a2 an) = a1g(a1 an) (2)

If i lt mminus1 then g isin IM [x1 xn] and so ω isin Ii+1M which is a contradiction Hence i = mminus1and so ω isin Imminus1M Again using (2) we see that f(x2 xn)minusx1g(x1 xn) isin IM [x1 xn]Since f does not involve x1 we have f isin IM [x1 xn] as required

The theorem shows that under the assumptions of (iii) any permutation of an M -regularsequence is M -regular

Corollary 65 Let A be a noetherian ring M a finitely generated A-module and let a1 anbe contained in the Jacobson radical of A Then a1 an is M -regular if and only if it is M -quasiregular In particular if a1 an is M -regular so is any permutation of the sequence

Proof From [AM69] we know that for any ideal I contained in the Jacobson radical the I-adictopology on any finitely generated A-module is separated

If A is a ring and M an A-module then any M -regular sequence a1 an isin A gives rise toa strictly increasing chain of submodules a1M (a1 a2)M (a1 an)M Hence the chain ofideals (a1) (a1 a2) (a1 an) must also be strictly increasing

Lemma 66 Let A be a noetherian ring and M an A-module Any M -regular sequence a1 anin an ideal I can be extended to a maximal M -regular sequence in I

Proof If a1 an is not maximal in I we can find an+1 isin I such that a1 an an+1 is anM -regular sequence Either this process terminates at a maximal M -regular sequence in I or itproduces a strictly ascending chain of ideals

(a1) sub (a1 a2) sub (a1 a2 a3) sub middot middot middot

Since A is noetherian we can exclude this latter possibility

Theorem 67 Let A be a noetherian ring M a finitely generated A-module and I an ideal of Awith IM 6= M Let n gt 0 be an integer Then the following are equivalent

(1) ExtiA(NM) = 0 for i lt n and every finitely generated A-module N with Supp(N) sube V (I)

(2) ExtiA(AIM) = 0 for i lt n

(3) There exists a finitely generated A-module N with Supp(N) = V (I) and ExtiA(NM) = 0for i lt n

(4) There exists an M -regular sequence a1 an of length n in I

Proof (1) rArr (2) rArr (3) is trivial With I fixed we show that (3) rArr (4) for every finitely generatedmodule M with IM 6= M by induction on n We have 0 = Ext0A(NM) sim= HomA(NM) SinceM is finitely generated and nonzero the set of associated primes of M is finite and nonempty Ifno elements of I are M -regular then I is contained in the union of these associated primes andhence I sube p for some p isin Ass(M) (see [AM69] for details) By definition there is a monomorphismof A-modules φ Ap minusrarrM There is an isomorphism of A-modules

(Ap)psim= ApotimesA Ap

sim= AppAp = k

It is not hard to check this is an isomorphism of Ap-modules as well Since φp is a monomorphismand k 6= 0 it follows that HomAp(kMp) 6= 0 Since p isin V (I) = Supp(N) we have Np 6= 0 andso the k-module NppNp 6= 0 is nonzero and therefore free so Homk(NppNp k) 6= 0 Sincek sim= (Ap)p as Ap-modules it follows that HomA(NAp)p

sim= HomAp(Np (Ap)p) 6= 0 Since Apis isomorphic to a submodule of M it follows that HomA(NM) 6= 0 which is a contradiction

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therefore there exists an M -regular element a1 isin I which takes care of the case n = 1 If n gt 1then put M1 = Ma1M From the exact sequence

0 Ma1 M M1

0 (3)

we get the long exact sequence

middot middot middot minusrarr ExtiA(NM) minusrarr ExtiA(NM1) minusrarr Exti+1A (NM) minusrarr middot middot middot

which shows that ExtiA(NM1) = 0 for 0 le i lt n minus 1 By the inductive hypothesis on n thereexists an M1-regular sequence a2 an in I The sequence a1 an is then an M -regularsequence in I

(4) rArr (1) By induction on n with I fixed For n = 1 we have a1 isin I regular on M and so (3)gives an exact sequence of R-modules

0 HomA(NM)a1 HomA(NM)

Where a1 denotes left multiplication by a1 Since Supp(N) = V (Ann(N)) sube V (I) it fol-lows that I sube r(Ann(N)) and so ar1N = 0 for some r gt 0 It follows that ar1 annihilatesHomA(NM) as well but since the action of ar1 on HomA(NM) gives an injective map it followsthat HomA(NM) = 0 Now assume n gt 1 and put M1 = Ma1M Then a2 an is anM1-regular sequence so by the inductive hypothesis ExtiA(NM1) = 0 for i lt nminus 1 So the longexact sequence corresponding to (3) gives an exact sequence for 0 le i lt n

0 ExtiA(NM)a1 ExtiA(NM)

Here a1 denotes left multiplication by a1 which is equal to ExtiA(αM) = Exti(N β) where α βare the endomorphisms given by left multiplication by a1 on NM respectively Assume that ar1annihilates N with r gt 0 Then αr is the zero map so ExtiA(αM)r = 0 and so ar1 also annihilatesExtiA(NM) Since the a1 is regular on this module it follows that ExtiA(NM) = 0 for i lt nas required

Corollary 68 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA with IM 6= M If a1 an a maximal M -regular sequence in I then ExtiA(AIM) = 0 fori lt n and ExtnA(AIM) 6= 0

Proof We already know that ExtiA(AIM) = 0 for i lt n so with I fixed we prove by inductionon n that ExtnA(AIM) 6= 0 for any finitely generated module M with IM 6= M admitting amaximal M -regular sequence of length n For n = 1 we have a1 isin I regular on M and an exactsequence (3) where M1 = Ma1M Part of the corresponding long exact sequence is

Ext0A(AIM) minusrarr Ext0A(AIM1) minusrarr Ext1A(AIM)

We know from the Theorem 67 that Ext0A(AIM) = 0 so it suffices to show that we haveHomA(AIM1) 6= 0 But if HomA(AIM1) = 0 then it follows from the proof of (3) rArr (4)above that there would be b isin I regular on M1 so a1 b is an M -regular sequence This is acontradiction since the sequence a1 was maximal so we conclude that Ext1A(AIM) 6= 0

Now assume n gt 1 and let a1 an be a maximal M -regular sequence in I Then a2 anis a maximal M1-regular sequence in I so by the inductive hypothesis Extnminus1

A (AIM1) 6= 0 Sofrom the long exact sequence for (3) we conclude that ExtnA(AIM) 6= 0 also

It follows that under the conditions of the Corollary every maximal M -regular sequence in Ihas a common length and you can find this length by looking at the sequence of abelian groups

HomA(AIM) Ext1A(AIM) Ext2A(AIM) ExtnA(AIM)

If there are M -regular sequences in I then this sequence will start off with n minus 1 zero groupswhere n ge 1 is the common length of every maximal M -regular sequence The nth group will

23

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

22 Going-up and Going-down

Definition 5 Let φ A minusrarr B be a morphism of rings We say that the going-up theorem holdsfor φ if the following condition is satisfied

(GU) For any p pprime isin Spec(A) such that p sub pprime and for any prime q isin Spec(B) lyingover p there exists qprime isin Spec(B) lying over pprime such that q sub qprime

Similarly we say that the going-down theorem holds for φ if the following condition is satisfied

(GD) For any p pprime isin Spec(A) such that p sub pprime and for any prime qprime isin Spec(B) lyingover pprime there exists q isin Spec(B) lying over p such that q sub qprime

Lemma 40 The condition (GD) is equivalent to the following condition (GDrsquo) For any p isinSpec(A) and any minimal prime overideal q of pB we have q capA = p

Proof (GD) rArr (GDrsquo) Clearly q cap A supe p If this inclusion is proper then by (GD) there exists aprime q1 of B with q1 sub q and q1 capA = p contradicting minimality of q (GDrsquo) rArr (GD) Supposeprimes p sub pprime of A are given and qprime cap A = pprime We can shrink qprime to a prime q minimal among allprime ideals containing pB and by assumption q capA = p which completes the proof

Let a be any proper radical ideal in a noetherian ring B Then a is the intersection of allits minimal primes p1 pn and the closed irreducible sets V (p1) sube V (a) are the irreduciblecomponents of the closed set V (a) in the noetherian space Spec(B)

Let φ A minusrarr B a morphism of rings put X = Spec(A) Y = Spec(B) and let Ψ Y minusrarr Xthe corresponding morphism of affine schemes and suppose B is noetherian Then (GDprime) can beformulated geometrically as follows let p isin X put X prime = V (p) sube X and let Y prime be an arbitraryirreducible component of Ψminus1(X prime) (which we assume is nonempty) Then Ψ maps Y prime genericallyonto X prime in the sense that the generic point of Y prime is mapped to the generic point p of X prime

Theorem 41 Let φ A minusrarr B be a flat morphism of rings Then the going-down theorem holdsfor φ

Proof Let pprime sub p be prime ideals of A and let q be a prime ideal of B lying over p Then Bq isflat over Ap by Lemma 33 hence faithfully flat since Ap minusrarr Bq is local Therefore Spec(Bq) minusrarrSpec(Ap) is surjective Let qprimeprime be a prime ideal of Bq lying over pprimeAp Then qprime = qprimeprimecapB is a primeideal of B lying over pprime and contained in q

Theorem 42 Let B be a ring and A a subring over which B is integral Then

(i) The canonical map Spec(B) minusrarr Spec(A) is surjective

(ii) If two prime ideals q sube qprime lie over the same prime ideal of A then they are equal

(iii) The going-up theorem holds for A sube B

(iv) If A is a local ring and m its maximal ideal then the prime ideals of B lying over m areprecisely the maximal ideals of B

(v) If A and B are integral domains and A is integrally closed then the going-down theoremholds for A sube B

Proof See [AM69] or [Mat80] Theorem 5

23 Constructible Sets

Definition 6 A topological space X is noetherian if the descending chain condition holds forthe closed sets in X The spectrum Spec(A) of a noetherian ring A is noetherian If a spaceis covered by a finite number of noetherian subspaces then it is noetherian Any subspace of anoetherian space is noetherian A noetherian space is quasi-compact In a noetherian space Xany nonempty closed set Z is uniquely decomposed into a finite number of irreducible closed setsZ = Z1 cup middot middot middot cupZn such that Zi Zj for i 6= j The Zi are called the irreducible components of Z

13

Lemma 43 (Noetherian Induction) Let X be a noetherian topological space and P a propertyof closed subsets of X Assume that for any closed subset Y of X if P holds for every properclosed subset of Y then P holds for Y (in particular P holds for the empty set) Then P holdsfor X

Proof Suppose that P does not hold for X and let Z be the set of all proper closed subsetsof X which do not satisfy P Then since X is noetherian Z has a minimal element Y SinceY is minimal every proper closed subset of Y must satisfy P and therefore Y satisfies Pcontradicting the fact that Y isin Z

Lemma 44 Let X be a noetherian topological space and P a property of general subsets of XAssume that for any subset Y of X if P holds for every proper subset Y prime of Y with Y prime sub Y thenP holds for Y (in particular P holds for the empty set) Then P holds for X

Proof Suppose that P does not hold for X and let Z be the set of all closures Q of proper subsetsQ of X with Q sub X and P not holding for Q Let Q be a minimal element of Z If Qprime is anyproper subset of Q with Qprime sub Q then Qprime must satisfy P otherwise Qprime would contradict minimalityof Q in Z But by assumption this implies that Q satisfies P which is a contradiction

Definition 7 Let X be a topological space and Z a subset of X We say Z is locally closed inX if it satisfies the following equivalent properties

(i) Every point z isin Z has an open neighborhood U in X such that U cap Z is closed in U

(ii) Z is the intersection of an open set in X and a closed set in X

(iii) Z is an open subset of its closure

Definition 8 Let X be a noetherian space We say a subset Z of X is a constructible set in Xif it is a finite union of locally closed sets in X so Z =

⋃mi=1(Ui capFi) with Ui open and Fi closed

The set F of all constructible subsets of X is the smallest collection of subsets of X containing allthe open sets which is closed with respect to the formation of finite intersections and complementsIt follows that all open and closed sets are constructible and F is also closed under finite unions

We say that a subset Z is pro-constructible (resp ind-constructible) if it is the intersection(resp union) of an arbitrary collection of constructible sets in X

Proposition 45 Let X be a noetherian space and Z a subset of X Then Z is constructible ifand only if the following condition is satisfied

(lowast) For each irreducible closed set X0 in X either X0capZ is not dense in X0 or X0capZcontains a nonempty open set of X0

Proof Assume that Z is constructible and Z cap X0 nonempty Then we can write X0 cap Z =⋃mi=1 Ui capFi for Ui open in X Fi closed and irreducible in X (by taking irreducible components)

and UicapFi nonempty for all i Then Ui cap Fi = Fi since Fi is irreducible therefore X0 cap Z =⋃i Fi

If X0 cap Z is dense in X0 we have X0 =⋃i Fi so that some Fi say F1 is equal to X0 Then

U1 capX0 = U1 cap F1 is a nonempty open subset of X0 contained in X0 cap ZNext we prove the converse We say that a subset T of X has the property P if whenever a

subset Z of T satisfies (lowast) it is constructible We need to show that X has the property P forwhich we use the form of noetherian induction given in Lemma 44 Suppose that Y is a subsetof X with P holding for every proper subset Y prime of Y with Y prime sub Y We need to show that Pholds for Y Let Z be a nonempty subset of Y satisfying (lowast) and let Z = F1 cup cup Fr be thedecomposition of Z into irreducible components Since Z = Z cap F1 cup middot middot middot cup Z cap Fr we have

F1 = F1 cap Z = F1 cap(Z cap F1 cup cup Z cap Fr

)= (F1 cap Z cap F1) cup middot middot middot cup (F1 cap Z cap Fr)

Since F1 is irreducible and not contained in any other Fi we must have F1 = Z cap F1 so F1 cap Zis dense in F1 whence by (lowast) there exists a proper closed subset F prime of F1 such that F1 F prime sube ZThen putting F lowast = F prime cupF2 cup middot middot middot cupFr we have Z = (F1 F prime)cup (Z capF lowast) The set F1 F prime is locally

14

closed in X so to complete the proof it suffices to show that Z cap F lowast is constructible in X SinceZ cap F lowast sube F lowast sub Z sube Y by the inductive hypothesis P holds for ZcapF lowast so it suffices to show thatZ cap F lowast satisfies (lowast) If X0 is irreducible and Z cap F lowast capX0 = X0 the closed set F lowast must containX0 and so Z cap F lowast capX0 = Z capX0 which contains a nonempty open subset of X0 since Z satisfies(lowast) and clearly Z capX0 is dense in X0

Lemma 46 Let φ A minusrarr B be a morphism of rings and f Spec(B) minusrarr Spec(A) the corre-sponding morphism of schemes Then f dominant if and only if Kerφ sube nil(A) In particular ifA is reduced the f dominant if and only if φ is injective

Proof Let X = Spec(A) and Y = Spec(B) The closure f(Y ) is the closed set V (I) defined bythe ideal I =

⋂pisinY φ

minus1p = φminus1⋂

pisinY p which is φminus1(nil(B)) Clearly Kerφ sube I Suppose thatf(Y ) is dense in X Then V (I) = X whence I = nil(A) and so Kerφ sube nil(A) Converselysuppose Kerφ sube nil(A) Then it is clear that I = φminus1(nil(B)) = nil(A) which means thatf(Y ) = V (I) = X

3 Associated Primes

This material can be found in [AM69] Chapter 11 webnotes of Robert Ash or in [Mat80] itselfThere is not much relevant to add here apart from a few small comments

Lemma 47 Let A be a ring and M an A-module Let a be an ideal in A that is maximal amongall annihilators of nonzero elements of M Then a is prime

Proof Say a = Ann(x) Given ab isin a we must show that a isin a or b isin a Assume a isin aThen ax 6= 0 We note that Ann(ax) supe a By hypothesis it cannot properly be larger HenceAnn(ax) = a Now b annihilates ax hence b isin a

Lemma 48 Let A be a noetherian ring and M an A-module If 0 6= a isin M then Ann(a) iscontained in an associated prime of M

Proposition 49 Let A be a noetherian ring and M a nonzero finitely generated A-module Amaximal ideal m is an associated prime of M if and only if no element of m is regular on M

Proof One implication is obvious If x isin m is not regular on M say x isin Ann(b) for some nonzerob then x is contained in an associated prime of M Thus m is contained in the finite union of theassociated primes of M and since m is maximal it must be one of them

Proposition 50 Let A be a nonzero noetherian ring I an ideal and M a nonzero finitelygenerated A-module If there exist elements x y isin I with x regular on A and y regular on M thenthere exists an element of I regular on both A and M

Proof Let p1 pn be the associated primes of A and q1 qm the associated primes of M By assumption I is not contained in any of these primes But if no element of I is regular on bothA and M then I is contained in the union p1 cup middot middot middot cup pn cup q1 cup middot middot middot cup qm and therefore containedin one of these primes which is a contradiction

4 Dimension

This is covered in [AM69] so we restrict ourselves here to mentioning some of the major pointsRecall that an ideal q sube R in a ring is primary if it is proper and if whenever xy isin q we haveeither x isin q or yn isin q for some n gt 0 Then the radical of q is a prime ideal p and we say q isa p-primary ideal If a is an ideal and b supe a is p-primary then in the ring Ra the ideal ba ispa-primary A minimal primary decomposition of an ideal b is an expression

b = q1 cap middot middot middot cap qn

15

where capj 6=iqj qi for all i and the primes pi = r(qi) are all distinct If a is an ideal contained inb then

ba = q1a cap middot middot middot cap qna

is a minimal primary decomposition of ba in AaLet A be a nonzero ring Recall that dimension of an A-module M is the Krull dimension of

the ring AAnn(M) and is defined for all modules M (minus1 if M = 0) The rank is defined for freeA-modules and is the common size of any basis (0 if M = 0) Throughout these notes dim(M)will denote the dimension not the rank

Definition 9 Let (Am k) be a noetherian local ring of dimension d An ideal of definition is anm-primary ideal Recall that the dimension of A is the size of the smallest collection of elementsof A which generates an m-primary ideal Recall that rankk(mm2) is equal to the size of thesmallest set of generators for m as an ideal so always d le rankk(mm2)

A system of parameters is a set of d elements generating an m-primary ideal If d = rankk(mm2)or equivalently there is a system of parameters generating m we say that A is a regular local ringand we call such a system of parameters a regular system of parameters

Proposition 51 Let (Am) be a noetherian local ring of dimension d ge 1 and let x1 xd bea system of parameters of A Then

dim(A(x1 xi)) = dminus i = dim(A)minus i

for each 1 le i le d

Proof Put A = A(x1 xi) If i = d then the zero ideal in A is an ideal of definition so clearlydim(A) = 0 If 1 le i lt d then dim(A) le dminus i since xi+1 xd generate an ideal of definition ofA Let dim(A) = p If p = 0 then (x1 xi) must be an ideal of definition contradicting i lt dSo p ge 1 and if y1 yp is a system of parameters of A then x1 xi y1 yp generate anideal of definition of A so that p+ i ge d That is p ge dminus i

Definition 10 Let A be a nonzero ring and I a proper ideal The height of I denoted htI isthe minimum of the heights of the prime ideals containing I

htI = infhtp | p supe I

This is a number in 0 1 2 infin Equivalently we can take the infimum over the heights ofprimes minimal over I Clearly ht0 = 0 and if I sube J are proper ideals then it is clear thathtI le htJ If I is a prime ideal then htI is the usual height of a prime ideal If A is a noetherianring then htI ltinfin for every proper ideal I since Ap is a local noetherian ring and htp = dim(Ap)

Lemma 52 Let A be a nonzero ring and I a proper ideal Then we have

htI = infhtIAm |m is a maximal ideal and I sube m

Lemma 53 Let A be a noetherian ring and suppose we have an exact sequence

0 minusrarrM prime minusrarrM minusrarrM primeprime minusrarr 0

in which M primeMM primeprime are nonzero and finitely generated Then dimM = maxdimM prime dimM primeprime

Proof We know that Supp(M) = Supp(M prime)cupSupp(M primeprime) and for all three modules the dimensionis the supremum of the coheights of prime ideals in the support So the result is straightforwardto check

16

41 Homomorphism and Dimension

Let φ A minusrarr B be a morphism of rings If p isin Spec(A) then put κ(p) = AppAp Let Bp denotethe ring Tminus1B where T = φ(Aminusp) There is an isomorphism of A-algebras Bp

sim= BotimesAAp Thereis a commutative diagram of rings

B

))RRRRRRRRRRRRRRRR Bp BppBp

B otimesA κ(p)

The vertical isomorphism is defined by bφ(s)+ pBp 7rarr botimes (1s+ pAp) We call Spec(BotimesA κ(p))the fibre over p of the map Φ Spec(B) minusrarr Spec(A) Since primes of BppBp clearly correspondto primes q of B with q cap A = p it is easy to see that the ring morphism B minusrarr B otimesA κ(p) givesrise to a continuous map Spec(B otimesA κ(p)) minusrarr SpecB which gives a homeomorphism betweenΦminus1(p) and Spec(B otimesA κ(p)) See [AM69] Chapter 3

Lemma 54 Let q be a prime ideal of B with q cap A = p and let P be the corresponding primeideal of B otimesA κ(p) Then there is an isomorphism of A-algebras

Bq otimesA κ(p) sim= (B otimesA κ(p))P

btotimes (as+ pAp) 7rarr (botimes (a1 + pAp))(totimes (s1 + pAp))

Proof It is not difficult to see that there is an isomorphism of rings Bqsim= (Bp)qBp defined by

bt 7rarr (b1)(t1) Consider the prime ideal qBppBp We know that there is a ring isomorphism

(B otimesA κ(p))Psim= (BppBp)qBppBp

sim= (Bp)qBpp(Bp)qBpsim= BqpBq

by the comments following Lemma 7 It is not hard to check there is a ring isomorphism BqpBqsim=

BqotimesA κ(p) defined by bt+ pBq 7rarr btotimes 1 (the inverse of btotimes (as+ pAp) is bφ(a)tφ(s)+ pBq)So by definition of P there is an isomorphism of rings (B otimesA κ(p))P

sim= Bq otimesA κ(p) and this isclearly an isomorphism of A-algebras

In particular if φ A minusrarr B is a ring morphism p isin Spec(A) and q isin Spec(B) such thatq capA = p then there is an isomorphism of rings (BpB)qpB

sim= BqpBq so we have ht(qpB) =dim(Bq otimesA κ(p))

Theorem 55 Let φ A minusrarr B be a morphism of noetherian rings Let q isin Spec(B) and putp = q capA Then

(1) htq le htp + ht(qpB) In other words dim(Bq) le dim(Ap) + dim(Bq otimesA κ(p))

(2) We have equality in (1) if the going-down theorem holds for φ (in particular if φ is flat)

(3) If Φ Spec(B) minusrarr Spec(A) is surjective and if the going-down theorem holds then we havedim(B) ge dim(A) and htI = ht(IB) for any proper ideal I of A

Proof (1) Replacing A by Ap and B by Bq we may suppose that (A p) and (B q) are noetherianlocal rings such that q cap A = p and we must show that dim(B) le dim(A) + dim(BpB) Let Ibe a p-primary ideal of A Then pn sube I for some n gt 0 so pnB sube IB sube pB Thus the ideals pBand IB have the same radical and so by definition dim(BpB) = dim(BIB) If dimA = 0 thenwe can take I = 0 and the result is trivial So assume dimA = r ge 1 and let I = (a1 ar) fora system of parameters a1 ar If dim(BIB) = 0 then IB is an q-primary ideal of B and sodim(B) le r as required Otherwise if dim(BIB) = s ge 1 let b1 + IB bs + IB be a systemof parameters of BIB Then b1 bs a1 ar generate an ideal of definition of B Hencedim(B) le r + s

(2) We use the same notation as above If ht(qpB) = s ge 0 then there exists a prime chainof length s q = q0 sup q1 sup middot middot middot sup qs such that qs supe pB As p = q cap A supe qi cap A supe p all the qi

17

lie over p If htp = r ge 0 then there exists a prime chain p = p0 sup p1 sup middot middot middot sup pr in A and bygoing-down there exists a prime chain qs = t0 sup t1 sup middot middot middot sup tr of B such that ti capA = pi Then

q = q0 sup middot middot middot sup qs sup t1 sup middot middot middot sup tr

is a prime chain of length r + s therefore htq ge r + s(3) (i) follows from (2) since dim(A) = suphtp | p isin Spec(A) (ii) Since Φ is surjective IB

is a proper ideal Let q be a minimal prime over IB such that htq = ht(IB) and put p = q capAThen ht(qpB) = 0 so by (2) we find that ht(IB) = htq = htp ge htI For the reverseinclusion let p be a minimal prime ideal over I such that htp = htI and take a prime q of Blying over p Replacing q if necessary we may assume that q is a minimal prime ideal over pBThen htI = htp = htq ge ht(IB)

Theorem 56 Let A be a nonzero subring of B and suppose that B is integral over A Then

(1) dim(A) = dim(B)

(2) Let q isin Spec(B) and set p = q capA Then we have cohtp = cohtq and htq le htp

(3) If the going-down theorem holds between A and B then for any ideal J of B with J capA 6= Awe have htJ = ht(J capA)

Proof (1) By Theorem 42 the going-up theorem holds for A sube B and Spec(B) minusrarr Spec(A) issurjective so we can lift any chain of prime ideals p0 sub p1 sub middot middot middot sub pn in A to a chain of primeideals q0 sub middot middot middot sub qn in B On the other hand if q sube qprime are prime ideals of B and q capA = qprime capAthen q = qprime so any chain of prime ideals in B restricts to a chain of the same length in A Hencedim(A) = dim(B)

(2) Since Bq is integral over Ap it is clear from (1) that cohtp = cohtq If q = q0 sup middot middot middot sup qnis a chain of prime ideals in B then intersecting with A gives a chain of length n descending fromp Hence htq le htp

(3) Given the going-down theorem it is clear that htq = htp in (2) Let J be a proper ideal ofB with J capA 6= A and let q be such that htJ = htq Then ht(J capA) le ht(qcapA) = htq = htJ On the other hand BJ is integral over BJ cap A so every prime ideal p of A containing J cap Acan be lifted to a prime ideal q of B containing J In particular we can lift a prime ideal p withht(J capA) = htp to see that htJ le htq = htp = ht(J capA) as required

5 Depth

Definition 11 Let A be a ring M an A-module and a1 ar a sequence of elements of AWe say a1 ar is an M -regular sequence (or simply M -sequence) if the following conditions aresatisfied

(1) For each 2 le i le r ai is regular on M(a1 aiminus1)M and a1 is regular on M

(2) M 6= (a1 an)M

If a1 ar is an M -regular sequence then so is a1 ai for any i le r When all ai belong to anideal I we say a1 ar is an M -regular sequence in I If moreover there is no b isin I such thata1 ar b is M -regular then a1 ar is said to be a maximal M -regular sequence in I Noticethat the notion of M -regular depends on the order of the elements in the sequence If MN areisomorphic A-modules then a sequence is regular on M iff it is regular on N

Lemma 57 A sequence a1 ar with r ge 2 is M -regular if and only if a1 is regular on Mand a2 ar is an Ma1M -regular sequence If the sequence a1 ar is a maximal M -regularsequence in I then a2 ar is a maximal Ma1M -regular sequence in I

18

Proof The key point is that for ideals a sube b there is a canonical isomorphism of A-modulesMbM sim= NbN where N = MaM If a1 ar is M -regular then a1 is regular on M a2 isregular on N = Ma1M and for 3 le i le r ai is regular on

M(a1 aiminus1)M sim= N(a2 aiminus1)N

Hence a2 ar is an N -regular sequence The converse follows from the same argument

More generally if a1 ar is an M -regular sequence and we set N = M(a1 ar)M andif b1 bs is an N -regular sequence then a1 ar b1 bs is an M -regular sequence

Lemma 58 If a1 ar is an A-regular sequence and M is a flat A-module then a1 ar isalso M -regular provided (a1 ar)M 6= M

Proof Left multiplication by a1 defines a monomorphism A minusrarr A since a1 is A-regular Tensoringwith M and using the fact that M is flat we see that left multiplication by a1 also gives amonomorphismM minusrarrM as required Similarly tensoring with the monomorphism a2 Aa1 minusrarrAa1 we get a monomorphism Ma1M minusrarrMa1M and so on

Lemma 59 Let A be a ring and M an A-module Given an integer n ge 1 a sequence a1 aris M -regular if and only if it is Mn-regular

Proof Suppose the sequence a1 ar is M -regular We prove it is Mn-regular by induction on rThe case r = 1 is trivial so assume r gt 1 By the inductive hypothesis the sequence a1 arminus1 isMn-regular Let L = (a1 arminus1)M Then (a1 arminus1)Mn = Ln and there is an isomorphismof A-modules MnLn sim= (ML)n So we need only show that ar is regular on (ML)n Since byassumption it is regular on ML this is not hard to check Clearly (a1 ar)Mn 6= Mn so thesequence a1 ar is Mn-regular as required The converse is similarly checked

Lemma 60 Let A be a nonzero ring M an A-module and a1 ar isin A If a1 ar isin Am isMm-regular for every maximal ideal m of A then the sequence a1 ar is M -regular

Proof This follows from the fact that given an A-module M an element a isin A is regular on M ifand only if its image in Am is regular on Mm for every maximal ideal m of A

Lemma 61 Suppose that a1 ar is M -regular and a1ξ1 + middot middot middot + arξr = 0 for ξi isin M Thenξi isin (a1 ar)M for all i

Proof By induction on r For r = 1 a1ξ1 = 0 implies that ξ1 = 0 Let r gt 1 Since ar is regularon M(a1 arminus1)M we have ξr =

sumrminus1i=1 aiηi so

sumrminus1i=1 ai(ξi + arηi) = 0 By the inductive

hypothesis for i lt r we have ξi + arηi isin (a1 arminus1)M so that ξi isin (a1 ar)M

Theorem 62 Let A be a ring and M an A-module and let a1 ar be an M -regular sequenceThen for every sequence n1 nr of integers gt 0 the sequence an1

1 anrr is M -regular

Proof Suppose we can prove the following statement

(lowast) Given an integer n gt 0 an A-module M and any M -regular sequence a1 arthe sequence an1 a2 ar is also M -regular

We prove the rest of the Theorem by induction on r For r = 1 this follows immediately from (lowast)Let r gt 1 and suppose a1 ar is M -regular Then by (lowast) an1

1 a2 ar is M -regular Hencea2 ar is Man1

1 M -regular By the inductive hypothesis an22 anr

r is Man11 M -regular and

therefore an11 anr

r is M -regular by Lemma 57So it only remains to prove (lowast) which we do by induction on n The case n = 1 is trivial so let

n gt 1 be given along with an A-moduleM and anM -regular sequence a1 ar By the inductivehypothesis anminus1

1 a2 ar is M -regular and clearly an1 is regular on M Since (an1 a2 ar) sube(anminus1

1 a2 ar) it is clear thatM 6= (an1 a2 ar)M Let i gt 1 and assume that an1 a2 aiminus1

is an M -regular sequence We need to show that ai is regular on M(an1 a2 aiminus1)M Suppose

19

that aiω = an1 ξ1 +a2ξ2 + middot middot middot+aiminus1ξiminus1 Then ω = anminus11 η1 +a2η2 + middot middot middot+aiminus1ηiminus1 by the inductive

hypothesis So

anminus11 (a1ξ1 minus aiη1) + a2(ξ2 minus aiη2) + middot middot middot+ aiminus1(ξiminus1 minus aiηiminus1) = 0

Hence a1ξ1minusaiη1 isin (anminus11 a2 aiminus1)M by Lemma 61 It follows that aiη1 isin (a1 a2 aiminus1)M

hence η1 isin (a1 aiminus1)M and so ω isin (an1 a2 aiminus1)M as required This proves (lowast) andtherefore completes the proof

Let A be a ring There is an isomorphism of A[x1 xn]-modules M otimesA A[x1 xn] sim=M [x1 xn] where the latter module consists of polynomials in x1 xn with coefficients in M(see our Polynomial Ring notes) For any f(x1 xn) isinM [x1 xn] and tuple (a1 an) isinAn we can define an element of M

f(a1 an) =sumα

aα11 middot middot middot aαn

n middot f(α)

For an element r isin R and h isinM [x1 xn]

(f + h)(a1 an) = f(a1 an) + h(a1 an)(r middot f)(a1 an) = r middot f(a1 an)

For an ideal I in R the R-submodule IM [x1 xn] consists of all polynomials whose coefficientsare in the R-submodule IM subeM

Let us review the definition of the associated graded ring and modules Let A be a ring and Ian ideal of A Then the abelian group

grI(A) = AI oplus II2 oplus I2I3 oplus middot middot middot

becomes a graded ring in a fairly obvious way For an A-module M we have the graded grI(A)-module

grI(M) = MIM oplus IMI2M oplus I2MI3M oplus middot middot middotIf A is noetherian and M is a finitely generated A-module then grI(A) is a noetherian ring andif grI(M) is a finitely generated grI(A)-module

Given elements a1 an isin A and I = (a1 an) we define a morphism of abelian groupsψ M [x1 xn] minusrarr grI(M) as follows if f is homogenous of degreem ge 0 define ψ(f) to be theimage of f(a1 an) in ImMIm+1M This defines a morphism of groups M [x1 xn]m minusrarrImMIm+1M and together these define the morphism of groups ψ Since ψ maps IM [x1 xn]to zero it induces a morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) and

Proposition 63 Let A be a ring and M an A-module Let a1 an isin A and set I =(a1 an) Then the following conditions are equivalent

(a) For every m gt 0 and for every homogenous polynomial f(x1 xn) isin M [x1 xn] ofdegree m such that f(a1 an) isin Im+1M we have f isin IM [x1 xn]

(b) If f(x1 xn) isin M [x1 xn] is homogenous and f(a1 an) = 0 then the coefficientsof f are in IM

(c) The morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) defined by mappinga homogenous polynomial f(x1 xn) of degree m to f(a1 an) isin ImMIm+1M is anisomorphism

Proof It is easy to see that (a) hArr (c) and (a) rArr (b) To show (b) rArr (a) let f isinM [x1 xn] bea homogenous polynomial of degree m gt 0 and suppose f(a1 an) isin Im+1M Any element ofIm+1M can be written as a sum of terms of the form aα1

1 middot middot middot aαnn middotm with

sumi αi = m+1 By shifting

one of the ai across we can write f(a1 an) = g(a1 an) for a homogenous polynomialg isinM [x1 xn] of degreem all of whose coefficients belong to IM Hence (fminusg)(a1 an) = 0so by (b) the coefficients of f minus g belong to IM and this implies implies that the coefficients of falso belong to IM as required

20

Definition 12 Let A be a ring and M an A-module A sequence a1 an isin A is M -quasiregularif it satisfies the equivalent conditions of the Proposition Obviously this concept does not dependon the order of the elements But a1 ai for i lt n need not be M -quasiregular

Recall that for an A-module M a submodule N sube M and x isin A the notation (N x) meansm isin M |xm isin N This is a submodule of M If A is a ring I an ideal and M an A-modulerecall that M is separated in the I-adic topology when

⋂n I

nM = 0

Theorem 64 Let A be a ring M a nonzero A-module a1 an isin A and I = (a1 an)Then

(i) If a1 an is M -quasiregular and x isin A is such that (IM x) = IM then (ImM x) =ImM for all m gt 0

(ii) If a1 an is M -regular then it is M -quasiregular

(iii) If MMa1MM(a1 a2)M M(a1 anminus1)M are separated in the I-adic topologythen the converse of (ii) is also true

Proof (i) By induction on m with the case m = 1 true by assumption Suppose m gt 1 andxξ isin ImM By the inductive hypothesis ξ isin Imminus1M Hence there exists a homogenous polynomialf isinM [x1 xn] of degree mminus 1 such that ξ = f(a1 an) Since xξ = xf(a1 an) isin ImMthe coefficients of f are in (IM x) = IM Therefore ξ = f(a1 an) isin ImM

(ii) By induction on n For n = 1 this is easy to check Let n gt 1 and suppose a1 an is M -regular Then by the induction hypothesis a1 anminus1 is M -quasiregular Let f isinM [x1 xn]be homogenous of degree m gt 0 such that f(a1 an) = 0 We prove that f isin IM [x1 xn]by induction on m (the case m = 0 being trivial) Write

f(x1 xn) = g(x1 xnminus1) + xnh(x1 xn)

Then g and h are homogenous of degrees m and mminus 1 respectively By (i) we have

h(a1 an) isin ((a1 anminus1)mM an) = (a1 anminus1)mM sube ImM

Since by assumption a1 an is regular on M so an is regular on M(a1 anminus1)M andhence ((a1 anminus1)M an) = (a1 anminus1)M So by the induction hypothesis on m we haveh isin IM [x1 xn] (by the argument of Proposition 63) Since h(a1 an) isin (a1 anminus1)mMthere exists H isin M [x1 xnminus1] which is homogenous of degree m such that h(a1 an) =H(a1 anminus1) Putting

G(x1 xnminus1) = g(x1 xnminus1) + anH(x1 xnminus1)

we have G(a1 anminus1) = 0 so by the inductive hypothesis on n we have G isin IM [x1 xn]hence g isin IM [x1 xn] and so f isin IM [x1 xn]

(iii) By induction on n ge 1 Assume that a1 an is M -quasiregular and the modulesMMa1M M(a1 anminus1)M are all separated in the I-adic topology If a1ξ = 0 thenξ isin IM hence ξ =

sumaiηi and

suma1aiηi = 0 hence ηi isin IM and so ξ isin I2M In this way we

see that ξ isin⋂t ItM = 0 Thus a1 is regular on M and this also takes care of the case n = 1

since M 6= IM by the separation condition So assume n gt 1 By Lemma 57 it suffices to showthat a2 an is an N -regular sequence where N = Ma1M Since there is an isomorphism ofA-modules for 2 le i le nminus 1

M(a1 ai)M sim= N(a2 ai)N

The modules NNa2N N(a2 anminus1)N are separated in the I-adic topology So by theinductive hypothesis it suffices to show that the sequence a2 an is N -quasiregular

It suffices to show that if f(x2 xn) isin M [x2 xn] is homogenous of degree m ge 1 withf(a2 an) isin a1M then the coefficients of f belong to IM Put f(a2 an) = a1ω We claim

21

that ω isin Imminus1M Let 0 le i le mminus1 be the largest integer with ω isin IiM Then ω = g(a1 an)for some homogenous polynomial of degree i and

f(a2 an) = a1g(a1 an) (2)

If i lt mminus1 then g isin IM [x1 xn] and so ω isin Ii+1M which is a contradiction Hence i = mminus1and so ω isin Imminus1M Again using (2) we see that f(x2 xn)minusx1g(x1 xn) isin IM [x1 xn]Since f does not involve x1 we have f isin IM [x1 xn] as required

The theorem shows that under the assumptions of (iii) any permutation of an M -regularsequence is M -regular

Corollary 65 Let A be a noetherian ring M a finitely generated A-module and let a1 anbe contained in the Jacobson radical of A Then a1 an is M -regular if and only if it is M -quasiregular In particular if a1 an is M -regular so is any permutation of the sequence

Proof From [AM69] we know that for any ideal I contained in the Jacobson radical the I-adictopology on any finitely generated A-module is separated

If A is a ring and M an A-module then any M -regular sequence a1 an isin A gives rise toa strictly increasing chain of submodules a1M (a1 a2)M (a1 an)M Hence the chain ofideals (a1) (a1 a2) (a1 an) must also be strictly increasing

Lemma 66 Let A be a noetherian ring and M an A-module Any M -regular sequence a1 anin an ideal I can be extended to a maximal M -regular sequence in I

Proof If a1 an is not maximal in I we can find an+1 isin I such that a1 an an+1 is anM -regular sequence Either this process terminates at a maximal M -regular sequence in I or itproduces a strictly ascending chain of ideals

(a1) sub (a1 a2) sub (a1 a2 a3) sub middot middot middot

Since A is noetherian we can exclude this latter possibility

Theorem 67 Let A be a noetherian ring M a finitely generated A-module and I an ideal of Awith IM 6= M Let n gt 0 be an integer Then the following are equivalent

(1) ExtiA(NM) = 0 for i lt n and every finitely generated A-module N with Supp(N) sube V (I)

(2) ExtiA(AIM) = 0 for i lt n

(3) There exists a finitely generated A-module N with Supp(N) = V (I) and ExtiA(NM) = 0for i lt n

(4) There exists an M -regular sequence a1 an of length n in I

Proof (1) rArr (2) rArr (3) is trivial With I fixed we show that (3) rArr (4) for every finitely generatedmodule M with IM 6= M by induction on n We have 0 = Ext0A(NM) sim= HomA(NM) SinceM is finitely generated and nonzero the set of associated primes of M is finite and nonempty Ifno elements of I are M -regular then I is contained in the union of these associated primes andhence I sube p for some p isin Ass(M) (see [AM69] for details) By definition there is a monomorphismof A-modules φ Ap minusrarrM There is an isomorphism of A-modules

(Ap)psim= ApotimesA Ap

sim= AppAp = k

It is not hard to check this is an isomorphism of Ap-modules as well Since φp is a monomorphismand k 6= 0 it follows that HomAp(kMp) 6= 0 Since p isin V (I) = Supp(N) we have Np 6= 0 andso the k-module NppNp 6= 0 is nonzero and therefore free so Homk(NppNp k) 6= 0 Sincek sim= (Ap)p as Ap-modules it follows that HomA(NAp)p

sim= HomAp(Np (Ap)p) 6= 0 Since Apis isomorphic to a submodule of M it follows that HomA(NM) 6= 0 which is a contradiction

22

therefore there exists an M -regular element a1 isin I which takes care of the case n = 1 If n gt 1then put M1 = Ma1M From the exact sequence

0 Ma1 M M1

0 (3)

we get the long exact sequence

middot middot middot minusrarr ExtiA(NM) minusrarr ExtiA(NM1) minusrarr Exti+1A (NM) minusrarr middot middot middot

which shows that ExtiA(NM1) = 0 for 0 le i lt n minus 1 By the inductive hypothesis on n thereexists an M1-regular sequence a2 an in I The sequence a1 an is then an M -regularsequence in I

(4) rArr (1) By induction on n with I fixed For n = 1 we have a1 isin I regular on M and so (3)gives an exact sequence of R-modules

0 HomA(NM)a1 HomA(NM)

Where a1 denotes left multiplication by a1 Since Supp(N) = V (Ann(N)) sube V (I) it fol-lows that I sube r(Ann(N)) and so ar1N = 0 for some r gt 0 It follows that ar1 annihilatesHomA(NM) as well but since the action of ar1 on HomA(NM) gives an injective map it followsthat HomA(NM) = 0 Now assume n gt 1 and put M1 = Ma1M Then a2 an is anM1-regular sequence so by the inductive hypothesis ExtiA(NM1) = 0 for i lt nminus 1 So the longexact sequence corresponding to (3) gives an exact sequence for 0 le i lt n

0 ExtiA(NM)a1 ExtiA(NM)

Here a1 denotes left multiplication by a1 which is equal to ExtiA(αM) = Exti(N β) where α βare the endomorphisms given by left multiplication by a1 on NM respectively Assume that ar1annihilates N with r gt 0 Then αr is the zero map so ExtiA(αM)r = 0 and so ar1 also annihilatesExtiA(NM) Since the a1 is regular on this module it follows that ExtiA(NM) = 0 for i lt nas required

Corollary 68 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA with IM 6= M If a1 an a maximal M -regular sequence in I then ExtiA(AIM) = 0 fori lt n and ExtnA(AIM) 6= 0

Proof We already know that ExtiA(AIM) = 0 for i lt n so with I fixed we prove by inductionon n that ExtnA(AIM) 6= 0 for any finitely generated module M with IM 6= M admitting amaximal M -regular sequence of length n For n = 1 we have a1 isin I regular on M and an exactsequence (3) where M1 = Ma1M Part of the corresponding long exact sequence is

Ext0A(AIM) minusrarr Ext0A(AIM1) minusrarr Ext1A(AIM)

We know from the Theorem 67 that Ext0A(AIM) = 0 so it suffices to show that we haveHomA(AIM1) 6= 0 But if HomA(AIM1) = 0 then it follows from the proof of (3) rArr (4)above that there would be b isin I regular on M1 so a1 b is an M -regular sequence This is acontradiction since the sequence a1 was maximal so we conclude that Ext1A(AIM) 6= 0

Now assume n gt 1 and let a1 an be a maximal M -regular sequence in I Then a2 anis a maximal M1-regular sequence in I so by the inductive hypothesis Extnminus1

A (AIM1) 6= 0 Sofrom the long exact sequence for (3) we conclude that ExtnA(AIM) 6= 0 also

It follows that under the conditions of the Corollary every maximal M -regular sequence in Ihas a common length and you can find this length by looking at the sequence of abelian groups

HomA(AIM) Ext1A(AIM) Ext2A(AIM) ExtnA(AIM)

If there are M -regular sequences in I then this sequence will start off with n minus 1 zero groupswhere n ge 1 is the common length of every maximal M -regular sequence The nth group will

23

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Lemma 43 (Noetherian Induction) Let X be a noetherian topological space and P a propertyof closed subsets of X Assume that for any closed subset Y of X if P holds for every properclosed subset of Y then P holds for Y (in particular P holds for the empty set) Then P holdsfor X

Proof Suppose that P does not hold for X and let Z be the set of all proper closed subsetsof X which do not satisfy P Then since X is noetherian Z has a minimal element Y SinceY is minimal every proper closed subset of Y must satisfy P and therefore Y satisfies Pcontradicting the fact that Y isin Z

Lemma 44 Let X be a noetherian topological space and P a property of general subsets of XAssume that for any subset Y of X if P holds for every proper subset Y prime of Y with Y prime sub Y thenP holds for Y (in particular P holds for the empty set) Then P holds for X

Proof Suppose that P does not hold for X and let Z be the set of all closures Q of proper subsetsQ of X with Q sub X and P not holding for Q Let Q be a minimal element of Z If Qprime is anyproper subset of Q with Qprime sub Q then Qprime must satisfy P otherwise Qprime would contradict minimalityof Q in Z But by assumption this implies that Q satisfies P which is a contradiction

Definition 7 Let X be a topological space and Z a subset of X We say Z is locally closed inX if it satisfies the following equivalent properties

(i) Every point z isin Z has an open neighborhood U in X such that U cap Z is closed in U

(ii) Z is the intersection of an open set in X and a closed set in X

(iii) Z is an open subset of its closure

Definition 8 Let X be a noetherian space We say a subset Z of X is a constructible set in Xif it is a finite union of locally closed sets in X so Z =

⋃mi=1(Ui capFi) with Ui open and Fi closed

The set F of all constructible subsets of X is the smallest collection of subsets of X containing allthe open sets which is closed with respect to the formation of finite intersections and complementsIt follows that all open and closed sets are constructible and F is also closed under finite unions

We say that a subset Z is pro-constructible (resp ind-constructible) if it is the intersection(resp union) of an arbitrary collection of constructible sets in X

Proposition 45 Let X be a noetherian space and Z a subset of X Then Z is constructible ifand only if the following condition is satisfied

(lowast) For each irreducible closed set X0 in X either X0capZ is not dense in X0 or X0capZcontains a nonempty open set of X0

Proof Assume that Z is constructible and Z cap X0 nonempty Then we can write X0 cap Z =⋃mi=1 Ui capFi for Ui open in X Fi closed and irreducible in X (by taking irreducible components)

and UicapFi nonempty for all i Then Ui cap Fi = Fi since Fi is irreducible therefore X0 cap Z =⋃i Fi

If X0 cap Z is dense in X0 we have X0 =⋃i Fi so that some Fi say F1 is equal to X0 Then

U1 capX0 = U1 cap F1 is a nonempty open subset of X0 contained in X0 cap ZNext we prove the converse We say that a subset T of X has the property P if whenever a

subset Z of T satisfies (lowast) it is constructible We need to show that X has the property P forwhich we use the form of noetherian induction given in Lemma 44 Suppose that Y is a subsetof X with P holding for every proper subset Y prime of Y with Y prime sub Y We need to show that Pholds for Y Let Z be a nonempty subset of Y satisfying (lowast) and let Z = F1 cup cup Fr be thedecomposition of Z into irreducible components Since Z = Z cap F1 cup middot middot middot cup Z cap Fr we have

F1 = F1 cap Z = F1 cap(Z cap F1 cup cup Z cap Fr

)= (F1 cap Z cap F1) cup middot middot middot cup (F1 cap Z cap Fr)

Since F1 is irreducible and not contained in any other Fi we must have F1 = Z cap F1 so F1 cap Zis dense in F1 whence by (lowast) there exists a proper closed subset F prime of F1 such that F1 F prime sube ZThen putting F lowast = F prime cupF2 cup middot middot middot cupFr we have Z = (F1 F prime)cup (Z capF lowast) The set F1 F prime is locally

14

closed in X so to complete the proof it suffices to show that Z cap F lowast is constructible in X SinceZ cap F lowast sube F lowast sub Z sube Y by the inductive hypothesis P holds for ZcapF lowast so it suffices to show thatZ cap F lowast satisfies (lowast) If X0 is irreducible and Z cap F lowast capX0 = X0 the closed set F lowast must containX0 and so Z cap F lowast capX0 = Z capX0 which contains a nonempty open subset of X0 since Z satisfies(lowast) and clearly Z capX0 is dense in X0

Lemma 46 Let φ A minusrarr B be a morphism of rings and f Spec(B) minusrarr Spec(A) the corre-sponding morphism of schemes Then f dominant if and only if Kerφ sube nil(A) In particular ifA is reduced the f dominant if and only if φ is injective

Proof Let X = Spec(A) and Y = Spec(B) The closure f(Y ) is the closed set V (I) defined bythe ideal I =

⋂pisinY φ

minus1p = φminus1⋂

pisinY p which is φminus1(nil(B)) Clearly Kerφ sube I Suppose thatf(Y ) is dense in X Then V (I) = X whence I = nil(A) and so Kerφ sube nil(A) Converselysuppose Kerφ sube nil(A) Then it is clear that I = φminus1(nil(B)) = nil(A) which means thatf(Y ) = V (I) = X

3 Associated Primes

This material can be found in [AM69] Chapter 11 webnotes of Robert Ash or in [Mat80] itselfThere is not much relevant to add here apart from a few small comments

Lemma 47 Let A be a ring and M an A-module Let a be an ideal in A that is maximal amongall annihilators of nonzero elements of M Then a is prime

Proof Say a = Ann(x) Given ab isin a we must show that a isin a or b isin a Assume a isin aThen ax 6= 0 We note that Ann(ax) supe a By hypothesis it cannot properly be larger HenceAnn(ax) = a Now b annihilates ax hence b isin a

Lemma 48 Let A be a noetherian ring and M an A-module If 0 6= a isin M then Ann(a) iscontained in an associated prime of M

Proposition 49 Let A be a noetherian ring and M a nonzero finitely generated A-module Amaximal ideal m is an associated prime of M if and only if no element of m is regular on M

Proof One implication is obvious If x isin m is not regular on M say x isin Ann(b) for some nonzerob then x is contained in an associated prime of M Thus m is contained in the finite union of theassociated primes of M and since m is maximal it must be one of them

Proposition 50 Let A be a nonzero noetherian ring I an ideal and M a nonzero finitelygenerated A-module If there exist elements x y isin I with x regular on A and y regular on M thenthere exists an element of I regular on both A and M

Proof Let p1 pn be the associated primes of A and q1 qm the associated primes of M By assumption I is not contained in any of these primes But if no element of I is regular on bothA and M then I is contained in the union p1 cup middot middot middot cup pn cup q1 cup middot middot middot cup qm and therefore containedin one of these primes which is a contradiction

4 Dimension

This is covered in [AM69] so we restrict ourselves here to mentioning some of the major pointsRecall that an ideal q sube R in a ring is primary if it is proper and if whenever xy isin q we haveeither x isin q or yn isin q for some n gt 0 Then the radical of q is a prime ideal p and we say q isa p-primary ideal If a is an ideal and b supe a is p-primary then in the ring Ra the ideal ba ispa-primary A minimal primary decomposition of an ideal b is an expression

b = q1 cap middot middot middot cap qn

15

where capj 6=iqj qi for all i and the primes pi = r(qi) are all distinct If a is an ideal contained inb then

ba = q1a cap middot middot middot cap qna

is a minimal primary decomposition of ba in AaLet A be a nonzero ring Recall that dimension of an A-module M is the Krull dimension of

the ring AAnn(M) and is defined for all modules M (minus1 if M = 0) The rank is defined for freeA-modules and is the common size of any basis (0 if M = 0) Throughout these notes dim(M)will denote the dimension not the rank

Definition 9 Let (Am k) be a noetherian local ring of dimension d An ideal of definition is anm-primary ideal Recall that the dimension of A is the size of the smallest collection of elementsof A which generates an m-primary ideal Recall that rankk(mm2) is equal to the size of thesmallest set of generators for m as an ideal so always d le rankk(mm2)

A system of parameters is a set of d elements generating an m-primary ideal If d = rankk(mm2)or equivalently there is a system of parameters generating m we say that A is a regular local ringand we call such a system of parameters a regular system of parameters

Proposition 51 Let (Am) be a noetherian local ring of dimension d ge 1 and let x1 xd bea system of parameters of A Then

dim(A(x1 xi)) = dminus i = dim(A)minus i

for each 1 le i le d

Proof Put A = A(x1 xi) If i = d then the zero ideal in A is an ideal of definition so clearlydim(A) = 0 If 1 le i lt d then dim(A) le dminus i since xi+1 xd generate an ideal of definition ofA Let dim(A) = p If p = 0 then (x1 xi) must be an ideal of definition contradicting i lt dSo p ge 1 and if y1 yp is a system of parameters of A then x1 xi y1 yp generate anideal of definition of A so that p+ i ge d That is p ge dminus i

Definition 10 Let A be a nonzero ring and I a proper ideal The height of I denoted htI isthe minimum of the heights of the prime ideals containing I

htI = infhtp | p supe I

This is a number in 0 1 2 infin Equivalently we can take the infimum over the heights ofprimes minimal over I Clearly ht0 = 0 and if I sube J are proper ideals then it is clear thathtI le htJ If I is a prime ideal then htI is the usual height of a prime ideal If A is a noetherianring then htI ltinfin for every proper ideal I since Ap is a local noetherian ring and htp = dim(Ap)

Lemma 52 Let A be a nonzero ring and I a proper ideal Then we have

htI = infhtIAm |m is a maximal ideal and I sube m

Lemma 53 Let A be a noetherian ring and suppose we have an exact sequence

0 minusrarrM prime minusrarrM minusrarrM primeprime minusrarr 0

in which M primeMM primeprime are nonzero and finitely generated Then dimM = maxdimM prime dimM primeprime

Proof We know that Supp(M) = Supp(M prime)cupSupp(M primeprime) and for all three modules the dimensionis the supremum of the coheights of prime ideals in the support So the result is straightforwardto check

16

41 Homomorphism and Dimension

Let φ A minusrarr B be a morphism of rings If p isin Spec(A) then put κ(p) = AppAp Let Bp denotethe ring Tminus1B where T = φ(Aminusp) There is an isomorphism of A-algebras Bp

sim= BotimesAAp Thereis a commutative diagram of rings

B

))RRRRRRRRRRRRRRRR Bp BppBp

B otimesA κ(p)

The vertical isomorphism is defined by bφ(s)+ pBp 7rarr botimes (1s+ pAp) We call Spec(BotimesA κ(p))the fibre over p of the map Φ Spec(B) minusrarr Spec(A) Since primes of BppBp clearly correspondto primes q of B with q cap A = p it is easy to see that the ring morphism B minusrarr B otimesA κ(p) givesrise to a continuous map Spec(B otimesA κ(p)) minusrarr SpecB which gives a homeomorphism betweenΦminus1(p) and Spec(B otimesA κ(p)) See [AM69] Chapter 3

Lemma 54 Let q be a prime ideal of B with q cap A = p and let P be the corresponding primeideal of B otimesA κ(p) Then there is an isomorphism of A-algebras

Bq otimesA κ(p) sim= (B otimesA κ(p))P

btotimes (as+ pAp) 7rarr (botimes (a1 + pAp))(totimes (s1 + pAp))

Proof It is not difficult to see that there is an isomorphism of rings Bqsim= (Bp)qBp defined by

bt 7rarr (b1)(t1) Consider the prime ideal qBppBp We know that there is a ring isomorphism

(B otimesA κ(p))Psim= (BppBp)qBppBp

sim= (Bp)qBpp(Bp)qBpsim= BqpBq

by the comments following Lemma 7 It is not hard to check there is a ring isomorphism BqpBqsim=

BqotimesA κ(p) defined by bt+ pBq 7rarr btotimes 1 (the inverse of btotimes (as+ pAp) is bφ(a)tφ(s)+ pBq)So by definition of P there is an isomorphism of rings (B otimesA κ(p))P

sim= Bq otimesA κ(p) and this isclearly an isomorphism of A-algebras

In particular if φ A minusrarr B is a ring morphism p isin Spec(A) and q isin Spec(B) such thatq capA = p then there is an isomorphism of rings (BpB)qpB

sim= BqpBq so we have ht(qpB) =dim(Bq otimesA κ(p))

Theorem 55 Let φ A minusrarr B be a morphism of noetherian rings Let q isin Spec(B) and putp = q capA Then

(1) htq le htp + ht(qpB) In other words dim(Bq) le dim(Ap) + dim(Bq otimesA κ(p))

(2) We have equality in (1) if the going-down theorem holds for φ (in particular if φ is flat)

(3) If Φ Spec(B) minusrarr Spec(A) is surjective and if the going-down theorem holds then we havedim(B) ge dim(A) and htI = ht(IB) for any proper ideal I of A

Proof (1) Replacing A by Ap and B by Bq we may suppose that (A p) and (B q) are noetherianlocal rings such that q cap A = p and we must show that dim(B) le dim(A) + dim(BpB) Let Ibe a p-primary ideal of A Then pn sube I for some n gt 0 so pnB sube IB sube pB Thus the ideals pBand IB have the same radical and so by definition dim(BpB) = dim(BIB) If dimA = 0 thenwe can take I = 0 and the result is trivial So assume dimA = r ge 1 and let I = (a1 ar) fora system of parameters a1 ar If dim(BIB) = 0 then IB is an q-primary ideal of B and sodim(B) le r as required Otherwise if dim(BIB) = s ge 1 let b1 + IB bs + IB be a systemof parameters of BIB Then b1 bs a1 ar generate an ideal of definition of B Hencedim(B) le r + s

(2) We use the same notation as above If ht(qpB) = s ge 0 then there exists a prime chainof length s q = q0 sup q1 sup middot middot middot sup qs such that qs supe pB As p = q cap A supe qi cap A supe p all the qi

17

lie over p If htp = r ge 0 then there exists a prime chain p = p0 sup p1 sup middot middot middot sup pr in A and bygoing-down there exists a prime chain qs = t0 sup t1 sup middot middot middot sup tr of B such that ti capA = pi Then

q = q0 sup middot middot middot sup qs sup t1 sup middot middot middot sup tr

is a prime chain of length r + s therefore htq ge r + s(3) (i) follows from (2) since dim(A) = suphtp | p isin Spec(A) (ii) Since Φ is surjective IB

is a proper ideal Let q be a minimal prime over IB such that htq = ht(IB) and put p = q capAThen ht(qpB) = 0 so by (2) we find that ht(IB) = htq = htp ge htI For the reverseinclusion let p be a minimal prime ideal over I such that htp = htI and take a prime q of Blying over p Replacing q if necessary we may assume that q is a minimal prime ideal over pBThen htI = htp = htq ge ht(IB)

Theorem 56 Let A be a nonzero subring of B and suppose that B is integral over A Then

(1) dim(A) = dim(B)

(2) Let q isin Spec(B) and set p = q capA Then we have cohtp = cohtq and htq le htp

(3) If the going-down theorem holds between A and B then for any ideal J of B with J capA 6= Awe have htJ = ht(J capA)

Proof (1) By Theorem 42 the going-up theorem holds for A sube B and Spec(B) minusrarr Spec(A) issurjective so we can lift any chain of prime ideals p0 sub p1 sub middot middot middot sub pn in A to a chain of primeideals q0 sub middot middot middot sub qn in B On the other hand if q sube qprime are prime ideals of B and q capA = qprime capAthen q = qprime so any chain of prime ideals in B restricts to a chain of the same length in A Hencedim(A) = dim(B)

(2) Since Bq is integral over Ap it is clear from (1) that cohtp = cohtq If q = q0 sup middot middot middot sup qnis a chain of prime ideals in B then intersecting with A gives a chain of length n descending fromp Hence htq le htp

(3) Given the going-down theorem it is clear that htq = htp in (2) Let J be a proper ideal ofB with J capA 6= A and let q be such that htJ = htq Then ht(J capA) le ht(qcapA) = htq = htJ On the other hand BJ is integral over BJ cap A so every prime ideal p of A containing J cap Acan be lifted to a prime ideal q of B containing J In particular we can lift a prime ideal p withht(J capA) = htp to see that htJ le htq = htp = ht(J capA) as required

5 Depth

Definition 11 Let A be a ring M an A-module and a1 ar a sequence of elements of AWe say a1 ar is an M -regular sequence (or simply M -sequence) if the following conditions aresatisfied

(1) For each 2 le i le r ai is regular on M(a1 aiminus1)M and a1 is regular on M

(2) M 6= (a1 an)M

If a1 ar is an M -regular sequence then so is a1 ai for any i le r When all ai belong to anideal I we say a1 ar is an M -regular sequence in I If moreover there is no b isin I such thata1 ar b is M -regular then a1 ar is said to be a maximal M -regular sequence in I Noticethat the notion of M -regular depends on the order of the elements in the sequence If MN areisomorphic A-modules then a sequence is regular on M iff it is regular on N

Lemma 57 A sequence a1 ar with r ge 2 is M -regular if and only if a1 is regular on Mand a2 ar is an Ma1M -regular sequence If the sequence a1 ar is a maximal M -regularsequence in I then a2 ar is a maximal Ma1M -regular sequence in I

18

Proof The key point is that for ideals a sube b there is a canonical isomorphism of A-modulesMbM sim= NbN where N = MaM If a1 ar is M -regular then a1 is regular on M a2 isregular on N = Ma1M and for 3 le i le r ai is regular on

M(a1 aiminus1)M sim= N(a2 aiminus1)N

Hence a2 ar is an N -regular sequence The converse follows from the same argument

More generally if a1 ar is an M -regular sequence and we set N = M(a1 ar)M andif b1 bs is an N -regular sequence then a1 ar b1 bs is an M -regular sequence

Lemma 58 If a1 ar is an A-regular sequence and M is a flat A-module then a1 ar isalso M -regular provided (a1 ar)M 6= M

Proof Left multiplication by a1 defines a monomorphism A minusrarr A since a1 is A-regular Tensoringwith M and using the fact that M is flat we see that left multiplication by a1 also gives amonomorphismM minusrarrM as required Similarly tensoring with the monomorphism a2 Aa1 minusrarrAa1 we get a monomorphism Ma1M minusrarrMa1M and so on

Lemma 59 Let A be a ring and M an A-module Given an integer n ge 1 a sequence a1 aris M -regular if and only if it is Mn-regular

Proof Suppose the sequence a1 ar is M -regular We prove it is Mn-regular by induction on rThe case r = 1 is trivial so assume r gt 1 By the inductive hypothesis the sequence a1 arminus1 isMn-regular Let L = (a1 arminus1)M Then (a1 arminus1)Mn = Ln and there is an isomorphismof A-modules MnLn sim= (ML)n So we need only show that ar is regular on (ML)n Since byassumption it is regular on ML this is not hard to check Clearly (a1 ar)Mn 6= Mn so thesequence a1 ar is Mn-regular as required The converse is similarly checked

Lemma 60 Let A be a nonzero ring M an A-module and a1 ar isin A If a1 ar isin Am isMm-regular for every maximal ideal m of A then the sequence a1 ar is M -regular

Proof This follows from the fact that given an A-module M an element a isin A is regular on M ifand only if its image in Am is regular on Mm for every maximal ideal m of A

Lemma 61 Suppose that a1 ar is M -regular and a1ξ1 + middot middot middot + arξr = 0 for ξi isin M Thenξi isin (a1 ar)M for all i

Proof By induction on r For r = 1 a1ξ1 = 0 implies that ξ1 = 0 Let r gt 1 Since ar is regularon M(a1 arminus1)M we have ξr =

sumrminus1i=1 aiηi so

sumrminus1i=1 ai(ξi + arηi) = 0 By the inductive

hypothesis for i lt r we have ξi + arηi isin (a1 arminus1)M so that ξi isin (a1 ar)M

Theorem 62 Let A be a ring and M an A-module and let a1 ar be an M -regular sequenceThen for every sequence n1 nr of integers gt 0 the sequence an1

1 anrr is M -regular

Proof Suppose we can prove the following statement

(lowast) Given an integer n gt 0 an A-module M and any M -regular sequence a1 arthe sequence an1 a2 ar is also M -regular

We prove the rest of the Theorem by induction on r For r = 1 this follows immediately from (lowast)Let r gt 1 and suppose a1 ar is M -regular Then by (lowast) an1

1 a2 ar is M -regular Hencea2 ar is Man1

1 M -regular By the inductive hypothesis an22 anr

r is Man11 M -regular and

therefore an11 anr

r is M -regular by Lemma 57So it only remains to prove (lowast) which we do by induction on n The case n = 1 is trivial so let

n gt 1 be given along with an A-moduleM and anM -regular sequence a1 ar By the inductivehypothesis anminus1

1 a2 ar is M -regular and clearly an1 is regular on M Since (an1 a2 ar) sube(anminus1

1 a2 ar) it is clear thatM 6= (an1 a2 ar)M Let i gt 1 and assume that an1 a2 aiminus1

is an M -regular sequence We need to show that ai is regular on M(an1 a2 aiminus1)M Suppose

19

that aiω = an1 ξ1 +a2ξ2 + middot middot middot+aiminus1ξiminus1 Then ω = anminus11 η1 +a2η2 + middot middot middot+aiminus1ηiminus1 by the inductive

hypothesis So

anminus11 (a1ξ1 minus aiη1) + a2(ξ2 minus aiη2) + middot middot middot+ aiminus1(ξiminus1 minus aiηiminus1) = 0

Hence a1ξ1minusaiη1 isin (anminus11 a2 aiminus1)M by Lemma 61 It follows that aiη1 isin (a1 a2 aiminus1)M

hence η1 isin (a1 aiminus1)M and so ω isin (an1 a2 aiminus1)M as required This proves (lowast) andtherefore completes the proof

Let A be a ring There is an isomorphism of A[x1 xn]-modules M otimesA A[x1 xn] sim=M [x1 xn] where the latter module consists of polynomials in x1 xn with coefficients in M(see our Polynomial Ring notes) For any f(x1 xn) isinM [x1 xn] and tuple (a1 an) isinAn we can define an element of M

f(a1 an) =sumα

aα11 middot middot middot aαn

n middot f(α)

For an element r isin R and h isinM [x1 xn]

(f + h)(a1 an) = f(a1 an) + h(a1 an)(r middot f)(a1 an) = r middot f(a1 an)

For an ideal I in R the R-submodule IM [x1 xn] consists of all polynomials whose coefficientsare in the R-submodule IM subeM

Let us review the definition of the associated graded ring and modules Let A be a ring and Ian ideal of A Then the abelian group

grI(A) = AI oplus II2 oplus I2I3 oplus middot middot middot

becomes a graded ring in a fairly obvious way For an A-module M we have the graded grI(A)-module

grI(M) = MIM oplus IMI2M oplus I2MI3M oplus middot middot middotIf A is noetherian and M is a finitely generated A-module then grI(A) is a noetherian ring andif grI(M) is a finitely generated grI(A)-module

Given elements a1 an isin A and I = (a1 an) we define a morphism of abelian groupsψ M [x1 xn] minusrarr grI(M) as follows if f is homogenous of degreem ge 0 define ψ(f) to be theimage of f(a1 an) in ImMIm+1M This defines a morphism of groups M [x1 xn]m minusrarrImMIm+1M and together these define the morphism of groups ψ Since ψ maps IM [x1 xn]to zero it induces a morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) and

Proposition 63 Let A be a ring and M an A-module Let a1 an isin A and set I =(a1 an) Then the following conditions are equivalent

(a) For every m gt 0 and for every homogenous polynomial f(x1 xn) isin M [x1 xn] ofdegree m such that f(a1 an) isin Im+1M we have f isin IM [x1 xn]

(b) If f(x1 xn) isin M [x1 xn] is homogenous and f(a1 an) = 0 then the coefficientsof f are in IM

(c) The morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) defined by mappinga homogenous polynomial f(x1 xn) of degree m to f(a1 an) isin ImMIm+1M is anisomorphism

Proof It is easy to see that (a) hArr (c) and (a) rArr (b) To show (b) rArr (a) let f isinM [x1 xn] bea homogenous polynomial of degree m gt 0 and suppose f(a1 an) isin Im+1M Any element ofIm+1M can be written as a sum of terms of the form aα1

1 middot middot middot aαnn middotm with

sumi αi = m+1 By shifting

one of the ai across we can write f(a1 an) = g(a1 an) for a homogenous polynomialg isinM [x1 xn] of degreem all of whose coefficients belong to IM Hence (fminusg)(a1 an) = 0so by (b) the coefficients of f minus g belong to IM and this implies implies that the coefficients of falso belong to IM as required

20

Definition 12 Let A be a ring and M an A-module A sequence a1 an isin A is M -quasiregularif it satisfies the equivalent conditions of the Proposition Obviously this concept does not dependon the order of the elements But a1 ai for i lt n need not be M -quasiregular

Recall that for an A-module M a submodule N sube M and x isin A the notation (N x) meansm isin M |xm isin N This is a submodule of M If A is a ring I an ideal and M an A-modulerecall that M is separated in the I-adic topology when

⋂n I

nM = 0

Theorem 64 Let A be a ring M a nonzero A-module a1 an isin A and I = (a1 an)Then

(i) If a1 an is M -quasiregular and x isin A is such that (IM x) = IM then (ImM x) =ImM for all m gt 0

(ii) If a1 an is M -regular then it is M -quasiregular

(iii) If MMa1MM(a1 a2)M M(a1 anminus1)M are separated in the I-adic topologythen the converse of (ii) is also true

Proof (i) By induction on m with the case m = 1 true by assumption Suppose m gt 1 andxξ isin ImM By the inductive hypothesis ξ isin Imminus1M Hence there exists a homogenous polynomialf isinM [x1 xn] of degree mminus 1 such that ξ = f(a1 an) Since xξ = xf(a1 an) isin ImMthe coefficients of f are in (IM x) = IM Therefore ξ = f(a1 an) isin ImM

(ii) By induction on n For n = 1 this is easy to check Let n gt 1 and suppose a1 an is M -regular Then by the induction hypothesis a1 anminus1 is M -quasiregular Let f isinM [x1 xn]be homogenous of degree m gt 0 such that f(a1 an) = 0 We prove that f isin IM [x1 xn]by induction on m (the case m = 0 being trivial) Write

f(x1 xn) = g(x1 xnminus1) + xnh(x1 xn)

Then g and h are homogenous of degrees m and mminus 1 respectively By (i) we have

h(a1 an) isin ((a1 anminus1)mM an) = (a1 anminus1)mM sube ImM

Since by assumption a1 an is regular on M so an is regular on M(a1 anminus1)M andhence ((a1 anminus1)M an) = (a1 anminus1)M So by the induction hypothesis on m we haveh isin IM [x1 xn] (by the argument of Proposition 63) Since h(a1 an) isin (a1 anminus1)mMthere exists H isin M [x1 xnminus1] which is homogenous of degree m such that h(a1 an) =H(a1 anminus1) Putting

G(x1 xnminus1) = g(x1 xnminus1) + anH(x1 xnminus1)

we have G(a1 anminus1) = 0 so by the inductive hypothesis on n we have G isin IM [x1 xn]hence g isin IM [x1 xn] and so f isin IM [x1 xn]

(iii) By induction on n ge 1 Assume that a1 an is M -quasiregular and the modulesMMa1M M(a1 anminus1)M are all separated in the I-adic topology If a1ξ = 0 thenξ isin IM hence ξ =

sumaiηi and

suma1aiηi = 0 hence ηi isin IM and so ξ isin I2M In this way we

see that ξ isin⋂t ItM = 0 Thus a1 is regular on M and this also takes care of the case n = 1

since M 6= IM by the separation condition So assume n gt 1 By Lemma 57 it suffices to showthat a2 an is an N -regular sequence where N = Ma1M Since there is an isomorphism ofA-modules for 2 le i le nminus 1

M(a1 ai)M sim= N(a2 ai)N

The modules NNa2N N(a2 anminus1)N are separated in the I-adic topology So by theinductive hypothesis it suffices to show that the sequence a2 an is N -quasiregular

It suffices to show that if f(x2 xn) isin M [x2 xn] is homogenous of degree m ge 1 withf(a2 an) isin a1M then the coefficients of f belong to IM Put f(a2 an) = a1ω We claim

21

that ω isin Imminus1M Let 0 le i le mminus1 be the largest integer with ω isin IiM Then ω = g(a1 an)for some homogenous polynomial of degree i and

f(a2 an) = a1g(a1 an) (2)

If i lt mminus1 then g isin IM [x1 xn] and so ω isin Ii+1M which is a contradiction Hence i = mminus1and so ω isin Imminus1M Again using (2) we see that f(x2 xn)minusx1g(x1 xn) isin IM [x1 xn]Since f does not involve x1 we have f isin IM [x1 xn] as required

The theorem shows that under the assumptions of (iii) any permutation of an M -regularsequence is M -regular

Corollary 65 Let A be a noetherian ring M a finitely generated A-module and let a1 anbe contained in the Jacobson radical of A Then a1 an is M -regular if and only if it is M -quasiregular In particular if a1 an is M -regular so is any permutation of the sequence

Proof From [AM69] we know that for any ideal I contained in the Jacobson radical the I-adictopology on any finitely generated A-module is separated

If A is a ring and M an A-module then any M -regular sequence a1 an isin A gives rise toa strictly increasing chain of submodules a1M (a1 a2)M (a1 an)M Hence the chain ofideals (a1) (a1 a2) (a1 an) must also be strictly increasing

Lemma 66 Let A be a noetherian ring and M an A-module Any M -regular sequence a1 anin an ideal I can be extended to a maximal M -regular sequence in I

Proof If a1 an is not maximal in I we can find an+1 isin I such that a1 an an+1 is anM -regular sequence Either this process terminates at a maximal M -regular sequence in I or itproduces a strictly ascending chain of ideals

(a1) sub (a1 a2) sub (a1 a2 a3) sub middot middot middot

Since A is noetherian we can exclude this latter possibility

Theorem 67 Let A be a noetherian ring M a finitely generated A-module and I an ideal of Awith IM 6= M Let n gt 0 be an integer Then the following are equivalent

(1) ExtiA(NM) = 0 for i lt n and every finitely generated A-module N with Supp(N) sube V (I)

(2) ExtiA(AIM) = 0 for i lt n

(3) There exists a finitely generated A-module N with Supp(N) = V (I) and ExtiA(NM) = 0for i lt n

(4) There exists an M -regular sequence a1 an of length n in I

Proof (1) rArr (2) rArr (3) is trivial With I fixed we show that (3) rArr (4) for every finitely generatedmodule M with IM 6= M by induction on n We have 0 = Ext0A(NM) sim= HomA(NM) SinceM is finitely generated and nonzero the set of associated primes of M is finite and nonempty Ifno elements of I are M -regular then I is contained in the union of these associated primes andhence I sube p for some p isin Ass(M) (see [AM69] for details) By definition there is a monomorphismof A-modules φ Ap minusrarrM There is an isomorphism of A-modules

(Ap)psim= ApotimesA Ap

sim= AppAp = k

It is not hard to check this is an isomorphism of Ap-modules as well Since φp is a monomorphismand k 6= 0 it follows that HomAp(kMp) 6= 0 Since p isin V (I) = Supp(N) we have Np 6= 0 andso the k-module NppNp 6= 0 is nonzero and therefore free so Homk(NppNp k) 6= 0 Sincek sim= (Ap)p as Ap-modules it follows that HomA(NAp)p

sim= HomAp(Np (Ap)p) 6= 0 Since Apis isomorphic to a submodule of M it follows that HomA(NM) 6= 0 which is a contradiction

22

therefore there exists an M -regular element a1 isin I which takes care of the case n = 1 If n gt 1then put M1 = Ma1M From the exact sequence

0 Ma1 M M1

0 (3)

we get the long exact sequence

middot middot middot minusrarr ExtiA(NM) minusrarr ExtiA(NM1) minusrarr Exti+1A (NM) minusrarr middot middot middot

which shows that ExtiA(NM1) = 0 for 0 le i lt n minus 1 By the inductive hypothesis on n thereexists an M1-regular sequence a2 an in I The sequence a1 an is then an M -regularsequence in I

(4) rArr (1) By induction on n with I fixed For n = 1 we have a1 isin I regular on M and so (3)gives an exact sequence of R-modules

0 HomA(NM)a1 HomA(NM)

Where a1 denotes left multiplication by a1 Since Supp(N) = V (Ann(N)) sube V (I) it fol-lows that I sube r(Ann(N)) and so ar1N = 0 for some r gt 0 It follows that ar1 annihilatesHomA(NM) as well but since the action of ar1 on HomA(NM) gives an injective map it followsthat HomA(NM) = 0 Now assume n gt 1 and put M1 = Ma1M Then a2 an is anM1-regular sequence so by the inductive hypothesis ExtiA(NM1) = 0 for i lt nminus 1 So the longexact sequence corresponding to (3) gives an exact sequence for 0 le i lt n

0 ExtiA(NM)a1 ExtiA(NM)

Here a1 denotes left multiplication by a1 which is equal to ExtiA(αM) = Exti(N β) where α βare the endomorphisms given by left multiplication by a1 on NM respectively Assume that ar1annihilates N with r gt 0 Then αr is the zero map so ExtiA(αM)r = 0 and so ar1 also annihilatesExtiA(NM) Since the a1 is regular on this module it follows that ExtiA(NM) = 0 for i lt nas required

Corollary 68 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA with IM 6= M If a1 an a maximal M -regular sequence in I then ExtiA(AIM) = 0 fori lt n and ExtnA(AIM) 6= 0

Proof We already know that ExtiA(AIM) = 0 for i lt n so with I fixed we prove by inductionon n that ExtnA(AIM) 6= 0 for any finitely generated module M with IM 6= M admitting amaximal M -regular sequence of length n For n = 1 we have a1 isin I regular on M and an exactsequence (3) where M1 = Ma1M Part of the corresponding long exact sequence is

Ext0A(AIM) minusrarr Ext0A(AIM1) minusrarr Ext1A(AIM)

We know from the Theorem 67 that Ext0A(AIM) = 0 so it suffices to show that we haveHomA(AIM1) 6= 0 But if HomA(AIM1) = 0 then it follows from the proof of (3) rArr (4)above that there would be b isin I regular on M1 so a1 b is an M -regular sequence This is acontradiction since the sequence a1 was maximal so we conclude that Ext1A(AIM) 6= 0

Now assume n gt 1 and let a1 an be a maximal M -regular sequence in I Then a2 anis a maximal M1-regular sequence in I so by the inductive hypothesis Extnminus1

A (AIM1) 6= 0 Sofrom the long exact sequence for (3) we conclude that ExtnA(AIM) 6= 0 also

It follows that under the conditions of the Corollary every maximal M -regular sequence in Ihas a common length and you can find this length by looking at the sequence of abelian groups

HomA(AIM) Ext1A(AIM) Ext2A(AIM) ExtnA(AIM)

If there are M -regular sequences in I then this sequence will start off with n minus 1 zero groupswhere n ge 1 is the common length of every maximal M -regular sequence The nth group will

23

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

closed in X so to complete the proof it suffices to show that Z cap F lowast is constructible in X SinceZ cap F lowast sube F lowast sub Z sube Y by the inductive hypothesis P holds for ZcapF lowast so it suffices to show thatZ cap F lowast satisfies (lowast) If X0 is irreducible and Z cap F lowast capX0 = X0 the closed set F lowast must containX0 and so Z cap F lowast capX0 = Z capX0 which contains a nonempty open subset of X0 since Z satisfies(lowast) and clearly Z capX0 is dense in X0

Lemma 46 Let φ A minusrarr B be a morphism of rings and f Spec(B) minusrarr Spec(A) the corre-sponding morphism of schemes Then f dominant if and only if Kerφ sube nil(A) In particular ifA is reduced the f dominant if and only if φ is injective

Proof Let X = Spec(A) and Y = Spec(B) The closure f(Y ) is the closed set V (I) defined bythe ideal I =

⋂pisinY φ

minus1p = φminus1⋂

pisinY p which is φminus1(nil(B)) Clearly Kerφ sube I Suppose thatf(Y ) is dense in X Then V (I) = X whence I = nil(A) and so Kerφ sube nil(A) Converselysuppose Kerφ sube nil(A) Then it is clear that I = φminus1(nil(B)) = nil(A) which means thatf(Y ) = V (I) = X

3 Associated Primes

This material can be found in [AM69] Chapter 11 webnotes of Robert Ash or in [Mat80] itselfThere is not much relevant to add here apart from a few small comments

Lemma 47 Let A be a ring and M an A-module Let a be an ideal in A that is maximal amongall annihilators of nonzero elements of M Then a is prime

Proof Say a = Ann(x) Given ab isin a we must show that a isin a or b isin a Assume a isin aThen ax 6= 0 We note that Ann(ax) supe a By hypothesis it cannot properly be larger HenceAnn(ax) = a Now b annihilates ax hence b isin a

Lemma 48 Let A be a noetherian ring and M an A-module If 0 6= a isin M then Ann(a) iscontained in an associated prime of M

Proposition 49 Let A be a noetherian ring and M a nonzero finitely generated A-module Amaximal ideal m is an associated prime of M if and only if no element of m is regular on M

Proof One implication is obvious If x isin m is not regular on M say x isin Ann(b) for some nonzerob then x is contained in an associated prime of M Thus m is contained in the finite union of theassociated primes of M and since m is maximal it must be one of them

Proposition 50 Let A be a nonzero noetherian ring I an ideal and M a nonzero finitelygenerated A-module If there exist elements x y isin I with x regular on A and y regular on M thenthere exists an element of I regular on both A and M

Proof Let p1 pn be the associated primes of A and q1 qm the associated primes of M By assumption I is not contained in any of these primes But if no element of I is regular on bothA and M then I is contained in the union p1 cup middot middot middot cup pn cup q1 cup middot middot middot cup qm and therefore containedin one of these primes which is a contradiction

4 Dimension

This is covered in [AM69] so we restrict ourselves here to mentioning some of the major pointsRecall that an ideal q sube R in a ring is primary if it is proper and if whenever xy isin q we haveeither x isin q or yn isin q for some n gt 0 Then the radical of q is a prime ideal p and we say q isa p-primary ideal If a is an ideal and b supe a is p-primary then in the ring Ra the ideal ba ispa-primary A minimal primary decomposition of an ideal b is an expression

b = q1 cap middot middot middot cap qn

15

where capj 6=iqj qi for all i and the primes pi = r(qi) are all distinct If a is an ideal contained inb then

ba = q1a cap middot middot middot cap qna

is a minimal primary decomposition of ba in AaLet A be a nonzero ring Recall that dimension of an A-module M is the Krull dimension of

the ring AAnn(M) and is defined for all modules M (minus1 if M = 0) The rank is defined for freeA-modules and is the common size of any basis (0 if M = 0) Throughout these notes dim(M)will denote the dimension not the rank

Definition 9 Let (Am k) be a noetherian local ring of dimension d An ideal of definition is anm-primary ideal Recall that the dimension of A is the size of the smallest collection of elementsof A which generates an m-primary ideal Recall that rankk(mm2) is equal to the size of thesmallest set of generators for m as an ideal so always d le rankk(mm2)

A system of parameters is a set of d elements generating an m-primary ideal If d = rankk(mm2)or equivalently there is a system of parameters generating m we say that A is a regular local ringand we call such a system of parameters a regular system of parameters

Proposition 51 Let (Am) be a noetherian local ring of dimension d ge 1 and let x1 xd bea system of parameters of A Then

dim(A(x1 xi)) = dminus i = dim(A)minus i

for each 1 le i le d

Proof Put A = A(x1 xi) If i = d then the zero ideal in A is an ideal of definition so clearlydim(A) = 0 If 1 le i lt d then dim(A) le dminus i since xi+1 xd generate an ideal of definition ofA Let dim(A) = p If p = 0 then (x1 xi) must be an ideal of definition contradicting i lt dSo p ge 1 and if y1 yp is a system of parameters of A then x1 xi y1 yp generate anideal of definition of A so that p+ i ge d That is p ge dminus i

Definition 10 Let A be a nonzero ring and I a proper ideal The height of I denoted htI isthe minimum of the heights of the prime ideals containing I

htI = infhtp | p supe I

This is a number in 0 1 2 infin Equivalently we can take the infimum over the heights ofprimes minimal over I Clearly ht0 = 0 and if I sube J are proper ideals then it is clear thathtI le htJ If I is a prime ideal then htI is the usual height of a prime ideal If A is a noetherianring then htI ltinfin for every proper ideal I since Ap is a local noetherian ring and htp = dim(Ap)

Lemma 52 Let A be a nonzero ring and I a proper ideal Then we have

htI = infhtIAm |m is a maximal ideal and I sube m

Lemma 53 Let A be a noetherian ring and suppose we have an exact sequence

0 minusrarrM prime minusrarrM minusrarrM primeprime minusrarr 0

in which M primeMM primeprime are nonzero and finitely generated Then dimM = maxdimM prime dimM primeprime

Proof We know that Supp(M) = Supp(M prime)cupSupp(M primeprime) and for all three modules the dimensionis the supremum of the coheights of prime ideals in the support So the result is straightforwardto check

16

41 Homomorphism and Dimension

Let φ A minusrarr B be a morphism of rings If p isin Spec(A) then put κ(p) = AppAp Let Bp denotethe ring Tminus1B where T = φ(Aminusp) There is an isomorphism of A-algebras Bp

sim= BotimesAAp Thereis a commutative diagram of rings

B

))RRRRRRRRRRRRRRRR Bp BppBp

B otimesA κ(p)

The vertical isomorphism is defined by bφ(s)+ pBp 7rarr botimes (1s+ pAp) We call Spec(BotimesA κ(p))the fibre over p of the map Φ Spec(B) minusrarr Spec(A) Since primes of BppBp clearly correspondto primes q of B with q cap A = p it is easy to see that the ring morphism B minusrarr B otimesA κ(p) givesrise to a continuous map Spec(B otimesA κ(p)) minusrarr SpecB which gives a homeomorphism betweenΦminus1(p) and Spec(B otimesA κ(p)) See [AM69] Chapter 3

Lemma 54 Let q be a prime ideal of B with q cap A = p and let P be the corresponding primeideal of B otimesA κ(p) Then there is an isomorphism of A-algebras

Bq otimesA κ(p) sim= (B otimesA κ(p))P

btotimes (as+ pAp) 7rarr (botimes (a1 + pAp))(totimes (s1 + pAp))

Proof It is not difficult to see that there is an isomorphism of rings Bqsim= (Bp)qBp defined by

bt 7rarr (b1)(t1) Consider the prime ideal qBppBp We know that there is a ring isomorphism

(B otimesA κ(p))Psim= (BppBp)qBppBp

sim= (Bp)qBpp(Bp)qBpsim= BqpBq

by the comments following Lemma 7 It is not hard to check there is a ring isomorphism BqpBqsim=

BqotimesA κ(p) defined by bt+ pBq 7rarr btotimes 1 (the inverse of btotimes (as+ pAp) is bφ(a)tφ(s)+ pBq)So by definition of P there is an isomorphism of rings (B otimesA κ(p))P

sim= Bq otimesA κ(p) and this isclearly an isomorphism of A-algebras

In particular if φ A minusrarr B is a ring morphism p isin Spec(A) and q isin Spec(B) such thatq capA = p then there is an isomorphism of rings (BpB)qpB

sim= BqpBq so we have ht(qpB) =dim(Bq otimesA κ(p))

Theorem 55 Let φ A minusrarr B be a morphism of noetherian rings Let q isin Spec(B) and putp = q capA Then

(1) htq le htp + ht(qpB) In other words dim(Bq) le dim(Ap) + dim(Bq otimesA κ(p))

(2) We have equality in (1) if the going-down theorem holds for φ (in particular if φ is flat)

(3) If Φ Spec(B) minusrarr Spec(A) is surjective and if the going-down theorem holds then we havedim(B) ge dim(A) and htI = ht(IB) for any proper ideal I of A

Proof (1) Replacing A by Ap and B by Bq we may suppose that (A p) and (B q) are noetherianlocal rings such that q cap A = p and we must show that dim(B) le dim(A) + dim(BpB) Let Ibe a p-primary ideal of A Then pn sube I for some n gt 0 so pnB sube IB sube pB Thus the ideals pBand IB have the same radical and so by definition dim(BpB) = dim(BIB) If dimA = 0 thenwe can take I = 0 and the result is trivial So assume dimA = r ge 1 and let I = (a1 ar) fora system of parameters a1 ar If dim(BIB) = 0 then IB is an q-primary ideal of B and sodim(B) le r as required Otherwise if dim(BIB) = s ge 1 let b1 + IB bs + IB be a systemof parameters of BIB Then b1 bs a1 ar generate an ideal of definition of B Hencedim(B) le r + s

(2) We use the same notation as above If ht(qpB) = s ge 0 then there exists a prime chainof length s q = q0 sup q1 sup middot middot middot sup qs such that qs supe pB As p = q cap A supe qi cap A supe p all the qi

17

lie over p If htp = r ge 0 then there exists a prime chain p = p0 sup p1 sup middot middot middot sup pr in A and bygoing-down there exists a prime chain qs = t0 sup t1 sup middot middot middot sup tr of B such that ti capA = pi Then

q = q0 sup middot middot middot sup qs sup t1 sup middot middot middot sup tr

is a prime chain of length r + s therefore htq ge r + s(3) (i) follows from (2) since dim(A) = suphtp | p isin Spec(A) (ii) Since Φ is surjective IB

is a proper ideal Let q be a minimal prime over IB such that htq = ht(IB) and put p = q capAThen ht(qpB) = 0 so by (2) we find that ht(IB) = htq = htp ge htI For the reverseinclusion let p be a minimal prime ideal over I such that htp = htI and take a prime q of Blying over p Replacing q if necessary we may assume that q is a minimal prime ideal over pBThen htI = htp = htq ge ht(IB)

Theorem 56 Let A be a nonzero subring of B and suppose that B is integral over A Then

(1) dim(A) = dim(B)

(2) Let q isin Spec(B) and set p = q capA Then we have cohtp = cohtq and htq le htp

(3) If the going-down theorem holds between A and B then for any ideal J of B with J capA 6= Awe have htJ = ht(J capA)

Proof (1) By Theorem 42 the going-up theorem holds for A sube B and Spec(B) minusrarr Spec(A) issurjective so we can lift any chain of prime ideals p0 sub p1 sub middot middot middot sub pn in A to a chain of primeideals q0 sub middot middot middot sub qn in B On the other hand if q sube qprime are prime ideals of B and q capA = qprime capAthen q = qprime so any chain of prime ideals in B restricts to a chain of the same length in A Hencedim(A) = dim(B)

(2) Since Bq is integral over Ap it is clear from (1) that cohtp = cohtq If q = q0 sup middot middot middot sup qnis a chain of prime ideals in B then intersecting with A gives a chain of length n descending fromp Hence htq le htp

(3) Given the going-down theorem it is clear that htq = htp in (2) Let J be a proper ideal ofB with J capA 6= A and let q be such that htJ = htq Then ht(J capA) le ht(qcapA) = htq = htJ On the other hand BJ is integral over BJ cap A so every prime ideal p of A containing J cap Acan be lifted to a prime ideal q of B containing J In particular we can lift a prime ideal p withht(J capA) = htp to see that htJ le htq = htp = ht(J capA) as required

5 Depth

Definition 11 Let A be a ring M an A-module and a1 ar a sequence of elements of AWe say a1 ar is an M -regular sequence (or simply M -sequence) if the following conditions aresatisfied

(1) For each 2 le i le r ai is regular on M(a1 aiminus1)M and a1 is regular on M

(2) M 6= (a1 an)M

If a1 ar is an M -regular sequence then so is a1 ai for any i le r When all ai belong to anideal I we say a1 ar is an M -regular sequence in I If moreover there is no b isin I such thata1 ar b is M -regular then a1 ar is said to be a maximal M -regular sequence in I Noticethat the notion of M -regular depends on the order of the elements in the sequence If MN areisomorphic A-modules then a sequence is regular on M iff it is regular on N

Lemma 57 A sequence a1 ar with r ge 2 is M -regular if and only if a1 is regular on Mand a2 ar is an Ma1M -regular sequence If the sequence a1 ar is a maximal M -regularsequence in I then a2 ar is a maximal Ma1M -regular sequence in I

18

Proof The key point is that for ideals a sube b there is a canonical isomorphism of A-modulesMbM sim= NbN where N = MaM If a1 ar is M -regular then a1 is regular on M a2 isregular on N = Ma1M and for 3 le i le r ai is regular on

M(a1 aiminus1)M sim= N(a2 aiminus1)N

Hence a2 ar is an N -regular sequence The converse follows from the same argument

More generally if a1 ar is an M -regular sequence and we set N = M(a1 ar)M andif b1 bs is an N -regular sequence then a1 ar b1 bs is an M -regular sequence

Lemma 58 If a1 ar is an A-regular sequence and M is a flat A-module then a1 ar isalso M -regular provided (a1 ar)M 6= M

Proof Left multiplication by a1 defines a monomorphism A minusrarr A since a1 is A-regular Tensoringwith M and using the fact that M is flat we see that left multiplication by a1 also gives amonomorphismM minusrarrM as required Similarly tensoring with the monomorphism a2 Aa1 minusrarrAa1 we get a monomorphism Ma1M minusrarrMa1M and so on

Lemma 59 Let A be a ring and M an A-module Given an integer n ge 1 a sequence a1 aris M -regular if and only if it is Mn-regular

Proof Suppose the sequence a1 ar is M -regular We prove it is Mn-regular by induction on rThe case r = 1 is trivial so assume r gt 1 By the inductive hypothesis the sequence a1 arminus1 isMn-regular Let L = (a1 arminus1)M Then (a1 arminus1)Mn = Ln and there is an isomorphismof A-modules MnLn sim= (ML)n So we need only show that ar is regular on (ML)n Since byassumption it is regular on ML this is not hard to check Clearly (a1 ar)Mn 6= Mn so thesequence a1 ar is Mn-regular as required The converse is similarly checked

Lemma 60 Let A be a nonzero ring M an A-module and a1 ar isin A If a1 ar isin Am isMm-regular for every maximal ideal m of A then the sequence a1 ar is M -regular

Proof This follows from the fact that given an A-module M an element a isin A is regular on M ifand only if its image in Am is regular on Mm for every maximal ideal m of A

Lemma 61 Suppose that a1 ar is M -regular and a1ξ1 + middot middot middot + arξr = 0 for ξi isin M Thenξi isin (a1 ar)M for all i

Proof By induction on r For r = 1 a1ξ1 = 0 implies that ξ1 = 0 Let r gt 1 Since ar is regularon M(a1 arminus1)M we have ξr =

sumrminus1i=1 aiηi so

sumrminus1i=1 ai(ξi + arηi) = 0 By the inductive

hypothesis for i lt r we have ξi + arηi isin (a1 arminus1)M so that ξi isin (a1 ar)M

Theorem 62 Let A be a ring and M an A-module and let a1 ar be an M -regular sequenceThen for every sequence n1 nr of integers gt 0 the sequence an1

1 anrr is M -regular

Proof Suppose we can prove the following statement

(lowast) Given an integer n gt 0 an A-module M and any M -regular sequence a1 arthe sequence an1 a2 ar is also M -regular

We prove the rest of the Theorem by induction on r For r = 1 this follows immediately from (lowast)Let r gt 1 and suppose a1 ar is M -regular Then by (lowast) an1

1 a2 ar is M -regular Hencea2 ar is Man1

1 M -regular By the inductive hypothesis an22 anr

r is Man11 M -regular and

therefore an11 anr

r is M -regular by Lemma 57So it only remains to prove (lowast) which we do by induction on n The case n = 1 is trivial so let

n gt 1 be given along with an A-moduleM and anM -regular sequence a1 ar By the inductivehypothesis anminus1

1 a2 ar is M -regular and clearly an1 is regular on M Since (an1 a2 ar) sube(anminus1

1 a2 ar) it is clear thatM 6= (an1 a2 ar)M Let i gt 1 and assume that an1 a2 aiminus1

is an M -regular sequence We need to show that ai is regular on M(an1 a2 aiminus1)M Suppose

19

that aiω = an1 ξ1 +a2ξ2 + middot middot middot+aiminus1ξiminus1 Then ω = anminus11 η1 +a2η2 + middot middot middot+aiminus1ηiminus1 by the inductive

hypothesis So

anminus11 (a1ξ1 minus aiη1) + a2(ξ2 minus aiη2) + middot middot middot+ aiminus1(ξiminus1 minus aiηiminus1) = 0

Hence a1ξ1minusaiη1 isin (anminus11 a2 aiminus1)M by Lemma 61 It follows that aiη1 isin (a1 a2 aiminus1)M

hence η1 isin (a1 aiminus1)M and so ω isin (an1 a2 aiminus1)M as required This proves (lowast) andtherefore completes the proof

Let A be a ring There is an isomorphism of A[x1 xn]-modules M otimesA A[x1 xn] sim=M [x1 xn] where the latter module consists of polynomials in x1 xn with coefficients in M(see our Polynomial Ring notes) For any f(x1 xn) isinM [x1 xn] and tuple (a1 an) isinAn we can define an element of M

f(a1 an) =sumα

aα11 middot middot middot aαn

n middot f(α)

For an element r isin R and h isinM [x1 xn]

(f + h)(a1 an) = f(a1 an) + h(a1 an)(r middot f)(a1 an) = r middot f(a1 an)

For an ideal I in R the R-submodule IM [x1 xn] consists of all polynomials whose coefficientsare in the R-submodule IM subeM

Let us review the definition of the associated graded ring and modules Let A be a ring and Ian ideal of A Then the abelian group

grI(A) = AI oplus II2 oplus I2I3 oplus middot middot middot

becomes a graded ring in a fairly obvious way For an A-module M we have the graded grI(A)-module

grI(M) = MIM oplus IMI2M oplus I2MI3M oplus middot middot middotIf A is noetherian and M is a finitely generated A-module then grI(A) is a noetherian ring andif grI(M) is a finitely generated grI(A)-module

Given elements a1 an isin A and I = (a1 an) we define a morphism of abelian groupsψ M [x1 xn] minusrarr grI(M) as follows if f is homogenous of degreem ge 0 define ψ(f) to be theimage of f(a1 an) in ImMIm+1M This defines a morphism of groups M [x1 xn]m minusrarrImMIm+1M and together these define the morphism of groups ψ Since ψ maps IM [x1 xn]to zero it induces a morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) and

Proposition 63 Let A be a ring and M an A-module Let a1 an isin A and set I =(a1 an) Then the following conditions are equivalent

(a) For every m gt 0 and for every homogenous polynomial f(x1 xn) isin M [x1 xn] ofdegree m such that f(a1 an) isin Im+1M we have f isin IM [x1 xn]

(b) If f(x1 xn) isin M [x1 xn] is homogenous and f(a1 an) = 0 then the coefficientsof f are in IM

(c) The morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) defined by mappinga homogenous polynomial f(x1 xn) of degree m to f(a1 an) isin ImMIm+1M is anisomorphism

Proof It is easy to see that (a) hArr (c) and (a) rArr (b) To show (b) rArr (a) let f isinM [x1 xn] bea homogenous polynomial of degree m gt 0 and suppose f(a1 an) isin Im+1M Any element ofIm+1M can be written as a sum of terms of the form aα1

1 middot middot middot aαnn middotm with

sumi αi = m+1 By shifting

one of the ai across we can write f(a1 an) = g(a1 an) for a homogenous polynomialg isinM [x1 xn] of degreem all of whose coefficients belong to IM Hence (fminusg)(a1 an) = 0so by (b) the coefficients of f minus g belong to IM and this implies implies that the coefficients of falso belong to IM as required

20

Definition 12 Let A be a ring and M an A-module A sequence a1 an isin A is M -quasiregularif it satisfies the equivalent conditions of the Proposition Obviously this concept does not dependon the order of the elements But a1 ai for i lt n need not be M -quasiregular

Recall that for an A-module M a submodule N sube M and x isin A the notation (N x) meansm isin M |xm isin N This is a submodule of M If A is a ring I an ideal and M an A-modulerecall that M is separated in the I-adic topology when

⋂n I

nM = 0

Theorem 64 Let A be a ring M a nonzero A-module a1 an isin A and I = (a1 an)Then

(i) If a1 an is M -quasiregular and x isin A is such that (IM x) = IM then (ImM x) =ImM for all m gt 0

(ii) If a1 an is M -regular then it is M -quasiregular

(iii) If MMa1MM(a1 a2)M M(a1 anminus1)M are separated in the I-adic topologythen the converse of (ii) is also true

Proof (i) By induction on m with the case m = 1 true by assumption Suppose m gt 1 andxξ isin ImM By the inductive hypothesis ξ isin Imminus1M Hence there exists a homogenous polynomialf isinM [x1 xn] of degree mminus 1 such that ξ = f(a1 an) Since xξ = xf(a1 an) isin ImMthe coefficients of f are in (IM x) = IM Therefore ξ = f(a1 an) isin ImM

(ii) By induction on n For n = 1 this is easy to check Let n gt 1 and suppose a1 an is M -regular Then by the induction hypothesis a1 anminus1 is M -quasiregular Let f isinM [x1 xn]be homogenous of degree m gt 0 such that f(a1 an) = 0 We prove that f isin IM [x1 xn]by induction on m (the case m = 0 being trivial) Write

f(x1 xn) = g(x1 xnminus1) + xnh(x1 xn)

Then g and h are homogenous of degrees m and mminus 1 respectively By (i) we have

h(a1 an) isin ((a1 anminus1)mM an) = (a1 anminus1)mM sube ImM

Since by assumption a1 an is regular on M so an is regular on M(a1 anminus1)M andhence ((a1 anminus1)M an) = (a1 anminus1)M So by the induction hypothesis on m we haveh isin IM [x1 xn] (by the argument of Proposition 63) Since h(a1 an) isin (a1 anminus1)mMthere exists H isin M [x1 xnminus1] which is homogenous of degree m such that h(a1 an) =H(a1 anminus1) Putting

G(x1 xnminus1) = g(x1 xnminus1) + anH(x1 xnminus1)

we have G(a1 anminus1) = 0 so by the inductive hypothesis on n we have G isin IM [x1 xn]hence g isin IM [x1 xn] and so f isin IM [x1 xn]

(iii) By induction on n ge 1 Assume that a1 an is M -quasiregular and the modulesMMa1M M(a1 anminus1)M are all separated in the I-adic topology If a1ξ = 0 thenξ isin IM hence ξ =

sumaiηi and

suma1aiηi = 0 hence ηi isin IM and so ξ isin I2M In this way we

see that ξ isin⋂t ItM = 0 Thus a1 is regular on M and this also takes care of the case n = 1

since M 6= IM by the separation condition So assume n gt 1 By Lemma 57 it suffices to showthat a2 an is an N -regular sequence where N = Ma1M Since there is an isomorphism ofA-modules for 2 le i le nminus 1

M(a1 ai)M sim= N(a2 ai)N

The modules NNa2N N(a2 anminus1)N are separated in the I-adic topology So by theinductive hypothesis it suffices to show that the sequence a2 an is N -quasiregular

It suffices to show that if f(x2 xn) isin M [x2 xn] is homogenous of degree m ge 1 withf(a2 an) isin a1M then the coefficients of f belong to IM Put f(a2 an) = a1ω We claim

21

that ω isin Imminus1M Let 0 le i le mminus1 be the largest integer with ω isin IiM Then ω = g(a1 an)for some homogenous polynomial of degree i and

f(a2 an) = a1g(a1 an) (2)

If i lt mminus1 then g isin IM [x1 xn] and so ω isin Ii+1M which is a contradiction Hence i = mminus1and so ω isin Imminus1M Again using (2) we see that f(x2 xn)minusx1g(x1 xn) isin IM [x1 xn]Since f does not involve x1 we have f isin IM [x1 xn] as required

The theorem shows that under the assumptions of (iii) any permutation of an M -regularsequence is M -regular

Corollary 65 Let A be a noetherian ring M a finitely generated A-module and let a1 anbe contained in the Jacobson radical of A Then a1 an is M -regular if and only if it is M -quasiregular In particular if a1 an is M -regular so is any permutation of the sequence

Proof From [AM69] we know that for any ideal I contained in the Jacobson radical the I-adictopology on any finitely generated A-module is separated

If A is a ring and M an A-module then any M -regular sequence a1 an isin A gives rise toa strictly increasing chain of submodules a1M (a1 a2)M (a1 an)M Hence the chain ofideals (a1) (a1 a2) (a1 an) must also be strictly increasing

Lemma 66 Let A be a noetherian ring and M an A-module Any M -regular sequence a1 anin an ideal I can be extended to a maximal M -regular sequence in I

Proof If a1 an is not maximal in I we can find an+1 isin I such that a1 an an+1 is anM -regular sequence Either this process terminates at a maximal M -regular sequence in I or itproduces a strictly ascending chain of ideals

(a1) sub (a1 a2) sub (a1 a2 a3) sub middot middot middot

Since A is noetherian we can exclude this latter possibility

Theorem 67 Let A be a noetherian ring M a finitely generated A-module and I an ideal of Awith IM 6= M Let n gt 0 be an integer Then the following are equivalent

(1) ExtiA(NM) = 0 for i lt n and every finitely generated A-module N with Supp(N) sube V (I)

(2) ExtiA(AIM) = 0 for i lt n

(3) There exists a finitely generated A-module N with Supp(N) = V (I) and ExtiA(NM) = 0for i lt n

(4) There exists an M -regular sequence a1 an of length n in I

Proof (1) rArr (2) rArr (3) is trivial With I fixed we show that (3) rArr (4) for every finitely generatedmodule M with IM 6= M by induction on n We have 0 = Ext0A(NM) sim= HomA(NM) SinceM is finitely generated and nonzero the set of associated primes of M is finite and nonempty Ifno elements of I are M -regular then I is contained in the union of these associated primes andhence I sube p for some p isin Ass(M) (see [AM69] for details) By definition there is a monomorphismof A-modules φ Ap minusrarrM There is an isomorphism of A-modules

(Ap)psim= ApotimesA Ap

sim= AppAp = k

It is not hard to check this is an isomorphism of Ap-modules as well Since φp is a monomorphismand k 6= 0 it follows that HomAp(kMp) 6= 0 Since p isin V (I) = Supp(N) we have Np 6= 0 andso the k-module NppNp 6= 0 is nonzero and therefore free so Homk(NppNp k) 6= 0 Sincek sim= (Ap)p as Ap-modules it follows that HomA(NAp)p

sim= HomAp(Np (Ap)p) 6= 0 Since Apis isomorphic to a submodule of M it follows that HomA(NM) 6= 0 which is a contradiction

22

therefore there exists an M -regular element a1 isin I which takes care of the case n = 1 If n gt 1then put M1 = Ma1M From the exact sequence

0 Ma1 M M1

0 (3)

we get the long exact sequence

middot middot middot minusrarr ExtiA(NM) minusrarr ExtiA(NM1) minusrarr Exti+1A (NM) minusrarr middot middot middot

which shows that ExtiA(NM1) = 0 for 0 le i lt n minus 1 By the inductive hypothesis on n thereexists an M1-regular sequence a2 an in I The sequence a1 an is then an M -regularsequence in I

(4) rArr (1) By induction on n with I fixed For n = 1 we have a1 isin I regular on M and so (3)gives an exact sequence of R-modules

0 HomA(NM)a1 HomA(NM)

Where a1 denotes left multiplication by a1 Since Supp(N) = V (Ann(N)) sube V (I) it fol-lows that I sube r(Ann(N)) and so ar1N = 0 for some r gt 0 It follows that ar1 annihilatesHomA(NM) as well but since the action of ar1 on HomA(NM) gives an injective map it followsthat HomA(NM) = 0 Now assume n gt 1 and put M1 = Ma1M Then a2 an is anM1-regular sequence so by the inductive hypothesis ExtiA(NM1) = 0 for i lt nminus 1 So the longexact sequence corresponding to (3) gives an exact sequence for 0 le i lt n

0 ExtiA(NM)a1 ExtiA(NM)

Here a1 denotes left multiplication by a1 which is equal to ExtiA(αM) = Exti(N β) where α βare the endomorphisms given by left multiplication by a1 on NM respectively Assume that ar1annihilates N with r gt 0 Then αr is the zero map so ExtiA(αM)r = 0 and so ar1 also annihilatesExtiA(NM) Since the a1 is regular on this module it follows that ExtiA(NM) = 0 for i lt nas required

Corollary 68 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA with IM 6= M If a1 an a maximal M -regular sequence in I then ExtiA(AIM) = 0 fori lt n and ExtnA(AIM) 6= 0

Proof We already know that ExtiA(AIM) = 0 for i lt n so with I fixed we prove by inductionon n that ExtnA(AIM) 6= 0 for any finitely generated module M with IM 6= M admitting amaximal M -regular sequence of length n For n = 1 we have a1 isin I regular on M and an exactsequence (3) where M1 = Ma1M Part of the corresponding long exact sequence is

Ext0A(AIM) minusrarr Ext0A(AIM1) minusrarr Ext1A(AIM)

We know from the Theorem 67 that Ext0A(AIM) = 0 so it suffices to show that we haveHomA(AIM1) 6= 0 But if HomA(AIM1) = 0 then it follows from the proof of (3) rArr (4)above that there would be b isin I regular on M1 so a1 b is an M -regular sequence This is acontradiction since the sequence a1 was maximal so we conclude that Ext1A(AIM) 6= 0

Now assume n gt 1 and let a1 an be a maximal M -regular sequence in I Then a2 anis a maximal M1-regular sequence in I so by the inductive hypothesis Extnminus1

A (AIM1) 6= 0 Sofrom the long exact sequence for (3) we conclude that ExtnA(AIM) 6= 0 also

It follows that under the conditions of the Corollary every maximal M -regular sequence in Ihas a common length and you can find this length by looking at the sequence of abelian groups

HomA(AIM) Ext1A(AIM) Ext2A(AIM) ExtnA(AIM)

If there are M -regular sequences in I then this sequence will start off with n minus 1 zero groupswhere n ge 1 is the common length of every maximal M -regular sequence The nth group will

23

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

where capj 6=iqj qi for all i and the primes pi = r(qi) are all distinct If a is an ideal contained inb then

ba = q1a cap middot middot middot cap qna

is a minimal primary decomposition of ba in AaLet A be a nonzero ring Recall that dimension of an A-module M is the Krull dimension of

the ring AAnn(M) and is defined for all modules M (minus1 if M = 0) The rank is defined for freeA-modules and is the common size of any basis (0 if M = 0) Throughout these notes dim(M)will denote the dimension not the rank

Definition 9 Let (Am k) be a noetherian local ring of dimension d An ideal of definition is anm-primary ideal Recall that the dimension of A is the size of the smallest collection of elementsof A which generates an m-primary ideal Recall that rankk(mm2) is equal to the size of thesmallest set of generators for m as an ideal so always d le rankk(mm2)

A system of parameters is a set of d elements generating an m-primary ideal If d = rankk(mm2)or equivalently there is a system of parameters generating m we say that A is a regular local ringand we call such a system of parameters a regular system of parameters

Proposition 51 Let (Am) be a noetherian local ring of dimension d ge 1 and let x1 xd bea system of parameters of A Then

dim(A(x1 xi)) = dminus i = dim(A)minus i

for each 1 le i le d

Proof Put A = A(x1 xi) If i = d then the zero ideal in A is an ideal of definition so clearlydim(A) = 0 If 1 le i lt d then dim(A) le dminus i since xi+1 xd generate an ideal of definition ofA Let dim(A) = p If p = 0 then (x1 xi) must be an ideal of definition contradicting i lt dSo p ge 1 and if y1 yp is a system of parameters of A then x1 xi y1 yp generate anideal of definition of A so that p+ i ge d That is p ge dminus i

Definition 10 Let A be a nonzero ring and I a proper ideal The height of I denoted htI isthe minimum of the heights of the prime ideals containing I

htI = infhtp | p supe I

This is a number in 0 1 2 infin Equivalently we can take the infimum over the heights ofprimes minimal over I Clearly ht0 = 0 and if I sube J are proper ideals then it is clear thathtI le htJ If I is a prime ideal then htI is the usual height of a prime ideal If A is a noetherianring then htI ltinfin for every proper ideal I since Ap is a local noetherian ring and htp = dim(Ap)

Lemma 52 Let A be a nonzero ring and I a proper ideal Then we have

htI = infhtIAm |m is a maximal ideal and I sube m

Lemma 53 Let A be a noetherian ring and suppose we have an exact sequence

0 minusrarrM prime minusrarrM minusrarrM primeprime minusrarr 0

in which M primeMM primeprime are nonzero and finitely generated Then dimM = maxdimM prime dimM primeprime

Proof We know that Supp(M) = Supp(M prime)cupSupp(M primeprime) and for all three modules the dimensionis the supremum of the coheights of prime ideals in the support So the result is straightforwardto check

16

41 Homomorphism and Dimension

Let φ A minusrarr B be a morphism of rings If p isin Spec(A) then put κ(p) = AppAp Let Bp denotethe ring Tminus1B where T = φ(Aminusp) There is an isomorphism of A-algebras Bp

sim= BotimesAAp Thereis a commutative diagram of rings

B

))RRRRRRRRRRRRRRRR Bp BppBp

B otimesA κ(p)

The vertical isomorphism is defined by bφ(s)+ pBp 7rarr botimes (1s+ pAp) We call Spec(BotimesA κ(p))the fibre over p of the map Φ Spec(B) minusrarr Spec(A) Since primes of BppBp clearly correspondto primes q of B with q cap A = p it is easy to see that the ring morphism B minusrarr B otimesA κ(p) givesrise to a continuous map Spec(B otimesA κ(p)) minusrarr SpecB which gives a homeomorphism betweenΦminus1(p) and Spec(B otimesA κ(p)) See [AM69] Chapter 3

Lemma 54 Let q be a prime ideal of B with q cap A = p and let P be the corresponding primeideal of B otimesA κ(p) Then there is an isomorphism of A-algebras

Bq otimesA κ(p) sim= (B otimesA κ(p))P

btotimes (as+ pAp) 7rarr (botimes (a1 + pAp))(totimes (s1 + pAp))

Proof It is not difficult to see that there is an isomorphism of rings Bqsim= (Bp)qBp defined by

bt 7rarr (b1)(t1) Consider the prime ideal qBppBp We know that there is a ring isomorphism

(B otimesA κ(p))Psim= (BppBp)qBppBp

sim= (Bp)qBpp(Bp)qBpsim= BqpBq

by the comments following Lemma 7 It is not hard to check there is a ring isomorphism BqpBqsim=

BqotimesA κ(p) defined by bt+ pBq 7rarr btotimes 1 (the inverse of btotimes (as+ pAp) is bφ(a)tφ(s)+ pBq)So by definition of P there is an isomorphism of rings (B otimesA κ(p))P

sim= Bq otimesA κ(p) and this isclearly an isomorphism of A-algebras

In particular if φ A minusrarr B is a ring morphism p isin Spec(A) and q isin Spec(B) such thatq capA = p then there is an isomorphism of rings (BpB)qpB

sim= BqpBq so we have ht(qpB) =dim(Bq otimesA κ(p))

Theorem 55 Let φ A minusrarr B be a morphism of noetherian rings Let q isin Spec(B) and putp = q capA Then

(1) htq le htp + ht(qpB) In other words dim(Bq) le dim(Ap) + dim(Bq otimesA κ(p))

(2) We have equality in (1) if the going-down theorem holds for φ (in particular if φ is flat)

(3) If Φ Spec(B) minusrarr Spec(A) is surjective and if the going-down theorem holds then we havedim(B) ge dim(A) and htI = ht(IB) for any proper ideal I of A

Proof (1) Replacing A by Ap and B by Bq we may suppose that (A p) and (B q) are noetherianlocal rings such that q cap A = p and we must show that dim(B) le dim(A) + dim(BpB) Let Ibe a p-primary ideal of A Then pn sube I for some n gt 0 so pnB sube IB sube pB Thus the ideals pBand IB have the same radical and so by definition dim(BpB) = dim(BIB) If dimA = 0 thenwe can take I = 0 and the result is trivial So assume dimA = r ge 1 and let I = (a1 ar) fora system of parameters a1 ar If dim(BIB) = 0 then IB is an q-primary ideal of B and sodim(B) le r as required Otherwise if dim(BIB) = s ge 1 let b1 + IB bs + IB be a systemof parameters of BIB Then b1 bs a1 ar generate an ideal of definition of B Hencedim(B) le r + s

(2) We use the same notation as above If ht(qpB) = s ge 0 then there exists a prime chainof length s q = q0 sup q1 sup middot middot middot sup qs such that qs supe pB As p = q cap A supe qi cap A supe p all the qi

17

lie over p If htp = r ge 0 then there exists a prime chain p = p0 sup p1 sup middot middot middot sup pr in A and bygoing-down there exists a prime chain qs = t0 sup t1 sup middot middot middot sup tr of B such that ti capA = pi Then

q = q0 sup middot middot middot sup qs sup t1 sup middot middot middot sup tr

is a prime chain of length r + s therefore htq ge r + s(3) (i) follows from (2) since dim(A) = suphtp | p isin Spec(A) (ii) Since Φ is surjective IB

is a proper ideal Let q be a minimal prime over IB such that htq = ht(IB) and put p = q capAThen ht(qpB) = 0 so by (2) we find that ht(IB) = htq = htp ge htI For the reverseinclusion let p be a minimal prime ideal over I such that htp = htI and take a prime q of Blying over p Replacing q if necessary we may assume that q is a minimal prime ideal over pBThen htI = htp = htq ge ht(IB)

Theorem 56 Let A be a nonzero subring of B and suppose that B is integral over A Then

(1) dim(A) = dim(B)

(2) Let q isin Spec(B) and set p = q capA Then we have cohtp = cohtq and htq le htp

(3) If the going-down theorem holds between A and B then for any ideal J of B with J capA 6= Awe have htJ = ht(J capA)

Proof (1) By Theorem 42 the going-up theorem holds for A sube B and Spec(B) minusrarr Spec(A) issurjective so we can lift any chain of prime ideals p0 sub p1 sub middot middot middot sub pn in A to a chain of primeideals q0 sub middot middot middot sub qn in B On the other hand if q sube qprime are prime ideals of B and q capA = qprime capAthen q = qprime so any chain of prime ideals in B restricts to a chain of the same length in A Hencedim(A) = dim(B)

(2) Since Bq is integral over Ap it is clear from (1) that cohtp = cohtq If q = q0 sup middot middot middot sup qnis a chain of prime ideals in B then intersecting with A gives a chain of length n descending fromp Hence htq le htp

(3) Given the going-down theorem it is clear that htq = htp in (2) Let J be a proper ideal ofB with J capA 6= A and let q be such that htJ = htq Then ht(J capA) le ht(qcapA) = htq = htJ On the other hand BJ is integral over BJ cap A so every prime ideal p of A containing J cap Acan be lifted to a prime ideal q of B containing J In particular we can lift a prime ideal p withht(J capA) = htp to see that htJ le htq = htp = ht(J capA) as required

5 Depth

Definition 11 Let A be a ring M an A-module and a1 ar a sequence of elements of AWe say a1 ar is an M -regular sequence (or simply M -sequence) if the following conditions aresatisfied

(1) For each 2 le i le r ai is regular on M(a1 aiminus1)M and a1 is regular on M

(2) M 6= (a1 an)M

If a1 ar is an M -regular sequence then so is a1 ai for any i le r When all ai belong to anideal I we say a1 ar is an M -regular sequence in I If moreover there is no b isin I such thata1 ar b is M -regular then a1 ar is said to be a maximal M -regular sequence in I Noticethat the notion of M -regular depends on the order of the elements in the sequence If MN areisomorphic A-modules then a sequence is regular on M iff it is regular on N

Lemma 57 A sequence a1 ar with r ge 2 is M -regular if and only if a1 is regular on Mand a2 ar is an Ma1M -regular sequence If the sequence a1 ar is a maximal M -regularsequence in I then a2 ar is a maximal Ma1M -regular sequence in I

18

Proof The key point is that for ideals a sube b there is a canonical isomorphism of A-modulesMbM sim= NbN where N = MaM If a1 ar is M -regular then a1 is regular on M a2 isregular on N = Ma1M and for 3 le i le r ai is regular on

M(a1 aiminus1)M sim= N(a2 aiminus1)N

Hence a2 ar is an N -regular sequence The converse follows from the same argument

More generally if a1 ar is an M -regular sequence and we set N = M(a1 ar)M andif b1 bs is an N -regular sequence then a1 ar b1 bs is an M -regular sequence

Lemma 58 If a1 ar is an A-regular sequence and M is a flat A-module then a1 ar isalso M -regular provided (a1 ar)M 6= M

Proof Left multiplication by a1 defines a monomorphism A minusrarr A since a1 is A-regular Tensoringwith M and using the fact that M is flat we see that left multiplication by a1 also gives amonomorphismM minusrarrM as required Similarly tensoring with the monomorphism a2 Aa1 minusrarrAa1 we get a monomorphism Ma1M minusrarrMa1M and so on

Lemma 59 Let A be a ring and M an A-module Given an integer n ge 1 a sequence a1 aris M -regular if and only if it is Mn-regular

Proof Suppose the sequence a1 ar is M -regular We prove it is Mn-regular by induction on rThe case r = 1 is trivial so assume r gt 1 By the inductive hypothesis the sequence a1 arminus1 isMn-regular Let L = (a1 arminus1)M Then (a1 arminus1)Mn = Ln and there is an isomorphismof A-modules MnLn sim= (ML)n So we need only show that ar is regular on (ML)n Since byassumption it is regular on ML this is not hard to check Clearly (a1 ar)Mn 6= Mn so thesequence a1 ar is Mn-regular as required The converse is similarly checked

Lemma 60 Let A be a nonzero ring M an A-module and a1 ar isin A If a1 ar isin Am isMm-regular for every maximal ideal m of A then the sequence a1 ar is M -regular

Proof This follows from the fact that given an A-module M an element a isin A is regular on M ifand only if its image in Am is regular on Mm for every maximal ideal m of A

Lemma 61 Suppose that a1 ar is M -regular and a1ξ1 + middot middot middot + arξr = 0 for ξi isin M Thenξi isin (a1 ar)M for all i

Proof By induction on r For r = 1 a1ξ1 = 0 implies that ξ1 = 0 Let r gt 1 Since ar is regularon M(a1 arminus1)M we have ξr =

sumrminus1i=1 aiηi so

sumrminus1i=1 ai(ξi + arηi) = 0 By the inductive

hypothesis for i lt r we have ξi + arηi isin (a1 arminus1)M so that ξi isin (a1 ar)M

Theorem 62 Let A be a ring and M an A-module and let a1 ar be an M -regular sequenceThen for every sequence n1 nr of integers gt 0 the sequence an1

1 anrr is M -regular

Proof Suppose we can prove the following statement

(lowast) Given an integer n gt 0 an A-module M and any M -regular sequence a1 arthe sequence an1 a2 ar is also M -regular

We prove the rest of the Theorem by induction on r For r = 1 this follows immediately from (lowast)Let r gt 1 and suppose a1 ar is M -regular Then by (lowast) an1

1 a2 ar is M -regular Hencea2 ar is Man1

1 M -regular By the inductive hypothesis an22 anr

r is Man11 M -regular and

therefore an11 anr

r is M -regular by Lemma 57So it only remains to prove (lowast) which we do by induction on n The case n = 1 is trivial so let

n gt 1 be given along with an A-moduleM and anM -regular sequence a1 ar By the inductivehypothesis anminus1

1 a2 ar is M -regular and clearly an1 is regular on M Since (an1 a2 ar) sube(anminus1

1 a2 ar) it is clear thatM 6= (an1 a2 ar)M Let i gt 1 and assume that an1 a2 aiminus1

is an M -regular sequence We need to show that ai is regular on M(an1 a2 aiminus1)M Suppose

19

that aiω = an1 ξ1 +a2ξ2 + middot middot middot+aiminus1ξiminus1 Then ω = anminus11 η1 +a2η2 + middot middot middot+aiminus1ηiminus1 by the inductive

hypothesis So

anminus11 (a1ξ1 minus aiη1) + a2(ξ2 minus aiη2) + middot middot middot+ aiminus1(ξiminus1 minus aiηiminus1) = 0

Hence a1ξ1minusaiη1 isin (anminus11 a2 aiminus1)M by Lemma 61 It follows that aiη1 isin (a1 a2 aiminus1)M

hence η1 isin (a1 aiminus1)M and so ω isin (an1 a2 aiminus1)M as required This proves (lowast) andtherefore completes the proof

Let A be a ring There is an isomorphism of A[x1 xn]-modules M otimesA A[x1 xn] sim=M [x1 xn] where the latter module consists of polynomials in x1 xn with coefficients in M(see our Polynomial Ring notes) For any f(x1 xn) isinM [x1 xn] and tuple (a1 an) isinAn we can define an element of M

f(a1 an) =sumα

aα11 middot middot middot aαn

n middot f(α)

For an element r isin R and h isinM [x1 xn]

(f + h)(a1 an) = f(a1 an) + h(a1 an)(r middot f)(a1 an) = r middot f(a1 an)

For an ideal I in R the R-submodule IM [x1 xn] consists of all polynomials whose coefficientsare in the R-submodule IM subeM

Let us review the definition of the associated graded ring and modules Let A be a ring and Ian ideal of A Then the abelian group

grI(A) = AI oplus II2 oplus I2I3 oplus middot middot middot

becomes a graded ring in a fairly obvious way For an A-module M we have the graded grI(A)-module

grI(M) = MIM oplus IMI2M oplus I2MI3M oplus middot middot middotIf A is noetherian and M is a finitely generated A-module then grI(A) is a noetherian ring andif grI(M) is a finitely generated grI(A)-module

Given elements a1 an isin A and I = (a1 an) we define a morphism of abelian groupsψ M [x1 xn] minusrarr grI(M) as follows if f is homogenous of degreem ge 0 define ψ(f) to be theimage of f(a1 an) in ImMIm+1M This defines a morphism of groups M [x1 xn]m minusrarrImMIm+1M and together these define the morphism of groups ψ Since ψ maps IM [x1 xn]to zero it induces a morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) and

Proposition 63 Let A be a ring and M an A-module Let a1 an isin A and set I =(a1 an) Then the following conditions are equivalent

(a) For every m gt 0 and for every homogenous polynomial f(x1 xn) isin M [x1 xn] ofdegree m such that f(a1 an) isin Im+1M we have f isin IM [x1 xn]

(b) If f(x1 xn) isin M [x1 xn] is homogenous and f(a1 an) = 0 then the coefficientsof f are in IM

(c) The morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) defined by mappinga homogenous polynomial f(x1 xn) of degree m to f(a1 an) isin ImMIm+1M is anisomorphism

Proof It is easy to see that (a) hArr (c) and (a) rArr (b) To show (b) rArr (a) let f isinM [x1 xn] bea homogenous polynomial of degree m gt 0 and suppose f(a1 an) isin Im+1M Any element ofIm+1M can be written as a sum of terms of the form aα1

1 middot middot middot aαnn middotm with

sumi αi = m+1 By shifting

one of the ai across we can write f(a1 an) = g(a1 an) for a homogenous polynomialg isinM [x1 xn] of degreem all of whose coefficients belong to IM Hence (fminusg)(a1 an) = 0so by (b) the coefficients of f minus g belong to IM and this implies implies that the coefficients of falso belong to IM as required

20

Definition 12 Let A be a ring and M an A-module A sequence a1 an isin A is M -quasiregularif it satisfies the equivalent conditions of the Proposition Obviously this concept does not dependon the order of the elements But a1 ai for i lt n need not be M -quasiregular

Recall that for an A-module M a submodule N sube M and x isin A the notation (N x) meansm isin M |xm isin N This is a submodule of M If A is a ring I an ideal and M an A-modulerecall that M is separated in the I-adic topology when

⋂n I

nM = 0

Theorem 64 Let A be a ring M a nonzero A-module a1 an isin A and I = (a1 an)Then

(i) If a1 an is M -quasiregular and x isin A is such that (IM x) = IM then (ImM x) =ImM for all m gt 0

(ii) If a1 an is M -regular then it is M -quasiregular

(iii) If MMa1MM(a1 a2)M M(a1 anminus1)M are separated in the I-adic topologythen the converse of (ii) is also true

Proof (i) By induction on m with the case m = 1 true by assumption Suppose m gt 1 andxξ isin ImM By the inductive hypothesis ξ isin Imminus1M Hence there exists a homogenous polynomialf isinM [x1 xn] of degree mminus 1 such that ξ = f(a1 an) Since xξ = xf(a1 an) isin ImMthe coefficients of f are in (IM x) = IM Therefore ξ = f(a1 an) isin ImM

(ii) By induction on n For n = 1 this is easy to check Let n gt 1 and suppose a1 an is M -regular Then by the induction hypothesis a1 anminus1 is M -quasiregular Let f isinM [x1 xn]be homogenous of degree m gt 0 such that f(a1 an) = 0 We prove that f isin IM [x1 xn]by induction on m (the case m = 0 being trivial) Write

f(x1 xn) = g(x1 xnminus1) + xnh(x1 xn)

Then g and h are homogenous of degrees m and mminus 1 respectively By (i) we have

h(a1 an) isin ((a1 anminus1)mM an) = (a1 anminus1)mM sube ImM

Since by assumption a1 an is regular on M so an is regular on M(a1 anminus1)M andhence ((a1 anminus1)M an) = (a1 anminus1)M So by the induction hypothesis on m we haveh isin IM [x1 xn] (by the argument of Proposition 63) Since h(a1 an) isin (a1 anminus1)mMthere exists H isin M [x1 xnminus1] which is homogenous of degree m such that h(a1 an) =H(a1 anminus1) Putting

G(x1 xnminus1) = g(x1 xnminus1) + anH(x1 xnminus1)

we have G(a1 anminus1) = 0 so by the inductive hypothesis on n we have G isin IM [x1 xn]hence g isin IM [x1 xn] and so f isin IM [x1 xn]

(iii) By induction on n ge 1 Assume that a1 an is M -quasiregular and the modulesMMa1M M(a1 anminus1)M are all separated in the I-adic topology If a1ξ = 0 thenξ isin IM hence ξ =

sumaiηi and

suma1aiηi = 0 hence ηi isin IM and so ξ isin I2M In this way we

see that ξ isin⋂t ItM = 0 Thus a1 is regular on M and this also takes care of the case n = 1

since M 6= IM by the separation condition So assume n gt 1 By Lemma 57 it suffices to showthat a2 an is an N -regular sequence where N = Ma1M Since there is an isomorphism ofA-modules for 2 le i le nminus 1

M(a1 ai)M sim= N(a2 ai)N

The modules NNa2N N(a2 anminus1)N are separated in the I-adic topology So by theinductive hypothesis it suffices to show that the sequence a2 an is N -quasiregular

It suffices to show that if f(x2 xn) isin M [x2 xn] is homogenous of degree m ge 1 withf(a2 an) isin a1M then the coefficients of f belong to IM Put f(a2 an) = a1ω We claim

21

that ω isin Imminus1M Let 0 le i le mminus1 be the largest integer with ω isin IiM Then ω = g(a1 an)for some homogenous polynomial of degree i and

f(a2 an) = a1g(a1 an) (2)

If i lt mminus1 then g isin IM [x1 xn] and so ω isin Ii+1M which is a contradiction Hence i = mminus1and so ω isin Imminus1M Again using (2) we see that f(x2 xn)minusx1g(x1 xn) isin IM [x1 xn]Since f does not involve x1 we have f isin IM [x1 xn] as required

The theorem shows that under the assumptions of (iii) any permutation of an M -regularsequence is M -regular

Corollary 65 Let A be a noetherian ring M a finitely generated A-module and let a1 anbe contained in the Jacobson radical of A Then a1 an is M -regular if and only if it is M -quasiregular In particular if a1 an is M -regular so is any permutation of the sequence

Proof From [AM69] we know that for any ideal I contained in the Jacobson radical the I-adictopology on any finitely generated A-module is separated

If A is a ring and M an A-module then any M -regular sequence a1 an isin A gives rise toa strictly increasing chain of submodules a1M (a1 a2)M (a1 an)M Hence the chain ofideals (a1) (a1 a2) (a1 an) must also be strictly increasing

Lemma 66 Let A be a noetherian ring and M an A-module Any M -regular sequence a1 anin an ideal I can be extended to a maximal M -regular sequence in I

Proof If a1 an is not maximal in I we can find an+1 isin I such that a1 an an+1 is anM -regular sequence Either this process terminates at a maximal M -regular sequence in I or itproduces a strictly ascending chain of ideals

(a1) sub (a1 a2) sub (a1 a2 a3) sub middot middot middot

Since A is noetherian we can exclude this latter possibility

Theorem 67 Let A be a noetherian ring M a finitely generated A-module and I an ideal of Awith IM 6= M Let n gt 0 be an integer Then the following are equivalent

(1) ExtiA(NM) = 0 for i lt n and every finitely generated A-module N with Supp(N) sube V (I)

(2) ExtiA(AIM) = 0 for i lt n

(3) There exists a finitely generated A-module N with Supp(N) = V (I) and ExtiA(NM) = 0for i lt n

(4) There exists an M -regular sequence a1 an of length n in I

Proof (1) rArr (2) rArr (3) is trivial With I fixed we show that (3) rArr (4) for every finitely generatedmodule M with IM 6= M by induction on n We have 0 = Ext0A(NM) sim= HomA(NM) SinceM is finitely generated and nonzero the set of associated primes of M is finite and nonempty Ifno elements of I are M -regular then I is contained in the union of these associated primes andhence I sube p for some p isin Ass(M) (see [AM69] for details) By definition there is a monomorphismof A-modules φ Ap minusrarrM There is an isomorphism of A-modules

(Ap)psim= ApotimesA Ap

sim= AppAp = k

It is not hard to check this is an isomorphism of Ap-modules as well Since φp is a monomorphismand k 6= 0 it follows that HomAp(kMp) 6= 0 Since p isin V (I) = Supp(N) we have Np 6= 0 andso the k-module NppNp 6= 0 is nonzero and therefore free so Homk(NppNp k) 6= 0 Sincek sim= (Ap)p as Ap-modules it follows that HomA(NAp)p

sim= HomAp(Np (Ap)p) 6= 0 Since Apis isomorphic to a submodule of M it follows that HomA(NM) 6= 0 which is a contradiction

22

therefore there exists an M -regular element a1 isin I which takes care of the case n = 1 If n gt 1then put M1 = Ma1M From the exact sequence

0 Ma1 M M1

0 (3)

we get the long exact sequence

middot middot middot minusrarr ExtiA(NM) minusrarr ExtiA(NM1) minusrarr Exti+1A (NM) minusrarr middot middot middot

which shows that ExtiA(NM1) = 0 for 0 le i lt n minus 1 By the inductive hypothesis on n thereexists an M1-regular sequence a2 an in I The sequence a1 an is then an M -regularsequence in I

(4) rArr (1) By induction on n with I fixed For n = 1 we have a1 isin I regular on M and so (3)gives an exact sequence of R-modules

0 HomA(NM)a1 HomA(NM)

Where a1 denotes left multiplication by a1 Since Supp(N) = V (Ann(N)) sube V (I) it fol-lows that I sube r(Ann(N)) and so ar1N = 0 for some r gt 0 It follows that ar1 annihilatesHomA(NM) as well but since the action of ar1 on HomA(NM) gives an injective map it followsthat HomA(NM) = 0 Now assume n gt 1 and put M1 = Ma1M Then a2 an is anM1-regular sequence so by the inductive hypothesis ExtiA(NM1) = 0 for i lt nminus 1 So the longexact sequence corresponding to (3) gives an exact sequence for 0 le i lt n

0 ExtiA(NM)a1 ExtiA(NM)

Here a1 denotes left multiplication by a1 which is equal to ExtiA(αM) = Exti(N β) where α βare the endomorphisms given by left multiplication by a1 on NM respectively Assume that ar1annihilates N with r gt 0 Then αr is the zero map so ExtiA(αM)r = 0 and so ar1 also annihilatesExtiA(NM) Since the a1 is regular on this module it follows that ExtiA(NM) = 0 for i lt nas required

Corollary 68 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA with IM 6= M If a1 an a maximal M -regular sequence in I then ExtiA(AIM) = 0 fori lt n and ExtnA(AIM) 6= 0

Proof We already know that ExtiA(AIM) = 0 for i lt n so with I fixed we prove by inductionon n that ExtnA(AIM) 6= 0 for any finitely generated module M with IM 6= M admitting amaximal M -regular sequence of length n For n = 1 we have a1 isin I regular on M and an exactsequence (3) where M1 = Ma1M Part of the corresponding long exact sequence is

Ext0A(AIM) minusrarr Ext0A(AIM1) minusrarr Ext1A(AIM)

We know from the Theorem 67 that Ext0A(AIM) = 0 so it suffices to show that we haveHomA(AIM1) 6= 0 But if HomA(AIM1) = 0 then it follows from the proof of (3) rArr (4)above that there would be b isin I regular on M1 so a1 b is an M -regular sequence This is acontradiction since the sequence a1 was maximal so we conclude that Ext1A(AIM) 6= 0

Now assume n gt 1 and let a1 an be a maximal M -regular sequence in I Then a2 anis a maximal M1-regular sequence in I so by the inductive hypothesis Extnminus1

A (AIM1) 6= 0 Sofrom the long exact sequence for (3) we conclude that ExtnA(AIM) 6= 0 also

It follows that under the conditions of the Corollary every maximal M -regular sequence in Ihas a common length and you can find this length by looking at the sequence of abelian groups

HomA(AIM) Ext1A(AIM) Ext2A(AIM) ExtnA(AIM)

If there are M -regular sequences in I then this sequence will start off with n minus 1 zero groupswhere n ge 1 is the common length of every maximal M -regular sequence The nth group will

23

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

41 Homomorphism and Dimension

Let φ A minusrarr B be a morphism of rings If p isin Spec(A) then put κ(p) = AppAp Let Bp denotethe ring Tminus1B where T = φ(Aminusp) There is an isomorphism of A-algebras Bp

sim= BotimesAAp Thereis a commutative diagram of rings

B

))RRRRRRRRRRRRRRRR Bp BppBp

B otimesA κ(p)

The vertical isomorphism is defined by bφ(s)+ pBp 7rarr botimes (1s+ pAp) We call Spec(BotimesA κ(p))the fibre over p of the map Φ Spec(B) minusrarr Spec(A) Since primes of BppBp clearly correspondto primes q of B with q cap A = p it is easy to see that the ring morphism B minusrarr B otimesA κ(p) givesrise to a continuous map Spec(B otimesA κ(p)) minusrarr SpecB which gives a homeomorphism betweenΦminus1(p) and Spec(B otimesA κ(p)) See [AM69] Chapter 3

Lemma 54 Let q be a prime ideal of B with q cap A = p and let P be the corresponding primeideal of B otimesA κ(p) Then there is an isomorphism of A-algebras

Bq otimesA κ(p) sim= (B otimesA κ(p))P

btotimes (as+ pAp) 7rarr (botimes (a1 + pAp))(totimes (s1 + pAp))

Proof It is not difficult to see that there is an isomorphism of rings Bqsim= (Bp)qBp defined by

bt 7rarr (b1)(t1) Consider the prime ideal qBppBp We know that there is a ring isomorphism

(B otimesA κ(p))Psim= (BppBp)qBppBp

sim= (Bp)qBpp(Bp)qBpsim= BqpBq

by the comments following Lemma 7 It is not hard to check there is a ring isomorphism BqpBqsim=

BqotimesA κ(p) defined by bt+ pBq 7rarr btotimes 1 (the inverse of btotimes (as+ pAp) is bφ(a)tφ(s)+ pBq)So by definition of P there is an isomorphism of rings (B otimesA κ(p))P

sim= Bq otimesA κ(p) and this isclearly an isomorphism of A-algebras

In particular if φ A minusrarr B is a ring morphism p isin Spec(A) and q isin Spec(B) such thatq capA = p then there is an isomorphism of rings (BpB)qpB

sim= BqpBq so we have ht(qpB) =dim(Bq otimesA κ(p))

Theorem 55 Let φ A minusrarr B be a morphism of noetherian rings Let q isin Spec(B) and putp = q capA Then

(1) htq le htp + ht(qpB) In other words dim(Bq) le dim(Ap) + dim(Bq otimesA κ(p))

(2) We have equality in (1) if the going-down theorem holds for φ (in particular if φ is flat)

(3) If Φ Spec(B) minusrarr Spec(A) is surjective and if the going-down theorem holds then we havedim(B) ge dim(A) and htI = ht(IB) for any proper ideal I of A

Proof (1) Replacing A by Ap and B by Bq we may suppose that (A p) and (B q) are noetherianlocal rings such that q cap A = p and we must show that dim(B) le dim(A) + dim(BpB) Let Ibe a p-primary ideal of A Then pn sube I for some n gt 0 so pnB sube IB sube pB Thus the ideals pBand IB have the same radical and so by definition dim(BpB) = dim(BIB) If dimA = 0 thenwe can take I = 0 and the result is trivial So assume dimA = r ge 1 and let I = (a1 ar) fora system of parameters a1 ar If dim(BIB) = 0 then IB is an q-primary ideal of B and sodim(B) le r as required Otherwise if dim(BIB) = s ge 1 let b1 + IB bs + IB be a systemof parameters of BIB Then b1 bs a1 ar generate an ideal of definition of B Hencedim(B) le r + s

(2) We use the same notation as above If ht(qpB) = s ge 0 then there exists a prime chainof length s q = q0 sup q1 sup middot middot middot sup qs such that qs supe pB As p = q cap A supe qi cap A supe p all the qi

17

lie over p If htp = r ge 0 then there exists a prime chain p = p0 sup p1 sup middot middot middot sup pr in A and bygoing-down there exists a prime chain qs = t0 sup t1 sup middot middot middot sup tr of B such that ti capA = pi Then

q = q0 sup middot middot middot sup qs sup t1 sup middot middot middot sup tr

is a prime chain of length r + s therefore htq ge r + s(3) (i) follows from (2) since dim(A) = suphtp | p isin Spec(A) (ii) Since Φ is surjective IB

is a proper ideal Let q be a minimal prime over IB such that htq = ht(IB) and put p = q capAThen ht(qpB) = 0 so by (2) we find that ht(IB) = htq = htp ge htI For the reverseinclusion let p be a minimal prime ideal over I such that htp = htI and take a prime q of Blying over p Replacing q if necessary we may assume that q is a minimal prime ideal over pBThen htI = htp = htq ge ht(IB)

Theorem 56 Let A be a nonzero subring of B and suppose that B is integral over A Then

(1) dim(A) = dim(B)

(2) Let q isin Spec(B) and set p = q capA Then we have cohtp = cohtq and htq le htp

(3) If the going-down theorem holds between A and B then for any ideal J of B with J capA 6= Awe have htJ = ht(J capA)

Proof (1) By Theorem 42 the going-up theorem holds for A sube B and Spec(B) minusrarr Spec(A) issurjective so we can lift any chain of prime ideals p0 sub p1 sub middot middot middot sub pn in A to a chain of primeideals q0 sub middot middot middot sub qn in B On the other hand if q sube qprime are prime ideals of B and q capA = qprime capAthen q = qprime so any chain of prime ideals in B restricts to a chain of the same length in A Hencedim(A) = dim(B)

(2) Since Bq is integral over Ap it is clear from (1) that cohtp = cohtq If q = q0 sup middot middot middot sup qnis a chain of prime ideals in B then intersecting with A gives a chain of length n descending fromp Hence htq le htp

(3) Given the going-down theorem it is clear that htq = htp in (2) Let J be a proper ideal ofB with J capA 6= A and let q be such that htJ = htq Then ht(J capA) le ht(qcapA) = htq = htJ On the other hand BJ is integral over BJ cap A so every prime ideal p of A containing J cap Acan be lifted to a prime ideal q of B containing J In particular we can lift a prime ideal p withht(J capA) = htp to see that htJ le htq = htp = ht(J capA) as required

5 Depth

Definition 11 Let A be a ring M an A-module and a1 ar a sequence of elements of AWe say a1 ar is an M -regular sequence (or simply M -sequence) if the following conditions aresatisfied

(1) For each 2 le i le r ai is regular on M(a1 aiminus1)M and a1 is regular on M

(2) M 6= (a1 an)M

If a1 ar is an M -regular sequence then so is a1 ai for any i le r When all ai belong to anideal I we say a1 ar is an M -regular sequence in I If moreover there is no b isin I such thata1 ar b is M -regular then a1 ar is said to be a maximal M -regular sequence in I Noticethat the notion of M -regular depends on the order of the elements in the sequence If MN areisomorphic A-modules then a sequence is regular on M iff it is regular on N

Lemma 57 A sequence a1 ar with r ge 2 is M -regular if and only if a1 is regular on Mand a2 ar is an Ma1M -regular sequence If the sequence a1 ar is a maximal M -regularsequence in I then a2 ar is a maximal Ma1M -regular sequence in I

18

Proof The key point is that for ideals a sube b there is a canonical isomorphism of A-modulesMbM sim= NbN where N = MaM If a1 ar is M -regular then a1 is regular on M a2 isregular on N = Ma1M and for 3 le i le r ai is regular on

M(a1 aiminus1)M sim= N(a2 aiminus1)N

Hence a2 ar is an N -regular sequence The converse follows from the same argument

More generally if a1 ar is an M -regular sequence and we set N = M(a1 ar)M andif b1 bs is an N -regular sequence then a1 ar b1 bs is an M -regular sequence

Lemma 58 If a1 ar is an A-regular sequence and M is a flat A-module then a1 ar isalso M -regular provided (a1 ar)M 6= M

Proof Left multiplication by a1 defines a monomorphism A minusrarr A since a1 is A-regular Tensoringwith M and using the fact that M is flat we see that left multiplication by a1 also gives amonomorphismM minusrarrM as required Similarly tensoring with the monomorphism a2 Aa1 minusrarrAa1 we get a monomorphism Ma1M minusrarrMa1M and so on

Lemma 59 Let A be a ring and M an A-module Given an integer n ge 1 a sequence a1 aris M -regular if and only if it is Mn-regular

Proof Suppose the sequence a1 ar is M -regular We prove it is Mn-regular by induction on rThe case r = 1 is trivial so assume r gt 1 By the inductive hypothesis the sequence a1 arminus1 isMn-regular Let L = (a1 arminus1)M Then (a1 arminus1)Mn = Ln and there is an isomorphismof A-modules MnLn sim= (ML)n So we need only show that ar is regular on (ML)n Since byassumption it is regular on ML this is not hard to check Clearly (a1 ar)Mn 6= Mn so thesequence a1 ar is Mn-regular as required The converse is similarly checked

Lemma 60 Let A be a nonzero ring M an A-module and a1 ar isin A If a1 ar isin Am isMm-regular for every maximal ideal m of A then the sequence a1 ar is M -regular

Proof This follows from the fact that given an A-module M an element a isin A is regular on M ifand only if its image in Am is regular on Mm for every maximal ideal m of A

Lemma 61 Suppose that a1 ar is M -regular and a1ξ1 + middot middot middot + arξr = 0 for ξi isin M Thenξi isin (a1 ar)M for all i

Proof By induction on r For r = 1 a1ξ1 = 0 implies that ξ1 = 0 Let r gt 1 Since ar is regularon M(a1 arminus1)M we have ξr =

sumrminus1i=1 aiηi so

sumrminus1i=1 ai(ξi + arηi) = 0 By the inductive

hypothesis for i lt r we have ξi + arηi isin (a1 arminus1)M so that ξi isin (a1 ar)M

Theorem 62 Let A be a ring and M an A-module and let a1 ar be an M -regular sequenceThen for every sequence n1 nr of integers gt 0 the sequence an1

1 anrr is M -regular

Proof Suppose we can prove the following statement

(lowast) Given an integer n gt 0 an A-module M and any M -regular sequence a1 arthe sequence an1 a2 ar is also M -regular

We prove the rest of the Theorem by induction on r For r = 1 this follows immediately from (lowast)Let r gt 1 and suppose a1 ar is M -regular Then by (lowast) an1

1 a2 ar is M -regular Hencea2 ar is Man1

1 M -regular By the inductive hypothesis an22 anr

r is Man11 M -regular and

therefore an11 anr

r is M -regular by Lemma 57So it only remains to prove (lowast) which we do by induction on n The case n = 1 is trivial so let

n gt 1 be given along with an A-moduleM and anM -regular sequence a1 ar By the inductivehypothesis anminus1

1 a2 ar is M -regular and clearly an1 is regular on M Since (an1 a2 ar) sube(anminus1

1 a2 ar) it is clear thatM 6= (an1 a2 ar)M Let i gt 1 and assume that an1 a2 aiminus1

is an M -regular sequence We need to show that ai is regular on M(an1 a2 aiminus1)M Suppose

19

that aiω = an1 ξ1 +a2ξ2 + middot middot middot+aiminus1ξiminus1 Then ω = anminus11 η1 +a2η2 + middot middot middot+aiminus1ηiminus1 by the inductive

hypothesis So

anminus11 (a1ξ1 minus aiη1) + a2(ξ2 minus aiη2) + middot middot middot+ aiminus1(ξiminus1 minus aiηiminus1) = 0

Hence a1ξ1minusaiη1 isin (anminus11 a2 aiminus1)M by Lemma 61 It follows that aiη1 isin (a1 a2 aiminus1)M

hence η1 isin (a1 aiminus1)M and so ω isin (an1 a2 aiminus1)M as required This proves (lowast) andtherefore completes the proof

Let A be a ring There is an isomorphism of A[x1 xn]-modules M otimesA A[x1 xn] sim=M [x1 xn] where the latter module consists of polynomials in x1 xn with coefficients in M(see our Polynomial Ring notes) For any f(x1 xn) isinM [x1 xn] and tuple (a1 an) isinAn we can define an element of M

f(a1 an) =sumα

aα11 middot middot middot aαn

n middot f(α)

For an element r isin R and h isinM [x1 xn]

(f + h)(a1 an) = f(a1 an) + h(a1 an)(r middot f)(a1 an) = r middot f(a1 an)

For an ideal I in R the R-submodule IM [x1 xn] consists of all polynomials whose coefficientsare in the R-submodule IM subeM

Let us review the definition of the associated graded ring and modules Let A be a ring and Ian ideal of A Then the abelian group

grI(A) = AI oplus II2 oplus I2I3 oplus middot middot middot

becomes a graded ring in a fairly obvious way For an A-module M we have the graded grI(A)-module

grI(M) = MIM oplus IMI2M oplus I2MI3M oplus middot middot middotIf A is noetherian and M is a finitely generated A-module then grI(A) is a noetherian ring andif grI(M) is a finitely generated grI(A)-module

Given elements a1 an isin A and I = (a1 an) we define a morphism of abelian groupsψ M [x1 xn] minusrarr grI(M) as follows if f is homogenous of degreem ge 0 define ψ(f) to be theimage of f(a1 an) in ImMIm+1M This defines a morphism of groups M [x1 xn]m minusrarrImMIm+1M and together these define the morphism of groups ψ Since ψ maps IM [x1 xn]to zero it induces a morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) and

Proposition 63 Let A be a ring and M an A-module Let a1 an isin A and set I =(a1 an) Then the following conditions are equivalent

(a) For every m gt 0 and for every homogenous polynomial f(x1 xn) isin M [x1 xn] ofdegree m such that f(a1 an) isin Im+1M we have f isin IM [x1 xn]

(b) If f(x1 xn) isin M [x1 xn] is homogenous and f(a1 an) = 0 then the coefficientsof f are in IM

(c) The morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) defined by mappinga homogenous polynomial f(x1 xn) of degree m to f(a1 an) isin ImMIm+1M is anisomorphism

Proof It is easy to see that (a) hArr (c) and (a) rArr (b) To show (b) rArr (a) let f isinM [x1 xn] bea homogenous polynomial of degree m gt 0 and suppose f(a1 an) isin Im+1M Any element ofIm+1M can be written as a sum of terms of the form aα1

1 middot middot middot aαnn middotm with

sumi αi = m+1 By shifting

one of the ai across we can write f(a1 an) = g(a1 an) for a homogenous polynomialg isinM [x1 xn] of degreem all of whose coefficients belong to IM Hence (fminusg)(a1 an) = 0so by (b) the coefficients of f minus g belong to IM and this implies implies that the coefficients of falso belong to IM as required

20

Definition 12 Let A be a ring and M an A-module A sequence a1 an isin A is M -quasiregularif it satisfies the equivalent conditions of the Proposition Obviously this concept does not dependon the order of the elements But a1 ai for i lt n need not be M -quasiregular

Recall that for an A-module M a submodule N sube M and x isin A the notation (N x) meansm isin M |xm isin N This is a submodule of M If A is a ring I an ideal and M an A-modulerecall that M is separated in the I-adic topology when

⋂n I

nM = 0

Theorem 64 Let A be a ring M a nonzero A-module a1 an isin A and I = (a1 an)Then

(i) If a1 an is M -quasiregular and x isin A is such that (IM x) = IM then (ImM x) =ImM for all m gt 0

(ii) If a1 an is M -regular then it is M -quasiregular

(iii) If MMa1MM(a1 a2)M M(a1 anminus1)M are separated in the I-adic topologythen the converse of (ii) is also true

Proof (i) By induction on m with the case m = 1 true by assumption Suppose m gt 1 andxξ isin ImM By the inductive hypothesis ξ isin Imminus1M Hence there exists a homogenous polynomialf isinM [x1 xn] of degree mminus 1 such that ξ = f(a1 an) Since xξ = xf(a1 an) isin ImMthe coefficients of f are in (IM x) = IM Therefore ξ = f(a1 an) isin ImM

(ii) By induction on n For n = 1 this is easy to check Let n gt 1 and suppose a1 an is M -regular Then by the induction hypothesis a1 anminus1 is M -quasiregular Let f isinM [x1 xn]be homogenous of degree m gt 0 such that f(a1 an) = 0 We prove that f isin IM [x1 xn]by induction on m (the case m = 0 being trivial) Write

f(x1 xn) = g(x1 xnminus1) + xnh(x1 xn)

Then g and h are homogenous of degrees m and mminus 1 respectively By (i) we have

h(a1 an) isin ((a1 anminus1)mM an) = (a1 anminus1)mM sube ImM

Since by assumption a1 an is regular on M so an is regular on M(a1 anminus1)M andhence ((a1 anminus1)M an) = (a1 anminus1)M So by the induction hypothesis on m we haveh isin IM [x1 xn] (by the argument of Proposition 63) Since h(a1 an) isin (a1 anminus1)mMthere exists H isin M [x1 xnminus1] which is homogenous of degree m such that h(a1 an) =H(a1 anminus1) Putting

G(x1 xnminus1) = g(x1 xnminus1) + anH(x1 xnminus1)

we have G(a1 anminus1) = 0 so by the inductive hypothesis on n we have G isin IM [x1 xn]hence g isin IM [x1 xn] and so f isin IM [x1 xn]

(iii) By induction on n ge 1 Assume that a1 an is M -quasiregular and the modulesMMa1M M(a1 anminus1)M are all separated in the I-adic topology If a1ξ = 0 thenξ isin IM hence ξ =

sumaiηi and

suma1aiηi = 0 hence ηi isin IM and so ξ isin I2M In this way we

see that ξ isin⋂t ItM = 0 Thus a1 is regular on M and this also takes care of the case n = 1

since M 6= IM by the separation condition So assume n gt 1 By Lemma 57 it suffices to showthat a2 an is an N -regular sequence where N = Ma1M Since there is an isomorphism ofA-modules for 2 le i le nminus 1

M(a1 ai)M sim= N(a2 ai)N

The modules NNa2N N(a2 anminus1)N are separated in the I-adic topology So by theinductive hypothesis it suffices to show that the sequence a2 an is N -quasiregular

It suffices to show that if f(x2 xn) isin M [x2 xn] is homogenous of degree m ge 1 withf(a2 an) isin a1M then the coefficients of f belong to IM Put f(a2 an) = a1ω We claim

21

that ω isin Imminus1M Let 0 le i le mminus1 be the largest integer with ω isin IiM Then ω = g(a1 an)for some homogenous polynomial of degree i and

f(a2 an) = a1g(a1 an) (2)

If i lt mminus1 then g isin IM [x1 xn] and so ω isin Ii+1M which is a contradiction Hence i = mminus1and so ω isin Imminus1M Again using (2) we see that f(x2 xn)minusx1g(x1 xn) isin IM [x1 xn]Since f does not involve x1 we have f isin IM [x1 xn] as required

The theorem shows that under the assumptions of (iii) any permutation of an M -regularsequence is M -regular

Corollary 65 Let A be a noetherian ring M a finitely generated A-module and let a1 anbe contained in the Jacobson radical of A Then a1 an is M -regular if and only if it is M -quasiregular In particular if a1 an is M -regular so is any permutation of the sequence

Proof From [AM69] we know that for any ideal I contained in the Jacobson radical the I-adictopology on any finitely generated A-module is separated

If A is a ring and M an A-module then any M -regular sequence a1 an isin A gives rise toa strictly increasing chain of submodules a1M (a1 a2)M (a1 an)M Hence the chain ofideals (a1) (a1 a2) (a1 an) must also be strictly increasing

Lemma 66 Let A be a noetherian ring and M an A-module Any M -regular sequence a1 anin an ideal I can be extended to a maximal M -regular sequence in I

Proof If a1 an is not maximal in I we can find an+1 isin I such that a1 an an+1 is anM -regular sequence Either this process terminates at a maximal M -regular sequence in I or itproduces a strictly ascending chain of ideals

(a1) sub (a1 a2) sub (a1 a2 a3) sub middot middot middot

Since A is noetherian we can exclude this latter possibility

Theorem 67 Let A be a noetherian ring M a finitely generated A-module and I an ideal of Awith IM 6= M Let n gt 0 be an integer Then the following are equivalent

(1) ExtiA(NM) = 0 for i lt n and every finitely generated A-module N with Supp(N) sube V (I)

(2) ExtiA(AIM) = 0 for i lt n

(3) There exists a finitely generated A-module N with Supp(N) = V (I) and ExtiA(NM) = 0for i lt n

(4) There exists an M -regular sequence a1 an of length n in I

Proof (1) rArr (2) rArr (3) is trivial With I fixed we show that (3) rArr (4) for every finitely generatedmodule M with IM 6= M by induction on n We have 0 = Ext0A(NM) sim= HomA(NM) SinceM is finitely generated and nonzero the set of associated primes of M is finite and nonempty Ifno elements of I are M -regular then I is contained in the union of these associated primes andhence I sube p for some p isin Ass(M) (see [AM69] for details) By definition there is a monomorphismof A-modules φ Ap minusrarrM There is an isomorphism of A-modules

(Ap)psim= ApotimesA Ap

sim= AppAp = k

It is not hard to check this is an isomorphism of Ap-modules as well Since φp is a monomorphismand k 6= 0 it follows that HomAp(kMp) 6= 0 Since p isin V (I) = Supp(N) we have Np 6= 0 andso the k-module NppNp 6= 0 is nonzero and therefore free so Homk(NppNp k) 6= 0 Sincek sim= (Ap)p as Ap-modules it follows that HomA(NAp)p

sim= HomAp(Np (Ap)p) 6= 0 Since Apis isomorphic to a submodule of M it follows that HomA(NM) 6= 0 which is a contradiction

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therefore there exists an M -regular element a1 isin I which takes care of the case n = 1 If n gt 1then put M1 = Ma1M From the exact sequence

0 Ma1 M M1

0 (3)

we get the long exact sequence

middot middot middot minusrarr ExtiA(NM) minusrarr ExtiA(NM1) minusrarr Exti+1A (NM) minusrarr middot middot middot

which shows that ExtiA(NM1) = 0 for 0 le i lt n minus 1 By the inductive hypothesis on n thereexists an M1-regular sequence a2 an in I The sequence a1 an is then an M -regularsequence in I

(4) rArr (1) By induction on n with I fixed For n = 1 we have a1 isin I regular on M and so (3)gives an exact sequence of R-modules

0 HomA(NM)a1 HomA(NM)

Where a1 denotes left multiplication by a1 Since Supp(N) = V (Ann(N)) sube V (I) it fol-lows that I sube r(Ann(N)) and so ar1N = 0 for some r gt 0 It follows that ar1 annihilatesHomA(NM) as well but since the action of ar1 on HomA(NM) gives an injective map it followsthat HomA(NM) = 0 Now assume n gt 1 and put M1 = Ma1M Then a2 an is anM1-regular sequence so by the inductive hypothesis ExtiA(NM1) = 0 for i lt nminus 1 So the longexact sequence corresponding to (3) gives an exact sequence for 0 le i lt n

0 ExtiA(NM)a1 ExtiA(NM)

Here a1 denotes left multiplication by a1 which is equal to ExtiA(αM) = Exti(N β) where α βare the endomorphisms given by left multiplication by a1 on NM respectively Assume that ar1annihilates N with r gt 0 Then αr is the zero map so ExtiA(αM)r = 0 and so ar1 also annihilatesExtiA(NM) Since the a1 is regular on this module it follows that ExtiA(NM) = 0 for i lt nas required

Corollary 68 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA with IM 6= M If a1 an a maximal M -regular sequence in I then ExtiA(AIM) = 0 fori lt n and ExtnA(AIM) 6= 0

Proof We already know that ExtiA(AIM) = 0 for i lt n so with I fixed we prove by inductionon n that ExtnA(AIM) 6= 0 for any finitely generated module M with IM 6= M admitting amaximal M -regular sequence of length n For n = 1 we have a1 isin I regular on M and an exactsequence (3) where M1 = Ma1M Part of the corresponding long exact sequence is

Ext0A(AIM) minusrarr Ext0A(AIM1) minusrarr Ext1A(AIM)

We know from the Theorem 67 that Ext0A(AIM) = 0 so it suffices to show that we haveHomA(AIM1) 6= 0 But if HomA(AIM1) = 0 then it follows from the proof of (3) rArr (4)above that there would be b isin I regular on M1 so a1 b is an M -regular sequence This is acontradiction since the sequence a1 was maximal so we conclude that Ext1A(AIM) 6= 0

Now assume n gt 1 and let a1 an be a maximal M -regular sequence in I Then a2 anis a maximal M1-regular sequence in I so by the inductive hypothesis Extnminus1

A (AIM1) 6= 0 Sofrom the long exact sequence for (3) we conclude that ExtnA(AIM) 6= 0 also

It follows that under the conditions of the Corollary every maximal M -regular sequence in Ihas a common length and you can find this length by looking at the sequence of abelian groups

HomA(AIM) Ext1A(AIM) Ext2A(AIM) ExtnA(AIM)

If there are M -regular sequences in I then this sequence will start off with n minus 1 zero groupswhere n ge 1 is the common length of every maximal M -regular sequence The nth group will

23

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

lie over p If htp = r ge 0 then there exists a prime chain p = p0 sup p1 sup middot middot middot sup pr in A and bygoing-down there exists a prime chain qs = t0 sup t1 sup middot middot middot sup tr of B such that ti capA = pi Then

q = q0 sup middot middot middot sup qs sup t1 sup middot middot middot sup tr

is a prime chain of length r + s therefore htq ge r + s(3) (i) follows from (2) since dim(A) = suphtp | p isin Spec(A) (ii) Since Φ is surjective IB

is a proper ideal Let q be a minimal prime over IB such that htq = ht(IB) and put p = q capAThen ht(qpB) = 0 so by (2) we find that ht(IB) = htq = htp ge htI For the reverseinclusion let p be a minimal prime ideal over I such that htp = htI and take a prime q of Blying over p Replacing q if necessary we may assume that q is a minimal prime ideal over pBThen htI = htp = htq ge ht(IB)

Theorem 56 Let A be a nonzero subring of B and suppose that B is integral over A Then

(1) dim(A) = dim(B)

(2) Let q isin Spec(B) and set p = q capA Then we have cohtp = cohtq and htq le htp

(3) If the going-down theorem holds between A and B then for any ideal J of B with J capA 6= Awe have htJ = ht(J capA)

Proof (1) By Theorem 42 the going-up theorem holds for A sube B and Spec(B) minusrarr Spec(A) issurjective so we can lift any chain of prime ideals p0 sub p1 sub middot middot middot sub pn in A to a chain of primeideals q0 sub middot middot middot sub qn in B On the other hand if q sube qprime are prime ideals of B and q capA = qprime capAthen q = qprime so any chain of prime ideals in B restricts to a chain of the same length in A Hencedim(A) = dim(B)

(2) Since Bq is integral over Ap it is clear from (1) that cohtp = cohtq If q = q0 sup middot middot middot sup qnis a chain of prime ideals in B then intersecting with A gives a chain of length n descending fromp Hence htq le htp

(3) Given the going-down theorem it is clear that htq = htp in (2) Let J be a proper ideal ofB with J capA 6= A and let q be such that htJ = htq Then ht(J capA) le ht(qcapA) = htq = htJ On the other hand BJ is integral over BJ cap A so every prime ideal p of A containing J cap Acan be lifted to a prime ideal q of B containing J In particular we can lift a prime ideal p withht(J capA) = htp to see that htJ le htq = htp = ht(J capA) as required

5 Depth

Definition 11 Let A be a ring M an A-module and a1 ar a sequence of elements of AWe say a1 ar is an M -regular sequence (or simply M -sequence) if the following conditions aresatisfied

(1) For each 2 le i le r ai is regular on M(a1 aiminus1)M and a1 is regular on M

(2) M 6= (a1 an)M

If a1 ar is an M -regular sequence then so is a1 ai for any i le r When all ai belong to anideal I we say a1 ar is an M -regular sequence in I If moreover there is no b isin I such thata1 ar b is M -regular then a1 ar is said to be a maximal M -regular sequence in I Noticethat the notion of M -regular depends on the order of the elements in the sequence If MN areisomorphic A-modules then a sequence is regular on M iff it is regular on N

Lemma 57 A sequence a1 ar with r ge 2 is M -regular if and only if a1 is regular on Mand a2 ar is an Ma1M -regular sequence If the sequence a1 ar is a maximal M -regularsequence in I then a2 ar is a maximal Ma1M -regular sequence in I

18

Proof The key point is that for ideals a sube b there is a canonical isomorphism of A-modulesMbM sim= NbN where N = MaM If a1 ar is M -regular then a1 is regular on M a2 isregular on N = Ma1M and for 3 le i le r ai is regular on

M(a1 aiminus1)M sim= N(a2 aiminus1)N

Hence a2 ar is an N -regular sequence The converse follows from the same argument

More generally if a1 ar is an M -regular sequence and we set N = M(a1 ar)M andif b1 bs is an N -regular sequence then a1 ar b1 bs is an M -regular sequence

Lemma 58 If a1 ar is an A-regular sequence and M is a flat A-module then a1 ar isalso M -regular provided (a1 ar)M 6= M

Proof Left multiplication by a1 defines a monomorphism A minusrarr A since a1 is A-regular Tensoringwith M and using the fact that M is flat we see that left multiplication by a1 also gives amonomorphismM minusrarrM as required Similarly tensoring with the monomorphism a2 Aa1 minusrarrAa1 we get a monomorphism Ma1M minusrarrMa1M and so on

Lemma 59 Let A be a ring and M an A-module Given an integer n ge 1 a sequence a1 aris M -regular if and only if it is Mn-regular

Proof Suppose the sequence a1 ar is M -regular We prove it is Mn-regular by induction on rThe case r = 1 is trivial so assume r gt 1 By the inductive hypothesis the sequence a1 arminus1 isMn-regular Let L = (a1 arminus1)M Then (a1 arminus1)Mn = Ln and there is an isomorphismof A-modules MnLn sim= (ML)n So we need only show that ar is regular on (ML)n Since byassumption it is regular on ML this is not hard to check Clearly (a1 ar)Mn 6= Mn so thesequence a1 ar is Mn-regular as required The converse is similarly checked

Lemma 60 Let A be a nonzero ring M an A-module and a1 ar isin A If a1 ar isin Am isMm-regular for every maximal ideal m of A then the sequence a1 ar is M -regular

Proof This follows from the fact that given an A-module M an element a isin A is regular on M ifand only if its image in Am is regular on Mm for every maximal ideal m of A

Lemma 61 Suppose that a1 ar is M -regular and a1ξ1 + middot middot middot + arξr = 0 for ξi isin M Thenξi isin (a1 ar)M for all i

Proof By induction on r For r = 1 a1ξ1 = 0 implies that ξ1 = 0 Let r gt 1 Since ar is regularon M(a1 arminus1)M we have ξr =

sumrminus1i=1 aiηi so

sumrminus1i=1 ai(ξi + arηi) = 0 By the inductive

hypothesis for i lt r we have ξi + arηi isin (a1 arminus1)M so that ξi isin (a1 ar)M

Theorem 62 Let A be a ring and M an A-module and let a1 ar be an M -regular sequenceThen for every sequence n1 nr of integers gt 0 the sequence an1

1 anrr is M -regular

Proof Suppose we can prove the following statement

(lowast) Given an integer n gt 0 an A-module M and any M -regular sequence a1 arthe sequence an1 a2 ar is also M -regular

We prove the rest of the Theorem by induction on r For r = 1 this follows immediately from (lowast)Let r gt 1 and suppose a1 ar is M -regular Then by (lowast) an1

1 a2 ar is M -regular Hencea2 ar is Man1

1 M -regular By the inductive hypothesis an22 anr

r is Man11 M -regular and

therefore an11 anr

r is M -regular by Lemma 57So it only remains to prove (lowast) which we do by induction on n The case n = 1 is trivial so let

n gt 1 be given along with an A-moduleM and anM -regular sequence a1 ar By the inductivehypothesis anminus1

1 a2 ar is M -regular and clearly an1 is regular on M Since (an1 a2 ar) sube(anminus1

1 a2 ar) it is clear thatM 6= (an1 a2 ar)M Let i gt 1 and assume that an1 a2 aiminus1

is an M -regular sequence We need to show that ai is regular on M(an1 a2 aiminus1)M Suppose

19

that aiω = an1 ξ1 +a2ξ2 + middot middot middot+aiminus1ξiminus1 Then ω = anminus11 η1 +a2η2 + middot middot middot+aiminus1ηiminus1 by the inductive

hypothesis So

anminus11 (a1ξ1 minus aiη1) + a2(ξ2 minus aiη2) + middot middot middot+ aiminus1(ξiminus1 minus aiηiminus1) = 0

Hence a1ξ1minusaiη1 isin (anminus11 a2 aiminus1)M by Lemma 61 It follows that aiη1 isin (a1 a2 aiminus1)M

hence η1 isin (a1 aiminus1)M and so ω isin (an1 a2 aiminus1)M as required This proves (lowast) andtherefore completes the proof

Let A be a ring There is an isomorphism of A[x1 xn]-modules M otimesA A[x1 xn] sim=M [x1 xn] where the latter module consists of polynomials in x1 xn with coefficients in M(see our Polynomial Ring notes) For any f(x1 xn) isinM [x1 xn] and tuple (a1 an) isinAn we can define an element of M

f(a1 an) =sumα

aα11 middot middot middot aαn

n middot f(α)

For an element r isin R and h isinM [x1 xn]

(f + h)(a1 an) = f(a1 an) + h(a1 an)(r middot f)(a1 an) = r middot f(a1 an)

For an ideal I in R the R-submodule IM [x1 xn] consists of all polynomials whose coefficientsare in the R-submodule IM subeM

Let us review the definition of the associated graded ring and modules Let A be a ring and Ian ideal of A Then the abelian group

grI(A) = AI oplus II2 oplus I2I3 oplus middot middot middot

becomes a graded ring in a fairly obvious way For an A-module M we have the graded grI(A)-module

grI(M) = MIM oplus IMI2M oplus I2MI3M oplus middot middot middotIf A is noetherian and M is a finitely generated A-module then grI(A) is a noetherian ring andif grI(M) is a finitely generated grI(A)-module

Given elements a1 an isin A and I = (a1 an) we define a morphism of abelian groupsψ M [x1 xn] minusrarr grI(M) as follows if f is homogenous of degreem ge 0 define ψ(f) to be theimage of f(a1 an) in ImMIm+1M This defines a morphism of groups M [x1 xn]m minusrarrImMIm+1M and together these define the morphism of groups ψ Since ψ maps IM [x1 xn]to zero it induces a morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) and

Proposition 63 Let A be a ring and M an A-module Let a1 an isin A and set I =(a1 an) Then the following conditions are equivalent

(a) For every m gt 0 and for every homogenous polynomial f(x1 xn) isin M [x1 xn] ofdegree m such that f(a1 an) isin Im+1M we have f isin IM [x1 xn]

(b) If f(x1 xn) isin M [x1 xn] is homogenous and f(a1 an) = 0 then the coefficientsof f are in IM

(c) The morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) defined by mappinga homogenous polynomial f(x1 xn) of degree m to f(a1 an) isin ImMIm+1M is anisomorphism

Proof It is easy to see that (a) hArr (c) and (a) rArr (b) To show (b) rArr (a) let f isinM [x1 xn] bea homogenous polynomial of degree m gt 0 and suppose f(a1 an) isin Im+1M Any element ofIm+1M can be written as a sum of terms of the form aα1

1 middot middot middot aαnn middotm with

sumi αi = m+1 By shifting

one of the ai across we can write f(a1 an) = g(a1 an) for a homogenous polynomialg isinM [x1 xn] of degreem all of whose coefficients belong to IM Hence (fminusg)(a1 an) = 0so by (b) the coefficients of f minus g belong to IM and this implies implies that the coefficients of falso belong to IM as required

20

Definition 12 Let A be a ring and M an A-module A sequence a1 an isin A is M -quasiregularif it satisfies the equivalent conditions of the Proposition Obviously this concept does not dependon the order of the elements But a1 ai for i lt n need not be M -quasiregular

Recall that for an A-module M a submodule N sube M and x isin A the notation (N x) meansm isin M |xm isin N This is a submodule of M If A is a ring I an ideal and M an A-modulerecall that M is separated in the I-adic topology when

⋂n I

nM = 0

Theorem 64 Let A be a ring M a nonzero A-module a1 an isin A and I = (a1 an)Then

(i) If a1 an is M -quasiregular and x isin A is such that (IM x) = IM then (ImM x) =ImM for all m gt 0

(ii) If a1 an is M -regular then it is M -quasiregular

(iii) If MMa1MM(a1 a2)M M(a1 anminus1)M are separated in the I-adic topologythen the converse of (ii) is also true

Proof (i) By induction on m with the case m = 1 true by assumption Suppose m gt 1 andxξ isin ImM By the inductive hypothesis ξ isin Imminus1M Hence there exists a homogenous polynomialf isinM [x1 xn] of degree mminus 1 such that ξ = f(a1 an) Since xξ = xf(a1 an) isin ImMthe coefficients of f are in (IM x) = IM Therefore ξ = f(a1 an) isin ImM

(ii) By induction on n For n = 1 this is easy to check Let n gt 1 and suppose a1 an is M -regular Then by the induction hypothesis a1 anminus1 is M -quasiregular Let f isinM [x1 xn]be homogenous of degree m gt 0 such that f(a1 an) = 0 We prove that f isin IM [x1 xn]by induction on m (the case m = 0 being trivial) Write

f(x1 xn) = g(x1 xnminus1) + xnh(x1 xn)

Then g and h are homogenous of degrees m and mminus 1 respectively By (i) we have

h(a1 an) isin ((a1 anminus1)mM an) = (a1 anminus1)mM sube ImM

Since by assumption a1 an is regular on M so an is regular on M(a1 anminus1)M andhence ((a1 anminus1)M an) = (a1 anminus1)M So by the induction hypothesis on m we haveh isin IM [x1 xn] (by the argument of Proposition 63) Since h(a1 an) isin (a1 anminus1)mMthere exists H isin M [x1 xnminus1] which is homogenous of degree m such that h(a1 an) =H(a1 anminus1) Putting

G(x1 xnminus1) = g(x1 xnminus1) + anH(x1 xnminus1)

we have G(a1 anminus1) = 0 so by the inductive hypothesis on n we have G isin IM [x1 xn]hence g isin IM [x1 xn] and so f isin IM [x1 xn]

(iii) By induction on n ge 1 Assume that a1 an is M -quasiregular and the modulesMMa1M M(a1 anminus1)M are all separated in the I-adic topology If a1ξ = 0 thenξ isin IM hence ξ =

sumaiηi and

suma1aiηi = 0 hence ηi isin IM and so ξ isin I2M In this way we

see that ξ isin⋂t ItM = 0 Thus a1 is regular on M and this also takes care of the case n = 1

since M 6= IM by the separation condition So assume n gt 1 By Lemma 57 it suffices to showthat a2 an is an N -regular sequence where N = Ma1M Since there is an isomorphism ofA-modules for 2 le i le nminus 1

M(a1 ai)M sim= N(a2 ai)N

The modules NNa2N N(a2 anminus1)N are separated in the I-adic topology So by theinductive hypothesis it suffices to show that the sequence a2 an is N -quasiregular

It suffices to show that if f(x2 xn) isin M [x2 xn] is homogenous of degree m ge 1 withf(a2 an) isin a1M then the coefficients of f belong to IM Put f(a2 an) = a1ω We claim

21

that ω isin Imminus1M Let 0 le i le mminus1 be the largest integer with ω isin IiM Then ω = g(a1 an)for some homogenous polynomial of degree i and

f(a2 an) = a1g(a1 an) (2)

If i lt mminus1 then g isin IM [x1 xn] and so ω isin Ii+1M which is a contradiction Hence i = mminus1and so ω isin Imminus1M Again using (2) we see that f(x2 xn)minusx1g(x1 xn) isin IM [x1 xn]Since f does not involve x1 we have f isin IM [x1 xn] as required

The theorem shows that under the assumptions of (iii) any permutation of an M -regularsequence is M -regular

Corollary 65 Let A be a noetherian ring M a finitely generated A-module and let a1 anbe contained in the Jacobson radical of A Then a1 an is M -regular if and only if it is M -quasiregular In particular if a1 an is M -regular so is any permutation of the sequence

Proof From [AM69] we know that for any ideal I contained in the Jacobson radical the I-adictopology on any finitely generated A-module is separated

If A is a ring and M an A-module then any M -regular sequence a1 an isin A gives rise toa strictly increasing chain of submodules a1M (a1 a2)M (a1 an)M Hence the chain ofideals (a1) (a1 a2) (a1 an) must also be strictly increasing

Lemma 66 Let A be a noetherian ring and M an A-module Any M -regular sequence a1 anin an ideal I can be extended to a maximal M -regular sequence in I

Proof If a1 an is not maximal in I we can find an+1 isin I such that a1 an an+1 is anM -regular sequence Either this process terminates at a maximal M -regular sequence in I or itproduces a strictly ascending chain of ideals

(a1) sub (a1 a2) sub (a1 a2 a3) sub middot middot middot

Since A is noetherian we can exclude this latter possibility

Theorem 67 Let A be a noetherian ring M a finitely generated A-module and I an ideal of Awith IM 6= M Let n gt 0 be an integer Then the following are equivalent

(1) ExtiA(NM) = 0 for i lt n and every finitely generated A-module N with Supp(N) sube V (I)

(2) ExtiA(AIM) = 0 for i lt n

(3) There exists a finitely generated A-module N with Supp(N) = V (I) and ExtiA(NM) = 0for i lt n

(4) There exists an M -regular sequence a1 an of length n in I

Proof (1) rArr (2) rArr (3) is trivial With I fixed we show that (3) rArr (4) for every finitely generatedmodule M with IM 6= M by induction on n We have 0 = Ext0A(NM) sim= HomA(NM) SinceM is finitely generated and nonzero the set of associated primes of M is finite and nonempty Ifno elements of I are M -regular then I is contained in the union of these associated primes andhence I sube p for some p isin Ass(M) (see [AM69] for details) By definition there is a monomorphismof A-modules φ Ap minusrarrM There is an isomorphism of A-modules

(Ap)psim= ApotimesA Ap

sim= AppAp = k

It is not hard to check this is an isomorphism of Ap-modules as well Since φp is a monomorphismand k 6= 0 it follows that HomAp(kMp) 6= 0 Since p isin V (I) = Supp(N) we have Np 6= 0 andso the k-module NppNp 6= 0 is nonzero and therefore free so Homk(NppNp k) 6= 0 Sincek sim= (Ap)p as Ap-modules it follows that HomA(NAp)p

sim= HomAp(Np (Ap)p) 6= 0 Since Apis isomorphic to a submodule of M it follows that HomA(NM) 6= 0 which is a contradiction

22

therefore there exists an M -regular element a1 isin I which takes care of the case n = 1 If n gt 1then put M1 = Ma1M From the exact sequence

0 Ma1 M M1

0 (3)

we get the long exact sequence

middot middot middot minusrarr ExtiA(NM) minusrarr ExtiA(NM1) minusrarr Exti+1A (NM) minusrarr middot middot middot

which shows that ExtiA(NM1) = 0 for 0 le i lt n minus 1 By the inductive hypothesis on n thereexists an M1-regular sequence a2 an in I The sequence a1 an is then an M -regularsequence in I

(4) rArr (1) By induction on n with I fixed For n = 1 we have a1 isin I regular on M and so (3)gives an exact sequence of R-modules

0 HomA(NM)a1 HomA(NM)

Where a1 denotes left multiplication by a1 Since Supp(N) = V (Ann(N)) sube V (I) it fol-lows that I sube r(Ann(N)) and so ar1N = 0 for some r gt 0 It follows that ar1 annihilatesHomA(NM) as well but since the action of ar1 on HomA(NM) gives an injective map it followsthat HomA(NM) = 0 Now assume n gt 1 and put M1 = Ma1M Then a2 an is anM1-regular sequence so by the inductive hypothesis ExtiA(NM1) = 0 for i lt nminus 1 So the longexact sequence corresponding to (3) gives an exact sequence for 0 le i lt n

0 ExtiA(NM)a1 ExtiA(NM)

Here a1 denotes left multiplication by a1 which is equal to ExtiA(αM) = Exti(N β) where α βare the endomorphisms given by left multiplication by a1 on NM respectively Assume that ar1annihilates N with r gt 0 Then αr is the zero map so ExtiA(αM)r = 0 and so ar1 also annihilatesExtiA(NM) Since the a1 is regular on this module it follows that ExtiA(NM) = 0 for i lt nas required

Corollary 68 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA with IM 6= M If a1 an a maximal M -regular sequence in I then ExtiA(AIM) = 0 fori lt n and ExtnA(AIM) 6= 0

Proof We already know that ExtiA(AIM) = 0 for i lt n so with I fixed we prove by inductionon n that ExtnA(AIM) 6= 0 for any finitely generated module M with IM 6= M admitting amaximal M -regular sequence of length n For n = 1 we have a1 isin I regular on M and an exactsequence (3) where M1 = Ma1M Part of the corresponding long exact sequence is

Ext0A(AIM) minusrarr Ext0A(AIM1) minusrarr Ext1A(AIM)

We know from the Theorem 67 that Ext0A(AIM) = 0 so it suffices to show that we haveHomA(AIM1) 6= 0 But if HomA(AIM1) = 0 then it follows from the proof of (3) rArr (4)above that there would be b isin I regular on M1 so a1 b is an M -regular sequence This is acontradiction since the sequence a1 was maximal so we conclude that Ext1A(AIM) 6= 0

Now assume n gt 1 and let a1 an be a maximal M -regular sequence in I Then a2 anis a maximal M1-regular sequence in I so by the inductive hypothesis Extnminus1

A (AIM1) 6= 0 Sofrom the long exact sequence for (3) we conclude that ExtnA(AIM) 6= 0 also

It follows that under the conditions of the Corollary every maximal M -regular sequence in Ihas a common length and you can find this length by looking at the sequence of abelian groups

HomA(AIM) Ext1A(AIM) Ext2A(AIM) ExtnA(AIM)

If there are M -regular sequences in I then this sequence will start off with n minus 1 zero groupswhere n ge 1 is the common length of every maximal M -regular sequence The nth group will

23

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Proof The key point is that for ideals a sube b there is a canonical isomorphism of A-modulesMbM sim= NbN where N = MaM If a1 ar is M -regular then a1 is regular on M a2 isregular on N = Ma1M and for 3 le i le r ai is regular on

M(a1 aiminus1)M sim= N(a2 aiminus1)N

Hence a2 ar is an N -regular sequence The converse follows from the same argument

More generally if a1 ar is an M -regular sequence and we set N = M(a1 ar)M andif b1 bs is an N -regular sequence then a1 ar b1 bs is an M -regular sequence

Lemma 58 If a1 ar is an A-regular sequence and M is a flat A-module then a1 ar isalso M -regular provided (a1 ar)M 6= M

Proof Left multiplication by a1 defines a monomorphism A minusrarr A since a1 is A-regular Tensoringwith M and using the fact that M is flat we see that left multiplication by a1 also gives amonomorphismM minusrarrM as required Similarly tensoring with the monomorphism a2 Aa1 minusrarrAa1 we get a monomorphism Ma1M minusrarrMa1M and so on

Lemma 59 Let A be a ring and M an A-module Given an integer n ge 1 a sequence a1 aris M -regular if and only if it is Mn-regular

Proof Suppose the sequence a1 ar is M -regular We prove it is Mn-regular by induction on rThe case r = 1 is trivial so assume r gt 1 By the inductive hypothesis the sequence a1 arminus1 isMn-regular Let L = (a1 arminus1)M Then (a1 arminus1)Mn = Ln and there is an isomorphismof A-modules MnLn sim= (ML)n So we need only show that ar is regular on (ML)n Since byassumption it is regular on ML this is not hard to check Clearly (a1 ar)Mn 6= Mn so thesequence a1 ar is Mn-regular as required The converse is similarly checked

Lemma 60 Let A be a nonzero ring M an A-module and a1 ar isin A If a1 ar isin Am isMm-regular for every maximal ideal m of A then the sequence a1 ar is M -regular

Proof This follows from the fact that given an A-module M an element a isin A is regular on M ifand only if its image in Am is regular on Mm for every maximal ideal m of A

Lemma 61 Suppose that a1 ar is M -regular and a1ξ1 + middot middot middot + arξr = 0 for ξi isin M Thenξi isin (a1 ar)M for all i

Proof By induction on r For r = 1 a1ξ1 = 0 implies that ξ1 = 0 Let r gt 1 Since ar is regularon M(a1 arminus1)M we have ξr =

sumrminus1i=1 aiηi so

sumrminus1i=1 ai(ξi + arηi) = 0 By the inductive

hypothesis for i lt r we have ξi + arηi isin (a1 arminus1)M so that ξi isin (a1 ar)M

Theorem 62 Let A be a ring and M an A-module and let a1 ar be an M -regular sequenceThen for every sequence n1 nr of integers gt 0 the sequence an1

1 anrr is M -regular

Proof Suppose we can prove the following statement

(lowast) Given an integer n gt 0 an A-module M and any M -regular sequence a1 arthe sequence an1 a2 ar is also M -regular

We prove the rest of the Theorem by induction on r For r = 1 this follows immediately from (lowast)Let r gt 1 and suppose a1 ar is M -regular Then by (lowast) an1

1 a2 ar is M -regular Hencea2 ar is Man1

1 M -regular By the inductive hypothesis an22 anr

r is Man11 M -regular and

therefore an11 anr

r is M -regular by Lemma 57So it only remains to prove (lowast) which we do by induction on n The case n = 1 is trivial so let

n gt 1 be given along with an A-moduleM and anM -regular sequence a1 ar By the inductivehypothesis anminus1

1 a2 ar is M -regular and clearly an1 is regular on M Since (an1 a2 ar) sube(anminus1

1 a2 ar) it is clear thatM 6= (an1 a2 ar)M Let i gt 1 and assume that an1 a2 aiminus1

is an M -regular sequence We need to show that ai is regular on M(an1 a2 aiminus1)M Suppose

19

that aiω = an1 ξ1 +a2ξ2 + middot middot middot+aiminus1ξiminus1 Then ω = anminus11 η1 +a2η2 + middot middot middot+aiminus1ηiminus1 by the inductive

hypothesis So

anminus11 (a1ξ1 minus aiη1) + a2(ξ2 minus aiη2) + middot middot middot+ aiminus1(ξiminus1 minus aiηiminus1) = 0

Hence a1ξ1minusaiη1 isin (anminus11 a2 aiminus1)M by Lemma 61 It follows that aiη1 isin (a1 a2 aiminus1)M

hence η1 isin (a1 aiminus1)M and so ω isin (an1 a2 aiminus1)M as required This proves (lowast) andtherefore completes the proof

Let A be a ring There is an isomorphism of A[x1 xn]-modules M otimesA A[x1 xn] sim=M [x1 xn] where the latter module consists of polynomials in x1 xn with coefficients in M(see our Polynomial Ring notes) For any f(x1 xn) isinM [x1 xn] and tuple (a1 an) isinAn we can define an element of M

f(a1 an) =sumα

aα11 middot middot middot aαn

n middot f(α)

For an element r isin R and h isinM [x1 xn]

(f + h)(a1 an) = f(a1 an) + h(a1 an)(r middot f)(a1 an) = r middot f(a1 an)

For an ideal I in R the R-submodule IM [x1 xn] consists of all polynomials whose coefficientsare in the R-submodule IM subeM

Let us review the definition of the associated graded ring and modules Let A be a ring and Ian ideal of A Then the abelian group

grI(A) = AI oplus II2 oplus I2I3 oplus middot middot middot

becomes a graded ring in a fairly obvious way For an A-module M we have the graded grI(A)-module

grI(M) = MIM oplus IMI2M oplus I2MI3M oplus middot middot middotIf A is noetherian and M is a finitely generated A-module then grI(A) is a noetherian ring andif grI(M) is a finitely generated grI(A)-module

Given elements a1 an isin A and I = (a1 an) we define a morphism of abelian groupsψ M [x1 xn] minusrarr grI(M) as follows if f is homogenous of degreem ge 0 define ψ(f) to be theimage of f(a1 an) in ImMIm+1M This defines a morphism of groups M [x1 xn]m minusrarrImMIm+1M and together these define the morphism of groups ψ Since ψ maps IM [x1 xn]to zero it induces a morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) and

Proposition 63 Let A be a ring and M an A-module Let a1 an isin A and set I =(a1 an) Then the following conditions are equivalent

(a) For every m gt 0 and for every homogenous polynomial f(x1 xn) isin M [x1 xn] ofdegree m such that f(a1 an) isin Im+1M we have f isin IM [x1 xn]

(b) If f(x1 xn) isin M [x1 xn] is homogenous and f(a1 an) = 0 then the coefficientsof f are in IM

(c) The morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) defined by mappinga homogenous polynomial f(x1 xn) of degree m to f(a1 an) isin ImMIm+1M is anisomorphism

Proof It is easy to see that (a) hArr (c) and (a) rArr (b) To show (b) rArr (a) let f isinM [x1 xn] bea homogenous polynomial of degree m gt 0 and suppose f(a1 an) isin Im+1M Any element ofIm+1M can be written as a sum of terms of the form aα1

1 middot middot middot aαnn middotm with

sumi αi = m+1 By shifting

one of the ai across we can write f(a1 an) = g(a1 an) for a homogenous polynomialg isinM [x1 xn] of degreem all of whose coefficients belong to IM Hence (fminusg)(a1 an) = 0so by (b) the coefficients of f minus g belong to IM and this implies implies that the coefficients of falso belong to IM as required

20

Definition 12 Let A be a ring and M an A-module A sequence a1 an isin A is M -quasiregularif it satisfies the equivalent conditions of the Proposition Obviously this concept does not dependon the order of the elements But a1 ai for i lt n need not be M -quasiregular

Recall that for an A-module M a submodule N sube M and x isin A the notation (N x) meansm isin M |xm isin N This is a submodule of M If A is a ring I an ideal and M an A-modulerecall that M is separated in the I-adic topology when

⋂n I

nM = 0

Theorem 64 Let A be a ring M a nonzero A-module a1 an isin A and I = (a1 an)Then

(i) If a1 an is M -quasiregular and x isin A is such that (IM x) = IM then (ImM x) =ImM for all m gt 0

(ii) If a1 an is M -regular then it is M -quasiregular

(iii) If MMa1MM(a1 a2)M M(a1 anminus1)M are separated in the I-adic topologythen the converse of (ii) is also true

Proof (i) By induction on m with the case m = 1 true by assumption Suppose m gt 1 andxξ isin ImM By the inductive hypothesis ξ isin Imminus1M Hence there exists a homogenous polynomialf isinM [x1 xn] of degree mminus 1 such that ξ = f(a1 an) Since xξ = xf(a1 an) isin ImMthe coefficients of f are in (IM x) = IM Therefore ξ = f(a1 an) isin ImM

(ii) By induction on n For n = 1 this is easy to check Let n gt 1 and suppose a1 an is M -regular Then by the induction hypothesis a1 anminus1 is M -quasiregular Let f isinM [x1 xn]be homogenous of degree m gt 0 such that f(a1 an) = 0 We prove that f isin IM [x1 xn]by induction on m (the case m = 0 being trivial) Write

f(x1 xn) = g(x1 xnminus1) + xnh(x1 xn)

Then g and h are homogenous of degrees m and mminus 1 respectively By (i) we have

h(a1 an) isin ((a1 anminus1)mM an) = (a1 anminus1)mM sube ImM

Since by assumption a1 an is regular on M so an is regular on M(a1 anminus1)M andhence ((a1 anminus1)M an) = (a1 anminus1)M So by the induction hypothesis on m we haveh isin IM [x1 xn] (by the argument of Proposition 63) Since h(a1 an) isin (a1 anminus1)mMthere exists H isin M [x1 xnminus1] which is homogenous of degree m such that h(a1 an) =H(a1 anminus1) Putting

G(x1 xnminus1) = g(x1 xnminus1) + anH(x1 xnminus1)

we have G(a1 anminus1) = 0 so by the inductive hypothesis on n we have G isin IM [x1 xn]hence g isin IM [x1 xn] and so f isin IM [x1 xn]

(iii) By induction on n ge 1 Assume that a1 an is M -quasiregular and the modulesMMa1M M(a1 anminus1)M are all separated in the I-adic topology If a1ξ = 0 thenξ isin IM hence ξ =

sumaiηi and

suma1aiηi = 0 hence ηi isin IM and so ξ isin I2M In this way we

see that ξ isin⋂t ItM = 0 Thus a1 is regular on M and this also takes care of the case n = 1

since M 6= IM by the separation condition So assume n gt 1 By Lemma 57 it suffices to showthat a2 an is an N -regular sequence where N = Ma1M Since there is an isomorphism ofA-modules for 2 le i le nminus 1

M(a1 ai)M sim= N(a2 ai)N

The modules NNa2N N(a2 anminus1)N are separated in the I-adic topology So by theinductive hypothesis it suffices to show that the sequence a2 an is N -quasiregular

It suffices to show that if f(x2 xn) isin M [x2 xn] is homogenous of degree m ge 1 withf(a2 an) isin a1M then the coefficients of f belong to IM Put f(a2 an) = a1ω We claim

21

that ω isin Imminus1M Let 0 le i le mminus1 be the largest integer with ω isin IiM Then ω = g(a1 an)for some homogenous polynomial of degree i and

f(a2 an) = a1g(a1 an) (2)

If i lt mminus1 then g isin IM [x1 xn] and so ω isin Ii+1M which is a contradiction Hence i = mminus1and so ω isin Imminus1M Again using (2) we see that f(x2 xn)minusx1g(x1 xn) isin IM [x1 xn]Since f does not involve x1 we have f isin IM [x1 xn] as required

The theorem shows that under the assumptions of (iii) any permutation of an M -regularsequence is M -regular

Corollary 65 Let A be a noetherian ring M a finitely generated A-module and let a1 anbe contained in the Jacobson radical of A Then a1 an is M -regular if and only if it is M -quasiregular In particular if a1 an is M -regular so is any permutation of the sequence

Proof From [AM69] we know that for any ideal I contained in the Jacobson radical the I-adictopology on any finitely generated A-module is separated

If A is a ring and M an A-module then any M -regular sequence a1 an isin A gives rise toa strictly increasing chain of submodules a1M (a1 a2)M (a1 an)M Hence the chain ofideals (a1) (a1 a2) (a1 an) must also be strictly increasing

Lemma 66 Let A be a noetherian ring and M an A-module Any M -regular sequence a1 anin an ideal I can be extended to a maximal M -regular sequence in I

Proof If a1 an is not maximal in I we can find an+1 isin I such that a1 an an+1 is anM -regular sequence Either this process terminates at a maximal M -regular sequence in I or itproduces a strictly ascending chain of ideals

(a1) sub (a1 a2) sub (a1 a2 a3) sub middot middot middot

Since A is noetherian we can exclude this latter possibility

Theorem 67 Let A be a noetherian ring M a finitely generated A-module and I an ideal of Awith IM 6= M Let n gt 0 be an integer Then the following are equivalent

(1) ExtiA(NM) = 0 for i lt n and every finitely generated A-module N with Supp(N) sube V (I)

(2) ExtiA(AIM) = 0 for i lt n

(3) There exists a finitely generated A-module N with Supp(N) = V (I) and ExtiA(NM) = 0for i lt n

(4) There exists an M -regular sequence a1 an of length n in I

Proof (1) rArr (2) rArr (3) is trivial With I fixed we show that (3) rArr (4) for every finitely generatedmodule M with IM 6= M by induction on n We have 0 = Ext0A(NM) sim= HomA(NM) SinceM is finitely generated and nonzero the set of associated primes of M is finite and nonempty Ifno elements of I are M -regular then I is contained in the union of these associated primes andhence I sube p for some p isin Ass(M) (see [AM69] for details) By definition there is a monomorphismof A-modules φ Ap minusrarrM There is an isomorphism of A-modules

(Ap)psim= ApotimesA Ap

sim= AppAp = k

It is not hard to check this is an isomorphism of Ap-modules as well Since φp is a monomorphismand k 6= 0 it follows that HomAp(kMp) 6= 0 Since p isin V (I) = Supp(N) we have Np 6= 0 andso the k-module NppNp 6= 0 is nonzero and therefore free so Homk(NppNp k) 6= 0 Sincek sim= (Ap)p as Ap-modules it follows that HomA(NAp)p

sim= HomAp(Np (Ap)p) 6= 0 Since Apis isomorphic to a submodule of M it follows that HomA(NM) 6= 0 which is a contradiction

22

therefore there exists an M -regular element a1 isin I which takes care of the case n = 1 If n gt 1then put M1 = Ma1M From the exact sequence

0 Ma1 M M1

0 (3)

we get the long exact sequence

middot middot middot minusrarr ExtiA(NM) minusrarr ExtiA(NM1) minusrarr Exti+1A (NM) minusrarr middot middot middot

which shows that ExtiA(NM1) = 0 for 0 le i lt n minus 1 By the inductive hypothesis on n thereexists an M1-regular sequence a2 an in I The sequence a1 an is then an M -regularsequence in I

(4) rArr (1) By induction on n with I fixed For n = 1 we have a1 isin I regular on M and so (3)gives an exact sequence of R-modules

0 HomA(NM)a1 HomA(NM)

Where a1 denotes left multiplication by a1 Since Supp(N) = V (Ann(N)) sube V (I) it fol-lows that I sube r(Ann(N)) and so ar1N = 0 for some r gt 0 It follows that ar1 annihilatesHomA(NM) as well but since the action of ar1 on HomA(NM) gives an injective map it followsthat HomA(NM) = 0 Now assume n gt 1 and put M1 = Ma1M Then a2 an is anM1-regular sequence so by the inductive hypothesis ExtiA(NM1) = 0 for i lt nminus 1 So the longexact sequence corresponding to (3) gives an exact sequence for 0 le i lt n

0 ExtiA(NM)a1 ExtiA(NM)

Here a1 denotes left multiplication by a1 which is equal to ExtiA(αM) = Exti(N β) where α βare the endomorphisms given by left multiplication by a1 on NM respectively Assume that ar1annihilates N with r gt 0 Then αr is the zero map so ExtiA(αM)r = 0 and so ar1 also annihilatesExtiA(NM) Since the a1 is regular on this module it follows that ExtiA(NM) = 0 for i lt nas required

Corollary 68 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA with IM 6= M If a1 an a maximal M -regular sequence in I then ExtiA(AIM) = 0 fori lt n and ExtnA(AIM) 6= 0

Proof We already know that ExtiA(AIM) = 0 for i lt n so with I fixed we prove by inductionon n that ExtnA(AIM) 6= 0 for any finitely generated module M with IM 6= M admitting amaximal M -regular sequence of length n For n = 1 we have a1 isin I regular on M and an exactsequence (3) where M1 = Ma1M Part of the corresponding long exact sequence is

Ext0A(AIM) minusrarr Ext0A(AIM1) minusrarr Ext1A(AIM)

We know from the Theorem 67 that Ext0A(AIM) = 0 so it suffices to show that we haveHomA(AIM1) 6= 0 But if HomA(AIM1) = 0 then it follows from the proof of (3) rArr (4)above that there would be b isin I regular on M1 so a1 b is an M -regular sequence This is acontradiction since the sequence a1 was maximal so we conclude that Ext1A(AIM) 6= 0

Now assume n gt 1 and let a1 an be a maximal M -regular sequence in I Then a2 anis a maximal M1-regular sequence in I so by the inductive hypothesis Extnminus1

A (AIM1) 6= 0 Sofrom the long exact sequence for (3) we conclude that ExtnA(AIM) 6= 0 also

It follows that under the conditions of the Corollary every maximal M -regular sequence in Ihas a common length and you can find this length by looking at the sequence of abelian groups

HomA(AIM) Ext1A(AIM) Ext2A(AIM) ExtnA(AIM)

If there are M -regular sequences in I then this sequence will start off with n minus 1 zero groupswhere n ge 1 is the common length of every maximal M -regular sequence The nth group will

23

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

that aiω = an1 ξ1 +a2ξ2 + middot middot middot+aiminus1ξiminus1 Then ω = anminus11 η1 +a2η2 + middot middot middot+aiminus1ηiminus1 by the inductive

hypothesis So

anminus11 (a1ξ1 minus aiη1) + a2(ξ2 minus aiη2) + middot middot middot+ aiminus1(ξiminus1 minus aiηiminus1) = 0

Hence a1ξ1minusaiη1 isin (anminus11 a2 aiminus1)M by Lemma 61 It follows that aiη1 isin (a1 a2 aiminus1)M

hence η1 isin (a1 aiminus1)M and so ω isin (an1 a2 aiminus1)M as required This proves (lowast) andtherefore completes the proof

Let A be a ring There is an isomorphism of A[x1 xn]-modules M otimesA A[x1 xn] sim=M [x1 xn] where the latter module consists of polynomials in x1 xn with coefficients in M(see our Polynomial Ring notes) For any f(x1 xn) isinM [x1 xn] and tuple (a1 an) isinAn we can define an element of M

f(a1 an) =sumα

aα11 middot middot middot aαn

n middot f(α)

For an element r isin R and h isinM [x1 xn]

(f + h)(a1 an) = f(a1 an) + h(a1 an)(r middot f)(a1 an) = r middot f(a1 an)

For an ideal I in R the R-submodule IM [x1 xn] consists of all polynomials whose coefficientsare in the R-submodule IM subeM

Let us review the definition of the associated graded ring and modules Let A be a ring and Ian ideal of A Then the abelian group

grI(A) = AI oplus II2 oplus I2I3 oplus middot middot middot

becomes a graded ring in a fairly obvious way For an A-module M we have the graded grI(A)-module

grI(M) = MIM oplus IMI2M oplus I2MI3M oplus middot middot middotIf A is noetherian and M is a finitely generated A-module then grI(A) is a noetherian ring andif grI(M) is a finitely generated grI(A)-module

Given elements a1 an isin A and I = (a1 an) we define a morphism of abelian groupsψ M [x1 xn] minusrarr grI(M) as follows if f is homogenous of degreem ge 0 define ψ(f) to be theimage of f(a1 an) in ImMIm+1M This defines a morphism of groups M [x1 xn]m minusrarrImMIm+1M and together these define the morphism of groups ψ Since ψ maps IM [x1 xn]to zero it induces a morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) and

Proposition 63 Let A be a ring and M an A-module Let a1 an isin A and set I =(a1 an) Then the following conditions are equivalent

(a) For every m gt 0 and for every homogenous polynomial f(x1 xn) isin M [x1 xn] ofdegree m such that f(a1 an) isin Im+1M we have f isin IM [x1 xn]

(b) If f(x1 xn) isin M [x1 xn] is homogenous and f(a1 an) = 0 then the coefficientsof f are in IM

(c) The morphism of abelian groups φ (MIM)[x1 xn] minusrarr grI(M) defined by mappinga homogenous polynomial f(x1 xn) of degree m to f(a1 an) isin ImMIm+1M is anisomorphism

Proof It is easy to see that (a) hArr (c) and (a) rArr (b) To show (b) rArr (a) let f isinM [x1 xn] bea homogenous polynomial of degree m gt 0 and suppose f(a1 an) isin Im+1M Any element ofIm+1M can be written as a sum of terms of the form aα1

1 middot middot middot aαnn middotm with

sumi αi = m+1 By shifting

one of the ai across we can write f(a1 an) = g(a1 an) for a homogenous polynomialg isinM [x1 xn] of degreem all of whose coefficients belong to IM Hence (fminusg)(a1 an) = 0so by (b) the coefficients of f minus g belong to IM and this implies implies that the coefficients of falso belong to IM as required

20

Definition 12 Let A be a ring and M an A-module A sequence a1 an isin A is M -quasiregularif it satisfies the equivalent conditions of the Proposition Obviously this concept does not dependon the order of the elements But a1 ai for i lt n need not be M -quasiregular

Recall that for an A-module M a submodule N sube M and x isin A the notation (N x) meansm isin M |xm isin N This is a submodule of M If A is a ring I an ideal and M an A-modulerecall that M is separated in the I-adic topology when

⋂n I

nM = 0

Theorem 64 Let A be a ring M a nonzero A-module a1 an isin A and I = (a1 an)Then

(i) If a1 an is M -quasiregular and x isin A is such that (IM x) = IM then (ImM x) =ImM for all m gt 0

(ii) If a1 an is M -regular then it is M -quasiregular

(iii) If MMa1MM(a1 a2)M M(a1 anminus1)M are separated in the I-adic topologythen the converse of (ii) is also true

Proof (i) By induction on m with the case m = 1 true by assumption Suppose m gt 1 andxξ isin ImM By the inductive hypothesis ξ isin Imminus1M Hence there exists a homogenous polynomialf isinM [x1 xn] of degree mminus 1 such that ξ = f(a1 an) Since xξ = xf(a1 an) isin ImMthe coefficients of f are in (IM x) = IM Therefore ξ = f(a1 an) isin ImM

(ii) By induction on n For n = 1 this is easy to check Let n gt 1 and suppose a1 an is M -regular Then by the induction hypothesis a1 anminus1 is M -quasiregular Let f isinM [x1 xn]be homogenous of degree m gt 0 such that f(a1 an) = 0 We prove that f isin IM [x1 xn]by induction on m (the case m = 0 being trivial) Write

f(x1 xn) = g(x1 xnminus1) + xnh(x1 xn)

Then g and h are homogenous of degrees m and mminus 1 respectively By (i) we have

h(a1 an) isin ((a1 anminus1)mM an) = (a1 anminus1)mM sube ImM

Since by assumption a1 an is regular on M so an is regular on M(a1 anminus1)M andhence ((a1 anminus1)M an) = (a1 anminus1)M So by the induction hypothesis on m we haveh isin IM [x1 xn] (by the argument of Proposition 63) Since h(a1 an) isin (a1 anminus1)mMthere exists H isin M [x1 xnminus1] which is homogenous of degree m such that h(a1 an) =H(a1 anminus1) Putting

G(x1 xnminus1) = g(x1 xnminus1) + anH(x1 xnminus1)

we have G(a1 anminus1) = 0 so by the inductive hypothesis on n we have G isin IM [x1 xn]hence g isin IM [x1 xn] and so f isin IM [x1 xn]

(iii) By induction on n ge 1 Assume that a1 an is M -quasiregular and the modulesMMa1M M(a1 anminus1)M are all separated in the I-adic topology If a1ξ = 0 thenξ isin IM hence ξ =

sumaiηi and

suma1aiηi = 0 hence ηi isin IM and so ξ isin I2M In this way we

see that ξ isin⋂t ItM = 0 Thus a1 is regular on M and this also takes care of the case n = 1

since M 6= IM by the separation condition So assume n gt 1 By Lemma 57 it suffices to showthat a2 an is an N -regular sequence where N = Ma1M Since there is an isomorphism ofA-modules for 2 le i le nminus 1

M(a1 ai)M sim= N(a2 ai)N

The modules NNa2N N(a2 anminus1)N are separated in the I-adic topology So by theinductive hypothesis it suffices to show that the sequence a2 an is N -quasiregular

It suffices to show that if f(x2 xn) isin M [x2 xn] is homogenous of degree m ge 1 withf(a2 an) isin a1M then the coefficients of f belong to IM Put f(a2 an) = a1ω We claim

21

that ω isin Imminus1M Let 0 le i le mminus1 be the largest integer with ω isin IiM Then ω = g(a1 an)for some homogenous polynomial of degree i and

f(a2 an) = a1g(a1 an) (2)

If i lt mminus1 then g isin IM [x1 xn] and so ω isin Ii+1M which is a contradiction Hence i = mminus1and so ω isin Imminus1M Again using (2) we see that f(x2 xn)minusx1g(x1 xn) isin IM [x1 xn]Since f does not involve x1 we have f isin IM [x1 xn] as required

The theorem shows that under the assumptions of (iii) any permutation of an M -regularsequence is M -regular

Corollary 65 Let A be a noetherian ring M a finitely generated A-module and let a1 anbe contained in the Jacobson radical of A Then a1 an is M -regular if and only if it is M -quasiregular In particular if a1 an is M -regular so is any permutation of the sequence

Proof From [AM69] we know that for any ideal I contained in the Jacobson radical the I-adictopology on any finitely generated A-module is separated

If A is a ring and M an A-module then any M -regular sequence a1 an isin A gives rise toa strictly increasing chain of submodules a1M (a1 a2)M (a1 an)M Hence the chain ofideals (a1) (a1 a2) (a1 an) must also be strictly increasing

Lemma 66 Let A be a noetherian ring and M an A-module Any M -regular sequence a1 anin an ideal I can be extended to a maximal M -regular sequence in I

Proof If a1 an is not maximal in I we can find an+1 isin I such that a1 an an+1 is anM -regular sequence Either this process terminates at a maximal M -regular sequence in I or itproduces a strictly ascending chain of ideals

(a1) sub (a1 a2) sub (a1 a2 a3) sub middot middot middot

Since A is noetherian we can exclude this latter possibility

Theorem 67 Let A be a noetherian ring M a finitely generated A-module and I an ideal of Awith IM 6= M Let n gt 0 be an integer Then the following are equivalent

(1) ExtiA(NM) = 0 for i lt n and every finitely generated A-module N with Supp(N) sube V (I)

(2) ExtiA(AIM) = 0 for i lt n

(3) There exists a finitely generated A-module N with Supp(N) = V (I) and ExtiA(NM) = 0for i lt n

(4) There exists an M -regular sequence a1 an of length n in I

Proof (1) rArr (2) rArr (3) is trivial With I fixed we show that (3) rArr (4) for every finitely generatedmodule M with IM 6= M by induction on n We have 0 = Ext0A(NM) sim= HomA(NM) SinceM is finitely generated and nonzero the set of associated primes of M is finite and nonempty Ifno elements of I are M -regular then I is contained in the union of these associated primes andhence I sube p for some p isin Ass(M) (see [AM69] for details) By definition there is a monomorphismof A-modules φ Ap minusrarrM There is an isomorphism of A-modules

(Ap)psim= ApotimesA Ap

sim= AppAp = k

It is not hard to check this is an isomorphism of Ap-modules as well Since φp is a monomorphismand k 6= 0 it follows that HomAp(kMp) 6= 0 Since p isin V (I) = Supp(N) we have Np 6= 0 andso the k-module NppNp 6= 0 is nonzero and therefore free so Homk(NppNp k) 6= 0 Sincek sim= (Ap)p as Ap-modules it follows that HomA(NAp)p

sim= HomAp(Np (Ap)p) 6= 0 Since Apis isomorphic to a submodule of M it follows that HomA(NM) 6= 0 which is a contradiction

22

therefore there exists an M -regular element a1 isin I which takes care of the case n = 1 If n gt 1then put M1 = Ma1M From the exact sequence

0 Ma1 M M1

0 (3)

we get the long exact sequence

middot middot middot minusrarr ExtiA(NM) minusrarr ExtiA(NM1) minusrarr Exti+1A (NM) minusrarr middot middot middot

which shows that ExtiA(NM1) = 0 for 0 le i lt n minus 1 By the inductive hypothesis on n thereexists an M1-regular sequence a2 an in I The sequence a1 an is then an M -regularsequence in I

(4) rArr (1) By induction on n with I fixed For n = 1 we have a1 isin I regular on M and so (3)gives an exact sequence of R-modules

0 HomA(NM)a1 HomA(NM)

Where a1 denotes left multiplication by a1 Since Supp(N) = V (Ann(N)) sube V (I) it fol-lows that I sube r(Ann(N)) and so ar1N = 0 for some r gt 0 It follows that ar1 annihilatesHomA(NM) as well but since the action of ar1 on HomA(NM) gives an injective map it followsthat HomA(NM) = 0 Now assume n gt 1 and put M1 = Ma1M Then a2 an is anM1-regular sequence so by the inductive hypothesis ExtiA(NM1) = 0 for i lt nminus 1 So the longexact sequence corresponding to (3) gives an exact sequence for 0 le i lt n

0 ExtiA(NM)a1 ExtiA(NM)

Here a1 denotes left multiplication by a1 which is equal to ExtiA(αM) = Exti(N β) where α βare the endomorphisms given by left multiplication by a1 on NM respectively Assume that ar1annihilates N with r gt 0 Then αr is the zero map so ExtiA(αM)r = 0 and so ar1 also annihilatesExtiA(NM) Since the a1 is regular on this module it follows that ExtiA(NM) = 0 for i lt nas required

Corollary 68 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA with IM 6= M If a1 an a maximal M -regular sequence in I then ExtiA(AIM) = 0 fori lt n and ExtnA(AIM) 6= 0

Proof We already know that ExtiA(AIM) = 0 for i lt n so with I fixed we prove by inductionon n that ExtnA(AIM) 6= 0 for any finitely generated module M with IM 6= M admitting amaximal M -regular sequence of length n For n = 1 we have a1 isin I regular on M and an exactsequence (3) where M1 = Ma1M Part of the corresponding long exact sequence is

Ext0A(AIM) minusrarr Ext0A(AIM1) minusrarr Ext1A(AIM)

We know from the Theorem 67 that Ext0A(AIM) = 0 so it suffices to show that we haveHomA(AIM1) 6= 0 But if HomA(AIM1) = 0 then it follows from the proof of (3) rArr (4)above that there would be b isin I regular on M1 so a1 b is an M -regular sequence This is acontradiction since the sequence a1 was maximal so we conclude that Ext1A(AIM) 6= 0

Now assume n gt 1 and let a1 an be a maximal M -regular sequence in I Then a2 anis a maximal M1-regular sequence in I so by the inductive hypothesis Extnminus1

A (AIM1) 6= 0 Sofrom the long exact sequence for (3) we conclude that ExtnA(AIM) 6= 0 also

It follows that under the conditions of the Corollary every maximal M -regular sequence in Ihas a common length and you can find this length by looking at the sequence of abelian groups

HomA(AIM) Ext1A(AIM) Ext2A(AIM) ExtnA(AIM)

If there are M -regular sequences in I then this sequence will start off with n minus 1 zero groupswhere n ge 1 is the common length of every maximal M -regular sequence The nth group will

23

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Definition 12 Let A be a ring and M an A-module A sequence a1 an isin A is M -quasiregularif it satisfies the equivalent conditions of the Proposition Obviously this concept does not dependon the order of the elements But a1 ai for i lt n need not be M -quasiregular

Recall that for an A-module M a submodule N sube M and x isin A the notation (N x) meansm isin M |xm isin N This is a submodule of M If A is a ring I an ideal and M an A-modulerecall that M is separated in the I-adic topology when

⋂n I

nM = 0

Theorem 64 Let A be a ring M a nonzero A-module a1 an isin A and I = (a1 an)Then

(i) If a1 an is M -quasiregular and x isin A is such that (IM x) = IM then (ImM x) =ImM for all m gt 0

(ii) If a1 an is M -regular then it is M -quasiregular

(iii) If MMa1MM(a1 a2)M M(a1 anminus1)M are separated in the I-adic topologythen the converse of (ii) is also true

Proof (i) By induction on m with the case m = 1 true by assumption Suppose m gt 1 andxξ isin ImM By the inductive hypothesis ξ isin Imminus1M Hence there exists a homogenous polynomialf isinM [x1 xn] of degree mminus 1 such that ξ = f(a1 an) Since xξ = xf(a1 an) isin ImMthe coefficients of f are in (IM x) = IM Therefore ξ = f(a1 an) isin ImM

(ii) By induction on n For n = 1 this is easy to check Let n gt 1 and suppose a1 an is M -regular Then by the induction hypothesis a1 anminus1 is M -quasiregular Let f isinM [x1 xn]be homogenous of degree m gt 0 such that f(a1 an) = 0 We prove that f isin IM [x1 xn]by induction on m (the case m = 0 being trivial) Write

f(x1 xn) = g(x1 xnminus1) + xnh(x1 xn)

Then g and h are homogenous of degrees m and mminus 1 respectively By (i) we have

h(a1 an) isin ((a1 anminus1)mM an) = (a1 anminus1)mM sube ImM

Since by assumption a1 an is regular on M so an is regular on M(a1 anminus1)M andhence ((a1 anminus1)M an) = (a1 anminus1)M So by the induction hypothesis on m we haveh isin IM [x1 xn] (by the argument of Proposition 63) Since h(a1 an) isin (a1 anminus1)mMthere exists H isin M [x1 xnminus1] which is homogenous of degree m such that h(a1 an) =H(a1 anminus1) Putting

G(x1 xnminus1) = g(x1 xnminus1) + anH(x1 xnminus1)

we have G(a1 anminus1) = 0 so by the inductive hypothesis on n we have G isin IM [x1 xn]hence g isin IM [x1 xn] and so f isin IM [x1 xn]

(iii) By induction on n ge 1 Assume that a1 an is M -quasiregular and the modulesMMa1M M(a1 anminus1)M are all separated in the I-adic topology If a1ξ = 0 thenξ isin IM hence ξ =

sumaiηi and

suma1aiηi = 0 hence ηi isin IM and so ξ isin I2M In this way we

see that ξ isin⋂t ItM = 0 Thus a1 is regular on M and this also takes care of the case n = 1

since M 6= IM by the separation condition So assume n gt 1 By Lemma 57 it suffices to showthat a2 an is an N -regular sequence where N = Ma1M Since there is an isomorphism ofA-modules for 2 le i le nminus 1

M(a1 ai)M sim= N(a2 ai)N

The modules NNa2N N(a2 anminus1)N are separated in the I-adic topology So by theinductive hypothesis it suffices to show that the sequence a2 an is N -quasiregular

It suffices to show that if f(x2 xn) isin M [x2 xn] is homogenous of degree m ge 1 withf(a2 an) isin a1M then the coefficients of f belong to IM Put f(a2 an) = a1ω We claim

21

that ω isin Imminus1M Let 0 le i le mminus1 be the largest integer with ω isin IiM Then ω = g(a1 an)for some homogenous polynomial of degree i and

f(a2 an) = a1g(a1 an) (2)

If i lt mminus1 then g isin IM [x1 xn] and so ω isin Ii+1M which is a contradiction Hence i = mminus1and so ω isin Imminus1M Again using (2) we see that f(x2 xn)minusx1g(x1 xn) isin IM [x1 xn]Since f does not involve x1 we have f isin IM [x1 xn] as required

The theorem shows that under the assumptions of (iii) any permutation of an M -regularsequence is M -regular

Corollary 65 Let A be a noetherian ring M a finitely generated A-module and let a1 anbe contained in the Jacobson radical of A Then a1 an is M -regular if and only if it is M -quasiregular In particular if a1 an is M -regular so is any permutation of the sequence

Proof From [AM69] we know that for any ideal I contained in the Jacobson radical the I-adictopology on any finitely generated A-module is separated

If A is a ring and M an A-module then any M -regular sequence a1 an isin A gives rise toa strictly increasing chain of submodules a1M (a1 a2)M (a1 an)M Hence the chain ofideals (a1) (a1 a2) (a1 an) must also be strictly increasing

Lemma 66 Let A be a noetherian ring and M an A-module Any M -regular sequence a1 anin an ideal I can be extended to a maximal M -regular sequence in I

Proof If a1 an is not maximal in I we can find an+1 isin I such that a1 an an+1 is anM -regular sequence Either this process terminates at a maximal M -regular sequence in I or itproduces a strictly ascending chain of ideals

(a1) sub (a1 a2) sub (a1 a2 a3) sub middot middot middot

Since A is noetherian we can exclude this latter possibility

Theorem 67 Let A be a noetherian ring M a finitely generated A-module and I an ideal of Awith IM 6= M Let n gt 0 be an integer Then the following are equivalent

(1) ExtiA(NM) = 0 for i lt n and every finitely generated A-module N with Supp(N) sube V (I)

(2) ExtiA(AIM) = 0 for i lt n

(3) There exists a finitely generated A-module N with Supp(N) = V (I) and ExtiA(NM) = 0for i lt n

(4) There exists an M -regular sequence a1 an of length n in I

Proof (1) rArr (2) rArr (3) is trivial With I fixed we show that (3) rArr (4) for every finitely generatedmodule M with IM 6= M by induction on n We have 0 = Ext0A(NM) sim= HomA(NM) SinceM is finitely generated and nonzero the set of associated primes of M is finite and nonempty Ifno elements of I are M -regular then I is contained in the union of these associated primes andhence I sube p for some p isin Ass(M) (see [AM69] for details) By definition there is a monomorphismof A-modules φ Ap minusrarrM There is an isomorphism of A-modules

(Ap)psim= ApotimesA Ap

sim= AppAp = k

It is not hard to check this is an isomorphism of Ap-modules as well Since φp is a monomorphismand k 6= 0 it follows that HomAp(kMp) 6= 0 Since p isin V (I) = Supp(N) we have Np 6= 0 andso the k-module NppNp 6= 0 is nonzero and therefore free so Homk(NppNp k) 6= 0 Sincek sim= (Ap)p as Ap-modules it follows that HomA(NAp)p

sim= HomAp(Np (Ap)p) 6= 0 Since Apis isomorphic to a submodule of M it follows that HomA(NM) 6= 0 which is a contradiction

22

therefore there exists an M -regular element a1 isin I which takes care of the case n = 1 If n gt 1then put M1 = Ma1M From the exact sequence

0 Ma1 M M1

0 (3)

we get the long exact sequence

middot middot middot minusrarr ExtiA(NM) minusrarr ExtiA(NM1) minusrarr Exti+1A (NM) minusrarr middot middot middot

which shows that ExtiA(NM1) = 0 for 0 le i lt n minus 1 By the inductive hypothesis on n thereexists an M1-regular sequence a2 an in I The sequence a1 an is then an M -regularsequence in I

(4) rArr (1) By induction on n with I fixed For n = 1 we have a1 isin I regular on M and so (3)gives an exact sequence of R-modules

0 HomA(NM)a1 HomA(NM)

Where a1 denotes left multiplication by a1 Since Supp(N) = V (Ann(N)) sube V (I) it fol-lows that I sube r(Ann(N)) and so ar1N = 0 for some r gt 0 It follows that ar1 annihilatesHomA(NM) as well but since the action of ar1 on HomA(NM) gives an injective map it followsthat HomA(NM) = 0 Now assume n gt 1 and put M1 = Ma1M Then a2 an is anM1-regular sequence so by the inductive hypothesis ExtiA(NM1) = 0 for i lt nminus 1 So the longexact sequence corresponding to (3) gives an exact sequence for 0 le i lt n

0 ExtiA(NM)a1 ExtiA(NM)

Here a1 denotes left multiplication by a1 which is equal to ExtiA(αM) = Exti(N β) where α βare the endomorphisms given by left multiplication by a1 on NM respectively Assume that ar1annihilates N with r gt 0 Then αr is the zero map so ExtiA(αM)r = 0 and so ar1 also annihilatesExtiA(NM) Since the a1 is regular on this module it follows that ExtiA(NM) = 0 for i lt nas required

Corollary 68 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA with IM 6= M If a1 an a maximal M -regular sequence in I then ExtiA(AIM) = 0 fori lt n and ExtnA(AIM) 6= 0

Proof We already know that ExtiA(AIM) = 0 for i lt n so with I fixed we prove by inductionon n that ExtnA(AIM) 6= 0 for any finitely generated module M with IM 6= M admitting amaximal M -regular sequence of length n For n = 1 we have a1 isin I regular on M and an exactsequence (3) where M1 = Ma1M Part of the corresponding long exact sequence is

Ext0A(AIM) minusrarr Ext0A(AIM1) minusrarr Ext1A(AIM)

We know from the Theorem 67 that Ext0A(AIM) = 0 so it suffices to show that we haveHomA(AIM1) 6= 0 But if HomA(AIM1) = 0 then it follows from the proof of (3) rArr (4)above that there would be b isin I regular on M1 so a1 b is an M -regular sequence This is acontradiction since the sequence a1 was maximal so we conclude that Ext1A(AIM) 6= 0

Now assume n gt 1 and let a1 an be a maximal M -regular sequence in I Then a2 anis a maximal M1-regular sequence in I so by the inductive hypothesis Extnminus1

A (AIM1) 6= 0 Sofrom the long exact sequence for (3) we conclude that ExtnA(AIM) 6= 0 also

It follows that under the conditions of the Corollary every maximal M -regular sequence in Ihas a common length and you can find this length by looking at the sequence of abelian groups

HomA(AIM) Ext1A(AIM) Ext2A(AIM) ExtnA(AIM)

If there are M -regular sequences in I then this sequence will start off with n minus 1 zero groupswhere n ge 1 is the common length of every maximal M -regular sequence The nth group will

23

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

that ω isin Imminus1M Let 0 le i le mminus1 be the largest integer with ω isin IiM Then ω = g(a1 an)for some homogenous polynomial of degree i and

f(a2 an) = a1g(a1 an) (2)

If i lt mminus1 then g isin IM [x1 xn] and so ω isin Ii+1M which is a contradiction Hence i = mminus1and so ω isin Imminus1M Again using (2) we see that f(x2 xn)minusx1g(x1 xn) isin IM [x1 xn]Since f does not involve x1 we have f isin IM [x1 xn] as required

The theorem shows that under the assumptions of (iii) any permutation of an M -regularsequence is M -regular

Corollary 65 Let A be a noetherian ring M a finitely generated A-module and let a1 anbe contained in the Jacobson radical of A Then a1 an is M -regular if and only if it is M -quasiregular In particular if a1 an is M -regular so is any permutation of the sequence

Proof From [AM69] we know that for any ideal I contained in the Jacobson radical the I-adictopology on any finitely generated A-module is separated

If A is a ring and M an A-module then any M -regular sequence a1 an isin A gives rise toa strictly increasing chain of submodules a1M (a1 a2)M (a1 an)M Hence the chain ofideals (a1) (a1 a2) (a1 an) must also be strictly increasing

Lemma 66 Let A be a noetherian ring and M an A-module Any M -regular sequence a1 anin an ideal I can be extended to a maximal M -regular sequence in I

Proof If a1 an is not maximal in I we can find an+1 isin I such that a1 an an+1 is anM -regular sequence Either this process terminates at a maximal M -regular sequence in I or itproduces a strictly ascending chain of ideals

(a1) sub (a1 a2) sub (a1 a2 a3) sub middot middot middot

Since A is noetherian we can exclude this latter possibility

Theorem 67 Let A be a noetherian ring M a finitely generated A-module and I an ideal of Awith IM 6= M Let n gt 0 be an integer Then the following are equivalent

(1) ExtiA(NM) = 0 for i lt n and every finitely generated A-module N with Supp(N) sube V (I)

(2) ExtiA(AIM) = 0 for i lt n

(3) There exists a finitely generated A-module N with Supp(N) = V (I) and ExtiA(NM) = 0for i lt n

(4) There exists an M -regular sequence a1 an of length n in I

Proof (1) rArr (2) rArr (3) is trivial With I fixed we show that (3) rArr (4) for every finitely generatedmodule M with IM 6= M by induction on n We have 0 = Ext0A(NM) sim= HomA(NM) SinceM is finitely generated and nonzero the set of associated primes of M is finite and nonempty Ifno elements of I are M -regular then I is contained in the union of these associated primes andhence I sube p for some p isin Ass(M) (see [AM69] for details) By definition there is a monomorphismof A-modules φ Ap minusrarrM There is an isomorphism of A-modules

(Ap)psim= ApotimesA Ap

sim= AppAp = k

It is not hard to check this is an isomorphism of Ap-modules as well Since φp is a monomorphismand k 6= 0 it follows that HomAp(kMp) 6= 0 Since p isin V (I) = Supp(N) we have Np 6= 0 andso the k-module NppNp 6= 0 is nonzero and therefore free so Homk(NppNp k) 6= 0 Sincek sim= (Ap)p as Ap-modules it follows that HomA(NAp)p

sim= HomAp(Np (Ap)p) 6= 0 Since Apis isomorphic to a submodule of M it follows that HomA(NM) 6= 0 which is a contradiction

22

therefore there exists an M -regular element a1 isin I which takes care of the case n = 1 If n gt 1then put M1 = Ma1M From the exact sequence

0 Ma1 M M1

0 (3)

we get the long exact sequence

middot middot middot minusrarr ExtiA(NM) minusrarr ExtiA(NM1) minusrarr Exti+1A (NM) minusrarr middot middot middot

which shows that ExtiA(NM1) = 0 for 0 le i lt n minus 1 By the inductive hypothesis on n thereexists an M1-regular sequence a2 an in I The sequence a1 an is then an M -regularsequence in I

(4) rArr (1) By induction on n with I fixed For n = 1 we have a1 isin I regular on M and so (3)gives an exact sequence of R-modules

0 HomA(NM)a1 HomA(NM)

Where a1 denotes left multiplication by a1 Since Supp(N) = V (Ann(N)) sube V (I) it fol-lows that I sube r(Ann(N)) and so ar1N = 0 for some r gt 0 It follows that ar1 annihilatesHomA(NM) as well but since the action of ar1 on HomA(NM) gives an injective map it followsthat HomA(NM) = 0 Now assume n gt 1 and put M1 = Ma1M Then a2 an is anM1-regular sequence so by the inductive hypothesis ExtiA(NM1) = 0 for i lt nminus 1 So the longexact sequence corresponding to (3) gives an exact sequence for 0 le i lt n

0 ExtiA(NM)a1 ExtiA(NM)

Here a1 denotes left multiplication by a1 which is equal to ExtiA(αM) = Exti(N β) where α βare the endomorphisms given by left multiplication by a1 on NM respectively Assume that ar1annihilates N with r gt 0 Then αr is the zero map so ExtiA(αM)r = 0 and so ar1 also annihilatesExtiA(NM) Since the a1 is regular on this module it follows that ExtiA(NM) = 0 for i lt nas required

Corollary 68 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA with IM 6= M If a1 an a maximal M -regular sequence in I then ExtiA(AIM) = 0 fori lt n and ExtnA(AIM) 6= 0

Proof We already know that ExtiA(AIM) = 0 for i lt n so with I fixed we prove by inductionon n that ExtnA(AIM) 6= 0 for any finitely generated module M with IM 6= M admitting amaximal M -regular sequence of length n For n = 1 we have a1 isin I regular on M and an exactsequence (3) where M1 = Ma1M Part of the corresponding long exact sequence is

Ext0A(AIM) minusrarr Ext0A(AIM1) minusrarr Ext1A(AIM)

We know from the Theorem 67 that Ext0A(AIM) = 0 so it suffices to show that we haveHomA(AIM1) 6= 0 But if HomA(AIM1) = 0 then it follows from the proof of (3) rArr (4)above that there would be b isin I regular on M1 so a1 b is an M -regular sequence This is acontradiction since the sequence a1 was maximal so we conclude that Ext1A(AIM) 6= 0

Now assume n gt 1 and let a1 an be a maximal M -regular sequence in I Then a2 anis a maximal M1-regular sequence in I so by the inductive hypothesis Extnminus1

A (AIM1) 6= 0 Sofrom the long exact sequence for (3) we conclude that ExtnA(AIM) 6= 0 also

It follows that under the conditions of the Corollary every maximal M -regular sequence in Ihas a common length and you can find this length by looking at the sequence of abelian groups

HomA(AIM) Ext1A(AIM) Ext2A(AIM) ExtnA(AIM)

If there are M -regular sequences in I then this sequence will start off with n minus 1 zero groupswhere n ge 1 is the common length of every maximal M -regular sequence The nth group will

23

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

therefore there exists an M -regular element a1 isin I which takes care of the case n = 1 If n gt 1then put M1 = Ma1M From the exact sequence

0 Ma1 M M1

0 (3)

we get the long exact sequence

middot middot middot minusrarr ExtiA(NM) minusrarr ExtiA(NM1) minusrarr Exti+1A (NM) minusrarr middot middot middot

which shows that ExtiA(NM1) = 0 for 0 le i lt n minus 1 By the inductive hypothesis on n thereexists an M1-regular sequence a2 an in I The sequence a1 an is then an M -regularsequence in I

(4) rArr (1) By induction on n with I fixed For n = 1 we have a1 isin I regular on M and so (3)gives an exact sequence of R-modules

0 HomA(NM)a1 HomA(NM)

Where a1 denotes left multiplication by a1 Since Supp(N) = V (Ann(N)) sube V (I) it fol-lows that I sube r(Ann(N)) and so ar1N = 0 for some r gt 0 It follows that ar1 annihilatesHomA(NM) as well but since the action of ar1 on HomA(NM) gives an injective map it followsthat HomA(NM) = 0 Now assume n gt 1 and put M1 = Ma1M Then a2 an is anM1-regular sequence so by the inductive hypothesis ExtiA(NM1) = 0 for i lt nminus 1 So the longexact sequence corresponding to (3) gives an exact sequence for 0 le i lt n

0 ExtiA(NM)a1 ExtiA(NM)

Here a1 denotes left multiplication by a1 which is equal to ExtiA(αM) = Exti(N β) where α βare the endomorphisms given by left multiplication by a1 on NM respectively Assume that ar1annihilates N with r gt 0 Then αr is the zero map so ExtiA(αM)r = 0 and so ar1 also annihilatesExtiA(NM) Since the a1 is regular on this module it follows that ExtiA(NM) = 0 for i lt nas required

Corollary 68 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA with IM 6= M If a1 an a maximal M -regular sequence in I then ExtiA(AIM) = 0 fori lt n and ExtnA(AIM) 6= 0

Proof We already know that ExtiA(AIM) = 0 for i lt n so with I fixed we prove by inductionon n that ExtnA(AIM) 6= 0 for any finitely generated module M with IM 6= M admitting amaximal M -regular sequence of length n For n = 1 we have a1 isin I regular on M and an exactsequence (3) where M1 = Ma1M Part of the corresponding long exact sequence is

Ext0A(AIM) minusrarr Ext0A(AIM1) minusrarr Ext1A(AIM)

We know from the Theorem 67 that Ext0A(AIM) = 0 so it suffices to show that we haveHomA(AIM1) 6= 0 But if HomA(AIM1) = 0 then it follows from the proof of (3) rArr (4)above that there would be b isin I regular on M1 so a1 b is an M -regular sequence This is acontradiction since the sequence a1 was maximal so we conclude that Ext1A(AIM) 6= 0

Now assume n gt 1 and let a1 an be a maximal M -regular sequence in I Then a2 anis a maximal M1-regular sequence in I so by the inductive hypothesis Extnminus1

A (AIM1) 6= 0 Sofrom the long exact sequence for (3) we conclude that ExtnA(AIM) 6= 0 also

It follows that under the conditions of the Corollary every maximal M -regular sequence in Ihas a common length and you can find this length by looking at the sequence of abelian groups

HomA(AIM) Ext1A(AIM) Ext2A(AIM) ExtnA(AIM)

If there are M -regular sequences in I then this sequence will start off with n minus 1 zero groupswhere n ge 1 is the common length of every maximal M -regular sequence The nth group will

23

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

be nonzero and we canrsquot necessarily say anything about the rest of the sequence Notice thatsince any M -regular sequence can be extended to a maximal one any M -regular sequence haslength le n There are no M -regular sequences in I if and only if the first term of this sequenceis nonzero

Definition 13 Let A be a noetherian ring M a finitely generated A-module and I an ideal ofA If IM 6= M then we define the I-depth of M to be

depthI(M) = infi |ExtiA(AIM) 6= 0

So depthI(M) = 0 if and only if there are no M -regular sequences in I and otherwise it is thecommon length of all maximal M -regular sequences in I or equivalently the supremum of thelengths of M -regular sequences in I We define depthI(M) = infin if IM = M In particulardepthI(0) = infin Isomorphic modules have the same I-depth When (Am) is a local ring we writedepth(M) or depthAM for depthm(M) and call it simply the depth of M Thus depth(M) = infiniff M = 0 and depth(M) = 0 iff m isin Ass(M)

Lemma 69 Let φ A minusrarr B be a surjective local morphism of local noetherian rings and let Mbe a finitely generated B-module Then depthA(M) = depthB(M)

Proof It is clear that depthA(M) = infin iff depthB(M) = infin so we may as well assume bothdepths are finite Given a sequence of elements a1 an isin mA it is clear that they are anM -regular sequence iff the images φ(a1) φ(an) isin mB are an M -regular sequence Given anM -regular sequence b1 bn in mB you can choose inverse images a1 an isin mA and theseform an M -regular sequence This makes it clear that depthA(M) = depthB(M)

Lemma 70 Let A be a noetherian ring and M a finitely generated A-module Then for any idealI and integer n ge 1 we have depthI(M) = depthI(Mn)

Proof We have IMn = (IM)n so depthI(M) = infin if and only if depthI(Mn) = infin In the finitecase the result follows immediately from Lemma 59

Lemma 71 Let A be a noetherian ring M a finitely generated A-module and p a prime idealThen depthAp(Mp) = 0 if and only if p isin AssA(M)

Proof We have depthAp(Mp) = 0 iff pAp isin AssAp(Mp) which by [Ash] Chapter 1 Lemma 142is iff p isin AssA(M) So the associated primes are precisely those with depthAp(Mp) = 0

Lemma 72 Let A be a noetherian ring and M a finitely generated A-module For any prime pwe have depthAp(Mp) ge depthp(M)

Proof If depthp(M) = infin then pM = M and this implies that (pAp)Mp = Mp so depthAp(M) =infin If depthAp(Mp) = 0 then pAp isin Ass(Mp) which can only occur if p isin Ass(M) and this impliesthat HomA(ApM) 6= 0 so depthp(M) = 0 (since we have already excluded the possibility ofpM = M) So we can reduce to the case where depthAp(Mp) = n with 0 lt n ltinfin and pM 6= M We have seen earlier in notes that there is an isomorphism of groups for i ge 0

ExtiAp((Ap)pMp) sim= ExtiA(ApM)p

As an Ap-module (Ap)psim= AppAp and by assumption ExtnAp

(AppApMp) 6= 0 so it followsthat ExtnA(ApM) 6= 0 and hence depthp(M) le n

Definition 14 Let A be a noetherian ring and M a finitely generated A-module Then wedefine the grade of M denoted grade(M) to be depthI(A) where I is the ideal Ann(M) Sograde(M) = infin if and only if M = 0 Isomorphic modules have the same grade

If I is an ideal of A then we call grade(AI) = depthI(A) the grade of I and denote it by G(I)So the grade of A is infin and the grade of any proper ideal I is the common length of the maximalA-regular sequences in I (zero if none exist)

24

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Lemma 73 Let A be a noetherian ring and M a nonzero finitely generated A-module Then

grade(M) = infi |Exti(MA) 6= 0

Proof Put I = Ann(M) Since M and AI are both finitely generated A-modules whose supportsare equal to V (I) it follows from Theorem 67 that for any n gt 0 we have Exti(AIA) = 0 forall i lt n if and only if Exti(MA) = 0 for all i lt n In particular Ext0(MA) 6= 0 if and only ifExt0(AIA) 6= 0 By definition

grade(M) = depthI(A) = infi |Exti(AIM) 6= 0

so the claim is straightforward to check

The following result is a generalisation of Krullrsquos Principal Ideal Theorem

Lemma 74 Let A be a noetherian ring and a1 ar an A-regular sequence Then every minimalprime over (a1 ar) has height r and in particular ht(a1 ar) = r

Proof By assumption I = (a1 ar) is a proper ideal If r = 1 then this is precisely KrullrsquosPID Theorem For r gt 1 we proceed by induction If a1 ar is an A-regular sequence thenset J = (a1 arminus1) Clearly ar + J is a regular element of RJ which is not a unit so everyminimal prime over (ar+J) in RJ has height 1 But these are precisely the primes in R minimalover I So if p is any prime ideal minimal over I there is a prime J sube q sub p with q minimal overJ By the inductive hypothesis htq = r minus 1 so htp ge r We know the height is le r by anotherresult of Krull

For any nonzero ring A the sequence x1 xn in A[x1 xn] is clearly a maximal A-regularsequence So in some sense regular sequences in a ring generalise the notion of independentvariables

Lemma 75 Let A be a noetherian ring M a nonzero finitely generated A-module and I a properideal Then grade(M) le projdimM and G(I) le htI

Proof For a nonzero module M the projective dimension is the largest i ge 0 for which thereexists a module N with Exti(MN) 6= 0 So clearly grade(M) le projdimM The second claimis trivial if G(I) = 0 and otherwise G(I) is the length r of a maximal A-regular sequence a1 arin I But then r = ht(a1 ar) le htI so the proof is complete

Proposition 76 Let A be a noetherian ring MN finitely generated A-modules with M nonzeroand suppose that grade(M) = k and projdimN = ` lt k Then

ExtiA(MN) = 0 (0 le i lt k minus `)

Proof Induction on ` If ` = minus1 then this is trivial If ` = 0 then is a direct summand of a freemodule Since our assertion holds for A by definition it holds for N also If ` gt 0 take an exactsequence 0 minusrarr N prime minusrarr L minusrarr N minusrarr 0 with L free Then projdimN prime = ` minus 1 and our assertionis proved by induction

Lemma 77 (Ischebeck) Let (Am) be a noetherian local ring and let MN be nonzero finitelygenerated A-modules Suppose that depth(M) = k dim(N) = r Then

ExtiA(NM) = 0 (0 le i lt k minus r)

Proof By induction on integers r with r lt k (we assume k gt 0) If r = 0 then Supp(N) = mso the assertion follows from Theorem 67 Let r gt 0 First we prove the result in the case whereN = Ap for a prime ideal p We can pick x isin m p and then the following sequence is exact

0 Nx N N prime 0

25

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

where N prime = A(p + Ax) has dimension lt r Then using the induction hypothesis we get exactsequences of A-modules

0 = Exti(N primeM) ExtiA(NM) x ExtiA(NM) Exti+1A (N primeM) = 0

for 0 le i lt kminus r and so ExtiA(NM) = 0 by Nakayama since the module ExtiA(NM) is finitelygenerated (see our Ext notes) This proves the result for modules of the form N = Ap

For general N we use know from [Ash] Chapter 1 Theorem 1510 that there is a chain ofmodules 0 = N0 sub middot middot middot sub Ns = N such that for 1 le j le s we have an isomorphism of A-modulesNjNjminus1

sim= Apj where the pj are prime ideals of A Lemma 53 shows that dimN1 le dimN2 lemiddot middot middot le dimN = r so since N1

sim= Ap1 we have already shown the result holds for N1 Considerthe exact sequence

0 minusrarr N1 minusrarr N2 minusrarr Ap2 minusrarr 0

By Lemma 53 we know that r ge dimAp2 so the result holds for Ap2 and the following piece ofthe long exact Ext sequence shows that the result is true for N2 as well

ExtiA(Ap2M) minusrarr ExtiA(N2M) minusrarr ExtiA(N1M)

Proceeding in this way proves the result for all Nj and hence for N completing the proof

Theorem 78 Let (Am) be a noetherian local ring and let M be a nonzero finitely generatedA-module Then depth(M) le dim(Ap) for every p isin Ass(M)

Proof If p isin Ass(M) then HomA(Apm) 6= 0 hence depth(M) le dim(Ap) by Lemma 77

Lemma 79 Let A be a ring and let EF be finitely generated A-modules Then Supp(EotimesAF ) =Supp(E) cap Supp(F )

Proof See [AM69] Chapter 3 Exercise 19

The Dimension Theorem for modules (see [AM69] Chapter 11) shows that for a nonzero finitelygenerated module M over a noetherian local ring A the dimension of M is zero iff M is of finitelength and otherwise dim(M) is the smallest r ge 1 for which there exists elements a1 ar isin mwith M(a1 an)M of finite length

Proposition 80 Let A be a noetherian local ring and M a finitely generated A-module Leta1 ar be an M -regular sequence Then

dimM(a1 ar)M = dimM minus r

In particular if a1 ar is an A-regular sequence then the dimension of the ring A(a1 ar)is dimAminus r

Proof Let N = M(a1 ar)M Then N is a nonzero finitely generated A-module so ifk = dim(N) then 0 le k lt infin If k = 0 then it is clear from the preceding comments thatdimM(a1 ar)M ge dimM minusr If k ge 1 and b1 bk isin m are elements such that the module

N(b1 bk)N sim= M(a1 ar b1 bk)M

is of finite length then since the ai all belong to m we conclude that dim(M) le r+k Hence we atleast have the inequality dimM(a1 an)M ge dim(M)minus r On the other hand suppose f isin mis an M -regular element We have Supp(MfM) = Supp(M) cap Supp(AfA) = Supp(M) cap V (f)by Lemma 79 and f is not in any minimal element of Supp(M) since these coincide with theminimal elements of Ass(M) and f is regular on M Since

dimM = supcohtp | p isin Supp(M)

it follows easily that dim(MfM) lt dimM Proceeding by induction on r we see that

dimM(a1 ar)M le dimM minus r

as required

26

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Corollary 81 Let (Am) be a noetherian local ring and M a nonzero finitely generated A-moduleThen depthM le dimM

Proof This is trivial if depthM = 0 Otherwise let a1 ar be a maximal M -regular sequencein m so depthM = r Then we know from Proposition 80 that r = dimMminusdimM(a1 ar)M so of course r le dimM

Lemma 82 Let A be a noetherian ring M a finitely generated A-module and I an ideal Leta1 ar be an M -regular sequence in I and assume IM 6= M Then

depthI(M(a1 ar)M) = depthI(M)minus r

Proof Let N = M(a1 ar)M It is clear that IM = M iff IN = N so both depths arefinite If depthI(N) = 0 then the sequence a1 ar must be a maximal M -regular sequence inI so depthI(M) = r and we are done Otherwise let b1 bs be a maximal N -regular sequencein I Then a1 ar b1 bs is a maximal M -regular sequence in I so depthI(M) = r + s =r + depthI(N) as required

Lemma 83 Let A be a noetherian local ring a1 ar an A-regular sequence If I = (a1 ar)then

depthAI(AI) = depthA(A)minus r

Proof A sequence b1 bs isin m is AI-regular iff b1 + I bs + I isin mI is AI-regular soit is clear that depthAI(AI) = depthA(AI) By Lemma 82 depthA(AI) = depthA(A)minus r asrequired

Proposition 84 Let A be a noetherian ring M a finitely generated A-module and I a properideal Then

depthI(M) = infdepthMp | p isin V (I)

Proof Let n denote the value of the right hand side If n = 0 then depthMp = 0 for some p supe Iand then I sube p isin Ass(M) since pAp isin Ass(Mp) implies p isin Ass(M) Thus depthI(M) = 0since there can be no M -regular sequences in p If 0 lt n lt infin then I is not contained in anyassociated prime of M and so it is not contained in their union which is the set of elements notregular on M Hence there exists a isin I regular on M Moreover IM 6= M since otherwise wewould have (pAp)Mp = Mp and hence depthMp = infin for any p supe I which would contradict thefact that n ltinfin Put M prime = MaM Then for any p supe I with Mp 6= 0 the element a1 isin Ap is anMp-regular sequence in pAp so

depthM primep = depthMpaMp = depthMp minus 1

and depthI(M prime) = depthI(M)minus1 by the Lemma 82 Therefore our assertion is proved by inductionon n

If n = infin then Mp = 0 for all p supe I If IM 6= M then Supp(MIM) is nonempty sinceAss(MIM) sube Supp(MIM) and Ass(MIM) = empty iff MIM = 0 If p isin Supp(MIM) =Supp(M) cap V (I) then (MIM)p 6= 0 and so MpIMp 6= 0 which is a contradiction HenceIM = M and therefore depthI(M) = infin

51 Cohen-Macaulay Rings

Definition 15 Let (Am) be a noetherian local ring and M a finitely generated A-module Weknow that depthM le dimM provided M is nonzero We say that M is Cohen-Macaulay if M = 0or if depthM = dimM If the noetherian local ring A is Cohen-Macaulay as an A-module thenwe call A a Cohen-Macaulay ring So a noetherian local ring is Cohen-Macaulay if its dimensionis equal to the common length of the maximal A-regular sequences in m The Cohen-Macaulayproperty is stable under isomorphisms of modules and rings

27

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Example 4 Let A be a noetherian local ring If dim(A) = 0 then A is Cohen-Macaulay sincem is an associated prime of A and therefore no element of m is regular If dim(A) = d ge 1 thenA is Cohen-Macaulay if and only if there is an A-regular sequence in m of length d

Recall that for a module M over a noetherian ring A the elements of Ass(M) which arenot minimal are called the embedded primes of M Since a noetherian ring has descending chaincondition on prime ideals every associated prime of M contains a minimal associated prime

Theorem 85 Let (Am) be a noetherian local ring and M a finitely generated A-module Then

(i) If M is a Cohen-Macaulay module and p isin Ass(M) then we have depthM = dim(Ap)Consequently M has no embedded primes

(ii) If a1 ar is an M -regular sequence in m and M prime = M(a1 ar)M then M is Cohen-Macaulay hArr M prime is Cohen-Macaulay

(iii) If M is Cohen-Macaulay then for every p isin Spec(A) the Ap-module Mp is Cohen-Macaulayand if Mp 6= 0 we have depthp(M) = depthAp(Mp)

Proof (i) Since Ass(M) 6= empty M is nonzero and so depthM = dimM Since p isin Supp(M) wehave p supe Ann(M) and therefore dimM ge dim(Ap) and dim(Ap) ge depthM by Theorem 78 Ifp isin Ass(M) were an embedded prime there would be a minimal prime q isin Ass(M) with q sub pBut since cohtp = cohtq are both finite this is impossible

(ii) By Nakayama we have M = 0 iff M prime = 0 Suppose M 6= 0 Then dimM prime = dimM minus r byProposition 80 and depthM prime = depthM minus r by Lemma 82

(iii) We may assume thatMp 6= 0 Hence p supe Ann(M) We know that dimMp ge depthApMp gedepthp(M) So we will prove depthp(M) = dimMp by induction on depthp(M) If depthp(M) = 0then no element of p is regular onM so by the usual argument p is contained in some pprime isin Ass(M)But Ann(M) sube p sube pprime and the associated primes of M are the minimal primes over the idealAnn(M) by (i) Hence p = pprime and so p is a minimal element of Supp(M) The dimension of Mp

is the length of the longest chain in Supp(Mp) If p0Ap sub middot middot middot sub psAp = pAp is a chain of lengths = dimMp then p0Ap is minimal and therefore p0 isin Ass(M) It follows that p0 = p and so s = 0as required

Now suppose depthp(M) gt 0 take an M -regular element a isin p and set M1 = MaM Theelement a1 isin Ap is then Mp-regular Therefore we have

dim(M1)p = dimMpaMp = dimMp minus 1

and depthp(M1) = depthp(M)minus1 Since M1 is Cohen-Macaulay by (ii) by the inductive hypoth-esis we have dim(M1)p = depthp(M1) which completes the proof

Corollary 86 Let (Am) be a noetherian local ring and a1 ar an A-regular sequence inm Let Aprime be the ring A(a1 ar) Then A is a Cohen-Macaulay ring if and only if Aprime is aCohen-Macaulay ring

Proof Let I = (a1 ar) It suffices to show that Aprime is a Cohen-Macaulay ring if and only if itis a Cohen-Macaulay module over A The dimension of Aprime as an A-module the Krull dimensionof Aprime and the dimension of Aprime as a module over itself are all equal So it suffices to observe thata sequence b1 bs isin m is an Aprime-regular sequence iff b1 + I bs + I isin mI is an Aprime-regularsequence so depthAAprime = depthAprimeA

prime

Corollary 87 Let A be a Cohen-Macaulay local ring and p a prime ideal Then Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A)

Proof This all follows immediately from Theorem 85 In the statement by dimAp we mean theKrull dimension of the ring

Lemma 88 Let A be a noetherian ring I a proper ideal and a isin I a regular element ThenhtI(a) = htI minus 1

28

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Proof The minimal primes over the ideal I(a) of A(a) correspond to the minimal primes over Iand we know from [Ash] Chapter 5 Corollary 548 that for any prime p containing a htp(a) =htpminus 1 so the proof is straightforward

Lemma 89 Let A be a Cohen-Macaulay local ring and I a proper ideal with htI = r ge 1 Thenwe can choose a1 ar isin I in such a way that ht(a1 ai) = i for 1 le i le r

Proof We claim that there exists a regular element a isin I Otherwise if every element of I was azero divisor on A then I would be contained in the union of the finite number of primes in Ass(A)and hence contained in some p isin Ass(M) By Theorem 85 (i) these primes are all minimal soI sube p implies htI = 0 a contradiction

Now we proceed by induction on r For r = 1 let a isin I be regular It follows from KrullrsquosPID Theorem that ht(a) = 1 Now assume r gt 1 and let a isin I be regular Then by Corollary 86the ring Aprime = A(a) is Cohen-Macaulay and by Lemma 88 htI(a) = r minus 1 so by the inductivehypothesis there are a1 arminus1 isin I with ht(a a1 ai)(a) = i for 1 le i le r minus 1 Hence

ht(a a1 ai) = i+ 1

for 1 le i le r minus 1 as required

Theorem 90 Let (Am) be a Cohen-Macaulay local ring Then

(i) For every proper ideal I of A we have

htI = depthI(A) = G(I)htI + dim(AI) = dimA

(ii) A is catenary

(iii) For every sequence a1 ar isin m the following conditions are equivalent

(1) The sequence a1 ar is A-regular

(2) ht(a1 ar) = i for 1 le i le r

(3) ht(a1 ar) = r

(4) There is a system of parameters of A containing a1 ar

Proof (iii) (1) rArr (2) is immediate by Lemma 74 (2) rArr (3) is trivial (3) rArr (4) If dimA = r ge 1then (a1 ar) must be m-primary so this is trivial If dimA gt r then m is not minimal over(a1 ar) so we can take ar+1 isin m which is not in any minimal prime ideal of (a1 ar)Then by construction ht(a1 ar+1) ge r+ 1 and therefore ht(a1 ar+1) = r+ 1 by KrullrsquosTheorem Continuing in this way we produce the desired system of parameters Note that theseimplications are true for any noetherian local ring (4) rArr (1) It suffices to show that every systemof parameters x1 xn of a Cohen-Macaulay ring A is an A-regular sequence which we do byinduction on n Let I = (x1 xn) and put Aprime = A(x1) If n = 1 and (x1) is m-primary thenit suffices to show that x1 is regular If not then x1 isin p for some p isin Ass(A) which impliesthat m = p is a minimal prime over 0 (since by Theorem 85 every prime of Ass(M) is minimal)contradicting the fact that dimA = 1 Now assume n gt 1 Since A is Cohen-Macaulay thedimensions dim(Ap) for p isin Ass(A) all agree and hence they are all equal to n = dimA Forany p isin Ass(A) the ideal p + I is m-primary since

r(I + p) = r(r(I) + r(p)) = r(m + p) = m

Thus p + Ip is an mp-primary ideal in the ring Ap which has dimension n so p + Ip cannotbe generated by fewer than n elements This shows that x1 isin p for any p isin Ass(A) and thereforex1 is A-regular Put Aprime = A(x1) By Corollary 86 Aprime is a Cohen-Macaulay ring and it hasdimension nminus1 by Proposition 80 The images of x2 xn in Aprime form a system of parameters forAprime Thus the residues x2+(x1) xn+(x1) form an Aprime-regular sequence (Aprime as an Aprime-module) by

29

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

the inductive hypothesis and therefore x2 xn is an Aprime-regular sequence (Aprime as an A-module)Hence x1 xn is an A-regular sequence and we are done

(i) Let I be a proper ideal of A If htI = 0 then there is a prime p minimal over I withhtp = 0 Since p is minimal over 0 we have p isin Ass(A) and every element of I annihilatessome nonzero element of A Therefore no A-regular sequence can exist in I and G(I) = 0 Nowassume htI = r with r ge 1 Using Lemma 89 we produce a1 ar isin I with ht(a1 ai) = ifor 1 le i le r Then the sequence a1 ar is A-regular by (iii) Hence r le G(I) Converselyif b1 bs is an A-regular sequence in I then ht(b1 bs) = s le htI by Lemma 74 HencehtI = G(I)

We first prove the second formula for prime ideals p Put dimA = depthA = n and htp = rIf r = 0 then dim(Ap) = depthA = n by Theorem 85 (i) If r ge 1 then since Ap is a Cohen-Macaulay local ring and htp = dimAp = depthp(A) we can find an A-regular sequence a1 arin p Then A(a1 ar) is a Cohen-Macaulay ring of dimension nminus r and p is a minimal primeof (a1 ar) Therefore dim(Ap) = nminus r by Theorem 85 (i) so the result is proved for primeideals Now let I be an arbitrary proper ideal with htI = r We have

dim(AI) = supdim(Ap) | p isin V (I)= supdimAminus htp | p isin V (I)

There exists a prime ideal p minimal over I with htp = r so it is clear that dim(AI) = dimAminusras required

(ii) If q sub p are prime ideals of A then since Ap is Cohen-Macaulay we have dimAp =htqAp + dimApqAp ie htpminus htq = ht(pq) Therefore A is catenary

Definition 16 We say a noetherian ring A is Cohen-Macaulay if Ap is a Cohen-Macaulay localring for every prime ideal of A A local noetherian ring is Cohen-Macaulay in this new sense iffit is Cohen-Macaulay in the original sense The Cohen-Macaulay property is stable under ringisomorphism

Lemma 91 Let A sube B be nonzero noetherian rings with B integral over A and suppose that Bis a flat A-module If A is Cohen-Macaulay then so is B

Proof Let q be a prime ideal of B and let p = q cap A By Lemma 33 Bq is flat over Ap andso using Lemma 58 it follows that depthBq(Bq) ge depthAp(Ap) = dim(Ap) By Theorem 56 wehave dim(Bq) le dim(Ap) and hence depthBq(Bq) ge dim(Bq) which shows that Bq is Cohen-Macaulay

Definition 17 Let A be a noetherian ring and I a proper ideal and let AssA(AI) = p1 psbe the associated primes of I We say that I is unmixed if htpi = htI for all i In that case all thepi are minimal and AI has no embedded primes We say that the unmixedness theorem holds inA if the following is true for r ge 0 if I is a proper ideal of height r generated by r elements thenI is unmixed Note that such an ideal is unmixed if and only if AI has no embedded primes andfor r = 0 the condition means that A has no embedded primes

Lemma 92 Let A be a noetherian ring If the unmixedness theorem holds in Am for everymaximal ideal m then the unmixedness theorem holds in A

Proof Let I be a proper ideal of height r generated by r elements with r ge 0 and let I = q1capmiddot middot middot qnbe a minimal primary decomposition with qi being pi-primary for 1 le i le n Assume that one ofthese associated primes say p1 is an embedded prime of I and let m be a maximal ideal containingp1 Arrange the qi so that the primes p1 ps are contained in m whereas ps+1 middot middot middot pn are notThen by [AM69] Proposition 49 the following is a minimal primary decomposition of the idealIAm

IAm = q1Am cap middot middot middot cap qsAm

So p1Am psAm are the associated primes of IAm Since p1 is embedded there is some1 le i le s with pi sub p1 and therefore piAm sub p1Am But this is a contradiction since IAm

has height r is generated by r elements and the unmixedness theorem holds in Am So theunmixedness theorem must hold in A

30

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Lemma 93 Let A be a noetherian ring and assume that the unmixedness theorem holds in A Ifa isin A is regular then the unmixedness theorem holds in A(a)

Proof Let I be a proper ideal of A containing a and supppose the ideal I(a) has height r andis generated by r elements in A(a) By Lemma 88 the ideal I has height r + 1 and is clearlygenerated by r+1 elements in A Therefore I is unmixed If p1 pn are the associated primesof I then the associated primes of I(a) are p1(a) pn(a) Since htp(a) = htp minus 1 =htI minus 1 = htI(a) it follows that I(a) is unmixed as required

Lemma 94 Let A be a noetherian ring and assume that the unmixedness theorem holds inA Then if I is a proper ideal with htI = r ge 1 we can choose a1 ar isin I such thatht(a1 ai) = i for 1 le i le r

Proof The proof is the same as Lemma 89 except we use the fact that 0 has no embedded primesto show I contains a regular element and we use Lemma 93

Theorem 95 Let A be a noetherian ring Then A is Cohen-Macaulay if and only if the unmixed-ness theorem holds in A

Proof Suppose the unmixedness theorem holds in A and let p be a prime ideal of height r ge 0We know that r = dimAp ge depth(Ap) ge depthpA by Lemma 72 If r = 0 then no regularelement can exist in p so depthpA = 0 and consequently dimAp = 0 = depth(Ap) so Ap is Cohen-Macaulay If r ge 1 then by Lemma 94 we can find a1 ar isin p such that ht(a1 ai) = i for1 le i le r The ideal (a1 ai) is unmixed by assumption so ai+1 lies in no associated primesof A(a1 ai) Thus a1 ar is an A-regular sequence in p so depthpA ge r and consequentlydimAp = r = depth(Ap) so again Ap is Cohen-Macaulay Hence A is a Cohen-Macaulay ring

Conversely suppose A is Cohen-Macaulay It suffices to show that the unmixedness theoremholds in Am for all maximal m so we can reduce to the case where A is a Cohen-Macaulay localring We know from Theorem 85 that 0 is unmixed Let (a1 ar) be an ideal of height r gt 0Then a1 ar is an A-regular sequence by Theorem 90 hence A(a1 ar) is Cohen-Macaulayand so (a1 ar) is unmixed

Corollary 96 A noetherian ring A is Cohen-Macaualy if and only if Am is a Cohen-Macaulaylocal ring for every maximal ideal m

Proof This follows immediately from Theorem 95 and Lemma 92

Corollary 97 Let A be a Cohen-Macaulay ring If a1 ar isin A are such that ht(a1 ai) = ifor 1 le i le r then a1 ar is an A-regular sequence

Theorem 98 Let A be a Cohen-Macaulay ring Then the polynomial ring A[x1 xn] is alsoCohen-Macaulay Hence any Cohen-Macaulay ring is universally catenary

Proof It is enough to consider the case n = 1 Let q be a prime ideal of B = A[x] and putp = q cap A We have to show that Bq is Cohen-Macaulay It follows from Lemma 10 that Bq isisomorphic to Ap[x]qAp[x] where qAp[x] is a prime ideal of Ap[x] contracting to pAp Since Ap

is Cohen-Macaulay we can reduce to showing Bq is Cohen-Macaulay in the case where A is aCohen-Macaulay local ring and p = q cap A is the maximal ideal Then BpB sim= k[x] where k is afield Therefore we have either q = pB or q = pB+fB where f isin B = A[x] is a monic polynomialof positive degree By Theorem 55 we have (Krull dimensions)

dim(Bq) = dim(A) + ht(qpB)

If q = pB then this implies that dim(Bq) = dim(A) So to show Bq is Cohen-Macaulay it sufficesto show that depthBq(Bq) ge dimA If dimA = 0 this is trivial so assume dimA = r ge 1 and leta1 ar be an A-regular sequence As B is flat over A so is Bq and therefore a1 ar is alsoa Bq-regular sequence by Lemma 58 It is then not difficult to check that the images of the ai inBq form a Bq-regular sequence so depthBq(Bq) ge r as required

31

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

If q = pB + fB then dim(Bq) = dim(A) + 1 (since every nonzero prime in k[x] has height1) and so it suffices to show that depthBq(Bq) ge dim(A) + 1 If dimA = 0 then since f ismonic it is clearly regular in B and therefore also in Bq which shows that depthBq(Bq) ge 1 IfdimA = r ge 1 let a1 ar be an A-regular sequence Since f is monic it follows that f is regularon B(a1 ar)B Therefore a1 ar f is a B-regular sequence Applying Lemma 58 we seethat this sequence is also Bq-regular and therefore the images in Bq form a Bq-regular sequenceThis shows that depthBq(Bq) ge r + 1 as required

It follows from Lemma 8 and Theorem 90 that any Cohen-Macaulay ring is catenary Thereforeif A is Cohen-Macaulay A[x1 xn] is catenary for n ge 1 and so any Cohen-Macaulay ring isuniversally catenary

Corollary 99 If k is a field then k[x1 xn] is Cohen-Macaulay and therefore universallycatenary for n ge 1

6 Normal and Regular Rings

61 Classical Theory

Definition 18 We say that an integral domain A is normal if it is integrally closed in its quotientfield The property of being normal is stable under ring isomorphism If an integral domain A isnormal then so is Sminus1A for any multiplicatively closed subset S of A not containing zero

Proposition 100 Let A be an integral domain Then the following are equivalent

(i) A is normal

(ii) Ap is normal for every prime ideal p

(iii) Am is normal for every maximal ideal m

Proof See [AM69] Proposition 513

Definition 19 Let A be an integral domain with quotient field K An element u isin K is almostintegral over A if there exists a nonzero element a isin A such that aun isin A for all n gt 0

Lemma 101 If u isin K is integral over A then it is almost integral over A The elements of Kalmost integral over A form a subring of K containing the integral closure of A If A is noetherianthen u isin K is integral if and only if it is almost integral

Proof It is clear that any element of A is almost integral over A Let u = bt isin K with b t isin Anonzero be integral over A and let

un + a1unminus1 + middot middot middot+ a1u+ a0 = 0

be an equation of integral dependence We claim that tnum isin A for any m gt 0 If m le n this istrivial and if m gt n then we can write um as an A-linear combination of strictly smaller powersof u so tnum isin A in this case as well It is easy to check that the almost integral elements forma subring of K

Now assume that A is noetherian and let u be almost integral over A If a is nonzero andaun isin A for n ge 1 then A[u] is a submodule of the finitely generated A-module aminus1A whenceA[u] itself is finitely generated over A and so u is integral over A

Definition 20 We say that an integral domain A is completely normal if every element u isin Kwhich is almost integral over A belongs to A Clearly a completely normal domain is normal andfor a noetherian ring domain normality and complete normality coincide The property of beingcompletely normal is stable under ring isomorphism

Example 5 Any field is completely normal and if k is a field then the domain k[x1 xn] iscompletely normal since it is noetherian and normal

32

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Definition 21 We say that a ring A is normal if Ap is a normal domain for every prime ideal pAn integral domain is normal in this new sense iff it is normal in the original sense The propertyof being normal is stable under ring isomorphism

Lemma 102 Let A be a ring and suppose that Ap is a domain for every prime ideal p Then Ais reduced In particular a normal ring is reduced

Proof Let a isin A be nilpotent For any prime ideal p we have a1 = 0 in Ap so ta = 0 for somet isin p Hence Ann(a) cannot be a proper ideal and so a = 0

Lemma 103 Let A1 An be normal domains Then A1 times middot middot middot timesAn is a normal ring

Proof Let A = A1 times middot middot middot times An A prime ideal p of A is A1 times middot middot middot times pi times middot middot middotAn for some 1 le i le nand prime ideal pi of Ai Moreover Ap

sim= (Ai)pi which by assumption is a normal domain Hence

A is a normal ring

Proposition 104 Let A be a completely normal domain Then a polynomial ring A[x1 xn]is also completely normal In particular k[x1 xn] is completely normal for any field k

Proof It is enough to treat the case n = 1 Let K denote the quotient field of A Then thecanonical injective ring morphism A[x] minusrarr K[x] induces an isomorphism between the quotientfield of A[x] and K(x) the quotient field of K[x] so we consider all our rings as subrings of K(x)Let 0 6= u isin K(x) be almost integral over A[x] Since A[x] sube K[x] and K[x] is completely normalthe element u must belong to K[x] Write

u = αrxr + αr+1x

r+1 + middot middot middot+ αdxd

for some r ge 0 and αr 6= 0 Let f(x) = bsxs+bs+1x

s+1+ middot middot middot+btxt isin A[x] with bs 6= 0 be such thatfun isin A[x] for all n gt 0 Then bsαnr isin A for all n so that αr isin A Then uminusαrxr = αr+1x

r+1+ middot middot middotis almost integral over A[x] so we get αr+1 isin A as before and so on Therefore u isin A[x]

Proposition 105 Let A be a normal ring Then A[x1 xn] is normal

Proof It suffices to consider the case n = 1 Let q be a prime ideal of A[x] and let p = q cap AThen A[x]q is a localisation of Ap[x] at a prime ideal and Ap is a normal domain So we reduceto the case where A is a normal domain with quotient field K As before we identify the quotientfield of A[x] with K(x) the quotient field of K[x] We have to prove that A[x] is integrally closedin K(x) Let u = p(x)q(x) with p q isin A[x] be a nonzero element of K(x) which is integral overA[x] Let

ud + f1(x)udminus1 + middot middot middot+ fd(x) = 0 fi isin A[x]

be an equation of integral dependence In order to prove that u isin A[x] consider the subring A0

of A generated by 1 and the coefficients of p q and all the fi Identify A0 A0[x] and the quotientfield of A0[x] with subrings of K(x) Then u is integral over A0[x] The proof of Proposition 104shows that u belongs to K[x] and moreover

u = αrxr + middot middot middot+ αdx

d

where each coefficient αi isin K is almost integral over A0 As A0 is noetherian αi is integral overA0 and therefore integral over A Therefore αi isin A which is what we wanted

Let A be a ring and I an ideal with⋂infinn=1 I

n = 0 Then for each nonzero a isin A there isan integer n ge 0 such that a isin In and a isin In+1 We then write n = ord(a) (or ordI(a))and call it the order of a with respect to I We have ord(a + b) ge minord(a) ord(b) andord(ab) ge ord(a) + ord(b) Put Aprime = grI(A) =

oplusnge0 I

nIn+1 For an element a of A withord(a) = n we call the sequence in Aprime with a single a in InIn+1 the leading form of a anddenote it by alowast Clearly alowast 6= 0 We define 0lowast = 0 The map a 7rarr alowast is in general not additiveor multiplicative but for nonzero a b if alowastblowast 6= 0 (ie if ord(ab) = ord(a) + ord(b)) then we have(ab)lowast = alowastblowast and if ord(a) = ord(b) and alowast + blowast 6= 0 then we have (a+ b)lowast = alowast + blowast

33

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Theorem 106 (Krull) Let A be a nonzero ring I an ideal and grI(A) the associated gradedring Then

(1) If⋂infinn=1 I

n = 0 and grI(A) is a domain so is A

(2) Suppose that A is noetherian and that I is contained in the Jacobson radical of A Then ifgrI(A) is a normal domain so is A

Proof We denote the ring grI(A) by Aprime for convenience (1) Let a b be nonzero elements of AThen alowast 6= 0 and blowast 6= 0 hence alowastblowast 6= 0 and therefore ab 6= 0

(2) Since I is contained in the Jacobson radical it is immediate that⋂infinn=1 I

n = 0 (see [AM69]Corollary 1019) and so by (1) the ring A is a domain Let K be the quotient field of A andsuppose we are given nonzero a b isin A with ab integral over A We have to prove that a isin bAThe A-module AbA is separated in the I-adic topology by Corollary 1019 of A amp M In otherwords

bA =infin⋂n=1

(bA+ In)

Therefore it suffices to prove the following for every n ge 1

(lowast) For nonzero a b isin A with ab integral over A if a isin bA+ Inminus1 then a isin bA+ In

Suppose that a isin bA + Inminus1 for some n ge 1 Then a = br + aprime with r isin A and aprime isin Inminus1 andaprimeb = abminus r is integral over A If aprime = 0 then a = br and we are done Otherwise we can reduceto proving (lowast) in the case where a isin Inminus1

So we are given an integer n ge 1 nonzero a b with a isin Inminus1 and ab integral over A and wehave to show that a isin bA + In Since ab is almost integral over A there exists nonzero c isin Asuch that cam isin bmA for all m gt 0 Since Aprime is a domain the map a 7rarr alowast is multiplicative hencewe have clowast(alowast)m isin (blowast)mAprime for all m and since Aprime is noetherian (see Proposition 1022 of A ampM) and normal we have alowast isin blowastAprime Therefore we can find d isin A with alowast = blowastdlowast If a isin In thenwe would be done so suppose a isin In and therefore ord(a) = n minus 1 Since alowast = blowastdlowast the residueof aminus bd in Inminus1In is zero and therefore aminus bd isin In Hence a isin bA+ In as required

Definition 22 Let (Am k) be a noetherian local ring of dimension d Recall that the ring Ais said to be regular if m can be generated by d elements or equivalently if rankkmm2 = dRegularity is stable under ring isomorphism

Recall that if k is a field a graded k-algebra is a k-algebra R which is also a graded ring insuch a way that the graded pieces Rd are k-submodules for every d ge 0 A morphism of gradedk-algebras is a morphism of graded rings which is also a morphism of k-modules

Theorem 107 Let (Am k) be a noetherian local ring of dimension d Then A is regular ifand only if the graded ring grm(A) =

oplusmnmn+1 is isomorphic as a graded k-algebra to the

polynomial ring k[x1 xd]

Proof The first summand in grm(A) is the field k = Am so this ring becomes a graded k-algebrain a canonical way For d = 0 we interpret the statement as saying A is regular iff grm(A) isisomorphic as a graded k-algebra to k itself See the section in [AM69] on regular local rings forthe proof

Theorem 108 Let A be a regular local ring of dimension d Then

(1) A is a normal domain

(2) A is a Cohen-Macaulay local ring

If d ge 1 and a1 ad is a regular system of parameters then

(3) a1 ad is an A-regular sequence

34

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

(4) pi = (a1 ai) is a prime ideal of height i for each 1 le i le d and Api is a regular localring of dimension dminus i

(5) Conversely if I is a proper ideal of A such that AI is regular and has dimension d minus ifor some 1 le i le d then there exists a regular system of parameters y1 yd such thatI = (y1 yi) In particular I is prime

Proof (1) Follows from Theorems 106 and 107(2) If d = 0 this is trivial and if d ge 1 this follows from (3) below(3) From the proof of [AM69] Theorem 1122 we know that there is an isomorphism of graded k-

algebras ϕ k[x1 xd] minusrarr grm(A) defined by xi 7rarr ai isin mm2 If f(x1 xd) is homogenousof degree m ge 0 then ϕ(f) is the element

sumα a

α11 middot middot middot aαn

n f(α) of mmmm+1 So ϕ agrees with themorphism of abelian groups defined in Proposition 63 (c) Thus a1 ad is an A-quasiregularsequence It then follows from Corollary 65 that a1 ad is an A-regular seqence

(4) We have dim(Api) = d minus i for 1 le i le d by Proposition 51 and hence htpi = i by (2)and Theorem 90 (i) The ring Apd is a field and therefore trivially a regular local ring of thecorrect dimension If i lt d then the maximal ideal mpi of Api is generated by d minus i elementsxi+1 xd Therefore Api is regular and hence pi is prime by (1)

(5) Let A = AI and put m = mI Then we can identify k with Am and there is clearly anisomorphism of k-modules

m2(m2 + I) sim= mm2

So we havedminus i = rankkmm

2 = rankkm(m2 + I)

Since I sube m the A-module (m2 + I)m2 is canonically a k-module and we have a short exactsequence of k-modules

0 minusrarr (m2 + I)m2 minusrarr mm2 minusrarr m(m2 + I) minusrarr 0

Consequently d minus i = rankkmm2 minus rankk(m2 + I)m2 and therefore rankk(m2 + I)m2 = i

Thus we can choose i elements y1 yi of I which span m2 + I mod m2 over k and d minus ielements yi+1 yd of m which together with y1 yi span m mod m2 over k (if i = d thenthe original y1 yi will do) Then y1 yd is a regular system of parameters of A so that(y1 yi) = p is a prime ideal of height i by (4) Since p sube I and dim(AI) = dim(Ap) = dminus iwe must have I = p

Let A be an integral domain with quotient field K A fractional ideal is an A-submodule of KIf MN are two fractional ideals then so is M middotN =

summini |mi isinMni isin N This product is

associative commutative and M middotA = M for any fractional ideal M For any nonzero ideal a of Awe put aminus1 = x isin K |xa sube A Then aminus1 is a fractional ideal and we have A sube aminus1 Moreovera middot aminus1 sube A is an ideal of A

Lemma 109 Let A be a noetherian domain with quotient field K Let a be a nonzero elementof A and p isin AssA(A(a)) Then pminus1 6= A

Proof By definition of associated primes there is b isin (a) with p = ((a) b) Then (ba)p sube A andba isin A

Lemma 110 Let (Am) be a noetherian local domain such that m 6= 0 and mmminus1 = A Then mis a principal ideal and so A is regular of dimension 1

Proof By assumption we have dimA ge 1 By [AM69] Proposition 86 it follows that m 6= m2Take a isin mminusm2 Then amminus1 sube A and if amminus1 sube m then (a) = amminus1m sube m2 contradicting thechoice of a Since amminus1 is an ideal we must have amminus1 = A that is (a) = amminus1m = m Usingthe dimension theory of noetherian local rings we see that dimA le 1 and therefore A is regularof dimension 1

Theorem 111 Let (Am) be a noetherian local ring of dimension 1 Then A is regular iff it isnormal

35

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Proof If A is regular then it is a normal domain by Theorem 108 Now suppose that A is normal(hence a domain since A sim= Am) By Lemma 110 to show A is regular it suffices to show thatmmminus1 = A Assume the contrary Then mmminus1 is a proper ideal and since 1 isin mminus1 we havem sube mmminus1 hence mmminus1 = m Let a1 an be generators for m (since dimA ge 1 we can assumeall ai 6= 0) and let a isin mminus1 Since aai isin m for all i we have coefficients rij isin A 1 le i j le n andequations aai = ri1a1 + middot middot middot+ rinan Collecting terms we have

0 = (r11 minus a)a1 + middot middot middot+ r1nan

0 = r21a1 + (r21 minus a)a2 + middot middot middot+ r2nan

0 = rn1a1 + middot middot middot+ (rnn minus a)an

The determinant of the coefficient matrix B = (rij minus δij middot a) must satisfy detB middot ai = 0 and thusdetB = 0 since A is a domain This gives an equation of integral dependence of a over A whencemminus1 = A since A is integrally closed But since dimA = 1 we have m isin Ass(A(b)) for anynonzero b isin m so that mminus1 6= A by Lemma 109 Thus mmminus1 = A cannot occur

Theorem 112 Let A be a noetherian normal domain Then as subrings of the quotient field Kof A we have

A =⋂

htp=1

Ap

Moreover any nonzero proper principal ideal in A is unmixed and if dim(A) le 2 then A isCohen-Macaulay

Proof Suppose 0 6= a is a nonunit of A and p isin Ass(A(a)) We claim that htp = 1 ReplacingA by Ap we may assume that A is local with maximal ideal p (since pAp = ((a1) (b1)) Thenwe have pminus1 6= A by Lemma 109 If htp gt 1 then ppminus1 = A since otherwise we can run the proofof Theorem 111 and obtain a contradiction (in that proof we only use dimA = 1 to show thatm 6= 0 and that m isin Ass(A(b)) for some nonzero b isin m) But then Lemma 110 implies that A isregular of dimension 1 contradicting the fact that htp gt 1 Hence htp = 1 which shows that theideal (a) is unmixed

Now suppose x isin Ap for all primes of height 1 and write x = ab We need to show that x isin Aso we can assume that b is not a unit and a isin (b) The ideal ((b) a) is the annihilator of thenonzero element a+ (b) of A(b) The set of annihilators of nonzero elements of A(b) containing((b) a) has a maximal element since A is noetherian and by Lemma 47 this maximal elementis a prime ideal p = ((b) h) for some h isin (b) By definition p isin Ass(A(b)) and thus htp = 1Since ab isin Ap we have ab = cs for some s isin p Then sa = bc isin (b) so s isin ((b) a) sube p whichis a contradiction Hence we must have had a isin (b) and thus x isin A to begin with

Now suppose that A is a noetherian normal domain with dim(A) le 2 By Theorem 95 itis enough to show that the unmixedness theorem holds in A Since A is a domain it is clearthat 0 has no embedded primes and we have just shown that every proper principal ideal ofheight 1 is unmixed If I = (a1 a2) is a proper ideal of height 2 then every associated primep of I has htp ge 2 but also htp le 2 since dim(A) = 2 Therefore I is unmixed and A isCohen-Macaulay

Definition 23 Let A be a nonzero noetherian ring Consider the following conditions about Afor k ge 0

(Sk) For every prime p of A we have depth(Ap) ge infk htp

(Rk) For every prime p of A if htp le k then Ap is regular

The condition (S0) is trivial and for every k ge 1 we have (Sk) rArr (Skminus1) and (Rk) rArr (Rkminus1)

For a nonzero noetherian ring A we can express (Sk) differently as follows for every primep if htp le k then depth(Ap) ge htp and otherwise depth(Ap) ge k We introduce the followingauxiliary condition for k ge 1

36

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

(Tk) For every prime p of A if htp ge k then depth(Ap) ge k

It is not hard to see that for k ge 1 the condition (Sk) is equivalent to (Ti) being satisfied for all1 le i le k

Proposition 113 Let A be a nonzero noetherian ring Then

(S1) hArr Ass(A) has no embedded primes hArr every prime p with htp ge 1 contains aregular element

(S2) hArr (S1) and Ass(AfA) has no embedded primes for any regular nonunit f isin A

The ring A is Cohen-Macaulay iff it satisfies (Sk) for all k ge 0

Proof For a noetherian ring A with prime ideal p we have depth(Ap) = 0 iff pAp isin Ass(Ap)which by [Ash] Chapter 1 Lemma 142 is iff p isin Ass(A) So the associated primes are preciselythose with depth(Ap) = 0 A prime p isin Ass(A) is embedded iff htp ge 1 so saying thatAss(A) has no embedded primes is equivalent to saying that if p isin Spec(A) and htp ge 1 thendepth(Ap) ge 1 Hence the first two statements are equivalent If Ass(A) has no embedded primesand htp ge 1 then p must contain a regular element since otherwise by [Ash] Chapter 1 Theorem136 p is contained in an associated prime of A and these all have height zero Conversely ifevery prime of height ge 1 contains a regular element then certainly no prime of height ge 1 canbe an associated prime of A so Ass(A) has no embedded primes

To prove the second statement we assume A is a nonzero noetherian ring satisfying (S1) andshow that (S2) is equivalent to Ass(AfA) having no embedded primes Suppose A satisfies (S2)and let a regular nonunit f be given If p isin Ass(AfA) then the following Lemma implies thathtp ge depth(Ap) = 1 and p is a minimal prime iff htp = 1 So the condition (T2) shows thatAss(AfA) can have no embedded primes Conversely suppose p is a prime ideal with htp ge 2not satisfying (T2) Since A has no embedded primes this can only happen if depth(Ap) = 1 Butthen by the following Lemma p isin Ass(AfA) for some regular f isin A Since htp ge 2 this is anembedded prime which is impossible

If A is Cohen-Macaulay then htp = depth(Ap) for every prime p so clearly (Sk) is satisfiedfor k ge 0 Conversely if A satisfies every (Sk) then by choosing k large enough we see thatdepth(Ap) ge htp for every prime p and hence A is Cohen-Macaulay

Lemma 114 Let A be a nonzero noetherian ring satisfying (S1) Then for a prime p the followingare equivalent

(i) depth(Ap) = 1

(ii) There exists a regular element f isin p with p isin Ass(AfA)

If f isin p is regular and p isin Ass(AfA) then p is a minimal prime of Ass(AfA) if and only ifhtp = 1

Proof Let f isin p be a regular element Then f1 isin Ap is regular and it is not hard to see thereis an isomorphism of Ap-modules ApfAp

sim= (AfA)p Note also that

depth(ApfAp) = depth(Ap)minus 1 (4)

(i) rArr (ii) Since htp = dim(Ap) ge depth(Ap) we have htp ge 1 and therefore since A satisfies (S1)there is a regular element f isin p The above shows that depth(ApfAp) = depth((AfA)p) = 0and therefore by Lemma 71 p isin Ass(AfA) as required (ii) rArr (i) follows from Lemma 71 and(4) If (i) is satisfied then the above proof shows that p is an associated prime of AfA for anyregular f isin p

Suppose p is a minimal prime of Ass(AfA) Then by (i) depth(Ap) = 1 and since p isa minimal prime over fA it follows from Krullrsquos PID Theorem that htp = 1 Conversely ifdepth(Ap) = htp = 1 then clearly p is minimal over fA

37

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Proposition 115 Let A be a nonzero noetherian ring Then A is reduced iff it satisfies (R0)and (S1)

Proof Suppose that A is reduced Then Lemma 13 shows that A satisfies (R0) Suppose that Adoes not satisfy (S1) Let p be an associated prime of A which is not minimal so htp ge 1 andp = Ann(b) for some nonzero b isin A Then Ap is a reduced noetherian ring in which every elementis either a unit or a zero-divisor so by Lemma 12 we must have dim(A) = 0 which contradictsthe fact that htp ge 1 Therefore A must satisfy (S1)

Now suppose that A satisfies (R0) and (S1) Let a isin A be nonzero and nilpotent By Lemma47 there is an associated prime p isin Ass(A) with Ann(a) sube p By (S1) we have htp = 0 andtherefore Ap is a field by (R0) Since a1 isin Ap is nilpotent we have ta = 0 for some t isin p whichis a contradiction Hence A is reduced

If A is a nonzero ring the set S of all regular elements is a multiplicatively closed subset LetΦA denote the localisation Sminus1A which we call the total quotient ring of A If A is a domainthis is clearly the quotient field

Theorem 116 (Criterion of Normality) A nonzero noetherian ring A is normal if and onlyif it satisfies (S2) and (R1)

Proof Let A be a nonzero noetherian ring Suppose first that A is normal and let p be a primeideal Then Ap is a field for htp = 0 and regular for htp = 1 by Theorem 111 hence the condition(R1) is satisfied Since A is normal it is reduced so it satisfies (S1) by Proposition 115 To showA satisfies (S2) it suffices by Proposition 113 to show that Ass(AfA) has no embedded primesfor any regular nonunit f Let f be a regular nonunit with associated primes

Ass(AfA) = p1 pn

Suppose wlog that p = p1 is an embedded prime and that p1 pi are the associated primescontained in p1 Since ApfAp

sim= (AfA)p we have

AssAp(ApfAp) = pAp p2Ap piAp

by [Ash] Chapter 1 Lemma 142 At least one of the pi is properly contained in p so pAp is anembedded prime of Ass(ApfAp) But since Ap is a noetherian normal domain this contradictsTheorem 112 Hence A satisfies (S2)

Next suppose that A satisfies (S2) and (R1) Then it also satisfies (R0) and (S1) so it isreduced Let p1 pr be the minimal prime ideals of A Then we have 0 = p1 cap middot middot middot cap pr LetS be the set of all regular elements in A Then by Proposition 113 the pi are precisely the primeideals of A avoiding S Therefore ΦA = Sminus1A is an artinian ring and since (Sminus1A)Sminus1pi

sim= Api

Proposition 14 gives an isomorphism of rings

θ ΦA minusrarrsprodj=1

Aqj as 7rarr (as as)

where for each i qi isin pi pr (some of the pi may more than once or not at all among theqj) Also note that by Lemma 13 for each j the ring Kj = Aqj

is a field For each j let Tjbe image of the ring morphism A minusrarr Kj Then taking the product gives a subring

prodj Tj ofprod

j Kj which contains the image of A under θ Let e1 es be the preimage in ΦA of the tuples(1 0 0) (0 0 1) These clearly form a family of orthogonal idempotents in ΦA

Suppose that we could show that A was integrally closed in ΦA For each j the element ejsatisfies e2jminusej = 0 so ej isin A We claim that θ identifies the subrings A and

prodj Tj It is enough to

show that θ maps the former subring onto the latter If a1 as isin A give a tuple (a11 as1)ofprodj Tj then since ej isin A we have e1a1 + middot middot middot + esas isin A and since θ(e1) = (1 0 0) and

similarly for the other ej it is clear that

θ(e1a1 + middot middot middot+ esas) = (a11 as1)

38

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

as required Since A is integrally closed in ΦA it is straightforward to check that each Tj isintegrally closed in Kj and is therefore a normal domain Hence A is isomorphic to a directproduct of normal domains so A is a normal ring by Lemma 103

So it only remains to show that A is integrally closed in ΦA Suppose we have an equation ofintegral dependence in ΦA

(ab)n + c1(ab)nminus1 + middot middot middot+ cn = 0

where a b and the ci are elements of A and b is A-regular Then an +sumni=1 cia

nminusibi = 0 Wewant to prove that a isin bA so we may assume b is a regular nonunit of A To show that a isin bAit suffices to show that ap isin bpAp for every associated prime p of bA (here ap denotes a1 isin Ap)Since bA is unmixed of height 1 by (S2) it suffices to prove this for primes p with htp = 1 By(R1) if htp = 1 then Ap is regular and therefore by Theorem 108 a normal domain But in thequotient field of Ap we have

anp +nsumi=1

(ci)panminusip bip = 0

If bp = 0 then clearly ap = 0 Otherwise apbp is integral over Ap and so ap isin bpAp asrequired

Corollary 117 A nonzero normal noetherian ring A is isomorphic to a finite direct product ofnormal domains

Theorem 118 Let A be a ring such that for every prime ideal p the localisation Ap is regularThen the polynomial ring A[x1 xn] over A has the same property

Proof As in the proof of Theorem 98 we reduce to the case where (A p) is a regular local ringn = 1 and q is a prime ideal of B = A[x] lying over p And we have to prove that Bq is regularWe have q supe pB and BpB sim= k[x] where k is a field Therefore either q = pB or q = pB + fBwhere f isin B = A[x] is a monic polynomial of positive degree Put dim(A) = d ge 0 Then p isgenerated by d elements so if q = pB then q is generated by d elements and by d + 1 elementsif q = pB + fB From [Ash] Chapter 5 we know that htpB = htp = d (use Propositions 563and 543) On the other hand if q = pB + fB then by Krullrsquos Theorem htq le d+ 1 and since qcontains p properly we must have htq = d+ 1 This shows that Bq is regular

Corollary 119 If k is a field then k[x1 xn]p is a regular local ring for every prime ideal pof k[x1 xn]

62 Homological Theory

The following results are proved in our Dimension notes

Proposition 120 Let A be a ring M an A-module Then

(i) M is projective iff Ext1A(MN) = 0 for all A-modules N

(ii) M is injective iff Ext1A(NM) = 0 for all A-modules N

(iii) M is injective iff Ext1A(AIM) = 0 for all ideals I of A

(iv) M is flat iff TorA1 (AIM) = 0 for all finitely generated ideals I

(v) M is flat iff TorA1 (NM) = 0 for all finitely generated A-modules N

So injectivity is characterised by vanishing of Ext1A(minusM) and we can restrict considerationto ideal quotients in the first variable Flatness is characterised by vanishing of Tor1A(minusM) (orequivalently Tor1A(Mminus)) and we can restrict consideration to ideal quotients or finitely generatedmodules The next result shows that the projectivity condition can also be restricted to a specialclass of modules

39

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Lemma 121 Let A be a noetherian ring and M a finitely generated A-module Then M isprojective if and only if Ext1A(MN) = 0 for all finitely generated A-modules N

Proof Take an exact sequence 0 minusrarr R minusrarr F minusrarr M minusrarr 0 with F finitely generated andfree Then R is finitely generated so by assumption Ext1A(MR) = 0 Thus the sequenceHom(FR) minusrarr Hom(RR) minusrarr 0 is exact It follows that R minusrarr F is a coretraction so that Mis a direct summand of a free module

If A is a nonzero ring then the global dimension of A denoted gldim(A) is the largestinteger n ge 0 for which there exists modules MN with ExtnA(MN) 6= 0 The Tor dimensionof A denoted tordim(A) is the largest integer n ge 0 for which there exists modules MN withTorAn (MN) 6= 0 We know from our Dimension notes that

gldim(A) = supprojdimM |M isin AMod= supinjdimM |M isin AMod= supprojdimAI | I a left ideal of A

and

tordim(A) = supflatdimM |M isin AMod= supflatdimAI | I is a left ideal of A

Proposition 122 Let A be a noetherian ring Then tordim(A) = gldim(A) and for everyfinitely generated A-module M flatdimM = projdimM

Proof See our Dimension notes

Lemma 123 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen for n ge 0

projdimM le n lArrrArr TorAn+1(Mk) = 0

In particular if M is nonzero then projdimM is the largest n ge 0 such that TorAn (Mk) 6= 0

Proof This is trivial if M = 0 so assume M is nonzero Since flatdimM le projdimM theimplication rArr is clear We prove the converse by induction on n Let m = rankk(MmM) Thenm ge 1 since M is nonzero and by Nakayama we can find elements u1 um which generate Mas an A-module and map to a k-basis in MmM Let ε Am minusrarr M be induced by the elementsui and let K be the kernel of ε which is finitely generated since A is noetherian So we have anexact sequence

0 minusrarr K minusrarr Am minusrarrM minusrarr 0

It follows that projdimM le projdimK + 1 If n gt 0 then using the long exact Tor sequencewe see that TorAn+1(Mk) sim= TorAn (K k) which proves the inductive step So it only remains toconsider the case n = 0 Then by assumption TorA1 (Mk) = 0 so the top row in the followingcommutative diagram of A-modules is exact

0 K otimesA k

Am otimesA kεotimes1

M otimesA k

0

0 KmK km MmM 0

By construction km minusrarrMmM is the morphism of k-modules corresponding to the basis definedby the ui so it is an isomorphism Hence KmK = 0 so K = 0 by Nakayamarsquos Lemma HenceM sim= Am and so projdimM = 0

40

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Remark 2 Let A be a ring and M an A-module By localising any finite projective resolutionof M we deduce that projdimApMp le projdimAM for any prime ideal p Given an Ap-moduleN we have N sim= Np as Ap-modules and it follows that gldim(Ap) le gldim(A)

Lemma 124 Let A be a nonzero noetherian ring and M a finitely generated A-module Then

(i) projdimM = supprojdimAmMm |m a maximal ideal of A

(ii) For n ge 0 projdimM le n if and only if TorAn+1(MAm) = 0 for every maximal ideal m

(iii) For every maximal ideal m gldim(Am) le gldim(A) Moreover

gldim(A) = supgldim(Am) |m a maximal ideal of A

Proof (i) This is trivial if M = 0 so assume M is nonzero For any module N and maximal idealm we know from Lemma 22 that there is an isomorphism of Am-modules for n ge 0

ExtnA(MN)msim= ExtnAm

(Mm Nm)

The module ExtnA(MN) is nonzero if and only if some ExtnAm(Mm Nm) is nonzero and projdimM

is the largest integer n ge 0 for which there exists a module N with ExtnA(MN) 6= 0 so the claimis easily checked

(ii) Let n ge 0 Then by (i) projdimM le n if and only if projdimAmMm le n for everymaximal ideal m Since AmmAm

sim= (Am)m as Am-modules we can use Lemma 22 and Lemma123 to see that this if and only if for every maximal ideal m

0 = TorAmn+1(Mm AmmAm) sim= TorAn+1(MAm)m

If m n are distinct maximal ideals then (Am)n = 0 so TorAn+1(MAm) = 0 if and only ifTorAn+1(MAm)m = 0 which completes the proof

(iii) For any maximal ideal m and Am-module N there is an isomorphism of Am-modulesN sim= Nm so using (i) and the fact that gldim(A) = supprojdimM the various claims are easyto check

Theorem 125 Let (Am k) be a noetherian local ring Then for n ge 0

gldim(A) le n lArrrArr TorAn+1(k k) = 0

Consequently we have gldim(A) = projdimA(k)

Proof Since tordim(A) = gldim(A) the implication rArr is immediate If TorAn+1(k k) = 0 thenprojdimA(k) le n by Lemma 123 Hence TorAn+1(Mk) = 0 for all modules M so by Lemma 123projdimM le n for every finitely generated module M Hence gldim(A) le n Using Lemma 123again we see that gldim(A) = projdimA(k)

Proposition 126 Let (Am k) be a noetherian local ring and M a nonzero finitely generatedA-module If projdimM = r ltinfin and if x isin m is M -regular then projdim(MxM) = r + 1

Proof By assumption the following sequence of A-modules is exact

0 Mx M MxM 0

Therefore the sequence 0 minusrarr TorAi (MxM k) minusrarr 0 is exact and so TorAi (MxM k) = 0 fori gt r + 1 The following sequence of A-modules is also exact

0 = TorAr+1(Mk) TorAr+1(MxM k) TorAr (Mk) x TorAr (Mk)

where x denotes left multiplication by x which is equal to TorAr (x k) and also TorAr (Mx) (see ourTor notes) Since k = Am is annihilated by x so is TorAr (Mk) Therefore TorAr+1(MxM k) sim=TorAr (Mk) 6= 0 and hence projdim(MxM) = r + 1 by Lemma 123

41

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Corollary 127 Let (Am k) be a noetherian local ring M a nonzero finitely generated A-moduleand a1 as an M -regular sequence If projdimM = r ltinfin then projdim(M(a1 as)) =r + s

Proof Since A is local and (a1 as)M 6= M we have ai isin m for each i We proceed by inductionon s The case s = 1 was handled by Proposition 126 If s gt 1 then set N = M(a1 asminus1)M Then as isin m is N -regular and by the inductive hypothesis projdimN = r + s minus 1 lt infin So bythe case s = 1 projdim(NasN) = r + s and NasN sim= M(a1 as)M so we are done

Theorem 128 Let (Am k) be a regular local ring of dimension d Then gldim(A) = d

Proof If d = 0 then A is a field and trivially gldim(A) = 0 Otherwise let a1 ad be aregular system of parameters Then the sequence a1 ad is A-regular by Theorem 108 and k =A(a1 ad) so projdimk = d by Corollary 127 Theorem 125 implies that gldim(A) = d

Among many other things Theorem 128 allows us to give a much stronger version of Lemma121 for regular local rings

Corollary 129 Let (Am k) be a regular local ring of dimension d and M a finitely generatedA-module Then

(i) M is projective if and only if Exti(MA) = 0 for i gt 0

(ii) For n ge 0 we have projdimM le n if and only if Exti(MA) = 0 for i gt n

Proof If M = 0 the result is trivial so assume otherwise (i) Suppose that Exti(MA) = 0 for alli gt 0 Since Exti(Mminus) is additive it follows that Exti(Mminus) vanishes on finite free A-modulesfor i gt 0 We show for 1 le j le d+ 1 that Extj(MN) = 0 for every finitely generated A-moduleN (we may assume d ge 1 since otherwise M is trivially projective)

Theorem 128 implies that projdimM le d and therefore Extd+1(Mminus) = 0 so this is atleast true for j = d + 1 Suppose that Extj(Mminus) vanishes on finitely generated modules andlet N be a finitely generated A-module We can find a short exact sequence of finitely generatedA-modules 0 minusrarr R minusrarr F minusrarr N minusrarr 0 with F a finite free A-module Since Extjminus1(MF ) = 0and Extj(MR) = 0 by the inductive hypothesis it follows from the long exact sequence thatExtjminus1(MN) = 0 as required The case j = 1 implies that M is projective using Lemma 121

(ii) The case n = 0 is (i) so assume n ge 1 If projdimM le n then by definition Exti(Mminus) =0 for i gt n so this direction is trivial For the converse suppose that Exti(MA) = 0 for i gt nWe can construct an exact sequence

0 minusrarr K minusrarr Pnminus1 minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

with K finitely generated and the Pi finitely generated projectives It suffices to show that Kis projective But by dimension shifting we have Exti(KA) sim= Exti+n(MA) = 0 for i gt 0Therefore by (i) K is projective and the proof is complete

Corollary 130 (Hilbertrsquos Syzygy Theorem) Let A = k[x1 xn] be a polynomial ring overa field k Then gldim(A) = n

Proof See our Dimension notes for another proof By Theorem 118 every local ring of A is regularSo if m is a maximal ideal then Am is regular of global dimension htm by Theorem 128 So byLemma 124 (iii) gldim(A) is the supremum of the heights of the maximal ideals in A which isclearly dim(A) = n

Theorem 131 Let (Am k) be a noetherian local ring and M a nonzero finitely generated A-module If projdim(M) ltinfin then

projdim(M) + depth(M) = depth(A)

42

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Proof By induction on depth(A) Let projdim(M) = n ge 0 If depth(A) = 0 then m isin Ass(A)This implies that there is a short exact sequence of A-modules

0 minusrarr k minusrarr A minusrarr C minusrarr 0

Thus we have an exact sequence

0 minusrarr TorAn+1(MC) minusrarr TorAn (Mk) minusrarr TorAn (MA)

By Proposition 122 flatdim(M) = n so TorAn+1(MC) = 0 But if n ge 1 then TorAn (MA) =0 and Lemma 123 yields TorAn+1(MC) sim= TorAn (Mk) 6= 0 which is a contradiction Henceprojdim(M) = 0 This means that M is projective and hence free by Proposition 24 Thus alsodepth(M) = 0 by Lemma 70 which completes the proof in the case depth(A) = 0

Now we fix a ring A with depth(A) gt 0 and proceed by induction on depth(M) First supposethat depth(M) = 0 Then m isin Ass(M) say m = Ann(y) with 0 6= y isin M Since depth(A) gt 0we can find a regular element x isin m Find an exact sequence

0 K Amε M 0

It follows from Lemma 70 that M cannot be free and hence by Proposition 24 cannot be projectiveeither Thus projdim(M) = projdim(K)+1 Choose u isin Am with ε(u) = y Clearly m sube (K u)and therefore xu isin K Since x is regular on Am and u isin K it follows that xu isin xK Butm sube (xK xu) so m isin Ass(KxK) and consequently depth(KxK) = 0 Since K is a submoduleof a free module x is regular on K By the third Change of Rings theorem for projective dimension(see our Dimension notes)

projdimAx(KxK) = projdimA(K) = projdimA(M)minus 1

By Lemma 83 depthAx(Ax) = depthA(A)minus 1 so using the inductive hypothesis (on A)

depthA(A) = 1 + depthAx(Ax)= 1 + depthAx(KxK) + projdimAx(KxK)= projdimA(M)

Finally we consider the case depth(M) gt 0 Let x isin m be regular on M By Lemma 82 we havedepth(MxM) = depth(M) minus 1 and by Proposition 126 projdim(MxM) = projdim(M) + 1Using the inductive hypothesis (for M) we have

depth(A) = depth(MxM) + projdim(MxM)= depth(M)minus 1 + projdim(M) + 1= depth(M) + projdim(M)

which completes the proof

Remark 3 If A is a regular local ring of dimension d then by Theorem 128 the global dimensionof A is d and for any A-module M we have projdimM le d We can now answer the questionhow big is the difference dminus projdimM

Corollary 132 Let A be a regular local ring of dimension d and M a nonzero finitely generatedA-module Then projdim(M) + depth(M) = d

Remark 4 With the notation of Corollary 132 the integer projdim(M) measures ldquohow projec-tiverdquo the module M is To be precise the closer projdim(M) is to zero the more projective M isUsing the Corollary we can rephrase this by saying that the projectivity of M is measured by thelargest number of ldquoindependent variablesrdquo in M The module M admits d independent variablesif and only if it is projective

43

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

63 Koszul Complexes

Throughout this section let A be a nonzero ring In this section a complex will mean a positivechain complex in AMod (notation of our Derived Functor notes) This is a sequence of A-modulesand module morphisms Mn dn Mn minusrarrMnminus1nisinZ with Mn = 0 for n lt 0 and dnminus1dn = 0 forall n Visually

middot middot middot Mndn Mnminus1

dnminus1 middot middot middot d1 M0d0 0 middot middot middot

We denote the complex by M and differentials dn by d where no confusion is likely Let C denotethe abelian category of all positive chain complexes in AMod (this is an abelian subcategoryof the category ChAMod of all chain complexes) If L is a complex then for k ge 0 let L[minus1]denote the complex obtained by shifting the objects and differentials one position left That isL[minus1]n = Lnminus1 Clearly if ϕ minusrarr Lprime is a morphism of complexes then ϕ[minus1]n = ϕnminus1 defines amorphism of complexes ϕ[minus1] L[minus1] minusrarr Lprime[minus1] This defines an exact functor T C minusrarr C andclearly T k shifts k positions left for k ge 1 If M is an A-module then we consider it as a complexconcentrated in degree 0 and denote this complex also by M

If L and M are two complexes we define a chain complex LotimesM by

(LotimesM)n =oplusp+q=n

Lp otimesAMq

= (L0 otimesAMn)oplus (L1 otimesAMnminus1)oplus middot middot middot oplus (Ln otimesAM0)

If x otimes y is an element of one of these summands then by abuse of notation we also use x otimes yto denote the image in (L otimesM)n For n ge 1 and integers p q ge 0 with p + q = n we induce amorphism κpq Lp otimesA Mq minusrarr (L otimesM)nminus1 of A-modules out of the tensor product using thefollowing formula

κpq(xotimes y) =

dL(x)otimes y + (minus1)pxotimes dM (y) p gt 0 q gt 0dL(x)otimes y q = 0(minus1)pxotimes dM (y) p = 0

Together these define a morphism of A-modules d (L otimesM)n minusrarr (L otimesM)nminus1 It is easy tocheck that this makes L otimes M into a complex of A-modules Given morphisms of complexesϕ L minusrarr Lprime and ψ M minusrarr M prime we obtain for each pair of integers p q ge 0 a morphism ofA-modules ϕp otimes ψq Lp otimesAMq minusrarr Lprimep otimesAM prime

q and these give rise to a morphism of complexes

ϕotimes ψ LotimesM minusrarr Lprime otimesM prime

(ϕotimes ψ)n = (ϕ0 otimes ψ0)oplus (ϕ1 otimes ψ1)oplus middot middot middot oplus (ϕn otimes ψn)

So the tensor product defines a covariant functor minus otimes minus C times C minusrarr C which is additive in eachvariable That is for any complex L the partial functors Lotimesminus and minusotimes L are additive

Proposition 133 For complexes LMN there is a canonical isomorphism

λLMN (LotimesM)otimesN minusrarr Lotimes (M otimesN)

which is natural in all three variables

44

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Proof For n ge 0 we have an isomorphism of A-modules

((LotimesM)otimesN)n =oplusp+q=n

(LotimesM)p otimesA Nq

=oplusp+q=n

( oplusr+s=p

Lr otimesAMs

)otimesA Nq

sim=oplus

r+s+q=n

(Lr otimesAMs)otimesA Nq

sim=oplus

r+s+q=n

Lr otimesA (Ms otimesA Nq)

sim=oplusp+q=n

Lp otimesA

( oplusr+s=q

Mr otimesA Ns

)= (Lotimes (M otimesN))n

Given integers with r+ s+ q = n and elements x isin Lr y isinMs z isin Nq we have xotimes y isin (LotimesM)pand this isomorphism sends (xotimes y)otimes z isin ((LotimesM)otimesN)n to xotimes (yotimes z) in (Lotimes (M otimesN))n It isstraightforward to check that this is an isomorphism of complexes natural in all three variables

Proposition 134 For any complex L the functors L otimes minus and minus otimes L are naturally equivalentand both are right exact The functor A otimes minus is naturally equivalent to the identity functor andA[minus1]otimesminus is naturally equivalent to T

Proof To show that L otimes minus and minus otimes L are naturally equivalent the only subtle point is that forp q ge 0 if ϕ Lp otimesMq

sim= Mq otimes Lp is the canonical isomorphism then we use the isomorphism(minus1)pqϕ in defining (LotimesM)n sim= (M otimesL)n Suppose we have a short exact sequence 0 minusrarr A minusrarrB minusrarr C minusrarr 0 in C Then for every j ge 0 the sequence of A-modules 0 minusrarr Aj minusrarr Bj minusrarrCj minusrarr 0 is exact and therefore

Li otimesAj minusrarr Li otimesBj minusrarr Li otimes Cj minusrarr 0

is also exact for any i ge 0 Coproducts are exact in AMod so for any n ge 0 the following sequenceis also exact oplus

i+j=n

Li otimesAj minusrarroplusi+j=n

Li otimesBj minusrarroplusi+j=n

Li otimes Cj minusrarr 0

But this is (L otimes A)n minusrarr (L otimes B)n minusrarr (L otimes C)n minusrarr 0 so the sequence L otimes A minusrarr L otimes B minusrarrL otimes C minusrarr 0 is pointwise exact and therefore exact Consider A as a complex concentrated indegree 0 For a complex M the natural isomorphism M sim= A otimes M is given pointwise by theisomorphism Mn

sim= AotimesMn There is also a natural isomorphism AotimesM sim= M given pointwise byAotimesMn

sim= Mn It is not hard to check that this is the same as M sim= AotimesM followed by the twistA otimesM sim= M otimes A The complex A[minus1] otimesM is isomorphic to M [minus1] but we have to be carefulsince the signs of the differentials in A[minus1]otimesM are the opposite of those in M [minus1] so we use theisomorphism M [minus1]n = Mnminus1

sim= A otimesMnminus1 given by (minus1)n+1ψ where ψ Mnminus1sim= A otimesMnminus1 is

canonical This isomorphism is clearly natural in M On the other hand there is a natural isomorphism M [minus1] sim= M otimes A[minus1] given pointwise by

M [minus1]n sim= Mnminus1 otimes A with no sign problems In fact this isomorphism is M [minus1] sim= A[minus1] otimesMfollowed by the twist A[minus1]otimesM sim= M otimesA[minus1]

In our Module Theory notes we define the exterior algebra andM associated to any A-moduleM It is a graded A-algebra and if M is free of rank n ge 1 with basis x1 xn then for0 le p le n andpM is free of rank

(np

)with basis xi1 and middot middot middot and xip indexed by strictly ascending

sequences i1 lt middot middot middot lt ip in the set 1 n For p gt n we have andpM = 0

45

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

Definition 24 Fix n ge 1 and let F = An be the canonical free A-module of rank n withcanonical basis x1 xn Suppose we are given elements a1 an isin A We define a complexof A-modules called the Koszul complex and denoted K(a1 an)

middot middot middotdp+1 andpF

dp middot middot middot d3 and2Fd2 and1F

d1 and0F 0

We identify and1F with F and and0F with A These modules become zero beyond andnF The map d1

is defined by d1(xi) = ai For p ge 2 with andpF 6= 0 we define

dp(xi1 and middot middot middot and xip) =psumr=1

(minus1)rminus1air (xi1 and middot middot middot and xir and middot middot middot and xip)

where xir indicates that we have omitted xir All other morphisms are zero It is not hard tocheck that dpdp+1 = 0 for all p ge 1 so this is actually a complex

Definition 25 Let a1 an isin A If C is a chain complex then we denote by C(a1 xn)the tensor product C otimesK(a1 xn) If M is an A-module then we consider it is as a complexconcentrated in degree 0 and denote by K(a1 anM) the complex M otimesK(a1 an) Thisis isomorphic to the complex

middot middot middot M otimes andpF middot middot middot M otimes and2F M otimes and1F M otimes and0F 0

Example 6 If a1 isin A then K(a1) is isomorphic to the complex

middot middot middot 0 Aa1 A 0

concentrated in degrees 0 and 1 where the morphism A minusrarr A is left multiplication by a1 ThenH0(K(a1)) = Aa1A and H1(K(a1)) = Ann(a1) as A-modules

Proposition 135 Let a1 an isin A and a multiplicatively closed set S sube A be given Then thereis a canonical isomorphism of complexes of Sminus1A-modules Sminus1K(a1 an) sim= K(a11 an1)

Proof There is a canonical isomorphism of Sminus1A-modules Sminus1F sim= (Sminus1A)n identifying xi1 withthe canonical ith basis element Using (TESCorollary 16) we have for each p ge 0 a canonicalisomorphism of Sminus1A-modules

Sminus1K(a1 an)p = Sminus1(pandA

F ) sim=pand

Sminus1A

Sminus1F sim=pand

Sminus1A

Gn = K(a11 an1)p

where G = (Sminus1A)n Together these isomorphisms form an isomorphism of complexes of Sminus1A-modules as required

Proposition 136 Let a1 an+1 isin A with n ge 1 Then there is a canonical isomorphism

K(a1 an)otimesK(an+1) sim= K(a1 an+1)

Proof Let T = K(a1 an) otimes K(an+1) Write F = An and let x1 xn be the canonicalbasis Let G = A with canonical basis xn+1 Then T0 = and0F otimes and0G sim= AotimesA sim= A and

T1 = (and0F otimes and1G)oplus (and1F otimes and0G) sim= and1Goplus and1F sim= An+1

For p ge 2 we have

Tp =oplusi+j=p

andiF otimes andjG sim= (andpminus1F otimes and1G)oplus (andpF otimes and0G) sim= andpminus1F oplus andpF

So for p gt n + 1 we have Tp = 0 and for p le n + 1 the A-module Tp is free of rank(n+1p

)

So at least the modules Tp are free of the same rank as Kp(a1 an+1) Let H = An+1 have

46

canonical basis e1 en+1 The isomorphism and0H sim= T0 sends 1 to 1 otimes 1 The isomorphismand1H sim= T1 sends e1 en to xi otimes 1 and en+1 to 1 otimes 1 For p ge 2 the action of isomorphismandpH sim= Tp on a basis element ei1 and middot middot middot and eip is described in two cases if ip le n then use the basiselement (xi1 and middot middot middot and xip) otimes 1 of andpF otimes and0G and otherwise if ip = n + 1 use the basis element(xi1 and middot middot middot and xipminus1)otimes 1 of andpminus1F otimesand1G One checks that these isomorphisms are compatible withthe differentials

For any a isin A we have an exact sequence of complexes

0 minusrarr A minusrarr K(a) minusrarr A[minus1] minusrarr 0

Let C be any complex Tensoring with C and using the natural isomorphisms C otimes A sim= C andC otimesA[minus1] sim= C[minus1] we have an exact sequence

0 minusrarr C minusrarr C(a) minusrarr C[minus1] minusrarr 0

For p isin Z we have Hp(C[minus1]) = Hpminus1(C) so the long exact homology sequence is

middot middot middot Hp+1(C) Hp+1(C(a)) Hp(C)δp Hp(C) middot middot middot

middot middot middot δ1 H1(C) H1(C(a)) H0(C)δ0 H0(C) H0(C(a)) 0

It is not difficult to check that the connecting morphism δp is multiplication by (minus1)pa Therefore

Lemma 137 If C is a complex with Hp(C) = 0 for p gt 0 then Hp(C(a)) = 0 for p gt 1 and thereis an exact sequence

0 H1(C(a)) H0(C) a H0(C) H0(C(a)) 0

If a is H0(C)-regular then we have Hp(C(a)) = 0 for all p gt 0 and H0(C(a)) sim= H0(C)aH0(C)

Theorem 138 Let A be a ring M an A-module and a1 an an M -regular sequence in AThen we have

Hp(K(a1 anM)) = 0 (p gt 0)H0(K(a1 anM)) sim= M(a1 an)M

Proof The last piece of Koszul complex K(a1 anM) is isomorphic to

middot middot middot minusrarrMn minusrarrM minusrarr 0

where the last map is (m1 mn) 7rarr (a1m1 anmn) So clearly there is an isomorphism of A-modules H0(K(a1 anM)) sim= M(a1 an)M We prove the other claim by induction on nhaving already proven the case n = 1 in Lemma 137 Let C be the complex K(a1 anminus1M)Then H0(C) sim= M(a1 anminus1)M so that an is H0(C)-regular By the inductive hypothesisHp(C) = 0 for p gt 0 and therefore by Lemma 137 Hp(C otimesK(an)) = 0 for p gt 0 But by Lemma136 and Proposition 133 there is an isomorphism CotimesK(an) sim= K(a1 anM) which completesthe proof

Remark 5 In other words for an M -regular sequence a1 an the corresponding Koszul com-plex K(a1 anM) gives a canonical resolution of the A-module M(a1 an)M That isthe following sequence is exact

middot middot middot minusrarrM otimes and2F minusrarrM otimes and1F minusrarrM otimes and0F minusrarrM(a1 an)M minusrarr 0

Taking M = A we see that the Koszul complex K(a1 an) gives a free resolution of the A-module A(a1 an) That is the following sequence is exact

0 minusrarr andnF minusrarr middot middot middot minusrarr and2F minusrarr and1F minusrarr and0F minusrarr A(a1 an) minusrarr 0 (5)

In particular we observe that projdimA(A(a1 an)) le n

47

Lemma 139 Let A be a ring and a1 an an A-regular sequence Then for any A-module Mthere is a canonical isomorphism of A-modules ExtnA(A(a1 an)M) sim= M(a1 an)M

Proof We have already observed that (5) is a projective resolution of A(a1 an) TakingHomA(minusM) the end of the complex we are interested in is

middot middot middot minusrarr HomA(andnminus1FM) minusrarr HomA(andnFM) minusrarr 0

Use the canonical bases to define isomorphisms andnminus1F sim= An and andnF sim= A Then we have acommutative diagram

HomA(andnminus1FM)

HomA(andnFM)

Mn

ψ M

where ψ(m1 mn) =sumnr=1(minus1)rminus1armr It is clear that Imψ = (a1 an)M so we have an

isomorphism of A-modules ExtnA(A(a1 an)M) sim= M(a1 an)M

Definition 26 Let (Am k) be a local ring and u M minusrarr N a morphism of A-modules Wesay that u is minimal if u otimes 1 M otimes k minusrarr N otimes k is an isomorphism Clearly any isomorphismM sim= N is minimal

Lemma 140 Let (Am k) be a local ring Then

(i) Let u M minusrarr N be a morphism of finitely generated A-modules Then u is minimal if andonly if it is surjective and Ker(u) sube mM

(ii) If M is a finitely generated A-module then there is a minimal morphism u F minusrarr M withF finite free and rankAF = rankk(M otimes k)

Proof (i) Suppose that u is minimal Let N prime be the image of M Then since MmM sim= NmNwe have N prime + mN = N and therefore N prime = N by Nakayama so u is surjective It is clear thatKer(u) sube mM Conversely suppose that u is surjective and Ker(u) sube mM Since u is surjectiveit follows that u(mM) = mN Therefore the morphism of A-modules M minusrarr N minusrarr NmNhas kernel mM and so MmM minusrarr NmN is an isomorphism as required (ii) If M = 0 thenthis is trivial since we can take F = 0 Otherwise let m1 mn be a minimal basis of M andu An minusrarrM the corresponding morphism This is clearly minimal

Let (Am k) be a noetherian local ring and M a finitely generated A-module A free resolution

L middot middot middot Lidi Liminus1

middot middot middot d1 L0d0 M 0

is called a minimal resolution if L0 minusrarrM is minimal and Li minusrarr Ker(diminus1) is minimal for eachi ge 1 Since Li+1 minusrarr Li minusrarr Ker(diminus1) = 0 for all i ge 1 it follows that in the complex ofA-modules Lotimes k

middot middot middot minusrarr Li otimes k minusrarr Liminus1 otimes k minusrarr middot middot middot minusrarr L0 otimes k minusrarr 0

the differentials are all zero Therefore we have TorAi (Mk) sim= Li otimes k as k-modules for all i ge 0Since Mk are finitely generated for i ge 0 the A-modules TorAi (Mk) and Li otimes k are finitelygenerated Hence Li otimes k is a finitely generated free k-module which shows that Li is a finitelygenerated A-module

Proposition 141 Let (Am k) be a noetherian local ring and M a finitely generated A-moduleThen a minimal free resolution of M exists and is unique up to a (non-canonical) isomorphism

48

Proof By Lemma 140 (ii) we can find a minimal morphism d0 L0 minusrarr M with L0 finite free ofrank rankk(M otimes k) Let K minusrarr L0 be the kernel of d0 Find a minimal morphism L1 minusrarr K withL1 finite free and so on This defines a minimal free resolution of M To prove the uniquenesslet ε L minusrarrM and εprime Lprime minusrarrM be two minimal free resolutions of M We can lift the identity1M to a morphism of chain complexes ϕ L minusrarr Lprime so we have a commutative diagram

middot middot middot L1

ϕ1

L0

ϕ0

ε

M

middot middot middot Lprime1 Lprime0

εprime M

Since ε εprime are minimal the map ϕ0 otimes 1 L0 otimes k minusrarr Lprime0 otimes k is an isomorphism of k-modules Inparticular we have

rankAL0 = rankk(L0 otimes k) = rankk(Lprime0 otimes k) = rankALprime0

So L0 Lprime0 are free of the same finite rank We claim that ϕ0 is an isomorphism This is trivial

if L0 = Lprime0 = 0 so assume they are both nonzero Then ϕ0 is described by a square matrixT isin Mn(A) If you take residues you get the matrix T prime isin Mn(k) of ϕ0 otimes 1 which has nonzerodeterminant since it is an isomorphism But it is clear that det(T ) + m = det(T prime) so det(T ) isin mTherefore ϕ0 itself is an isomorphism

Since ϕ0 is an isomorphism so the induced morphism on the kernels Ker(ε) minusrarr Ker(εprime) andwe can repeat the same argument to see that ϕ1 is an isomorphism and similarly to show thatall the ϕi are isomorphisms

Lemma 142 Let (Am k) be a noetherian local ring and u F minusrarr G a morphism of finitelygenerated free A-modules Then u is minimal if and only if it is an isomorphism

Definition 27 Let (Am k) be a noetherian local ring and M a finitely generated A-moduleChoose a minimal free resolution of M Then for i ge 0 the integer bi = rankALi ge 0 iscalled the i-th Betti number of M It is independent of the chosen resolution and moreoverrankkTor

Ai (Mk) = bi

Example 7 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen

(i) Proposition 141 shows that b0 = rankk(M otimes k)

(ii) If M = 0 then the zero complex is a minimal free resolution of M so bi = 0 for i ge 0

(iii) If M is flat then TorAi (Mk) = 0 for all i ge 1 so bi = 0 for i ge 1 In particular this is trueif M is free or projective

(iv) If M is free of finite rank s ge 1 then M otimes k is a free k-module of rank s so b0 = s

Lemma 143 Let (Am k) be a noetherian local ring and M a finitely generated A-moduleSuppose that we have two complexes LF together with morphisms ε εprime such that the followingsequences are exact in the last two nonzero positions

L middot middot middot Lidi Liminus1

middot middot middot d1 L0ε M 0

F middot middot middot Fidprimei Fiminus1

middot middot middotdprime1 F0

εprime M 0

Assume the following

(i) L is a minimal free resolution of M

(ii) Each Fi is a finitely generated free A-module

49

(iii) εprime otimes 1 F0 otimes k minusrarrM otimes k is injective

(iv) For each i ge 0 dprimei+1(Fi+1) sube mFi and the induced morphism Fi+1mFi+1 minusrarr mFim2Fi is

an injection

Then there exists a morphism of complexes f F minusrarr L lifting the identity 1M such that for fimaps Fi isomorphically onto a direct summand of Li Consequently we have

rankAFi le rankALi = rankkTorAi (Mk)

Proof Both LF are positive chain complexes with F projective and L acyclic so by our DerivedFunctor notes there is a morphism f F minusrarr L of chain complexes giving a commutative diagram

middot middot middot F0

f0

εprime M

0

middot middot middot L0 ε M 0

We have to prove that for each i ge 0 the morphism fi Fi minusrarr Li is a coretractionWe claim thatfi is a coretraction iff fi otimes 1 Fi otimes k minusrarr Li otimes k is an injective morphism of k-modules Oneimplication is clear So assume that fi otimes 1 is injective The claim is trivial if either of Fi Li arezero so assume they are both of nonzero finite rank Pick bases for Fi Li (which are obviouslyminimal bases) and use the fact that fi otimes 1 is a coretraction to define a morphism ϕ Li minusrarr Fisuch that (ϕfi) otimes 1 Fi otimes k minusrarr Fi otimes k is the identity By Lemma 142 it follows that ϕfi is anisomorphism and therefore clearly fi is a coretraction

We prove by induction that fi otimes 1 is injective for all i ge 0 By assumptions (i) (iii) it is clearthat f0 otimes 1 is injective We have the following commutative diagram

F1

β FFFF

FFFF

F

f1

dprime1 F0

f0

εprime M

Kerεprime

xxxxxxxxx

α

Kerε

FFFF

FFFF

F

L1

γxxxxxxxxx

d1

L0 ε M

By assumption γ otimes 1 is an isomorphism So to show f1 otimes 1 is injective it suffices to show thatαotimes 1 β otimes 1 are injective or equivalently that αminus1(mKerε) = mKerεprime and βminus1(mKerεprime) = mF1Suppose that a isin F1 and dprime1(a) isin mKerεprime Since εprime otimes 1 is injective we have mKerεprime sube m2F0Hence dprime1(a) isin m2F0 and therefore by (iv) a isin mF1 as required

Now suppose that a isin Kerεprime and f0(a) isin mKerε Let g be such that gf0 = 1 Then f0(a) isinm2L0 and therefore a = gf0(a) isin g(m2L0) sube m2F0 Let b isin F1 be such that dprime1(b) = a isin m2F0Then (iv) implies that b isin mF1 and therefore a = β(b) isin mKerεprime as required This shows thatf1 otimes 1 is injective

Suppose that fi otimes 1 is injective for some i ge 1 Then we show fi+1 otimes 1 is injective using asimilar setup We replace Kerεprime by Imdprimei+1 (in the case i = 0 they are equal) and use (iv) toshow that Kerdprimei sube mFi and (i) to show that Kerdi sube mLi The proof that β otimes 1 is injective isstraightforward For αotimes 1 let a isin Imdprimei+1 be such that fi(a) isin mKerdi As before we find thatfi(a) isin m2Li and hence a = gfi(a) isin m2Fi Let b isin Fi+1 be such that a = dprimei+1(b) Then by (iv)b isin mFi+1 and therefore a isin mImdprimei+1 as required This proves that fi is a coretraction for i ge 0and the rank claim follows from the fact that rankAFi = rankk(Fi otimes k) le rankk(Li otimes k)

50

Lemma 144 Let A be a ring with maximal ideal m If s isin m and a isin A then sa isin mk impliesa isin mk for any k ge 1

Theorem 145 Let (Am k) be a noetherian local ring and let s = rankkmm2 Then we have

rankkTorAi (k k) ge

(s

i

)0 le i le s

Here rankkTorAi (k k) is the i-th Betti number of the A-module k

Proof We have TorA0 (k k) sim= k as k-modules so rankkTorA0 (k k) = rankkk = 1 which takes

care of the case s = 0 So assume that s ge 1 and let a1 as be a minimal basis of m withassociated Koszul complex F = K(a1 as) The canonical morphism εprime F0

sim= A minusrarr k gives acomplex exact in the last two nonzero places

middot middot middot minusrarr Fi minusrarr Fiminus1 minusrarr middot middot middot minusrarr F1 minusrarr F0 minusrarr k minusrarr 0

We claim this complex satisfies the conditions of Lemma 143 It clearly satisfies (ii) and (iii)It only remains to check condition (iii) By the definition of dp+1 Fp+1 minusrarr Fp it is clear thatdp+1(Fp+1) sube mFp for p ge 0 We also have to show that dminus1

p+1(m2Fp) sube mFp+1 This is trivial

if p + 1 gt s and also if p = 0 since a1 as is a minimal basis So assume 0 lt p le s minus 1Assume that

dp+1

sumi1ltmiddotmiddotmiddotltip+1

mi1middotmiddotmiddotip+1(xi1 and middot middot middot and xip+1)

=

sumi1ltmiddotmiddotmiddotltip+1

psumr=1

(minus1)rminus1airmi1middotmiddotmiddotip+1(xi1 and middot middot middot and xir and middot middot middot and xip+1) isin m2Fp

Then collecting terms we obtain a number of equations of the formsum

(minus1)etatmt isin m2 where atis one of the air and mt one of the mi1middotmiddotmiddotip+1 Since the residues of the ai give a basis of mm2

over k it follows that mt isin m which completes the proof that F satisfies all the conditions ofLemma 143 Choosing any minimal free resolution L of M and applying Lemma 143 we see thatfor 0 le i le s (

s

i

)= rankAFi le rankkTor

Ai (k k)

as required

Theorem 146 (Serre) Let (Am k) be a noetherian local ring Then A is regular if and only ifthe global dimension of A is finite

Proof We have already proved one part in Theorem 128 So suppose that gldim(A) lt infinThen TorAs (k k) 6= 0 by Theorem 145 hence gldim(A) ge rankkmm

2 since by Proposition122 we have tordim(A) = gldim(A) On the other hand it follows from Theorem 125 thatprojdim(k) = gldim(A) ltinfin so by Theorem 131 we have gldim(A) = projdim(k) = depth(A)Therefore we get

dim(A) le rankkmm2 le gldim(A) = depth(A) le dim(A)

Whence dim(A) = rankkmm2 and A is regular

Corollary 147 If A is a regular local ring then Ap is regular for any p isin Spec(A)

Proof Let M be a nonzero Ap-module Then considering M as an A-module there is an exactsequence of finite length n le gldim(A) with all Pi projective

0 minusrarr Pn minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

51

Since Ap is flat the following sequence is also exact

0 minusrarr (Pn)p minusrarr middot middot middot minusrarr (P0)p minusrarrMp minusrarr 0

The modules (Pi)p are projective Ap-modules and M sim= Mp as Ap-modules so it follows thatgldim(Ap) le gldim(A) ltinfin

Definition 28 A ring A is called a regular ring if Ap is a regular local ring for every prime idealp of A Note that A is not required to be noetherian Regularity is stable under ring isomorphismA noetherian local ring A is regular in this sense if and only if it is regular in the normal sense

It follows from Theorem 108 that any regular ring is normal and a noetherian regular ring isCohen-Macaulay It follows from Lemma 8 and Theorem 90 that a regular ring is catenary

Lemma 148 A ring A is regular if and only if Am is regular for all maximal ideals m

Proof One implication is clear For the other given a prime ideal p find a maximal ideal m withp sube m Then Ap

sim= (Am)pAm so Ap is a regular local ring

Lemma 149 If A is a regular ring and S sube A is multiplicatively closed then Sminus1A is a regularring

Lemma 150 If A is a regular ring then so is A[x1 xn] In particular k[x1 xn] is aregular ring for any field k

Proof This follows immediately from Theorem 118

Theorem 151 Let A be a regular local ring which is a subring of a domain B and suppose thatB is a finitely generated A-module Then B is flat (equivalently free) over A if and only if B isCohen-Macaulay In particular if B is regular then it is a free A-module

Proof Since B is a finitely generated A-module it is integral over A and so by Lemma 91 if B isflat it is Cohen-Macaulay Conversely suppose that B is Cohen-Macaulay If dim(A) = 0 then Ais a field so B is trivially flat so throughout we may assume dim(A) ge 1 Since A is normal thegoing-down theorem holds between A and B by Theorem 42 so by Theorem 55 (3) for any properideal I of A IB is proper and htI = htIB We claim that depthA(A) = depthA(B) Notice thatdepthA(B) is finite since otherwise m isin AssA(B) and hence dim(A) = 0

Firstly we prove the inequality le Since A is regular it is Cohen-Macaulay so depthA(A) =dim(A) Set s = dim(A) and let a1 as be a regular system of parameters Then ht(a1 ai)A =i and therefore ht(a1 ai)B = i for all 1 le i le s by Theorem 108 It follows from Corollary97 that a1 as is a B-regular sequence and therefore depthA(B) ge s

To prove the reverse inequality set d = depthA(B) and let a1 ad isin m be a maximal B-regular sequence Then as elements of B the sequence a1 ad is B-regular so by Lemma 74 wehave ht(a1 ad) = d But (a1 ad) sube mB and htmB = htm = dim(A) so d le dim(A) asrequired

Since gldim(A) lt infin we have projdimA(B) lt infin so we can apply Theorem 131 to see thatprojdimA(B) + depthA(B) = depthA(A) so projdimA(B) = 0 and therefore B is projectiveSince A is local and B finitely generated projective hArr free hArr flat so the proof is complete

64 Unique Factorisation

Recall that if A is a ring two elements p q isin A are said to be associates if p = uq for some unitu isin A This is an equivalence relation on the elements of A

Definition 29 Let A be an integral domain An element of A is irreducible if it is a nonzerononunit which cannot be written as the product of two nonunits An element p isin A is prime if it isa nonzero nonunit with the property that if p|ab then p|a or p|b Equivalently p is prime iff (p) isa nonzero prime ideal We say A is a unique factorisation domain if every nonzero nonunit a isin Acan be written essentially uniquely as a = up1 middot middot middot pr where u is a unit and each pi is irreducible

52

Essentially uniquely means that if a = vq1 middot middot middot qs where v is a unit and the qj irreducible thenr = s and after reordering (if necessary) qi is an associate of pi The property of being a UFD isstable under ring isomorphism

Theorem 152 A noetherian domain A is a UFD if and only if every prime ideal of height 1 isprincipal

Lemma 153 Let A be a noetherian domain and let x isin A be prime Then A is a UFD if andonly if Ax is

Proof By assumption (x) is a prime ideal of height 1 If p is a prime ideal of height 1 then eitherx isin p in which case p = (x) or x isin p and these primes are in bijection with the primes of Ax Sousing Theorem 152 it is clear that if A is a UFD so is Ax Suppose that Ax is a UFD and let p bea prime ideal of height 1 in A We can assume that x isin p Let a isin p be such that pAx = a1AxBy [AM69] Corollary 1018 we have capi(xi) = 0 so if x|a there is a largest integer n ge 1 with xn|aWrite a = cxn Since x isin p we have c isin p so by replacing a with c we can assume pAx = a1Axwith a isin (x) Then it is clear that p = (a) as required

Definition 30 Let R be an integral domain If M is a torsion-free R-module then the rank ofM is the maximum number of linearly independent elements in M rank(M) isin 0 1 infin

Proposition 154 Let R be an integral domain and M a torsion-free R-module If T sube Ris multiplicatively closed then Tminus1M is a torsion-free Tminus1R-module and rankTminus1R(Tminus1M) =rankR(M)

Proof If rank(M) = 0 this is trivial so assumeM is nonzero It is clear that Tminus1M is torsion-freeIf rankR(M) = r and x1 xr isin M are linearly independent then x11 xr1 isin Tminus1M arelinearly independent over Tminus1R Similarly if x1s1 xnsn isin Tminus1M are linearly independentin Tminus1M then x1 xn are linearly independent in M So the result is clear

Corollary 155 Let R be an integral domain with quotient field K Then

(i) If M is a torsion-free R-module rankR(M) = dimK(M otimesK)

(ii) If MN are two torsion-free R-modules of finite rank then rankR(M oplusN) = rankR(M) +rankR(N)

In particular if M is a free R-module then the rank just defined is equal to the normal free rankand we can write rank(M) without confusion

Let R be a noetherian domain and suppose a1 an isin R are linearly independent elementswhich do not generate R Then a1 ar is an R-regular sequence so by Lemma 74 the ideals(a1 ai) have height i for 1 le i le n So it follows immediately that

Lemma 156 Let R be a noetherian domain and I an ideal Then rank(I) le htI

Lemma 157 Let R be a domain and M a finitely generated projective R-module of rank 1 ThenandiM = 0 for i gt 1

Proof By localisation If p is a prime ideal then Mp is a finitely generated projective module overthe local ring Rp so Mp is free of rank 1 and Mp

sim= Rp Hence for i gt 1

(andiM)psim= andiMp

sim= andiRq = 0

as required

Theorem 158 (Auslander-Buchsbaum) A regular local ring (Am) is UFD

53

Proof We use induction on dimA If dimA = 0 then A is a field and if dimA = 1 then A is aprincipal ideal domain Suppose dimA gt 1 and let a1 ad be a regular system of parametersThen x = a1 is prime by Theorem 108 so it suffices by Lemma 153 to show that Ax is UFDLet q be a prime ideal of height 1 in Ax and put p = q cap A so q = pAx By Theorem 128gldimA = dimA lt infin so we can produce an exact sequence of A-modules with all Fi finitelygenerated free

0 minusrarr Fn minusrarr Fnminus1 minusrarr middot middot middot minusrarr F0 minusrarr p minusrarr 0 (6)

Maximal ideals of Ax correspond to primes of A maximal among those not containing x Theseprimes must all be properly contained in m so if PAx is a maximal ideal then htP lt dimATherefore (Ax)PAx

sim= AP is UFD by the inductive assumption and so q(Ax)n is either principalor zero for every maximal n of Ax Then by Lemma 124 we have projdimAx(q) = 0 and thereforeq is projective Localising (6) with respect to S = 1 x x2 we see that the following sequenceof Ax-modules is exact

0 minusrarr F primen minusrarr F primenminus1 minusrarr middot middot middot minusrarr F prime0 minusrarr q minusrarr 0 (7)

where F primei = Fi otimesAx are finitely generated and free over Ax If we decompose (7) into short exactsequences

0 minusrarr K prime0 minusrarrF prime0 minusrarr q minusrarr 0

0 minusrarr K prime1 minusrarrF prime1 minusrarr K prime

0 minusrarr 0

0 minusrarr F primen minusrarrF primenminus1 minusrarr K primenminus2 minusrarr 0

(8)

then the first sequence splits since q is projective Hence K prime0 must be projective and in this way

we show that all the sequences split and all the K primei are projective It follows thatoplus

i even

F primeisim=oplusi odd

F primei oplus q

Thus we have finite free Ax-modules FG such that F sim= Goplus q Since Ax is a noetherian domainand q a nonzero ideal of height 1 it follows from Lemma 156 that rank(q) = 1 If rank(G) = rthen rank(F ) = r + 1

So to show q is principal and complete the proof it suffices to show that q is free But by ournotes on Tensor Symmetric and Exterior algebras we have

Ax sim=r+1and

F sim=r+1and

(Goplus q) sim=oplus

i+j=r+1

(andiG)otimes (andjq) sim= (andr+1Gotimes and0q)oplus (andrGotimes and1q) sim= q

Since andr+1G = 0andrG sim= Ax and andiq = 0 for i gt 1 by Lemma 157

References

[AM69] M F Atiyah and I G Macdonald Introduction to commutative algebra Addison-WesleyPublishing Co Reading Mass-London-Don Mills Ont 1969 MR MR0242802 (394129)

[Ash] Robert Ash A course in commutative algebra httpwwwmathuiucedusimr-ashComAlghtml

[Eis95] David Eisenbud Commutative algebra Graduate Texts in Mathematics vol 150Springer-Verlag New York 1995 With a view toward algebraic geometry MRMR1322960 (97a13001)

[Mat80] Hideyuki Matsumura Commutative algebra second ed Mathematics Lecture Note Se-ries vol 56 BenjaminCummings Publishing Co Inc Reading Mass 1980 MRMR575344 (82i13003)

54

• General Rings
• Flatness
• • Faithful Flatness
  • Going-up and Going-down
  • Constructible Sets
  • • Associated Primes
    • Dimension
    • • Homomorphism and Dimension
      • • Depth
        • • Cohen-Macaulay Rings
          • • Normal and Regular Rings
            • • Classical Theory
              • Homological Theory
              • Koszul Complexes
              • Unique Factorisation

Lemma 139 Let A be a ring and a1 an an A-regular sequence Then for any A-module Mthere is a canonical isomorphism of A-modules ExtnA(A(a1 an)M) sim= M(a1 an)M

Proof We have already observed that (5) is a projective resolution of A(a1 an) TakingHomA(minusM) the end of the complex we are interested in is

middot middot middot minusrarr HomA(andnminus1FM) minusrarr HomA(andnFM) minusrarr 0

Use the canonical bases to define isomorphisms andnminus1F sim= An and andnF sim= A Then we have acommutative diagram

HomA(andnminus1FM)

HomA(andnFM)

Mn

ψ M

where ψ(m1 mn) =sumnr=1(minus1)rminus1armr It is clear that Imψ = (a1 an)M so we have an

isomorphism of A-modules ExtnA(A(a1 an)M) sim= M(a1 an)M

Definition 26 Let (Am k) be a local ring and u M minusrarr N a morphism of A-modules Wesay that u is minimal if u otimes 1 M otimes k minusrarr N otimes k is an isomorphism Clearly any isomorphismM sim= N is minimal

Lemma 140 Let (Am k) be a local ring Then

(i) Let u M minusrarr N be a morphism of finitely generated A-modules Then u is minimal if andonly if it is surjective and Ker(u) sube mM

(ii) If M is a finitely generated A-module then there is a minimal morphism u F minusrarr M withF finite free and rankAF = rankk(M otimes k)

Proof (i) Suppose that u is minimal Let N prime be the image of M Then since MmM sim= NmNwe have N prime + mN = N and therefore N prime = N by Nakayama so u is surjective It is clear thatKer(u) sube mM Conversely suppose that u is surjective and Ker(u) sube mM Since u is surjectiveit follows that u(mM) = mN Therefore the morphism of A-modules M minusrarr N minusrarr NmNhas kernel mM and so MmM minusrarr NmN is an isomorphism as required (ii) If M = 0 thenthis is trivial since we can take F = 0 Otherwise let m1 mn be a minimal basis of M andu An minusrarrM the corresponding morphism This is clearly minimal

Let (Am k) be a noetherian local ring and M a finitely generated A-module A free resolution

L middot middot middot Lidi Liminus1

middot middot middot d1 L0d0 M 0

is called a minimal resolution if L0 minusrarrM is minimal and Li minusrarr Ker(diminus1) is minimal for eachi ge 1 Since Li+1 minusrarr Li minusrarr Ker(diminus1) = 0 for all i ge 1 it follows that in the complex ofA-modules Lotimes k

middot middot middot minusrarr Li otimes k minusrarr Liminus1 otimes k minusrarr middot middot middot minusrarr L0 otimes k minusrarr 0

the differentials are all zero Therefore we have TorAi (Mk) sim= Li otimes k as k-modules for all i ge 0Since Mk are finitely generated for i ge 0 the A-modules TorAi (Mk) and Li otimes k are finitelygenerated Hence Li otimes k is a finitely generated free k-module which shows that Li is a finitelygenerated A-module

Proposition 141 Let (Am k) be a noetherian local ring and M a finitely generated A-moduleThen a minimal free resolution of M exists and is unique up to a (non-canonical) isomorphism

48

Proof By Lemma 140 (ii) we can find a minimal morphism d0 L0 minusrarr M with L0 finite free ofrank rankk(M otimes k) Let K minusrarr L0 be the kernel of d0 Find a minimal morphism L1 minusrarr K withL1 finite free and so on This defines a minimal free resolution of M To prove the uniquenesslet ε L minusrarrM and εprime Lprime minusrarrM be two minimal free resolutions of M We can lift the identity1M to a morphism of chain complexes ϕ L minusrarr Lprime so we have a commutative diagram

middot middot middot L1

ϕ1

L0

ϕ0

ε

M

middot middot middot Lprime1 Lprime0

εprime M

Since ε εprime are minimal the map ϕ0 otimes 1 L0 otimes k minusrarr Lprime0 otimes k is an isomorphism of k-modules Inparticular we have

rankAL0 = rankk(L0 otimes k) = rankk(Lprime0 otimes k) = rankALprime0

So L0 Lprime0 are free of the same finite rank We claim that ϕ0 is an isomorphism This is trivial

if L0 = Lprime0 = 0 so assume they are both nonzero Then ϕ0 is described by a square matrixT isin Mn(A) If you take residues you get the matrix T prime isin Mn(k) of ϕ0 otimes 1 which has nonzerodeterminant since it is an isomorphism But it is clear that det(T ) + m = det(T prime) so det(T ) isin mTherefore ϕ0 itself is an isomorphism

Since ϕ0 is an isomorphism so the induced morphism on the kernels Ker(ε) minusrarr Ker(εprime) andwe can repeat the same argument to see that ϕ1 is an isomorphism and similarly to show thatall the ϕi are isomorphisms

Lemma 142 Let (Am k) be a noetherian local ring and u F minusrarr G a morphism of finitelygenerated free A-modules Then u is minimal if and only if it is an isomorphism

Definition 27 Let (Am k) be a noetherian local ring and M a finitely generated A-moduleChoose a minimal free resolution of M Then for i ge 0 the integer bi = rankALi ge 0 iscalled the i-th Betti number of M It is independent of the chosen resolution and moreoverrankkTor

Ai (Mk) = bi

Example 7 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen

(i) Proposition 141 shows that b0 = rankk(M otimes k)

(ii) If M = 0 then the zero complex is a minimal free resolution of M so bi = 0 for i ge 0

(iii) If M is flat then TorAi (Mk) = 0 for all i ge 1 so bi = 0 for i ge 1 In particular this is trueif M is free or projective

(iv) If M is free of finite rank s ge 1 then M otimes k is a free k-module of rank s so b0 = s

Lemma 143 Let (Am k) be a noetherian local ring and M a finitely generated A-moduleSuppose that we have two complexes LF together with morphisms ε εprime such that the followingsequences are exact in the last two nonzero positions

L middot middot middot Lidi Liminus1

middot middot middot d1 L0ε M 0

F middot middot middot Fidprimei Fiminus1

middot middot middotdprime1 F0

εprime M 0

Assume the following

(i) L is a minimal free resolution of M

(ii) Each Fi is a finitely generated free A-module

49

(iii) εprime otimes 1 F0 otimes k minusrarrM otimes k is injective

(iv) For each i ge 0 dprimei+1(Fi+1) sube mFi and the induced morphism Fi+1mFi+1 minusrarr mFim2Fi is

an injection

Then there exists a morphism of complexes f F minusrarr L lifting the identity 1M such that for fimaps Fi isomorphically onto a direct summand of Li Consequently we have

rankAFi le rankALi = rankkTorAi (Mk)

Proof Both LF are positive chain complexes with F projective and L acyclic so by our DerivedFunctor notes there is a morphism f F minusrarr L of chain complexes giving a commutative diagram

middot middot middot F0

f0

εprime M

0

middot middot middot L0 ε M 0

We have to prove that for each i ge 0 the morphism fi Fi minusrarr Li is a coretractionWe claim thatfi is a coretraction iff fi otimes 1 Fi otimes k minusrarr Li otimes k is an injective morphism of k-modules Oneimplication is clear So assume that fi otimes 1 is injective The claim is trivial if either of Fi Li arezero so assume they are both of nonzero finite rank Pick bases for Fi Li (which are obviouslyminimal bases) and use the fact that fi otimes 1 is a coretraction to define a morphism ϕ Li minusrarr Fisuch that (ϕfi) otimes 1 Fi otimes k minusrarr Fi otimes k is the identity By Lemma 142 it follows that ϕfi is anisomorphism and therefore clearly fi is a coretraction

We prove by induction that fi otimes 1 is injective for all i ge 0 By assumptions (i) (iii) it is clearthat f0 otimes 1 is injective We have the following commutative diagram

F1

β FFFF

FFFF

F

f1

dprime1 F0

f0

εprime M

Kerεprime

xxxxxxxxx

α

Kerε

FFFF

FFFF

F

L1

γxxxxxxxxx

d1

L0 ε M

By assumption γ otimes 1 is an isomorphism So to show f1 otimes 1 is injective it suffices to show thatαotimes 1 β otimes 1 are injective or equivalently that αminus1(mKerε) = mKerεprime and βminus1(mKerεprime) = mF1Suppose that a isin F1 and dprime1(a) isin mKerεprime Since εprime otimes 1 is injective we have mKerεprime sube m2F0Hence dprime1(a) isin m2F0 and therefore by (iv) a isin mF1 as required

Now suppose that a isin Kerεprime and f0(a) isin mKerε Let g be such that gf0 = 1 Then f0(a) isinm2L0 and therefore a = gf0(a) isin g(m2L0) sube m2F0 Let b isin F1 be such that dprime1(b) = a isin m2F0Then (iv) implies that b isin mF1 and therefore a = β(b) isin mKerεprime as required This shows thatf1 otimes 1 is injective

Suppose that fi otimes 1 is injective for some i ge 1 Then we show fi+1 otimes 1 is injective using asimilar setup We replace Kerεprime by Imdprimei+1 (in the case i = 0 they are equal) and use (iv) toshow that Kerdprimei sube mFi and (i) to show that Kerdi sube mLi The proof that β otimes 1 is injective isstraightforward For αotimes 1 let a isin Imdprimei+1 be such that fi(a) isin mKerdi As before we find thatfi(a) isin m2Li and hence a = gfi(a) isin m2Fi Let b isin Fi+1 be such that a = dprimei+1(b) Then by (iv)b isin mFi+1 and therefore a isin mImdprimei+1 as required This proves that fi is a coretraction for i ge 0and the rank claim follows from the fact that rankAFi = rankk(Fi otimes k) le rankk(Li otimes k)

50

Lemma 144 Let A be a ring with maximal ideal m If s isin m and a isin A then sa isin mk impliesa isin mk for any k ge 1

Theorem 145 Let (Am k) be a noetherian local ring and let s = rankkmm2 Then we have

rankkTorAi (k k) ge

(s

i

)0 le i le s

Here rankkTorAi (k k) is the i-th Betti number of the A-module k

Proof We have TorA0 (k k) sim= k as k-modules so rankkTorA0 (k k) = rankkk = 1 which takes

care of the case s = 0 So assume that s ge 1 and let a1 as be a minimal basis of m withassociated Koszul complex F = K(a1 as) The canonical morphism εprime F0

sim= A minusrarr k gives acomplex exact in the last two nonzero places

middot middot middot minusrarr Fi minusrarr Fiminus1 minusrarr middot middot middot minusrarr F1 minusrarr F0 minusrarr k minusrarr 0

We claim this complex satisfies the conditions of Lemma 143 It clearly satisfies (ii) and (iii)It only remains to check condition (iii) By the definition of dp+1 Fp+1 minusrarr Fp it is clear thatdp+1(Fp+1) sube mFp for p ge 0 We also have to show that dminus1

p+1(m2Fp) sube mFp+1 This is trivial

if p + 1 gt s and also if p = 0 since a1 as is a minimal basis So assume 0 lt p le s minus 1Assume that

dp+1

sumi1ltmiddotmiddotmiddotltip+1

mi1middotmiddotmiddotip+1(xi1 and middot middot middot and xip+1)

=

sumi1ltmiddotmiddotmiddotltip+1

psumr=1

(minus1)rminus1airmi1middotmiddotmiddotip+1(xi1 and middot middot middot and xir and middot middot middot and xip+1) isin m2Fp

Then collecting terms we obtain a number of equations of the formsum

(minus1)etatmt isin m2 where atis one of the air and mt one of the mi1middotmiddotmiddotip+1 Since the residues of the ai give a basis of mm2

over k it follows that mt isin m which completes the proof that F satisfies all the conditions ofLemma 143 Choosing any minimal free resolution L of M and applying Lemma 143 we see thatfor 0 le i le s (

s

i

)= rankAFi le rankkTor

Ai (k k)

as required

Theorem 146 (Serre) Let (Am k) be a noetherian local ring Then A is regular if and only ifthe global dimension of A is finite

Proof We have already proved one part in Theorem 128 So suppose that gldim(A) lt infinThen TorAs (k k) 6= 0 by Theorem 145 hence gldim(A) ge rankkmm

2 since by Proposition122 we have tordim(A) = gldim(A) On the other hand it follows from Theorem 125 thatprojdim(k) = gldim(A) ltinfin so by Theorem 131 we have gldim(A) = projdim(k) = depth(A)Therefore we get

dim(A) le rankkmm2 le gldim(A) = depth(A) le dim(A)

Whence dim(A) = rankkmm2 and A is regular

Corollary 147 If A is a regular local ring then Ap is regular for any p isin Spec(A)

Proof Let M be a nonzero Ap-module Then considering M as an A-module there is an exactsequence of finite length n le gldim(A) with all Pi projective

0 minusrarr Pn minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

51

Since Ap is flat the following sequence is also exact

0 minusrarr (Pn)p minusrarr middot middot middot minusrarr (P0)p minusrarrMp minusrarr 0

The modules (Pi)p are projective Ap-modules and M sim= Mp as Ap-modules so it follows thatgldim(Ap) le gldim(A) ltinfin

Definition 28 A ring A is called a regular ring if Ap is a regular local ring for every prime idealp of A Note that A is not required to be noetherian Regularity is stable under ring isomorphismA noetherian local ring A is regular in this sense if and only if it is regular in the normal sense

It follows from Theorem 108 that any regular ring is normal and a noetherian regular ring isCohen-Macaulay It follows from Lemma 8 and Theorem 90 that a regular ring is catenary

Lemma 148 A ring A is regular if and only if Am is regular for all maximal ideals m

Proof One implication is clear For the other given a prime ideal p find a maximal ideal m withp sube m Then Ap

sim= (Am)pAm so Ap is a regular local ring

Lemma 149 If A is a regular ring and S sube A is multiplicatively closed then Sminus1A is a regularring

Lemma 150 If A is a regular ring then so is A[x1 xn] In particular k[x1 xn] is aregular ring for any field k

Proof This follows immediately from Theorem 118

Theorem 151 Let A be a regular local ring which is a subring of a domain B and suppose thatB is a finitely generated A-module Then B is flat (equivalently free) over A if and only if B isCohen-Macaulay In particular if B is regular then it is a free A-module

Proof Since B is a finitely generated A-module it is integral over A and so by Lemma 91 if B isflat it is Cohen-Macaulay Conversely suppose that B is Cohen-Macaulay If dim(A) = 0 then Ais a field so B is trivially flat so throughout we may assume dim(A) ge 1 Since A is normal thegoing-down theorem holds between A and B by Theorem 42 so by Theorem 55 (3) for any properideal I of A IB is proper and htI = htIB We claim that depthA(A) = depthA(B) Notice thatdepthA(B) is finite since otherwise m isin AssA(B) and hence dim(A) = 0

Firstly we prove the inequality le Since A is regular it is Cohen-Macaulay so depthA(A) =dim(A) Set s = dim(A) and let a1 as be a regular system of parameters Then ht(a1 ai)A =i and therefore ht(a1 ai)B = i for all 1 le i le s by Theorem 108 It follows from Corollary97 that a1 as is a B-regular sequence and therefore depthA(B) ge s

To prove the reverse inequality set d = depthA(B) and let a1 ad isin m be a maximal B-regular sequence Then as elements of B the sequence a1 ad is B-regular so by Lemma 74 wehave ht(a1 ad) = d But (a1 ad) sube mB and htmB = htm = dim(A) so d le dim(A) asrequired

Since gldim(A) lt infin we have projdimA(B) lt infin so we can apply Theorem 131 to see thatprojdimA(B) + depthA(B) = depthA(A) so projdimA(B) = 0 and therefore B is projectiveSince A is local and B finitely generated projective hArr free hArr flat so the proof is complete

64 Unique Factorisation

Recall that if A is a ring two elements p q isin A are said to be associates if p = uq for some unitu isin A This is an equivalence relation on the elements of A

Definition 29 Let A be an integral domain An element of A is irreducible if it is a nonzerononunit which cannot be written as the product of two nonunits An element p isin A is prime if it isa nonzero nonunit with the property that if p|ab then p|a or p|b Equivalently p is prime iff (p) isa nonzero prime ideal We say A is a unique factorisation domain if every nonzero nonunit a isin Acan be written essentially uniquely as a = up1 middot middot middot pr where u is a unit and each pi is irreducible

52

Essentially uniquely means that if a = vq1 middot middot middot qs where v is a unit and the qj irreducible thenr = s and after reordering (if necessary) qi is an associate of pi The property of being a UFD isstable under ring isomorphism

Theorem 152 A noetherian domain A is a UFD if and only if every prime ideal of height 1 isprincipal

Lemma 153 Let A be a noetherian domain and let x isin A be prime Then A is a UFD if andonly if Ax is

Proof By assumption (x) is a prime ideal of height 1 If p is a prime ideal of height 1 then eitherx isin p in which case p = (x) or x isin p and these primes are in bijection with the primes of Ax Sousing Theorem 152 it is clear that if A is a UFD so is Ax Suppose that Ax is a UFD and let p bea prime ideal of height 1 in A We can assume that x isin p Let a isin p be such that pAx = a1AxBy [AM69] Corollary 1018 we have capi(xi) = 0 so if x|a there is a largest integer n ge 1 with xn|aWrite a = cxn Since x isin p we have c isin p so by replacing a with c we can assume pAx = a1Axwith a isin (x) Then it is clear that p = (a) as required

Definition 30 Let R be an integral domain If M is a torsion-free R-module then the rank ofM is the maximum number of linearly independent elements in M rank(M) isin 0 1 infin

Proposition 154 Let R be an integral domain and M a torsion-free R-module If T sube Ris multiplicatively closed then Tminus1M is a torsion-free Tminus1R-module and rankTminus1R(Tminus1M) =rankR(M)

Proof If rank(M) = 0 this is trivial so assumeM is nonzero It is clear that Tminus1M is torsion-freeIf rankR(M) = r and x1 xr isin M are linearly independent then x11 xr1 isin Tminus1M arelinearly independent over Tminus1R Similarly if x1s1 xnsn isin Tminus1M are linearly independentin Tminus1M then x1 xn are linearly independent in M So the result is clear

Corollary 155 Let R be an integral domain with quotient field K Then

(i) If M is a torsion-free R-module rankR(M) = dimK(M otimesK)

(ii) If MN are two torsion-free R-modules of finite rank then rankR(M oplusN) = rankR(M) +rankR(N)

In particular if M is a free R-module then the rank just defined is equal to the normal free rankand we can write rank(M) without confusion

Let R be a noetherian domain and suppose a1 an isin R are linearly independent elementswhich do not generate R Then a1 ar is an R-regular sequence so by Lemma 74 the ideals(a1 ai) have height i for 1 le i le n So it follows immediately that

Lemma 156 Let R be a noetherian domain and I an ideal Then rank(I) le htI

Lemma 157 Let R be a domain and M a finitely generated projective R-module of rank 1 ThenandiM = 0 for i gt 1

Proof By localisation If p is a prime ideal then Mp is a finitely generated projective module overthe local ring Rp so Mp is free of rank 1 and Mp

sim= Rp Hence for i gt 1

(andiM)psim= andiMp

sim= andiRq = 0

as required

Theorem 158 (Auslander-Buchsbaum) A regular local ring (Am) is UFD

53

Proof We use induction on dimA If dimA = 0 then A is a field and if dimA = 1 then A is aprincipal ideal domain Suppose dimA gt 1 and let a1 ad be a regular system of parametersThen x = a1 is prime by Theorem 108 so it suffices by Lemma 153 to show that Ax is UFDLet q be a prime ideal of height 1 in Ax and put p = q cap A so q = pAx By Theorem 128gldimA = dimA lt infin so we can produce an exact sequence of A-modules with all Fi finitelygenerated free

0 minusrarr Fn minusrarr Fnminus1 minusrarr middot middot middot minusrarr F0 minusrarr p minusrarr 0 (6)

Maximal ideals of Ax correspond to primes of A maximal among those not containing x Theseprimes must all be properly contained in m so if PAx is a maximal ideal then htP lt dimATherefore (Ax)PAx

sim= AP is UFD by the inductive assumption and so q(Ax)n is either principalor zero for every maximal n of Ax Then by Lemma 124 we have projdimAx(q) = 0 and thereforeq is projective Localising (6) with respect to S = 1 x x2 we see that the following sequenceof Ax-modules is exact

0 minusrarr F primen minusrarr F primenminus1 minusrarr middot middot middot minusrarr F prime0 minusrarr q minusrarr 0 (7)

where F primei = Fi otimesAx are finitely generated and free over Ax If we decompose (7) into short exactsequences

0 minusrarr K prime0 minusrarrF prime0 minusrarr q minusrarr 0

0 minusrarr K prime1 minusrarrF prime1 minusrarr K prime

0 minusrarr 0

0 minusrarr F primen minusrarrF primenminus1 minusrarr K primenminus2 minusrarr 0

(8)

then the first sequence splits since q is projective Hence K prime0 must be projective and in this way

we show that all the sequences split and all the K primei are projective It follows thatoplus

i even

F primeisim=oplusi odd

F primei oplus q

Thus we have finite free Ax-modules FG such that F sim= Goplus q Since Ax is a noetherian domainand q a nonzero ideal of height 1 it follows from Lemma 156 that rank(q) = 1 If rank(G) = rthen rank(F ) = r + 1

So to show q is principal and complete the proof it suffices to show that q is free But by ournotes on Tensor Symmetric and Exterior algebras we have

Ax sim=r+1and

F sim=r+1and

(Goplus q) sim=oplus

i+j=r+1

(andiG)otimes (andjq) sim= (andr+1Gotimes and0q)oplus (andrGotimes and1q) sim= q

Since andr+1G = 0andrG sim= Ax and andiq = 0 for i gt 1 by Lemma 157

References

[AM69] M F Atiyah and I G Macdonald Introduction to commutative algebra Addison-WesleyPublishing Co Reading Mass-London-Don Mills Ont 1969 MR MR0242802 (394129)

[Ash] Robert Ash A course in commutative algebra httpwwwmathuiucedusimr-ashComAlghtml

[Eis95] David Eisenbud Commutative algebra Graduate Texts in Mathematics vol 150Springer-Verlag New York 1995 With a view toward algebraic geometry MRMR1322960 (97a13001)

[Mat80] Hideyuki Matsumura Commutative algebra second ed Mathematics Lecture Note Se-ries vol 56 BenjaminCummings Publishing Co Inc Reading Mass 1980 MRMR575344 (82i13003)

54

• General Rings
• Flatness
• • Faithful Flatness
  • Going-up and Going-down
  • Constructible Sets
  • • Associated Primes
    • Dimension
    • • Homomorphism and Dimension
      • • Depth
        • • Cohen-Macaulay Rings
          • • Normal and Regular Rings
            • • Classical Theory
              • Homological Theory
              • Koszul Complexes
              • Unique Factorisation

Proof By Lemma 140 (ii) we can find a minimal morphism d0 L0 minusrarr M with L0 finite free ofrank rankk(M otimes k) Let K minusrarr L0 be the kernel of d0 Find a minimal morphism L1 minusrarr K withL1 finite free and so on This defines a minimal free resolution of M To prove the uniquenesslet ε L minusrarrM and εprime Lprime minusrarrM be two minimal free resolutions of M We can lift the identity1M to a morphism of chain complexes ϕ L minusrarr Lprime so we have a commutative diagram

middot middot middot L1

ϕ1

L0

ϕ0

ε

M

middot middot middot Lprime1 Lprime0

εprime M

Since ε εprime are minimal the map ϕ0 otimes 1 L0 otimes k minusrarr Lprime0 otimes k is an isomorphism of k-modules Inparticular we have

rankAL0 = rankk(L0 otimes k) = rankk(Lprime0 otimes k) = rankALprime0

So L0 Lprime0 are free of the same finite rank We claim that ϕ0 is an isomorphism This is trivial

if L0 = Lprime0 = 0 so assume they are both nonzero Then ϕ0 is described by a square matrixT isin Mn(A) If you take residues you get the matrix T prime isin Mn(k) of ϕ0 otimes 1 which has nonzerodeterminant since it is an isomorphism But it is clear that det(T ) + m = det(T prime) so det(T ) isin mTherefore ϕ0 itself is an isomorphism

Since ϕ0 is an isomorphism so the induced morphism on the kernels Ker(ε) minusrarr Ker(εprime) andwe can repeat the same argument to see that ϕ1 is an isomorphism and similarly to show thatall the ϕi are isomorphisms

Lemma 142 Let (Am k) be a noetherian local ring and u F minusrarr G a morphism of finitelygenerated free A-modules Then u is minimal if and only if it is an isomorphism

Definition 27 Let (Am k) be a noetherian local ring and M a finitely generated A-moduleChoose a minimal free resolution of M Then for i ge 0 the integer bi = rankALi ge 0 iscalled the i-th Betti number of M It is independent of the chosen resolution and moreoverrankkTor

Ai (Mk) = bi

Example 7 Let (Am k) be a noetherian local ring and let M be a finitely generated A-moduleThen

(i) Proposition 141 shows that b0 = rankk(M otimes k)

(ii) If M = 0 then the zero complex is a minimal free resolution of M so bi = 0 for i ge 0

(iii) If M is flat then TorAi (Mk) = 0 for all i ge 1 so bi = 0 for i ge 1 In particular this is trueif M is free or projective

(iv) If M is free of finite rank s ge 1 then M otimes k is a free k-module of rank s so b0 = s

Lemma 143 Let (Am k) be a noetherian local ring and M a finitely generated A-moduleSuppose that we have two complexes LF together with morphisms ε εprime such that the followingsequences are exact in the last two nonzero positions

L middot middot middot Lidi Liminus1

middot middot middot d1 L0ε M 0

F middot middot middot Fidprimei Fiminus1

middot middot middotdprime1 F0

εprime M 0

Assume the following

(i) L is a minimal free resolution of M

(ii) Each Fi is a finitely generated free A-module

49

(iii) εprime otimes 1 F0 otimes k minusrarrM otimes k is injective

(iv) For each i ge 0 dprimei+1(Fi+1) sube mFi and the induced morphism Fi+1mFi+1 minusrarr mFim2Fi is

an injection

Then there exists a morphism of complexes f F minusrarr L lifting the identity 1M such that for fimaps Fi isomorphically onto a direct summand of Li Consequently we have

rankAFi le rankALi = rankkTorAi (Mk)

Proof Both LF are positive chain complexes with F projective and L acyclic so by our DerivedFunctor notes there is a morphism f F minusrarr L of chain complexes giving a commutative diagram

middot middot middot F0

f0

εprime M

0

middot middot middot L0 ε M 0

We have to prove that for each i ge 0 the morphism fi Fi minusrarr Li is a coretractionWe claim thatfi is a coretraction iff fi otimes 1 Fi otimes k minusrarr Li otimes k is an injective morphism of k-modules Oneimplication is clear So assume that fi otimes 1 is injective The claim is trivial if either of Fi Li arezero so assume they are both of nonzero finite rank Pick bases for Fi Li (which are obviouslyminimal bases) and use the fact that fi otimes 1 is a coretraction to define a morphism ϕ Li minusrarr Fisuch that (ϕfi) otimes 1 Fi otimes k minusrarr Fi otimes k is the identity By Lemma 142 it follows that ϕfi is anisomorphism and therefore clearly fi is a coretraction

We prove by induction that fi otimes 1 is injective for all i ge 0 By assumptions (i) (iii) it is clearthat f0 otimes 1 is injective We have the following commutative diagram

F1

β FFFF

FFFF

F

f1

dprime1 F0

f0

εprime M

Kerεprime

xxxxxxxxx

α

Kerε

FFFF

FFFF

F

L1

γxxxxxxxxx

d1

L0 ε M

By assumption γ otimes 1 is an isomorphism So to show f1 otimes 1 is injective it suffices to show thatαotimes 1 β otimes 1 are injective or equivalently that αminus1(mKerε) = mKerεprime and βminus1(mKerεprime) = mF1Suppose that a isin F1 and dprime1(a) isin mKerεprime Since εprime otimes 1 is injective we have mKerεprime sube m2F0Hence dprime1(a) isin m2F0 and therefore by (iv) a isin mF1 as required

Now suppose that a isin Kerεprime and f0(a) isin mKerε Let g be such that gf0 = 1 Then f0(a) isinm2L0 and therefore a = gf0(a) isin g(m2L0) sube m2F0 Let b isin F1 be such that dprime1(b) = a isin m2F0Then (iv) implies that b isin mF1 and therefore a = β(b) isin mKerεprime as required This shows thatf1 otimes 1 is injective

Suppose that fi otimes 1 is injective for some i ge 1 Then we show fi+1 otimes 1 is injective using asimilar setup We replace Kerεprime by Imdprimei+1 (in the case i = 0 they are equal) and use (iv) toshow that Kerdprimei sube mFi and (i) to show that Kerdi sube mLi The proof that β otimes 1 is injective isstraightforward For αotimes 1 let a isin Imdprimei+1 be such that fi(a) isin mKerdi As before we find thatfi(a) isin m2Li and hence a = gfi(a) isin m2Fi Let b isin Fi+1 be such that a = dprimei+1(b) Then by (iv)b isin mFi+1 and therefore a isin mImdprimei+1 as required This proves that fi is a coretraction for i ge 0and the rank claim follows from the fact that rankAFi = rankk(Fi otimes k) le rankk(Li otimes k)

50

Lemma 144 Let A be a ring with maximal ideal m If s isin m and a isin A then sa isin mk impliesa isin mk for any k ge 1

Theorem 145 Let (Am k) be a noetherian local ring and let s = rankkmm2 Then we have

rankkTorAi (k k) ge

(s

i

)0 le i le s

Here rankkTorAi (k k) is the i-th Betti number of the A-module k

Proof We have TorA0 (k k) sim= k as k-modules so rankkTorA0 (k k) = rankkk = 1 which takes

care of the case s = 0 So assume that s ge 1 and let a1 as be a minimal basis of m withassociated Koszul complex F = K(a1 as) The canonical morphism εprime F0

sim= A minusrarr k gives acomplex exact in the last two nonzero places

middot middot middot minusrarr Fi minusrarr Fiminus1 minusrarr middot middot middot minusrarr F1 minusrarr F0 minusrarr k minusrarr 0

We claim this complex satisfies the conditions of Lemma 143 It clearly satisfies (ii) and (iii)It only remains to check condition (iii) By the definition of dp+1 Fp+1 minusrarr Fp it is clear thatdp+1(Fp+1) sube mFp for p ge 0 We also have to show that dminus1

p+1(m2Fp) sube mFp+1 This is trivial

if p + 1 gt s and also if p = 0 since a1 as is a minimal basis So assume 0 lt p le s minus 1Assume that

dp+1

sumi1ltmiddotmiddotmiddotltip+1

mi1middotmiddotmiddotip+1(xi1 and middot middot middot and xip+1)

=

sumi1ltmiddotmiddotmiddotltip+1

psumr=1

(minus1)rminus1airmi1middotmiddotmiddotip+1(xi1 and middot middot middot and xir and middot middot middot and xip+1) isin m2Fp

Then collecting terms we obtain a number of equations of the formsum

(minus1)etatmt isin m2 where atis one of the air and mt one of the mi1middotmiddotmiddotip+1 Since the residues of the ai give a basis of mm2

over k it follows that mt isin m which completes the proof that F satisfies all the conditions ofLemma 143 Choosing any minimal free resolution L of M and applying Lemma 143 we see thatfor 0 le i le s (

s

i

)= rankAFi le rankkTor

Ai (k k)

as required

Theorem 146 (Serre) Let (Am k) be a noetherian local ring Then A is regular if and only ifthe global dimension of A is finite

Proof We have already proved one part in Theorem 128 So suppose that gldim(A) lt infinThen TorAs (k k) 6= 0 by Theorem 145 hence gldim(A) ge rankkmm

2 since by Proposition122 we have tordim(A) = gldim(A) On the other hand it follows from Theorem 125 thatprojdim(k) = gldim(A) ltinfin so by Theorem 131 we have gldim(A) = projdim(k) = depth(A)Therefore we get

dim(A) le rankkmm2 le gldim(A) = depth(A) le dim(A)

Whence dim(A) = rankkmm2 and A is regular

Corollary 147 If A is a regular local ring then Ap is regular for any p isin Spec(A)

Proof Let M be a nonzero Ap-module Then considering M as an A-module there is an exactsequence of finite length n le gldim(A) with all Pi projective

0 minusrarr Pn minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

51

Since Ap is flat the following sequence is also exact

0 minusrarr (Pn)p minusrarr middot middot middot minusrarr (P0)p minusrarrMp minusrarr 0

The modules (Pi)p are projective Ap-modules and M sim= Mp as Ap-modules so it follows thatgldim(Ap) le gldim(A) ltinfin

Definition 28 A ring A is called a regular ring if Ap is a regular local ring for every prime idealp of A Note that A is not required to be noetherian Regularity is stable under ring isomorphismA noetherian local ring A is regular in this sense if and only if it is regular in the normal sense

It follows from Theorem 108 that any regular ring is normal and a noetherian regular ring isCohen-Macaulay It follows from Lemma 8 and Theorem 90 that a regular ring is catenary

Lemma 148 A ring A is regular if and only if Am is regular for all maximal ideals m

Proof One implication is clear For the other given a prime ideal p find a maximal ideal m withp sube m Then Ap

sim= (Am)pAm so Ap is a regular local ring

Lemma 149 If A is a regular ring and S sube A is multiplicatively closed then Sminus1A is a regularring

Lemma 150 If A is a regular ring then so is A[x1 xn] In particular k[x1 xn] is aregular ring for any field k

Proof This follows immediately from Theorem 118

Theorem 151 Let A be a regular local ring which is a subring of a domain B and suppose thatB is a finitely generated A-module Then B is flat (equivalently free) over A if and only if B isCohen-Macaulay In particular if B is regular then it is a free A-module

Proof Since B is a finitely generated A-module it is integral over A and so by Lemma 91 if B isflat it is Cohen-Macaulay Conversely suppose that B is Cohen-Macaulay If dim(A) = 0 then Ais a field so B is trivially flat so throughout we may assume dim(A) ge 1 Since A is normal thegoing-down theorem holds between A and B by Theorem 42 so by Theorem 55 (3) for any properideal I of A IB is proper and htI = htIB We claim that depthA(A) = depthA(B) Notice thatdepthA(B) is finite since otherwise m isin AssA(B) and hence dim(A) = 0

Firstly we prove the inequality le Since A is regular it is Cohen-Macaulay so depthA(A) =dim(A) Set s = dim(A) and let a1 as be a regular system of parameters Then ht(a1 ai)A =i and therefore ht(a1 ai)B = i for all 1 le i le s by Theorem 108 It follows from Corollary97 that a1 as is a B-regular sequence and therefore depthA(B) ge s

To prove the reverse inequality set d = depthA(B) and let a1 ad isin m be a maximal B-regular sequence Then as elements of B the sequence a1 ad is B-regular so by Lemma 74 wehave ht(a1 ad) = d But (a1 ad) sube mB and htmB = htm = dim(A) so d le dim(A) asrequired

Since gldim(A) lt infin we have projdimA(B) lt infin so we can apply Theorem 131 to see thatprojdimA(B) + depthA(B) = depthA(A) so projdimA(B) = 0 and therefore B is projectiveSince A is local and B finitely generated projective hArr free hArr flat so the proof is complete

64 Unique Factorisation

Recall that if A is a ring two elements p q isin A are said to be associates if p = uq for some unitu isin A This is an equivalence relation on the elements of A

Definition 29 Let A be an integral domain An element of A is irreducible if it is a nonzerononunit which cannot be written as the product of two nonunits An element p isin A is prime if it isa nonzero nonunit with the property that if p|ab then p|a or p|b Equivalently p is prime iff (p) isa nonzero prime ideal We say A is a unique factorisation domain if every nonzero nonunit a isin Acan be written essentially uniquely as a = up1 middot middot middot pr where u is a unit and each pi is irreducible

52

Essentially uniquely means that if a = vq1 middot middot middot qs where v is a unit and the qj irreducible thenr = s and after reordering (if necessary) qi is an associate of pi The property of being a UFD isstable under ring isomorphism

Theorem 152 A noetherian domain A is a UFD if and only if every prime ideal of height 1 isprincipal

Lemma 153 Let A be a noetherian domain and let x isin A be prime Then A is a UFD if andonly if Ax is

Proof By assumption (x) is a prime ideal of height 1 If p is a prime ideal of height 1 then eitherx isin p in which case p = (x) or x isin p and these primes are in bijection with the primes of Ax Sousing Theorem 152 it is clear that if A is a UFD so is Ax Suppose that Ax is a UFD and let p bea prime ideal of height 1 in A We can assume that x isin p Let a isin p be such that pAx = a1AxBy [AM69] Corollary 1018 we have capi(xi) = 0 so if x|a there is a largest integer n ge 1 with xn|aWrite a = cxn Since x isin p we have c isin p so by replacing a with c we can assume pAx = a1Axwith a isin (x) Then it is clear that p = (a) as required

Definition 30 Let R be an integral domain If M is a torsion-free R-module then the rank ofM is the maximum number of linearly independent elements in M rank(M) isin 0 1 infin

Proposition 154 Let R be an integral domain and M a torsion-free R-module If T sube Ris multiplicatively closed then Tminus1M is a torsion-free Tminus1R-module and rankTminus1R(Tminus1M) =rankR(M)

Proof If rank(M) = 0 this is trivial so assumeM is nonzero It is clear that Tminus1M is torsion-freeIf rankR(M) = r and x1 xr isin M are linearly independent then x11 xr1 isin Tminus1M arelinearly independent over Tminus1R Similarly if x1s1 xnsn isin Tminus1M are linearly independentin Tminus1M then x1 xn are linearly independent in M So the result is clear

Corollary 155 Let R be an integral domain with quotient field K Then

(i) If M is a torsion-free R-module rankR(M) = dimK(M otimesK)

(ii) If MN are two torsion-free R-modules of finite rank then rankR(M oplusN) = rankR(M) +rankR(N)

In particular if M is a free R-module then the rank just defined is equal to the normal free rankand we can write rank(M) without confusion

Let R be a noetherian domain and suppose a1 an isin R are linearly independent elementswhich do not generate R Then a1 ar is an R-regular sequence so by Lemma 74 the ideals(a1 ai) have height i for 1 le i le n So it follows immediately that

Lemma 156 Let R be a noetherian domain and I an ideal Then rank(I) le htI

Lemma 157 Let R be a domain and M a finitely generated projective R-module of rank 1 ThenandiM = 0 for i gt 1

Proof By localisation If p is a prime ideal then Mp is a finitely generated projective module overthe local ring Rp so Mp is free of rank 1 and Mp

sim= Rp Hence for i gt 1

(andiM)psim= andiMp

sim= andiRq = 0

as required

Theorem 158 (Auslander-Buchsbaum) A regular local ring (Am) is UFD

53

Proof We use induction on dimA If dimA = 0 then A is a field and if dimA = 1 then A is aprincipal ideal domain Suppose dimA gt 1 and let a1 ad be a regular system of parametersThen x = a1 is prime by Theorem 108 so it suffices by Lemma 153 to show that Ax is UFDLet q be a prime ideal of height 1 in Ax and put p = q cap A so q = pAx By Theorem 128gldimA = dimA lt infin so we can produce an exact sequence of A-modules with all Fi finitelygenerated free

0 minusrarr Fn minusrarr Fnminus1 minusrarr middot middot middot minusrarr F0 minusrarr p minusrarr 0 (6)

Maximal ideals of Ax correspond to primes of A maximal among those not containing x Theseprimes must all be properly contained in m so if PAx is a maximal ideal then htP lt dimATherefore (Ax)PAx

sim= AP is UFD by the inductive assumption and so q(Ax)n is either principalor zero for every maximal n of Ax Then by Lemma 124 we have projdimAx(q) = 0 and thereforeq is projective Localising (6) with respect to S = 1 x x2 we see that the following sequenceof Ax-modules is exact

0 minusrarr F primen minusrarr F primenminus1 minusrarr middot middot middot minusrarr F prime0 minusrarr q minusrarr 0 (7)

where F primei = Fi otimesAx are finitely generated and free over Ax If we decompose (7) into short exactsequences

0 minusrarr K prime0 minusrarrF prime0 minusrarr q minusrarr 0

0 minusrarr K prime1 minusrarrF prime1 minusrarr K prime

0 minusrarr 0

0 minusrarr F primen minusrarrF primenminus1 minusrarr K primenminus2 minusrarr 0

(8)

then the first sequence splits since q is projective Hence K prime0 must be projective and in this way

we show that all the sequences split and all the K primei are projective It follows thatoplus

i even

F primeisim=oplusi odd

F primei oplus q

Thus we have finite free Ax-modules FG such that F sim= Goplus q Since Ax is a noetherian domainand q a nonzero ideal of height 1 it follows from Lemma 156 that rank(q) = 1 If rank(G) = rthen rank(F ) = r + 1

So to show q is principal and complete the proof it suffices to show that q is free But by ournotes on Tensor Symmetric and Exterior algebras we have

Ax sim=r+1and

F sim=r+1and

(Goplus q) sim=oplus

i+j=r+1

(andiG)otimes (andjq) sim= (andr+1Gotimes and0q)oplus (andrGotimes and1q) sim= q

Since andr+1G = 0andrG sim= Ax and andiq = 0 for i gt 1 by Lemma 157

References

[AM69] M F Atiyah and I G Macdonald Introduction to commutative algebra Addison-WesleyPublishing Co Reading Mass-London-Don Mills Ont 1969 MR MR0242802 (394129)

[Ash] Robert Ash A course in commutative algebra httpwwwmathuiucedusimr-ashComAlghtml

[Eis95] David Eisenbud Commutative algebra Graduate Texts in Mathematics vol 150Springer-Verlag New York 1995 With a view toward algebraic geometry MRMR1322960 (97a13001)

[Mat80] Hideyuki Matsumura Commutative algebra second ed Mathematics Lecture Note Se-ries vol 56 BenjaminCummings Publishing Co Inc Reading Mass 1980 MRMR575344 (82i13003)

54

• General Rings
• Flatness
• • Faithful Flatness
  • Going-up and Going-down
  • Constructible Sets
  • • Associated Primes
    • Dimension
    • • Homomorphism and Dimension
      • • Depth
        • • Cohen-Macaulay Rings
          • • Normal and Regular Rings
            • • Classical Theory
              • Homological Theory
              • Koszul Complexes
              • Unique Factorisation

(iii) εprime otimes 1 F0 otimes k minusrarrM otimes k is injective

(iv) For each i ge 0 dprimei+1(Fi+1) sube mFi and the induced morphism Fi+1mFi+1 minusrarr mFim2Fi is

an injection

Then there exists a morphism of complexes f F minusrarr L lifting the identity 1M such that for fimaps Fi isomorphically onto a direct summand of Li Consequently we have

rankAFi le rankALi = rankkTorAi (Mk)

Proof Both LF are positive chain complexes with F projective and L acyclic so by our DerivedFunctor notes there is a morphism f F minusrarr L of chain complexes giving a commutative diagram

middot middot middot F0

f0

εprime M

0

middot middot middot L0 ε M 0

We have to prove that for each i ge 0 the morphism fi Fi minusrarr Li is a coretractionWe claim thatfi is a coretraction iff fi otimes 1 Fi otimes k minusrarr Li otimes k is an injective morphism of k-modules Oneimplication is clear So assume that fi otimes 1 is injective The claim is trivial if either of Fi Li arezero so assume they are both of nonzero finite rank Pick bases for Fi Li (which are obviouslyminimal bases) and use the fact that fi otimes 1 is a coretraction to define a morphism ϕ Li minusrarr Fisuch that (ϕfi) otimes 1 Fi otimes k minusrarr Fi otimes k is the identity By Lemma 142 it follows that ϕfi is anisomorphism and therefore clearly fi is a coretraction

We prove by induction that fi otimes 1 is injective for all i ge 0 By assumptions (i) (iii) it is clearthat f0 otimes 1 is injective We have the following commutative diagram

F1

β FFFF

FFFF

F

f1

dprime1 F0

f0

εprime M

Kerεprime

xxxxxxxxx

α

Kerε

FFFF

FFFF

F

L1

γxxxxxxxxx

d1

L0 ε M

By assumption γ otimes 1 is an isomorphism So to show f1 otimes 1 is injective it suffices to show thatαotimes 1 β otimes 1 are injective or equivalently that αminus1(mKerε) = mKerεprime and βminus1(mKerεprime) = mF1Suppose that a isin F1 and dprime1(a) isin mKerεprime Since εprime otimes 1 is injective we have mKerεprime sube m2F0Hence dprime1(a) isin m2F0 and therefore by (iv) a isin mF1 as required

Now suppose that a isin Kerεprime and f0(a) isin mKerε Let g be such that gf0 = 1 Then f0(a) isinm2L0 and therefore a = gf0(a) isin g(m2L0) sube m2F0 Let b isin F1 be such that dprime1(b) = a isin m2F0Then (iv) implies that b isin mF1 and therefore a = β(b) isin mKerεprime as required This shows thatf1 otimes 1 is injective

Suppose that fi otimes 1 is injective for some i ge 1 Then we show fi+1 otimes 1 is injective using asimilar setup We replace Kerεprime by Imdprimei+1 (in the case i = 0 they are equal) and use (iv) toshow that Kerdprimei sube mFi and (i) to show that Kerdi sube mLi The proof that β otimes 1 is injective isstraightforward For αotimes 1 let a isin Imdprimei+1 be such that fi(a) isin mKerdi As before we find thatfi(a) isin m2Li and hence a = gfi(a) isin m2Fi Let b isin Fi+1 be such that a = dprimei+1(b) Then by (iv)b isin mFi+1 and therefore a isin mImdprimei+1 as required This proves that fi is a coretraction for i ge 0and the rank claim follows from the fact that rankAFi = rankk(Fi otimes k) le rankk(Li otimes k)

50

Lemma 144 Let A be a ring with maximal ideal m If s isin m and a isin A then sa isin mk impliesa isin mk for any k ge 1

Theorem 145 Let (Am k) be a noetherian local ring and let s = rankkmm2 Then we have

rankkTorAi (k k) ge

(s

i

)0 le i le s

Here rankkTorAi (k k) is the i-th Betti number of the A-module k

Proof We have TorA0 (k k) sim= k as k-modules so rankkTorA0 (k k) = rankkk = 1 which takes

care of the case s = 0 So assume that s ge 1 and let a1 as be a minimal basis of m withassociated Koszul complex F = K(a1 as) The canonical morphism εprime F0

sim= A minusrarr k gives acomplex exact in the last two nonzero places

middot middot middot minusrarr Fi minusrarr Fiminus1 minusrarr middot middot middot minusrarr F1 minusrarr F0 minusrarr k minusrarr 0

We claim this complex satisfies the conditions of Lemma 143 It clearly satisfies (ii) and (iii)It only remains to check condition (iii) By the definition of dp+1 Fp+1 minusrarr Fp it is clear thatdp+1(Fp+1) sube mFp for p ge 0 We also have to show that dminus1

p+1(m2Fp) sube mFp+1 This is trivial

if p + 1 gt s and also if p = 0 since a1 as is a minimal basis So assume 0 lt p le s minus 1Assume that

dp+1

sumi1ltmiddotmiddotmiddotltip+1

mi1middotmiddotmiddotip+1(xi1 and middot middot middot and xip+1)

=

sumi1ltmiddotmiddotmiddotltip+1

psumr=1

(minus1)rminus1airmi1middotmiddotmiddotip+1(xi1 and middot middot middot and xir and middot middot middot and xip+1) isin m2Fp

Then collecting terms we obtain a number of equations of the formsum

(minus1)etatmt isin m2 where atis one of the air and mt one of the mi1middotmiddotmiddotip+1 Since the residues of the ai give a basis of mm2

over k it follows that mt isin m which completes the proof that F satisfies all the conditions ofLemma 143 Choosing any minimal free resolution L of M and applying Lemma 143 we see thatfor 0 le i le s (

s

i

)= rankAFi le rankkTor

Ai (k k)

as required

Theorem 146 (Serre) Let (Am k) be a noetherian local ring Then A is regular if and only ifthe global dimension of A is finite

Proof We have already proved one part in Theorem 128 So suppose that gldim(A) lt infinThen TorAs (k k) 6= 0 by Theorem 145 hence gldim(A) ge rankkmm

2 since by Proposition122 we have tordim(A) = gldim(A) On the other hand it follows from Theorem 125 thatprojdim(k) = gldim(A) ltinfin so by Theorem 131 we have gldim(A) = projdim(k) = depth(A)Therefore we get

dim(A) le rankkmm2 le gldim(A) = depth(A) le dim(A)

Whence dim(A) = rankkmm2 and A is regular

Corollary 147 If A is a regular local ring then Ap is regular for any p isin Spec(A)

Proof Let M be a nonzero Ap-module Then considering M as an A-module there is an exactsequence of finite length n le gldim(A) with all Pi projective

0 minusrarr Pn minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

51

Since Ap is flat the following sequence is also exact

0 minusrarr (Pn)p minusrarr middot middot middot minusrarr (P0)p minusrarrMp minusrarr 0

The modules (Pi)p are projective Ap-modules and M sim= Mp as Ap-modules so it follows thatgldim(Ap) le gldim(A) ltinfin

Definition 28 A ring A is called a regular ring if Ap is a regular local ring for every prime idealp of A Note that A is not required to be noetherian Regularity is stable under ring isomorphismA noetherian local ring A is regular in this sense if and only if it is regular in the normal sense

It follows from Theorem 108 that any regular ring is normal and a noetherian regular ring isCohen-Macaulay It follows from Lemma 8 and Theorem 90 that a regular ring is catenary

Lemma 148 A ring A is regular if and only if Am is regular for all maximal ideals m

Proof One implication is clear For the other given a prime ideal p find a maximal ideal m withp sube m Then Ap

sim= (Am)pAm so Ap is a regular local ring

Lemma 149 If A is a regular ring and S sube A is multiplicatively closed then Sminus1A is a regularring

Lemma 150 If A is a regular ring then so is A[x1 xn] In particular k[x1 xn] is aregular ring for any field k

Proof This follows immediately from Theorem 118

Theorem 151 Let A be a regular local ring which is a subring of a domain B and suppose thatB is a finitely generated A-module Then B is flat (equivalently free) over A if and only if B isCohen-Macaulay In particular if B is regular then it is a free A-module

Proof Since B is a finitely generated A-module it is integral over A and so by Lemma 91 if B isflat it is Cohen-Macaulay Conversely suppose that B is Cohen-Macaulay If dim(A) = 0 then Ais a field so B is trivially flat so throughout we may assume dim(A) ge 1 Since A is normal thegoing-down theorem holds between A and B by Theorem 42 so by Theorem 55 (3) for any properideal I of A IB is proper and htI = htIB We claim that depthA(A) = depthA(B) Notice thatdepthA(B) is finite since otherwise m isin AssA(B) and hence dim(A) = 0

Firstly we prove the inequality le Since A is regular it is Cohen-Macaulay so depthA(A) =dim(A) Set s = dim(A) and let a1 as be a regular system of parameters Then ht(a1 ai)A =i and therefore ht(a1 ai)B = i for all 1 le i le s by Theorem 108 It follows from Corollary97 that a1 as is a B-regular sequence and therefore depthA(B) ge s

To prove the reverse inequality set d = depthA(B) and let a1 ad isin m be a maximal B-regular sequence Then as elements of B the sequence a1 ad is B-regular so by Lemma 74 wehave ht(a1 ad) = d But (a1 ad) sube mB and htmB = htm = dim(A) so d le dim(A) asrequired

Since gldim(A) lt infin we have projdimA(B) lt infin so we can apply Theorem 131 to see thatprojdimA(B) + depthA(B) = depthA(A) so projdimA(B) = 0 and therefore B is projectiveSince A is local and B finitely generated projective hArr free hArr flat so the proof is complete

64 Unique Factorisation

Recall that if A is a ring two elements p q isin A are said to be associates if p = uq for some unitu isin A This is an equivalence relation on the elements of A

Definition 29 Let A be an integral domain An element of A is irreducible if it is a nonzerononunit which cannot be written as the product of two nonunits An element p isin A is prime if it isa nonzero nonunit with the property that if p|ab then p|a or p|b Equivalently p is prime iff (p) isa nonzero prime ideal We say A is a unique factorisation domain if every nonzero nonunit a isin Acan be written essentially uniquely as a = up1 middot middot middot pr where u is a unit and each pi is irreducible

52

Essentially uniquely means that if a = vq1 middot middot middot qs where v is a unit and the qj irreducible thenr = s and after reordering (if necessary) qi is an associate of pi The property of being a UFD isstable under ring isomorphism

Theorem 152 A noetherian domain A is a UFD if and only if every prime ideal of height 1 isprincipal

Lemma 153 Let A be a noetherian domain and let x isin A be prime Then A is a UFD if andonly if Ax is

Proof By assumption (x) is a prime ideal of height 1 If p is a prime ideal of height 1 then eitherx isin p in which case p = (x) or x isin p and these primes are in bijection with the primes of Ax Sousing Theorem 152 it is clear that if A is a UFD so is Ax Suppose that Ax is a UFD and let p bea prime ideal of height 1 in A We can assume that x isin p Let a isin p be such that pAx = a1AxBy [AM69] Corollary 1018 we have capi(xi) = 0 so if x|a there is a largest integer n ge 1 with xn|aWrite a = cxn Since x isin p we have c isin p so by replacing a with c we can assume pAx = a1Axwith a isin (x) Then it is clear that p = (a) as required

Definition 30 Let R be an integral domain If M is a torsion-free R-module then the rank ofM is the maximum number of linearly independent elements in M rank(M) isin 0 1 infin

Proposition 154 Let R be an integral domain and M a torsion-free R-module If T sube Ris multiplicatively closed then Tminus1M is a torsion-free Tminus1R-module and rankTminus1R(Tminus1M) =rankR(M)

Proof If rank(M) = 0 this is trivial so assumeM is nonzero It is clear that Tminus1M is torsion-freeIf rankR(M) = r and x1 xr isin M are linearly independent then x11 xr1 isin Tminus1M arelinearly independent over Tminus1R Similarly if x1s1 xnsn isin Tminus1M are linearly independentin Tminus1M then x1 xn are linearly independent in M So the result is clear

Corollary 155 Let R be an integral domain with quotient field K Then

(i) If M is a torsion-free R-module rankR(M) = dimK(M otimesK)

(ii) If MN are two torsion-free R-modules of finite rank then rankR(M oplusN) = rankR(M) +rankR(N)

In particular if M is a free R-module then the rank just defined is equal to the normal free rankand we can write rank(M) without confusion

Let R be a noetherian domain and suppose a1 an isin R are linearly independent elementswhich do not generate R Then a1 ar is an R-regular sequence so by Lemma 74 the ideals(a1 ai) have height i for 1 le i le n So it follows immediately that

Lemma 156 Let R be a noetherian domain and I an ideal Then rank(I) le htI

Lemma 157 Let R be a domain and M a finitely generated projective R-module of rank 1 ThenandiM = 0 for i gt 1

Proof By localisation If p is a prime ideal then Mp is a finitely generated projective module overthe local ring Rp so Mp is free of rank 1 and Mp

sim= Rp Hence for i gt 1

(andiM)psim= andiMp

sim= andiRq = 0

as required

Theorem 158 (Auslander-Buchsbaum) A regular local ring (Am) is UFD

53

Proof We use induction on dimA If dimA = 0 then A is a field and if dimA = 1 then A is aprincipal ideal domain Suppose dimA gt 1 and let a1 ad be a regular system of parametersThen x = a1 is prime by Theorem 108 so it suffices by Lemma 153 to show that Ax is UFDLet q be a prime ideal of height 1 in Ax and put p = q cap A so q = pAx By Theorem 128gldimA = dimA lt infin so we can produce an exact sequence of A-modules with all Fi finitelygenerated free

0 minusrarr Fn minusrarr Fnminus1 minusrarr middot middot middot minusrarr F0 minusrarr p minusrarr 0 (6)

Maximal ideals of Ax correspond to primes of A maximal among those not containing x Theseprimes must all be properly contained in m so if PAx is a maximal ideal then htP lt dimATherefore (Ax)PAx

sim= AP is UFD by the inductive assumption and so q(Ax)n is either principalor zero for every maximal n of Ax Then by Lemma 124 we have projdimAx(q) = 0 and thereforeq is projective Localising (6) with respect to S = 1 x x2 we see that the following sequenceof Ax-modules is exact

0 minusrarr F primen minusrarr F primenminus1 minusrarr middot middot middot minusrarr F prime0 minusrarr q minusrarr 0 (7)

where F primei = Fi otimesAx are finitely generated and free over Ax If we decompose (7) into short exactsequences

0 minusrarr K prime0 minusrarrF prime0 minusrarr q minusrarr 0

0 minusrarr K prime1 minusrarrF prime1 minusrarr K prime

0 minusrarr 0

0 minusrarr F primen minusrarrF primenminus1 minusrarr K primenminus2 minusrarr 0

(8)

then the first sequence splits since q is projective Hence K prime0 must be projective and in this way

we show that all the sequences split and all the K primei are projective It follows thatoplus

i even

F primeisim=oplusi odd

F primei oplus q

Thus we have finite free Ax-modules FG such that F sim= Goplus q Since Ax is a noetherian domainand q a nonzero ideal of height 1 it follows from Lemma 156 that rank(q) = 1 If rank(G) = rthen rank(F ) = r + 1

So to show q is principal and complete the proof it suffices to show that q is free But by ournotes on Tensor Symmetric and Exterior algebras we have

Ax sim=r+1and

F sim=r+1and

(Goplus q) sim=oplus

i+j=r+1

(andiG)otimes (andjq) sim= (andr+1Gotimes and0q)oplus (andrGotimes and1q) sim= q

Since andr+1G = 0andrG sim= Ax and andiq = 0 for i gt 1 by Lemma 157

References

[AM69] M F Atiyah and I G Macdonald Introduction to commutative algebra Addison-WesleyPublishing Co Reading Mass-London-Don Mills Ont 1969 MR MR0242802 (394129)

[Ash] Robert Ash A course in commutative algebra httpwwwmathuiucedusimr-ashComAlghtml

[Eis95] David Eisenbud Commutative algebra Graduate Texts in Mathematics vol 150Springer-Verlag New York 1995 With a view toward algebraic geometry MRMR1322960 (97a13001)

[Mat80] Hideyuki Matsumura Commutative algebra second ed Mathematics Lecture Note Se-ries vol 56 BenjaminCummings Publishing Co Inc Reading Mass 1980 MRMR575344 (82i13003)

54

• General Rings
• Flatness
• • Faithful Flatness
  • Going-up and Going-down
  • Constructible Sets
  • • Associated Primes
    • Dimension
    • • Homomorphism and Dimension
      • • Depth
        • • Cohen-Macaulay Rings
          • • Normal and Regular Rings
            • • Classical Theory
              • Homological Theory
              • Koszul Complexes
              • Unique Factorisation

Lemma 144 Let A be a ring with maximal ideal m If s isin m and a isin A then sa isin mk impliesa isin mk for any k ge 1

Theorem 145 Let (Am k) be a noetherian local ring and let s = rankkmm2 Then we have

rankkTorAi (k k) ge

(s

i

)0 le i le s

Here rankkTorAi (k k) is the i-th Betti number of the A-module k

Proof We have TorA0 (k k) sim= k as k-modules so rankkTorA0 (k k) = rankkk = 1 which takes

care of the case s = 0 So assume that s ge 1 and let a1 as be a minimal basis of m withassociated Koszul complex F = K(a1 as) The canonical morphism εprime F0

sim= A minusrarr k gives acomplex exact in the last two nonzero places

middot middot middot minusrarr Fi minusrarr Fiminus1 minusrarr middot middot middot minusrarr F1 minusrarr F0 minusrarr k minusrarr 0

We claim this complex satisfies the conditions of Lemma 143 It clearly satisfies (ii) and (iii)It only remains to check condition (iii) By the definition of dp+1 Fp+1 minusrarr Fp it is clear thatdp+1(Fp+1) sube mFp for p ge 0 We also have to show that dminus1

p+1(m2Fp) sube mFp+1 This is trivial

if p + 1 gt s and also if p = 0 since a1 as is a minimal basis So assume 0 lt p le s minus 1Assume that

dp+1

sumi1ltmiddotmiddotmiddotltip+1

mi1middotmiddotmiddotip+1(xi1 and middot middot middot and xip+1)

=

sumi1ltmiddotmiddotmiddotltip+1

psumr=1

(minus1)rminus1airmi1middotmiddotmiddotip+1(xi1 and middot middot middot and xir and middot middot middot and xip+1) isin m2Fp

Then collecting terms we obtain a number of equations of the formsum

(minus1)etatmt isin m2 where atis one of the air and mt one of the mi1middotmiddotmiddotip+1 Since the residues of the ai give a basis of mm2

over k it follows that mt isin m which completes the proof that F satisfies all the conditions ofLemma 143 Choosing any minimal free resolution L of M and applying Lemma 143 we see thatfor 0 le i le s (

s

i

)= rankAFi le rankkTor

Ai (k k)

as required

Theorem 146 (Serre) Let (Am k) be a noetherian local ring Then A is regular if and only ifthe global dimension of A is finite

Proof We have already proved one part in Theorem 128 So suppose that gldim(A) lt infinThen TorAs (k k) 6= 0 by Theorem 145 hence gldim(A) ge rankkmm

2 since by Proposition122 we have tordim(A) = gldim(A) On the other hand it follows from Theorem 125 thatprojdim(k) = gldim(A) ltinfin so by Theorem 131 we have gldim(A) = projdim(k) = depth(A)Therefore we get

dim(A) le rankkmm2 le gldim(A) = depth(A) le dim(A)

Whence dim(A) = rankkmm2 and A is regular

Corollary 147 If A is a regular local ring then Ap is regular for any p isin Spec(A)

Proof Let M be a nonzero Ap-module Then considering M as an A-module there is an exactsequence of finite length n le gldim(A) with all Pi projective

0 minusrarr Pn minusrarr middot middot middot minusrarr P0 minusrarrM minusrarr 0

51

Since Ap is flat the following sequence is also exact

0 minusrarr (Pn)p minusrarr middot middot middot minusrarr (P0)p minusrarrMp minusrarr 0

The modules (Pi)p are projective Ap-modules and M sim= Mp as Ap-modules so it follows thatgldim(Ap) le gldim(A) ltinfin

Definition 28 A ring A is called a regular ring if Ap is a regular local ring for every prime idealp of A Note that A is not required to be noetherian Regularity is stable under ring isomorphismA noetherian local ring A is regular in this sense if and only if it is regular in the normal sense

It follows from Theorem 108 that any regular ring is normal and a noetherian regular ring isCohen-Macaulay It follows from Lemma 8 and Theorem 90 that a regular ring is catenary

Lemma 148 A ring A is regular if and only if Am is regular for all maximal ideals m

Proof One implication is clear For the other given a prime ideal p find a maximal ideal m withp sube m Then Ap

sim= (Am)pAm so Ap is a regular local ring

Lemma 149 If A is a regular ring and S sube A is multiplicatively closed then Sminus1A is a regularring

Lemma 150 If A is a regular ring then so is A[x1 xn] In particular k[x1 xn] is aregular ring for any field k

Proof This follows immediately from Theorem 118

Theorem 151 Let A be a regular local ring which is a subring of a domain B and suppose thatB is a finitely generated A-module Then B is flat (equivalently free) over A if and only if B isCohen-Macaulay In particular if B is regular then it is a free A-module

Proof Since B is a finitely generated A-module it is integral over A and so by Lemma 91 if B isflat it is Cohen-Macaulay Conversely suppose that B is Cohen-Macaulay If dim(A) = 0 then Ais a field so B is trivially flat so throughout we may assume dim(A) ge 1 Since A is normal thegoing-down theorem holds between A and B by Theorem 42 so by Theorem 55 (3) for any properideal I of A IB is proper and htI = htIB We claim that depthA(A) = depthA(B) Notice thatdepthA(B) is finite since otherwise m isin AssA(B) and hence dim(A) = 0

Firstly we prove the inequality le Since A is regular it is Cohen-Macaulay so depthA(A) =dim(A) Set s = dim(A) and let a1 as be a regular system of parameters Then ht(a1 ai)A =i and therefore ht(a1 ai)B = i for all 1 le i le s by Theorem 108 It follows from Corollary97 that a1 as is a B-regular sequence and therefore depthA(B) ge s

To prove the reverse inequality set d = depthA(B) and let a1 ad isin m be a maximal B-regular sequence Then as elements of B the sequence a1 ad is B-regular so by Lemma 74 wehave ht(a1 ad) = d But (a1 ad) sube mB and htmB = htm = dim(A) so d le dim(A) asrequired

Since gldim(A) lt infin we have projdimA(B) lt infin so we can apply Theorem 131 to see thatprojdimA(B) + depthA(B) = depthA(A) so projdimA(B) = 0 and therefore B is projectiveSince A is local and B finitely generated projective hArr free hArr flat so the proof is complete

64 Unique Factorisation

Recall that if A is a ring two elements p q isin A are said to be associates if p = uq for some unitu isin A This is an equivalence relation on the elements of A

Definition 29 Let A be an integral domain An element of A is irreducible if it is a nonzerononunit which cannot be written as the product of two nonunits An element p isin A is prime if it isa nonzero nonunit with the property that if p|ab then p|a or p|b Equivalently p is prime iff (p) isa nonzero prime ideal We say A is a unique factorisation domain if every nonzero nonunit a isin Acan be written essentially uniquely as a = up1 middot middot middot pr where u is a unit and each pi is irreducible

52

Essentially uniquely means that if a = vq1 middot middot middot qs where v is a unit and the qj irreducible thenr = s and after reordering (if necessary) qi is an associate of pi The property of being a UFD isstable under ring isomorphism

Theorem 152 A noetherian domain A is a UFD if and only if every prime ideal of height 1 isprincipal

Lemma 153 Let A be a noetherian domain and let x isin A be prime Then A is a UFD if andonly if Ax is

Proof By assumption (x) is a prime ideal of height 1 If p is a prime ideal of height 1 then eitherx isin p in which case p = (x) or x isin p and these primes are in bijection with the primes of Ax Sousing Theorem 152 it is clear that if A is a UFD so is Ax Suppose that Ax is a UFD and let p bea prime ideal of height 1 in A We can assume that x isin p Let a isin p be such that pAx = a1AxBy [AM69] Corollary 1018 we have capi(xi) = 0 so if x|a there is a largest integer n ge 1 with xn|aWrite a = cxn Since x isin p we have c isin p so by replacing a with c we can assume pAx = a1Axwith a isin (x) Then it is clear that p = (a) as required

Definition 30 Let R be an integral domain If M is a torsion-free R-module then the rank ofM is the maximum number of linearly independent elements in M rank(M) isin 0 1 infin

Proposition 154 Let R be an integral domain and M a torsion-free R-module If T sube Ris multiplicatively closed then Tminus1M is a torsion-free Tminus1R-module and rankTminus1R(Tminus1M) =rankR(M)

Proof If rank(M) = 0 this is trivial so assumeM is nonzero It is clear that Tminus1M is torsion-freeIf rankR(M) = r and x1 xr isin M are linearly independent then x11 xr1 isin Tminus1M arelinearly independent over Tminus1R Similarly if x1s1 xnsn isin Tminus1M are linearly independentin Tminus1M then x1 xn are linearly independent in M So the result is clear

Corollary 155 Let R be an integral domain with quotient field K Then

(i) If M is a torsion-free R-module rankR(M) = dimK(M otimesK)

(ii) If MN are two torsion-free R-modules of finite rank then rankR(M oplusN) = rankR(M) +rankR(N)

In particular if M is a free R-module then the rank just defined is equal to the normal free rankand we can write rank(M) without confusion

Let R be a noetherian domain and suppose a1 an isin R are linearly independent elementswhich do not generate R Then a1 ar is an R-regular sequence so by Lemma 74 the ideals(a1 ai) have height i for 1 le i le n So it follows immediately that

Lemma 156 Let R be a noetherian domain and I an ideal Then rank(I) le htI

Lemma 157 Let R be a domain and M a finitely generated projective R-module of rank 1 ThenandiM = 0 for i gt 1

Proof By localisation If p is a prime ideal then Mp is a finitely generated projective module overthe local ring Rp so Mp is free of rank 1 and Mp

sim= Rp Hence for i gt 1

(andiM)psim= andiMp

sim= andiRq = 0

as required

Theorem 158 (Auslander-Buchsbaum) A regular local ring (Am) is UFD

53

Proof We use induction on dimA If dimA = 0 then A is a field and if dimA = 1 then A is aprincipal ideal domain Suppose dimA gt 1 and let a1 ad be a regular system of parametersThen x = a1 is prime by Theorem 108 so it suffices by Lemma 153 to show that Ax is UFDLet q be a prime ideal of height 1 in Ax and put p = q cap A so q = pAx By Theorem 128gldimA = dimA lt infin so we can produce an exact sequence of A-modules with all Fi finitelygenerated free

0 minusrarr Fn minusrarr Fnminus1 minusrarr middot middot middot minusrarr F0 minusrarr p minusrarr 0 (6)

Maximal ideals of Ax correspond to primes of A maximal among those not containing x Theseprimes must all be properly contained in m so if PAx is a maximal ideal then htP lt dimATherefore (Ax)PAx

sim= AP is UFD by the inductive assumption and so q(Ax)n is either principalor zero for every maximal n of Ax Then by Lemma 124 we have projdimAx(q) = 0 and thereforeq is projective Localising (6) with respect to S = 1 x x2 we see that the following sequenceof Ax-modules is exact

0 minusrarr F primen minusrarr F primenminus1 minusrarr middot middot middot minusrarr F prime0 minusrarr q minusrarr 0 (7)

where F primei = Fi otimesAx are finitely generated and free over Ax If we decompose (7) into short exactsequences

0 minusrarr K prime0 minusrarrF prime0 minusrarr q minusrarr 0

0 minusrarr K prime1 minusrarrF prime1 minusrarr K prime

0 minusrarr 0

0 minusrarr F primen minusrarrF primenminus1 minusrarr K primenminus2 minusrarr 0

(8)

then the first sequence splits since q is projective Hence K prime0 must be projective and in this way

we show that all the sequences split and all the K primei are projective It follows thatoplus

i even

F primeisim=oplusi odd

F primei oplus q

Thus we have finite free Ax-modules FG such that F sim= Goplus q Since Ax is a noetherian domainand q a nonzero ideal of height 1 it follows from Lemma 156 that rank(q) = 1 If rank(G) = rthen rank(F ) = r + 1

So to show q is principal and complete the proof it suffices to show that q is free But by ournotes on Tensor Symmetric and Exterior algebras we have

Ax sim=r+1and

F sim=r+1and

(Goplus q) sim=oplus

i+j=r+1

(andiG)otimes (andjq) sim= (andr+1Gotimes and0q)oplus (andrGotimes and1q) sim= q

Since andr+1G = 0andrG sim= Ax and andiq = 0 for i gt 1 by Lemma 157

References

[AM69] M F Atiyah and I G Macdonald Introduction to commutative algebra Addison-WesleyPublishing Co Reading Mass-London-Don Mills Ont 1969 MR MR0242802 (394129)

[Ash] Robert Ash A course in commutative algebra httpwwwmathuiucedusimr-ashComAlghtml

[Eis95] David Eisenbud Commutative algebra Graduate Texts in Mathematics vol 150Springer-Verlag New York 1995 With a view toward algebraic geometry MRMR1322960 (97a13001)

[Mat80] Hideyuki Matsumura Commutative algebra second ed Mathematics Lecture Note Se-ries vol 56 BenjaminCummings Publishing Co Inc Reading Mass 1980 MRMR575344 (82i13003)

54

• General Rings
• Flatness
• • Faithful Flatness
  • Going-up and Going-down
  • Constructible Sets
  • • Associated Primes
    • Dimension
    • • Homomorphism and Dimension
      • • Depth
        • • Cohen-Macaulay Rings
          • • Normal and Regular Rings
            • • Classical Theory
              • Homological Theory
              • Koszul Complexes
              • Unique Factorisation

Since Ap is flat the following sequence is also exact

0 minusrarr (Pn)p minusrarr middot middot middot minusrarr (P0)p minusrarrMp minusrarr 0

The modules (Pi)p are projective Ap-modules and M sim= Mp as Ap-modules so it follows thatgldim(Ap) le gldim(A) ltinfin

Definition 28 A ring A is called a regular ring if Ap is a regular local ring for every prime idealp of A Note that A is not required to be noetherian Regularity is stable under ring isomorphismA noetherian local ring A is regular in this sense if and only if it is regular in the normal sense

It follows from Theorem 108 that any regular ring is normal and a noetherian regular ring isCohen-Macaulay It follows from Lemma 8 and Theorem 90 that a regular ring is catenary

Lemma 148 A ring A is regular if and only if Am is regular for all maximal ideals m

Proof One implication is clear For the other given a prime ideal p find a maximal ideal m withp sube m Then Ap

sim= (Am)pAm so Ap is a regular local ring

Lemma 149 If A is a regular ring and S sube A is multiplicatively closed then Sminus1A is a regularring

Lemma 150 If A is a regular ring then so is A[x1 xn] In particular k[x1 xn] is aregular ring for any field k

Proof This follows immediately from Theorem 118

Theorem 151 Let A be a regular local ring which is a subring of a domain B and suppose thatB is a finitely generated A-module Then B is flat (equivalently free) over A if and only if B isCohen-Macaulay In particular if B is regular then it is a free A-module

Proof Since B is a finitely generated A-module it is integral over A and so by Lemma 91 if B isflat it is Cohen-Macaulay Conversely suppose that B is Cohen-Macaulay If dim(A) = 0 then Ais a field so B is trivially flat so throughout we may assume dim(A) ge 1 Since A is normal thegoing-down theorem holds between A and B by Theorem 42 so by Theorem 55 (3) for any properideal I of A IB is proper and htI = htIB We claim that depthA(A) = depthA(B) Notice thatdepthA(B) is finite since otherwise m isin AssA(B) and hence dim(A) = 0

Firstly we prove the inequality le Since A is regular it is Cohen-Macaulay so depthA(A) =dim(A) Set s = dim(A) and let a1 as be a regular system of parameters Then ht(a1 ai)A =i and therefore ht(a1 ai)B = i for all 1 le i le s by Theorem 108 It follows from Corollary97 that a1 as is a B-regular sequence and therefore depthA(B) ge s

To prove the reverse inequality set d = depthA(B) and let a1 ad isin m be a maximal B-regular sequence Then as elements of B the sequence a1 ad is B-regular so by Lemma 74 wehave ht(a1 ad) = d But (a1 ad) sube mB and htmB = htm = dim(A) so d le dim(A) asrequired

Since gldim(A) lt infin we have projdimA(B) lt infin so we can apply Theorem 131 to see thatprojdimA(B) + depthA(B) = depthA(A) so projdimA(B) = 0 and therefore B is projectiveSince A is local and B finitely generated projective hArr free hArr flat so the proof is complete

64 Unique Factorisation

Recall that if A is a ring two elements p q isin A are said to be associates if p = uq for some unitu isin A This is an equivalence relation on the elements of A

Definition 29 Let A be an integral domain An element of A is irreducible if it is a nonzerononunit which cannot be written as the product of two nonunits An element p isin A is prime if it isa nonzero nonunit with the property that if p|ab then p|a or p|b Equivalently p is prime iff (p) isa nonzero prime ideal We say A is a unique factorisation domain if every nonzero nonunit a isin Acan be written essentially uniquely as a = up1 middot middot middot pr where u is a unit and each pi is irreducible

52

Essentially uniquely means that if a = vq1 middot middot middot qs where v is a unit and the qj irreducible thenr = s and after reordering (if necessary) qi is an associate of pi The property of being a UFD isstable under ring isomorphism

Theorem 152 A noetherian domain A is a UFD if and only if every prime ideal of height 1 isprincipal

Lemma 153 Let A be a noetherian domain and let x isin A be prime Then A is a UFD if andonly if Ax is

Proof By assumption (x) is a prime ideal of height 1 If p is a prime ideal of height 1 then eitherx isin p in which case p = (x) or x isin p and these primes are in bijection with the primes of Ax Sousing Theorem 152 it is clear that if A is a UFD so is Ax Suppose that Ax is a UFD and let p bea prime ideal of height 1 in A We can assume that x isin p Let a isin p be such that pAx = a1AxBy [AM69] Corollary 1018 we have capi(xi) = 0 so if x|a there is a largest integer n ge 1 with xn|aWrite a = cxn Since x isin p we have c isin p so by replacing a with c we can assume pAx = a1Axwith a isin (x) Then it is clear that p = (a) as required

Definition 30 Let R be an integral domain If M is a torsion-free R-module then the rank ofM is the maximum number of linearly independent elements in M rank(M) isin 0 1 infin

Proposition 154 Let R be an integral domain and M a torsion-free R-module If T sube Ris multiplicatively closed then Tminus1M is a torsion-free Tminus1R-module and rankTminus1R(Tminus1M) =rankR(M)

Proof If rank(M) = 0 this is trivial so assumeM is nonzero It is clear that Tminus1M is torsion-freeIf rankR(M) = r and x1 xr isin M are linearly independent then x11 xr1 isin Tminus1M arelinearly independent over Tminus1R Similarly if x1s1 xnsn isin Tminus1M are linearly independentin Tminus1M then x1 xn are linearly independent in M So the result is clear

Corollary 155 Let R be an integral domain with quotient field K Then

(i) If M is a torsion-free R-module rankR(M) = dimK(M otimesK)

(ii) If MN are two torsion-free R-modules of finite rank then rankR(M oplusN) = rankR(M) +rankR(N)

In particular if M is a free R-module then the rank just defined is equal to the normal free rankand we can write rank(M) without confusion

Let R be a noetherian domain and suppose a1 an isin R are linearly independent elementswhich do not generate R Then a1 ar is an R-regular sequence so by Lemma 74 the ideals(a1 ai) have height i for 1 le i le n So it follows immediately that

Lemma 156 Let R be a noetherian domain and I an ideal Then rank(I) le htI

Lemma 157 Let R be a domain and M a finitely generated projective R-module of rank 1 ThenandiM = 0 for i gt 1

Proof By localisation If p is a prime ideal then Mp is a finitely generated projective module overthe local ring Rp so Mp is free of rank 1 and Mp

sim= Rp Hence for i gt 1

(andiM)psim= andiMp

sim= andiRq = 0

as required

Theorem 158 (Auslander-Buchsbaum) A regular local ring (Am) is UFD

53

Proof We use induction on dimA If dimA = 0 then A is a field and if dimA = 1 then A is aprincipal ideal domain Suppose dimA gt 1 and let a1 ad be a regular system of parametersThen x = a1 is prime by Theorem 108 so it suffices by Lemma 153 to show that Ax is UFDLet q be a prime ideal of height 1 in Ax and put p = q cap A so q = pAx By Theorem 128gldimA = dimA lt infin so we can produce an exact sequence of A-modules with all Fi finitelygenerated free

0 minusrarr Fn minusrarr Fnminus1 minusrarr middot middot middot minusrarr F0 minusrarr p minusrarr 0 (6)

Maximal ideals of Ax correspond to primes of A maximal among those not containing x Theseprimes must all be properly contained in m so if PAx is a maximal ideal then htP lt dimATherefore (Ax)PAx

sim= AP is UFD by the inductive assumption and so q(Ax)n is either principalor zero for every maximal n of Ax Then by Lemma 124 we have projdimAx(q) = 0 and thereforeq is projective Localising (6) with respect to S = 1 x x2 we see that the following sequenceof Ax-modules is exact

0 minusrarr F primen minusrarr F primenminus1 minusrarr middot middot middot minusrarr F prime0 minusrarr q minusrarr 0 (7)

where F primei = Fi otimesAx are finitely generated and free over Ax If we decompose (7) into short exactsequences

0 minusrarr K prime0 minusrarrF prime0 minusrarr q minusrarr 0

0 minusrarr K prime1 minusrarrF prime1 minusrarr K prime

0 minusrarr 0

0 minusrarr F primen minusrarrF primenminus1 minusrarr K primenminus2 minusrarr 0

(8)

then the first sequence splits since q is projective Hence K prime0 must be projective and in this way

we show that all the sequences split and all the K primei are projective It follows thatoplus

i even

F primeisim=oplusi odd

F primei oplus q

Thus we have finite free Ax-modules FG such that F sim= Goplus q Since Ax is a noetherian domainand q a nonzero ideal of height 1 it follows from Lemma 156 that rank(q) = 1 If rank(G) = rthen rank(F ) = r + 1

So to show q is principal and complete the proof it suffices to show that q is free But by ournotes on Tensor Symmetric and Exterior algebras we have

Ax sim=r+1and

F sim=r+1and

(Goplus q) sim=oplus

i+j=r+1

(andiG)otimes (andjq) sim= (andr+1Gotimes and0q)oplus (andrGotimes and1q) sim= q

Since andr+1G = 0andrG sim= Ax and andiq = 0 for i gt 1 by Lemma 157

References

[AM69] M F Atiyah and I G Macdonald Introduction to commutative algebra Addison-WesleyPublishing Co Reading Mass-London-Don Mills Ont 1969 MR MR0242802 (394129)

[Ash] Robert Ash A course in commutative algebra httpwwwmathuiucedusimr-ashComAlghtml

[Eis95] David Eisenbud Commutative algebra Graduate Texts in Mathematics vol 150Springer-Verlag New York 1995 With a view toward algebraic geometry MRMR1322960 (97a13001)

[Mat80] Hideyuki Matsumura Commutative algebra second ed Mathematics Lecture Note Se-ries vol 56 BenjaminCummings Publishing Co Inc Reading Mass 1980 MRMR575344 (82i13003)

54

• General Rings
• Flatness
• • Faithful Flatness
  • Going-up and Going-down
  • Constructible Sets
  • • Associated Primes
    • Dimension
    • • Homomorphism and Dimension
      • • Depth
        • • Cohen-Macaulay Rings
          • • Normal and Regular Rings
            • • Classical Theory
              • Homological Theory
              • Koszul Complexes
              • Unique Factorisation

Essentially uniquely means that if a = vq1 middot middot middot qs where v is a unit and the qj irreducible thenr = s and after reordering (if necessary) qi is an associate of pi The property of being a UFD isstable under ring isomorphism

Theorem 152 A noetherian domain A is a UFD if and only if every prime ideal of height 1 isprincipal

Lemma 153 Let A be a noetherian domain and let x isin A be prime Then A is a UFD if andonly if Ax is

Proof By assumption (x) is a prime ideal of height 1 If p is a prime ideal of height 1 then eitherx isin p in which case p = (x) or x isin p and these primes are in bijection with the primes of Ax Sousing Theorem 152 it is clear that if A is a UFD so is Ax Suppose that Ax is a UFD and let p bea prime ideal of height 1 in A We can assume that x isin p Let a isin p be such that pAx = a1AxBy [AM69] Corollary 1018 we have capi(xi) = 0 so if x|a there is a largest integer n ge 1 with xn|aWrite a = cxn Since x isin p we have c isin p so by replacing a with c we can assume pAx = a1Axwith a isin (x) Then it is clear that p = (a) as required

Definition 30 Let R be an integral domain If M is a torsion-free R-module then the rank ofM is the maximum number of linearly independent elements in M rank(M) isin 0 1 infin

Proposition 154 Let R be an integral domain and M a torsion-free R-module If T sube Ris multiplicatively closed then Tminus1M is a torsion-free Tminus1R-module and rankTminus1R(Tminus1M) =rankR(M)

Proof If rank(M) = 0 this is trivial so assumeM is nonzero It is clear that Tminus1M is torsion-freeIf rankR(M) = r and x1 xr isin M are linearly independent then x11 xr1 isin Tminus1M arelinearly independent over Tminus1R Similarly if x1s1 xnsn isin Tminus1M are linearly independentin Tminus1M then x1 xn are linearly independent in M So the result is clear

Corollary 155 Let R be an integral domain with quotient field K Then

(i) If M is a torsion-free R-module rankR(M) = dimK(M otimesK)

(ii) If MN are two torsion-free R-modules of finite rank then rankR(M oplusN) = rankR(M) +rankR(N)

In particular if M is a free R-module then the rank just defined is equal to the normal free rankand we can write rank(M) without confusion

Let R be a noetherian domain and suppose a1 an isin R are linearly independent elementswhich do not generate R Then a1 ar is an R-regular sequence so by Lemma 74 the ideals(a1 ai) have height i for 1 le i le n So it follows immediately that

Lemma 156 Let R be a noetherian domain and I an ideal Then rank(I) le htI

Lemma 157 Let R be a domain and M a finitely generated projective R-module of rank 1 ThenandiM = 0 for i gt 1

Proof By localisation If p is a prime ideal then Mp is a finitely generated projective module overthe local ring Rp so Mp is free of rank 1 and Mp

sim= Rp Hence for i gt 1

(andiM)psim= andiMp

sim= andiRq = 0

as required

Theorem 158 (Auslander-Buchsbaum) A regular local ring (Am) is UFD

53

Proof We use induction on dimA If dimA = 0 then A is a field and if dimA = 1 then A is aprincipal ideal domain Suppose dimA gt 1 and let a1 ad be a regular system of parametersThen x = a1 is prime by Theorem 108 so it suffices by Lemma 153 to show that Ax is UFDLet q be a prime ideal of height 1 in Ax and put p = q cap A so q = pAx By Theorem 128gldimA = dimA lt infin so we can produce an exact sequence of A-modules with all Fi finitelygenerated free

0 minusrarr Fn minusrarr Fnminus1 minusrarr middot middot middot minusrarr F0 minusrarr p minusrarr 0 (6)

Maximal ideals of Ax correspond to primes of A maximal among those not containing x Theseprimes must all be properly contained in m so if PAx is a maximal ideal then htP lt dimATherefore (Ax)PAx

sim= AP is UFD by the inductive assumption and so q(Ax)n is either principalor zero for every maximal n of Ax Then by Lemma 124 we have projdimAx(q) = 0 and thereforeq is projective Localising (6) with respect to S = 1 x x2 we see that the following sequenceof Ax-modules is exact

0 minusrarr F primen minusrarr F primenminus1 minusrarr middot middot middot minusrarr F prime0 minusrarr q minusrarr 0 (7)

where F primei = Fi otimesAx are finitely generated and free over Ax If we decompose (7) into short exactsequences

0 minusrarr K prime0 minusrarrF prime0 minusrarr q minusrarr 0

0 minusrarr K prime1 minusrarrF prime1 minusrarr K prime

0 minusrarr 0

0 minusrarr F primen minusrarrF primenminus1 minusrarr K primenminus2 minusrarr 0

(8)

then the first sequence splits since q is projective Hence K prime0 must be projective and in this way

we show that all the sequences split and all the K primei are projective It follows thatoplus

i even

F primeisim=oplusi odd

F primei oplus q

Thus we have finite free Ax-modules FG such that F sim= Goplus q Since Ax is a noetherian domainand q a nonzero ideal of height 1 it follows from Lemma 156 that rank(q) = 1 If rank(G) = rthen rank(F ) = r + 1

So to show q is principal and complete the proof it suffices to show that q is free But by ournotes on Tensor Symmetric and Exterior algebras we have

Ax sim=r+1and

F sim=r+1and

(Goplus q) sim=oplus

i+j=r+1

(andiG)otimes (andjq) sim= (andr+1Gotimes and0q)oplus (andrGotimes and1q) sim= q

Since andr+1G = 0andrG sim= Ax and andiq = 0 for i gt 1 by Lemma 157

References

[AM69] M F Atiyah and I G Macdonald Introduction to commutative algebra Addison-WesleyPublishing Co Reading Mass-London-Don Mills Ont 1969 MR MR0242802 (394129)

[Ash] Robert Ash A course in commutative algebra httpwwwmathuiucedusimr-ashComAlghtml

[Eis95] David Eisenbud Commutative algebra Graduate Texts in Mathematics vol 150Springer-Verlag New York 1995 With a view toward algebraic geometry MRMR1322960 (97a13001)

[Mat80] Hideyuki Matsumura Commutative algebra second ed Mathematics Lecture Note Se-ries vol 56 BenjaminCummings Publishing Co Inc Reading Mass 1980 MRMR575344 (82i13003)

54

• General Rings
• Flatness
• • Faithful Flatness
  • Going-up and Going-down
  • Constructible Sets
  • • Associated Primes
    • Dimension
    • • Homomorphism and Dimension
      • • Depth
        • • Cohen-Macaulay Rings
          • • Normal and Regular Rings
            • • Classical Theory
              • Homological Theory
              • Koszul Complexes
              • Unique Factorisation

Proof We use induction on dimA If dimA = 0 then A is a field and if dimA = 1 then A is aprincipal ideal domain Suppose dimA gt 1 and let a1 ad be a regular system of parametersThen x = a1 is prime by Theorem 108 so it suffices by Lemma 153 to show that Ax is UFDLet q be a prime ideal of height 1 in Ax and put p = q cap A so q = pAx By Theorem 128gldimA = dimA lt infin so we can produce an exact sequence of A-modules with all Fi finitelygenerated free

0 minusrarr Fn minusrarr Fnminus1 minusrarr middot middot middot minusrarr F0 minusrarr p minusrarr 0 (6)

Maximal ideals of Ax correspond to primes of A maximal among those not containing x Theseprimes must all be properly contained in m so if PAx is a maximal ideal then htP lt dimATherefore (Ax)PAx

sim= AP is UFD by the inductive assumption and so q(Ax)n is either principalor zero for every maximal n of Ax Then by Lemma 124 we have projdimAx(q) = 0 and thereforeq is projective Localising (6) with respect to S = 1 x x2 we see that the following sequenceof Ax-modules is exact

0 minusrarr F primen minusrarr F primenminus1 minusrarr middot middot middot minusrarr F prime0 minusrarr q minusrarr 0 (7)

where F primei = Fi otimesAx are finitely generated and free over Ax If we decompose (7) into short exactsequences

0 minusrarr K prime0 minusrarrF prime0 minusrarr q minusrarr 0

0 minusrarr K prime1 minusrarrF prime1 minusrarr K prime

0 minusrarr 0

0 minusrarr F primen minusrarrF primenminus1 minusrarr K primenminus2 minusrarr 0

(8)

then the first sequence splits since q is projective Hence K prime0 must be projective and in this way

we show that all the sequences split and all the K primei are projective It follows thatoplus

i even

F primeisim=oplusi odd

F primei oplus q

Thus we have finite free Ax-modules FG such that F sim= Goplus q Since Ax is a noetherian domainand q a nonzero ideal of height 1 it follows from Lemma 156 that rank(q) = 1 If rank(G) = rthen rank(F ) = r + 1

So to show q is principal and complete the proof it suffices to show that q is free But by ournotes on Tensor Symmetric and Exterior algebras we have

Ax sim=r+1and

F sim=r+1and

(Goplus q) sim=oplus

i+j=r+1

(andiG)otimes (andjq) sim= (andr+1Gotimes and0q)oplus (andrGotimes and1q) sim= q

Since andr+1G = 0andrG sim= Ax and andiq = 0 for i gt 1 by Lemma 157

References

[AM69] M F Atiyah and I G Macdonald Introduction to commutative algebra Addison-WesleyPublishing Co Reading Mass-London-Don Mills Ont 1969 MR MR0242802 (394129)

[Ash] Robert Ash A course in commutative algebra httpwwwmathuiucedusimr-ashComAlghtml

[Eis95] David Eisenbud Commutative algebra Graduate Texts in Mathematics vol 150Springer-Verlag New York 1995 With a view toward algebraic geometry MRMR1322960 (97a13001)

[Mat80] Hideyuki Matsumura Commutative algebra second ed Mathematics Lecture Note Se-ries vol 56 BenjaminCummings Publishing Co Inc Reading Mass 1980 MRMR575344 (82i13003)

54

• General Rings
• Flatness
• • Faithful Flatness
  • Going-up and Going-down
  • Constructible Sets
  • • Associated Primes
    • Dimension
    • • Homomorphism and Dimension
      • • Depth
        • • Cohen-Macaulay Rings
          • • Normal and Regular Rings
            • • Classical Theory
              • Homological Theory
              • Koszul Complexes
              • Unique Factorisation