math 221 (commutative algebra) notes, 2nd halfpeople.math.harvard.edu/~gaitsgde/comalg_2008/... ·...

Math 221 (Commutative Algebra) Notes, 2nd Half

Live [Locally] Free or Die.

This is Math 221: Commutative Algebra as taught by Prof. Gaitsgory at Harvard duringthe Fall Term of 2008, and as understood by yours truly. There are probably typos andmistakes (which are all mine); use at your own risk. If you find the aforementioned mistakesor typos, and are feeling in a generous mood, then you may consider sharing this informationwith me.1

-S. Gong

[email protected]

1

CONTENTS

Contents

1 Divisors 21.1 Discrete Valuation Rings (DVRs), Krull Dimension 1 . . . . . . . . . . . . . 21.2 Dedekind Domains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.3 Divisors and Fractional Ideals . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4 Dedekind Domains and Integral Closedness . . . . . . . . . . . . . . . . . . . 111.5 The Picard Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2 Dimension via Hilbert Functions 212.1 Filtrations and Gradations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.2 Dimension of Modules over a Polynomial Algebra . . . . . . . . . . . . . . . 242.3 For an arbitrary finitely generated algebra . . . . . . . . . . . . . . . . . . . 27

3 Dimension via Transcendency Degree 313.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 313.2 Agreement with Hilbert dimension . . . . . . . . . . . . . . . . . . . . . . . 32

4 Dimension via Chains of Primes 344.1 Definition via chains of primes and Noether Normalization . . . . . . . . . . 344.2 Agreement with Hilbert dimension . . . . . . . . . . . . . . . . . . . . . . . 35

5 Kahler Differentials 37

6 Completions 426.1 Introduction to inverse limits and completions . . . . . . . . . . . . . . . . . 426.2 Artin Rees Pattern . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

7 Local Rings and Other Notions of Dimension 507.1 Dimension theory for Local Noetherian Rings . . . . . . . . . . . . . . . . . 50

8 Smoothness and Regularity 538.1 Regular Local Rings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 538.2 Smoothness/Regularity for algebraically closed fields . . . . . . . . . . . . . 558.3 A taste of smoothness and regularity for non-algebraically closed fields . . . 60

9 Appendix 62

1

1 Divisors

1.1 Discrete Valuation Rings (DVRs), Krull Dimension 1

Let A be a Noetherian local domain with maximal ideal m.

Definition 1.1. We say that A is a discrete valuation ring, or DVR, if m = (π), that is, ifthe maximal ideal is principal.

Lemma 1.2. If A is a DVR, then dimA/m(m/m2) = 1.

Proof. Note that the image of π generates m/m2 over A/m, so the dimension is at most 1.However, by Nakayama’s Lemma, m/m2 is not the zero module, so the dimension is 1.

Lemma 1.3. For A a DVR, for any a ∈ A, there is a unique n ∈ Z≥0 such that a = πna0

for a0 invertible. We call this n the order of a.

Proof. Uniqueness is clear by cancellation, so we seek to establish existence. Let n be themaximal power of π that divides a. This is finite because ∩mk = 0. Then if a = πna0,then a0 6∈ m, so a0 is invertible.

Corollary 1.4. Any DVR is a PID (principal ideal domain).

Proof. For any ideal I, choose an element a ∈ I of minimal order. Then if a = πna0 for a0

invertible, then πn divides all elements of I, but πn ∈ I, so I = (πn), as desired.

The above corollary leads directly to the following one:

Corollary 1.5. If A is a DVR with maximal ideal m, then the only prime ideals in A are m

and (0).

Now why are they called DVRs? Here is the apparent reason:

Proposition 1.6. DVRs are in bijection with fields K, equipped with a map ν : K →Z ∪ ∞ satisfying the following properties:

1. ν(a+ b) ≥ min(ν(a), ν(b)) for any a, b ∈ K

2. ν(ab) = ν(a) + ν(b) for any a, b ∈ K∗.

3. ν(x) =∞ if and only if x = 0

This map is called a valuation.

Proof. Given a discrete valuation ring A, take K = Frac(A) and take ν(πna0) = n where a0

is invertible. Conversely, for K take A to be its ring of integers, i.e, A = a ∈ K|ν(a) ≥ 0.One may check that this is a DVR with m = a ∈ K|ν(a) > 0, because it is clear that anyelement with valuation 0 is invertible in A and therefore any element of valuation 1 generatesm.

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Definition 1.7. Let M be a module over a domain A. We say that M is torsion-free, if forany nonzero a ∈ A, a : M → M is injective. We say that M is torsion if for any m ∈ M ,there is nonzero a ∈ A such that am = 0.

Lemma 1.8. For any module finitely generated module M over a Noetherian domain A,there is a short exact sequence

0 - Mtors- M - Mtors−free - 0

where Mtors is killed by an element of A and Mtors−free is torsion-free.

Proof. This is because we may take Mtors to be all the elements which are killed by a non-zero element of A. Then this is clearly a sub-module. Since A is Noetherian, it is finitelygenerated, which means that it can be killed by one element of A (take the product of theelements that kill the generators). Then it is easy to check that the quotient M/Mtors istorsion-free.

Lemma 1.9. For R a PID, a module M over R is flat if and only if it is torsion-free.

Proof. This is the content of Problem 2 on the Midterm.

Using this, we will classify modules over DVRs.

Proposition 1.10. let M be a finitely generated module over a DVR R. Then

M = Mtors ⊕R⊕n,

i.e, where Mtors can be annihilated by πn for some n.

Proof. Set Mtors ⊂ M be as in Lemma 1.8 so that M/Mtors is torsion-free. Therefore, byCorollary 1.4 and Lemma 1.9 we see that it is flat. But it is over a local ring, so that meansthat it is free. So we have M/Mtors = R⊕n for some n. Furthermore, since R⊕n is free, it isadditionally projective, so the above sequence splits, so

M = Mtors ⊕R⊕n

as desired.

There is nothing more to say about the free part, so let us discuss the torsion part inmore detail.

Lemma 1.11. Any finitely generated torsion module over a DVR is⊕R/πnR.

Before we prove this, let us give two examples:

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1. Take R = k[[t]], which is a DVR with maximal ideal (t). Thus, by the lemma, for afinitely generated torsion module M , t : M → M is a nilpotent operator. However,k[[t]]/tn is a Jordan block so we are exactly saying that linear transformations can bewritten in Jordan block form.

2. Let R = Zp. Here the lemma implies that finitely generated torsion modules over Zp

can be written as a direct sum of p-groups. Miaaaoow??

Now let us proceed with the proof of the lemma.

Proof of Lemma 1.11. Let n be the minimal integer such that πn kills M . This means thatM is a module over Rn = R/πnR, and also there is an element m ∈ M , and an injectivemap Rn →M , because we may choose m to be an element which is not annihilated by πn−1,and then take the map to be 1 7→ m.

Proceeding by induction, it suffices to show that the above map Rn →M splits. But forthis it suffices that Rn is an injective module over itself. This property of rings is called theFrobenius property, and it is very rare. We will write this as a lemma.

Lemma 1.12. Rn is injective as a module over itself.

Proof of Lemma 1.12. Note that a module M over a ring R is injective if and only if for anyideal I ⊂ R, Ext1(R/I,M) = 0. This was shown on Problem Set 8, Problem 2a.

Thus we wish to show that for any ideal I, Ext1Rn(Rn/I,Rn) = 0. Note that since R is a

DVR, we know that it is a PID, and also any element has the form r = πkr0 for some k ≥ 0and some r0 invertible. Then all ideals in R are of the form (πk) for some k, so all ideals inRn are also of this form. Therefore, Rn/I = Rm for some m ≤ n, so it suffices to show thatfor m ≤ n, Ext1

Rn(Rm, Rn) = 0.But note that we have short exact sequence

0 - Rn−mπm·- Rn

- Rm- 0

which gives a corresponding long exact sequence of Exts

0 - HomRn(Rm, Rn) - HomRn(Rn, Rn)♥- HomRn(Rn−m, Rn)

- Ext1Rn(Rm, Rn) - Ext1

Rn(Rn, Rn) - · · ·But note that any map of Rn modules, Rn−m → Rn, must map 1 ∈ Rn−m to an elementwhich is killed by πn−m, which means it must be a multiple of πm, so say is is πma. Thenthe map is

r 7→ πmar,

which is the image of the map

[r 7→ ar] ∈ HomRn(Rn, Rn).

Thus, ♥ is surjective. Also note that Rn is projective over itself, so Ext1Rn(Rn, Rn) = 0.

This, along with the surjectivity of ♥ shows that

Ext1Rn(Rm, Rn) = 0

as desired.

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As mentioned earlier, this lemma concludes our proof of Lemma 1.11 as well.

Definition 1.13. We say that a Noetherian domain A is of Krull dimension 1, if any nonzeroprime ideal is maximal.

Lemma 1.14. For A a Noetherian domain of Krull dimension 1, for any I ⊂ A an non-zeroideal, A/I is Artinian.

Proof. Note that any prime ideal of A/I is maximal, so Spec(A/I) has finitely many points,andA/I is therefore Artinian.

Lemma 1.15. Let A be a Noetherian domain of Krull dimension 1 and let M be a finitelygenerated A module. Then the following are equivalent:

1. M is of finite length

2. M is torsion

3. supp(M) = V (Ann(M)) ( Spec(A).

Proof. (iii) ⇒ (ii) This is clear.(ii) ⇒ (iii) Take the elements of A that kill the generators of M , and multiply them. Thenthis is in the annihilator of M .(iii) ⇒ (i) M is a module over A/Ann(M), which is Artinian by Lemma 1.14. Thereforeby a Problem Set 8, Problem 5, M is of finite length.(i) ⇒ (iii) Again, by Problem Set 8, Problem 5, since M is of finite length A/Ann(M) isArtinian. So this means that Ann(M) is non-zero, but A is a domain, so this means thatV (Ann(M)) 6= Spec(A), as desired.

The above lemma would not work out without the condition of Krull dimension 1. Forexample, if we take the ring to be C[x, y] and the module to be C[x] = C[x, y]/y. Then thismodule is torsion, but it is not of finite length.

The following proposition will be very useful later in the development of the theory ofdivisors.

Proposition 1.16. For a Noetherian domain of Krull dimension 1, nonzero ideals 0 6= I ⊂ Aare in bijection with a collection of nonzero ideals in Ap for each p, almost all of which areequal to Ap, i.e.,

I ⊂ A←→ Ip ⊂ Ap|Ip = Ap for all but finitely many p.

Proof. To get from an ideal to a set of Ip for each p, we just localize. We need to checkthat this actually gives us a collection satisfying the desired properties, namely that theyare nonzero and Ip = Ap for all but finitely many p. But note that by the flatness of Ap,we have Ap/Ip = (A/I)p, and (A/I)p 6= 0 only when p is in the support of (A/I), but byLemma 1.14, we see that this only happens for finitely many p. So (A/I)p = 0 for all but

5

1.2 Dedekind Domains

finitely many p. We now just need that they are nonzero. But this is true, because since Iis an ideal, the annihilator of I is 0, so the supp(I) = Spec(A), so none of the localizationsare 0.

Now let us construct a map in the other direction. For a collection of nonzero idealsIp ⊂ Ap, almost all Ap, and we claim that the natural map

A -⊕

p

(Ap/Ip)

is surjective. This is because if we consider it locally, then it is

Aq-⊕

p

(Ap/Ip)q

but then note that Aq → (Aq/Iq)q = (Aq/Iq) is surjective, and for p 6= q, regarding Ap/Ip asan A module, we have (Ap/Ip)q = 0, as was shown in Problem Set 9, problem 4b.

Now let us take I to be the kernel of the above map, so that we have

0 - I - A -⊕

p

(Ap/Ip) - 0

and we map our collection Ip to I.It is easy to see that the maps we have constructed in the two directions are inverse to

each-other.


Definition 1.17. We say that a ring A is a Dedekind Domain if

1. A is Noetherian of Krull dimension 1

2. For any non-zero prime p ∈ Spec(A), Ap is a DVR.

For example, note that Z and k[t] are Dedekind. In fact, any PID is Dedekind: it is easyto check that a PID is Noetherian; it has Krull dimension 1 because every prime ideal is (a)for some a irreducible, so if (a) ⊂ (b) for some other ideal b, then b|a, so if b is not a unitthen it is off from a by a unit; lastly, a localization of a principal ideal domain A at primeideal (a) is such that the maximal ideal (a)(a) is generated by the image of a, so the uniquemaximal ideal is principal, so A(a) is a DVR, as desired.

So we have that all PIDs are Dedekind domains. However, not all Dedekind domains arePIDs, as we shall later see.

Let us now consider some properties of Dedekind domains.

Proposition 1.18. Let A be a Dedekind domain, and let M be a finitely generated Amodule. Then M is projective (or equivalently flat, or locally free) if and only if it is torsion-free.

6


Proof. If M is projective, then it is a direct summand of a free module, so it is torsion-free.So we need to show that if M is torsion-free, then it is projective. Recall that to show M isprojective, it suffices to show that Mp is projective for any prime p ⊂M . But note that Ap

is a PID so a module over it is torsion free if and only if it is flat, by Lemma 1.9. However,it is also a local Noetherian ring, so a module is flat if and only if it is projective. So Mp isprojective if and only if it is torsion-free, so it now suffices to show that it is torsion-free.

However for any multiplicative set S ⊂ A, if M is torsion-free then MS is also torsion-free.This is because if

a

s′· ms

= 0

then there is t such that tam = 0, as desired.

Proposition 1.19. Let A be a Dedekind domain. Then any finitely generated module Mover it has (not canonically) a decomposition M = M tors ⊕M tors−free.

Proof. Note that by Lemma 1.8, we have a short exact sequence

0 - M tors - M - M tors−free - 0

but by proposition 1.18 the torsion free part is projective, so M can be split, not necessarilycanonically as M tors ⊕M tors−free, as desired.

Note that we may give further information about the torsion free part of the module:

M tors =⊕

p

M torsp

First note that there is a map

M tors →⊕

p

M torsp

because M is torsion, every element is supported at finitely many points, so the image of f inM tors

p is only nonzero for finitely many p. It is an isomorphism, because it is an isomorphismafter every localization.

So we have pretty much specified what the torsion part is. We can in fact also classifythe torsion free part; in particular, we have

M tors−free ' ⊕L

where L are locally free modules of rank 1. This is because we know from above that thetorsion free module is projective, we may apply Problem Set 10, Problem 12, and then sinceL is a line bundle, and I−D is also, L⊗ I−D is a line bundle, and then M/L⊗ I−D is flat, soit is projective, so we may split it off.

MEEOOOWWW!

Lemma 1.20. For A a Dedekind Domain, and I ⊂ A an ideal, then I is a locally freemodule of rank 1.

7

1.3 Divisors and Fractional Ideals

Proof. First note that I is torsion-free and therefore projective by 1.18, and it is also finitelygenerated, because A is Noetherian. But for a finitely generated module over a Noetherianring, we know that it is projective if and only if it is locally free, so we have shown that itis locally free.

Also recall that for a module which is locally free, the rank is well defined, i.e, anylocalization which makes it free makes it free of the same rank. So to test the rank, itsuffices to show that if we tensor with K = Frac(A), it is free of rank 1. But note that sinceK, being a localization of A is flat over A so we have short exact sequence

0 - I ⊗AK - A⊗

AK - (A/I)⊗

AK - 0

However, note that supp(A/I) = V (Ann(A/I)) = V (I), and the prime (0) is not in V (I),so A/I ⊗

AK, which is the localization of A/I at (0) vanishes, so we have

I ⊗AK ' A⊗

AK

but this is one-dimensional as a free K module, so the rank is 1, as desired.


Definition 1.21. For a ring A, a divisor is a formal linear combination of the prime idealsof A.

D =∑

p

np · p

where np ∈ Z, and almost all np are zero. They clearly form an abelian group. Thisgroup of divisors is called Div(A).

An effective divisor is a divisor∑npp such that for all p ∈ Spec(A), np ≥ 0. The set of

effective divisors is denoted Div+(A).

Lemma 1.22. To specify a non-zero ideal in a Dedekind domain is the same as to specifyan effective divisor.

Proof. To specify a non-zero ideal, by Proposition 1.16, is the same as to specify somecollection of non-zero ideals Ip ⊂ Ap, where all but finitely many of them are Ap. But notethat Ap is a DVR, so Ip = m

npp for some np, where the np are clearly non-negative, and almost

all the np are zero, since almost all the Ip are Ap. Therefore we may consider mapping theideal to

∑npp and it is easy to see that this is a bijection between ideals of A and effective

divisors.

Remark 1.23. Note that if A is a Dedekind domain, then Ap is a DVR, so we may considerthe corresponding valuation,

νp : Frac(Ap)→ Z

but note that Frac(Ap) = Frac(A), so we can see it as νp : K → Z. where K = Frac(A).

8


Lemma 1.24. With νp : K∗ → Z defined as in the above remark, A = a ∈ K|νp(a) ≥ 0.Also, the ideal in A corresponding to the effective divisor D =

∑npp in Lemma 1.22 is

actually the idealID = a ∈ A|νp(a) ≥ np.

Proof. Both parts follow from Problem Set 9, problem 1a.

So we have shown (Lemma 1.24) that effective divisors are in bijection with ideals in A,with the ideal corresponding to

∑npp being a ∈ K|νp(a) ≥ np, where the effectiveness of

the divisor ensures that this submodule of K is actually an ideal of A. This suggests thatwe may want to consider relating divisors directly with submodules of K, so let us prepareto do this.

Definition 1.25. Given a Dedekind domain A, fractional ideal is a nonzero A-submoduleL ⊂ K, which is finitely generated as an A module.

Lemma 1.26. Any fractional ideal L is locally free over A of rank 1.

Proof. Note that L, which sits in K, is clearly torsion-free, so by Proposition 1.18 it is locallyfree. We want to show that it is rank 1. So we just need to show that localized at (0), it isa 1-dimensional vector space over K. Consider the inclusion

L ⊂inc- K

and note that if we tensor with K, it is still an inclusion, as K is flat over A

L⊗AK ⊂

inc- K ⊗AK = K

but this means the dimension of L⊗AK is at most 1. However, if it were 0, then L would be

locally free of rank 0, which would mean that it is 0, which it is not.

Note that for a fractional ideal L we have shown that it is a locally free module of rank1, equipped with an isomorphism

L⊗AK

∼- A⊗AK.

However, the following lemma says that this may be used as an alternate definition offractional ideals:

Lemma 1.27. Fractional ideals are in bijection with locally free module L of rank 1,equipped with an isomorphism

L⊗AK

∼- A⊗AK

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and where two such are considered isomorphic if there is an isomorphism φ : L1 → L2

between them, such that

L1 ⊗AK

φ- L2 ⊗

AK

K

α1

?

α 2

commutes.

Proof. Taking the proof of Lemma 1.26 into account, we just need to see how to get from alocally free module of rank 1 with the aforementioned map to a fractional ideal. To do thisconsider

L∼- L⊗

AA ⊂ - L⊗

AK

∼- A⊗AK

∼- K

which gives us a way of seeing L as a submodule of K, and it is easy to check that this isthe corresponding fractional ideal.

Now, as promised before we will show the correspondence between divisors and fractionalideals

Theorem 1.28. There is a canonical bijection of sets between

fractional ideals divisors

Proof.(←−) To get from a divisor to a fractional ideal: For a divisor D =

∑npp, consider the set

ID = f ∈ K|νp(f) ≥ np

(we are using the first definition of fractional ideals). This ID is clearly a submodule. Wejust need to show that it is finitely generated. Note that we may write D = D1 − D2 forD1, D2 ∈ Div+.

Then note that ID ⊂ I−D2 , so it suffices to show that for any effective divisor D, I−D isfinitely generated.

Then we may choose a ∈ A such that νp(a) ≥ D(p), because there are only finitelymany p such that D(p) 6= 0, and calling these p1, . . . ps, we may consider πi in A such that

νpi(πi) ≥ 1, and since πi ∈ A, νpj(πi) ≥ 0, so that the product∏πD(pi)i works.

Now note that I−D ⊂ a−1A ⊂ K as an ideal, and a−1 is finitely generated, so I−D isfinitely generated, as desired.(−→) Now we start with a fractional ideal and we want to recover the corresponding divisor.So consider a finitely generated submodule L ⊂ K. Then note that we may consider

Lp → Kp = K.

10

1.4 Dedekind Domains and Integral Closedness

but then note that Lp is a finitely generated Ap submodule of K, and Ap is a DVR. Let πbe the element generating the maximal ideal of Ap and let np be the maximal integer suchthat π−np ∈ Lp, (which exists because Lp is finitely generated, so the smallest −np could beis the minimal valuation of the generators).

So then we will consider mapping L to∑np defined as above. We need to show that this

is a divisor. That is, we want to show that there are only finitely many p such that np = 0.Consider f1, . . . fn the generators of L. Then, as mentioned earlier, we have that

np ≤ −min(νp(f1), νp(f2), . . . νp(fn))

and for each f , all but finitely many of νp(f) are zero. This is because if f = xy

then x and yare only in finitely many primes, and for all primes such that x, y 6∈ p, we have f ∈ Ap andf is invertible in Ap, so νp(f) = 0.


DG: Another question you may ask is “why the hell are we talking about this?”This section will show why it is interesting to consider Dedekind domains– we will show

that many rings that people might be interested in are Dedekind.

Definition 1.29. For A a domain and K its field of fractions, the integral closure of A inK, denoted Aint is defined to be the subalgebra of K formed by elements which are integralover A. More generally for A ⊂ K ⊂ K ′ for any other field K ′, the integral closure of Ain K ′ is defined to be the set of elements of K ′ which are integral over A. (Recall that forA → B, an element of b ∈ B is integral over A if b satisfies a monic polynomial in A, orequivalently, if the image of A[x]→ B given by mapping x to b is finite as an A module).

Definition 1.30. We say that A is integrally closed if A is its own integral closure in K.

Theorem 1.31. Let A be a local Noetherian domain of Krull dimension 1. Then A is aDVR if and only if it is integrally closed.

Proof.

DVR =⇒ Integrally closed. Suppose we have some x ∈ K such that

xn + an−1xn−1 + · · ·+ a0 = 0.

Then,−xn = an−1x

n−1 + · · ·+ a0

and then we may compute the valuation of both sides. Suppose for the sake of contradictionthat the valuation of x is −a, with a > 0. Then

ν(−xn) = −na

butν(an−1x

n−1 + · · ·+ a0) ≥ −(n− 1)(a)

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a contradiction. So ν(x) ≥ 0, so x ∈ A.Integrally closed =⇒ DVR We want to show that the maximal ideal m is principal. Take

a ∈ m. Then by Lemma 1.14 A/(a)A is Artinian, so there is some k such that mk ⊂ (a).Let k be the minimum such integer. Consider b ∈ mk−1 such that b 6∈ (a). Then considery = b

a6∈ A.

We shall show that y−1 ∈ m generates m. This would prove that m is principal.Consider ym ⊂ K, and note that ym ⊂ A. This is because bm ⊂ (a), so b

am ⊂ (1) = A.

ym is clearly an A submodule of K, so since it is in A, it is an ideal of A. Note that if wecould show ym 6⊂ m, then ym is an ideal that is not contained in the maximal ideal, so itmust be A, so ym = A, so m = y−1A, and we would be done.

Thus, to prove this theorem, it now suffices to show ym 6⊂ m. We proceed by contra-diction: if ym ⊂ m, then apply the next lemma, Lemma 1.32, where B = A, M = m andT = y, and we see that there are a0, . . . an−1 ∈ A such that

yn + an−1yn−1 + · · ·+ a0 = 0.

But then since A is integrally closed, y ∈ A, a contradiction.

Lemma 1.32. Let M be a finitely generated module over a Noetherian domain B, and letT : M →M be a map of modules. Then T satisfies a monic polynomial.

Proof. Consider f : Bn M for some n, which we may because M is finitely generated.Then the map T : M →M lifts to a map Bn → Bn, by lifting setting T ′(bi) to be some liftof T (f(bi)), for bi the generators of Bn. So we have diagram

Bn f- M

Bn

T ′

? f- M

T

?

where T ′ is a matrix with coefficients in B. Consider P the characteristic polynomial of T ′,which is monic by definition. Then P (T ′) = 0, but we have commutative diagram

Bn f- M

Bn

P (T ′)

? f- M

P (T )

?

but f is surjective, so P (T ) = 0, as desired.

We would now like to establish a global version of Theorem 1.31. To do this, let usfirst study the behaviour of integral closures under localization. In particular consider thefollowing lemma:

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Lemma 1.33. Let A ⊂ B be an inclusion of rings and let Aint be the integral closure of Ain B. Let S be a multiplicative subset of A. Then

(AS)int = (Aint)S.

where (AS)int is the integral closure of AS in BS.

Proof. To do this we will show inclusions in both directions:(AS)int ⊃ (Aint)S: Take b satisfying monic polynomial

bn + an−1bn−1 + · · ·+ a0 = 0

and s ∈ S, and we want to show bs∈ (AS)int.

But note that (b

s

)n+an−1

s

(b

s

)n−1

+ · · ·+ a0

sn= 0

so bs

satisfies a monic polynomial with coefficients in AS, as desired.(AS)int ⊂ (Aint)S: Let b′ ∈ BS be in the integral closure of AS, ie there are a′0, . . . a

′n−1 ∈ AS

such that(b′)n + a′n−1(b′)n−1 + · · ·+ a′0 = 0

then let a′i = aisi

for ai ∈ A. Then take b = b′∏si, and plugging into the above equation and

multiplying both sides by (∏si)

n, gives a monic polynomial for b with coefficients in A, sob ∈ Aint. Then

b′ =b∏si∈ (Aint)S

Now let us proceed to the global version of Theorem 1.31.

Theorem 1.34. Let A be a Noetherian domain of Krull dimension 1. Then A is Dedekindif and only if it is integrally closed.

Proof. Suppose that A is Dedekind. We want to show it is integrally closed. ConsiderA ⊂ Aint ⊂ K, and we want to show that A ⊂ Aint is an isomorphism. It suffices to showthat it is an isomorphism after all localizations at prime ideals. So it suffices to show thatAp = (Aint)p. By Lemma 1.33, (Aint)p = (Ap)int. But note that since A is Dedekind, Ap is aDVR, so applying Theorem 1.31, it is integrally closed, so Ap = (Ap)int, as desired.

Now suppose A is integrally closed. We want to show it is Dedekind, which means weneed to show that its localizations are DVRs. By Theorem 1.31 it suffices to show that thelocalizations are integrally closed. But this also follows from the above lemma.

——————–Here we get to the crux of this section: showing that many rings one may be interested

in are Dedekind.First let us recall that a field extension k′ of k is called separable if k ⊗

kk′ is a reduced

algebra over k, where k is the algebraic closure of k. In the ramblings around Theorem 5.11,we see that all field extensions of a field of characteristic 0 are separable.

13


Theorem 1.35. Let A be a Dedekind domain, and let K be its field of fractions. Let K ′

be a finite separable field extension . Let A′ be the integral closure of A in K ′. Then A′ isDedekind.

Before we prove the theorem, we provide some applications

1. Taking the case A = Z, so K = Q, we see that for any field extension K ′ of Q, thering of integers in it is a Dedekind domain.

2. Taking A = k[x] for k of characteristic 0, and a field extension K ′ ⊃ k(x), the ring ofintegers in it is Dedekind.

Using the above, we may construct examples of Dedekind domains which are not PIDs.Let us present a lemma that we shall use to prove the theorem:

Lemma 1.36. In the set-up of Theorem 1.35, A′ is finitely generated as an A module.

We will prove the theorem from the lemma:

Proof that Lemma ⇒ Theorem. Note that by the Hilbert basis theorem, we get that A′ isNoetherian.

This also gives us that A′ is integrally closed: for b ∈ K ′ if b is integral over A′, then bis integral over A. This is because A′[b] is finite as a module over A′, and A′ is finite as amodule over A (by the lemma), so A′[b] is finite as a module over A.

We will now show that A′ has Krull dimension 1. Consider a nonzero prime p′ ∈ Spec(A′).Then we have ψ : A → A′ an inclusion, which induces Ψ : Spec(A′) → Spec(A). Letp ∈ Spec(A) be the image of p′ in Spec(A). We claim that p is not (0). To show this werecall that for a map φ : A→ B and Φ : Spec(A)→ Spec(B), from Problem Set 9 Problem9 that Φ−1(p) = Spec(B ⊗

AAp/pp). So we have

Φ−1((0)) = Spec(A′ ⊗AK) = Spec(K ′)

where we have A′ ⊗AK = K ′ (see footnote for proof).2

2Note that there is clearly map A′ ⊗K → K ′ given by viewing A′ and K both in K ′ and multiplying.We will show that this map is surjective and injective. To show it is surjective, consider x ∈ K ′. Then sinceK ′ is finite dimensional, x satisfies a polynomial with coefficients in K, and we can scale the coefficients sothey are in A. So say we have b0, . . . bn such that

bnxn + · · ·+ b0 = 0.

Then this means(bnxn)n + bn−1(bnxn)n−1 + bn−2bn(bnxn)n−2 + · · ·+ b0b

n−1n = 0

so bnxn satisfies a monic polynomial with coefficients in A. So bnxn ∈ A′. and x is the image of bnxn ⊗ 1bn

,establishing surjectivity.

To establish injectivity, note that all elements in A′ ⊗AK can be written as pure tensors, because for

x, y ∈ A′, α, β, γ, δ ∈ A, we have

x⊗ α

β+ y ⊗ γ

δ= (xαδ + yγβ)⊗ 1

βδ

14


But K ′ is a field so Spec(K ′) is a point, so there is only one pre-image to the prime (0),but we know (0) ∈ Spec(A′) maps to (0) ∈ Spec(A), so this must be the one pre-image, sop′ is not a pre-image of (0).

So we have p 6= (0). But recall that A is of Krull dimension 1, so this means that p ismaximal. Now we want to show that p′ is also maximal. To do this, note that applyingProblem Set 9, Problem 9 again, we see that

Φ−1(p) = Spec(A′ ⊗AAp/pp) = Spec(A′ ⊗

AA/p) = Spec(A′/pA′)

where pA′ is an ideal in A′. But A′/pA′ is an algebra over the field A/pA, but since A′ isfinitely generated as an A module, A′/pA′ is a finite dimensional A/pA algebra. Thus it isArtinian, so p′ ∈ Spec(A′/pA′) is a closed point. But Spec(A′/pA′) ⊂ Spec(A′) is a closedsubset, so p′ is also a closed point in Spec(A′), which means p′ is maximal, as desired.

So we have shown that A′ is Noetherian, integrally closed, and of Krull dimension 1, soapplying 1.34 we see that A′ is Dedekind, as desired.

Now we have shown how the theorem results from the lemma, so let us prove the lemma

Proof of Lemma. First let us give an alternate definition to our definition of “separable”.For a finite field extension k′ of k, we may consider the bilinear pairing k′⊗

kk′ → k given by

x, y 7→ Trk′/k(xy). Which is to say xy ∈ k′ can be seen as a k-linear map of finite dimensionalvector spaces k′ → k′, and we are considering the trace of this map. Then we claim that k′

is separable if and only if the bilinear pairing k′ × k′ → k is non-degenerate.To show the above claim, first note that the pairing is non-degenerate if and only if it

is non-degenerate after tensoring with the algebraic closure. This is because if Tr(xy) = 0for all y ∈ k′, then Tr((x ⊗ 1k)y) = 0 for all y ∈ k′ ⊗

kk, which we may see to be true by

decomposing into pure tensors. The other direction is obtained by selecting a basis of k overk, and then noting that for yi basis elements, if Tr(

∑xyi) = 0 then Tr(xyi) = 0 for each i.

So now we just need to show that X = k′⊗kk is reduced if and only if the map X⊗

k

X → k

given by a⊗b 7→ Tr(ab) is non-degenerate. To do this, we show that elements of the kernel ofthe bilinear map are exactly the nilpotents. But note that X is a finite dimensional algebraover k, and we may elements as matrices. Then if Tr(AB) = 0 for all B if and only ifTr(PAP−1PBP−1) = 0 for all B, so we may assume A is in Upper Triangular Form. Fromthis, the claim becomes clear.

Now, pick a basis ei of K ′ over K. Let us scale each ei to make it in A′. (We may dothis because K ′ = A′⊗

AK. Then the non-degenerate bilinear map 〈−,−〉 given by considering

the trace gives a dual basis di, where 〈di, ej〉 = δij. Then note that

l =∑〈l, ei〉di

and it is easy to show that nonzero pure tensors cannot map to zero.An alternate, more abstract method of establishing injectivity goes as follows. We have

A′ ⊗AK - K ′ ⊗

AK - K ′

and the first map is injective because K is a flat A module, and the second is injective because everythingin A acts invertibly on K ′ as an A module, so localizing again has no effect.

15

1.5 The Picard Group

so we have for b ∈ A′,b =

∑〈b, ei〉di

so it suffices to show that 〈b, ei〉 ∈ A, because this would mean that the di generate A′ as anA module. Note that

〈b, ei〉 = Tr(bei)

and bei ∈ A′, so it suffices to show that for any element in A′, its trace as a mapK ′ → K ′ of K vector spaces is in A.

Well consider α ∈ A′, and we want to show that the trace of α : K ′ → K ′ is in A. Bydefinition of A′ there are an−1, . . . a0 ∈ A, such that

P (α) = αn + an−1αn−1 + · · ·+ a0 = 0.

Now consider α as a matrix. Then let us consider α : K ′⊗K → K ′⊗K of K vector spaces.Then it still satisfies P (β) = 0.

Let β be an upper triangular matrix which is conjugate to α as a map of K vector spaces.Then note that β satisfies the same polynomial as α, and the trace of β is equal to that ofα, so we now just need to show that the trace of β is in A. But since β is upper triangular,every diagonal entry of β also satisfies P , so they are all in the integral closure of A in K.Thus, their sum is also in this integral closure. But their sum is the trace of α, which weknow is in K, so it is in the intersection of K with the integral closure of A in K, but sinceA is Dedekind, this means that it is in A, as desired.


Definition 1.37. For A a Dedekind domain, define Pic(A) to be the set of isomorphismclasses of finitely generated locally free A modules of rank 1.

Note that Pic(A) forms a semigroup with the multiplication given by

(L1,L2) 7→ L1 ⊗A

L2.

With this multiplication, A, viewed as a module over itself, is the identity.This semi-group is in fact a group. To show this, it suffices to show that HomA(L, A) is

the inverse to L, which is to say, to show that

L⊗A

HomA(L, A) ' A

which was problem set 10, problem 1a.

Definition 1.38. For A a Dedekind domain, denote Pic(A) to be the set of isomorphismclasses of (L, α) where L is a finitely generated locally free A module of rank 1 and α is an

16


isomorphism α : L ⊗AK ' K, and where two such are considered isomorphic if there is an

isomorphism φ : L1 → L2 between them, such that

L1 ⊗AK

φ- L2 ⊗

AK

K

α1

?

α 2

commutes. Alternately, similarly to in Lemma 1.27 we may define Pic(A) to be the iso-morphism classes (L, i) where L is a finitely generated A sub-module, and i is an injectioni : L→ K, where (L1, i1) and (L2, i2) are considered the same if we have map φ : L1 → L2,such that the diagram

L1

φ- L2

K

i1

?

i 2

commutes.

Remark 1.39. Note that we may define a multiplication structure on Pic(A) also, given by

(L1, α1) · (L2, α2) = (L1 ⊗A

L2, α)

where α : K ⊗A

L1 ⊗A

L2 → K is α2 (α1 ⊗ IdL2).

Here, A with A ⊗AK → K, is again the identity, as was checked in Problem Set 10,

problem 1b and inverse of (L, φ) is given by

(HomA(L, A), ψ)

where ψ is the composed map

K ⊗HomA(L, A) - HomK(L⊗K,K) −α−1- HomK(K,K) - K

where the first map is given by Problem Set 5, Problem 2, since tensoring K is the sameas localizing at (0), and it is an isomorphism by Problem Set 5, Problem 2c. The last mapis just the canonical map. We check that this is the inverse of (L, φ) in Problem Set 10,Problem 1b.

Lemma 1.40. There is an isomorphism of groups

Div(A)∼- Pic(A).

17


Proof. The bijection of sets was shown in Theorem 1.28, since Pic(A) is just the set offractional ideals, with a group structure placed on it. That the isomorphism is a grouphomomorphism is checked in Problem Set 10, Problem 2.

Note that there is naturally a map

Pic(A) - Pic(A)

given by forgetting the map L ⊗AK → K. We claim that the above map is a surjection; to

show this, we need only show that for every finitely generated locally free A module L of

rank 1, there is an isomorphism L⊗AK

∼- K, but this is true because it is of rank 1, so for

some localization AS there is an isomorphism L⊗AAS

∼- AS, and then we can just further

localize.Now let us explore the what the kernel of this map could be. Note that we may map

K∗ to Pic(A) such that the map lands in the kernel. To do this, let us map f ∈ K∗ to

(A,A⊗AK

×f- K). It is clear that the image of this map K∗ → Pic(A) lies in the kernel of

the map Pic(A)→ Pic(A), since (A,A⊗AK

×f- K) maps to A, which is the unit of Pic(A).

We shall (falsely) propose that this is gives us the kernel of Pic(A)→ Pic(A).

False Proposition 1.41. There is a short exact sequence

0 - K∗ - Pic - Pic - 0

Proof. (Keeping in mind the lack of veracity of the proposition.)

It suffices to show that every (L, i : L⊗AK

∼- K) that goes to a unit comes from some

f ∈ K∗, i.e, there is a unique f such that we may say

(L, i : L⊗AK - K) = (A,A⊗

AK

×f- K).

Well, to construct the f , let us note that since (L, i : L⊗AK

∼- K) maps to the unit,

we have that there is some isomorphism φ : A→ L. Then, this gives us a map A⊗AK → K

such that the following commutes:

L⊗AK

i- K

A⊗AK

φ⊗ id6

ψ

-

but note that ψ : A ⊗AK → K is an isomorphism of A modules, but A ⊗

AK = K, and the

only isomorphisms of A modules from K to K are multiplications by elements of K∗.

18


So this gives us that (L, i : L ⊗AK

∼- K) comes from some f . However, it does not

establish uniqueness, and herein lies the problem with this proof and this proposition; wehave shown that the map K∗ → Pic(A) surjects to the kernel of Pic(A) → Pic(A), but the

map K∗ → Pic(A) may not be injective– there may be multiple f that map to the same(L, i).

Let us trace exactly how this issue may arise– when there are two f that map to thesame (L, i). This happens when φ1 : A → L and φ2 : A → L give different f . When this

happens, note we may consider the isomorphism A∼- A that is the composition φ−1

1 φ2

and we get

L⊗AK

f1 - K

A⊗AK

φ−11 φ2

6f 2

-

but note that φ−11 φ2 is an isomorphism of A modules from A to A, so it is an invertible

element of A. Thus we have shown that for any f1 and f2 that map to the same L, f1 = f2afor a some invertible element of A. Also it is easy to see that the converse also holds; iff1 = f2a for some a invertible in A, then f1 and f2 map to the same (L, i), so we have thatthe true statement behind the false proposition is this true proposition:

Proposition 1.42. There is a short exact sequence

0 - K∗/A∗ - Pic(A) - Pic(A) - 0

where A∗ denotes the subset of A consisting of invertible elements.

Proof. The proof of this lies in the pre-amble.

Recall that we had a group isomorphism Div(A) ' Pic(A). So we may consider the abovemaps in terms of Div(A).

Lemma 1.43. There is a commutative diagram

0 - K∗/A∗ - Pic(A) - Pic(A) - 0

‖ ‖

0 - K∗/A∗α- Div(A)

∼6

AJ- Pic(A) - 0

where α is given by α(f) =∑νp(f)p where you will note that if a ∈ A∗ then νp(a) = 0, so

this α can be seen as a map α : K∗/A∗ → Div(A). Also AJ is given by taking D =∑npp

toK ⊃ ID = f ∈ K|νp(f) ≥ np

19


which was shown to be finitely generated in Theorem 1.28, and by Lemma 1.26, we knowthat it is locally free of rank 1, since it is a fractional ideal, although we are viewing it inPic here, so we are forgetting the map to K.

Proof. That the right square commutes is clear by chasing through our map Div(A) →Pic(A) from Lemma 1.40. That the right square commutes is the content of problem Set 10,Problem 3.

Definition 1.44. Divisors in the image of α in the above lemma are called principal divisors.

It is shown in Problem Set 10, Problem 2 that the Dedekind domain A is a PrincipalIdeal Domain if and only if α is surjective, i.e, if all divisors are principal.

20

2 Dimension via Hilbert Functions

2.1 Filtrations and Gradations

2.1.1 Filtered and Graded Abelian Groups

Let M be an abelian group.

Definition 2.1. A filtration Fil(M) on M is a sequences of submodules

· · · ⊆Mi−1 ⊆Mi ⊆Mi+1 ⊆ · · ·

of M which satisfy:

1. ∪Mi = M

2. There exists i0 such that if i < i0 then Mi = 0. (We shall henceforth denote “Thereexists i0 such that if i < i0 then P holds” by “For i 0, P holds).

Definition 2.2. Given filtrations on · · · ⊆Mi−1 ⊆Mi ⊆Mi+1 ⊆ · · · of M and · · · ⊆ Ni−1 ⊆Ni ⊆ Ni+1 ⊆ · · · of N a map φ : M → N is called compatible with filtrations if for anym ∈Mi, φ(m) ∈ Ni. We may alternately call φ a map of filtered abelian groups.

Definition 2.3. Induced Filtrations:Given a filtration · · · ⊆ Mi−1 ⊆ Mi ⊆ Mi+1 ⊆ · · · on M and an abelian group N ⊂ M ,

the induced filtration on N is defined by Ni = Mi ∩N .Given a filtration · · · ⊆ Mi−1 ⊆ Mi ⊆ Mi+1 ⊆ · · · on M , an abelian group N and a

surjection φ : M → N , the induced filtration on N is defined to be Ni = φ(Mi).It is easy to check that these are filtrations.

Definition 2.4. A graded abelian group is M = ⊕i∈ZM i where M i = 0 for i 0.

To a graded abelian group we may associate a filtered abelian group by Mi = Mj≤i

j.

To a filtered abelian group we may consider the associated graded abelian group givenby

gri(M) = Mi/Mi−1.

Note that for a map of filtered abelian groups φ : M → N , since φ(Mi) ⊂ Ni andφ(Mi−1) ⊂ Ni−1, we have φ induces a map Mi/Mi−1 → Ni/Ni−1. Thus we may considergr(φ) : gr(M)→ gr(N).

Here is a useful lemma relating behavior of φ to behavior of gr(φ):

Lemma 2.5. Let φ : M → N be a map of filtered abelian groups. Suppose that gr(φ) :gr(M)→ gr(N) is injective (resp. surjective). Then φ is injective (resp. surjective).

Proof. Injectivity:Suppose, for the sake of contradiction that the kernel of φ is non-empty. Consider the

minimal i such that Mi contains a nonzero element of the kernel. Then consider m ∈ker(φ) ∩Mi. Then note that m 6= 0 ∈ Mi/Mi−1 because Mi−1 does not contain an nonzeroelement of the kernel. But by definition gr(φ)(m) = 0, so m ∈ ker(gr(φ)).

21


Surjectivity:We proceed by induction to show that in fact Mi Ni for all i. Suppose that for all

i′ < i, Mi′ Ni′ .Take n ∈ Ni. Then note that n ∈ gri(N) is in the image of φ. So there issome m ∈Mi such that φ(m) = n. Thus n− φ(m) ∈ Ni−1, so we are done by induction.

While the converse of the above statement is not true (See Problem Set 10 problem 5 inthe attached problems at the end), it is true in the specific case of induced filtrations:

For N ⊂ M with the induced filtration on N , we have gr(N) → gr(M). To show this,we need that for each n ∈ Ni, with n 6= 0, the image of n in M does not belong to Mi−1,but this is just definition.

We may similarly show that if we have a filtered abelian group M and φ : M N theinduced map gr(φ) : gr(M)→ gr(N) with the induced filtration on N is a surjection.

2.1.2 Filtered and Graded Rings and Modules

Definition 2.6. A graded ring is a ring A that may be written as A = ⊕i∈Zi≥0

Ai such that the

multiplication map sends Ai × Aj → Ai+j.

Definition 2.7. A filtered ring is a ring A with a filtration of abelian groups A0 ⊆ A1 ⊆ · · ·such that multiplication sends Ai × Aj → Ai+j

Note that for a filtered ring we may consider the associated graded as abelian groups.This is naturally a graded ring because we may consider the map Ai/Ai−1 ⊗ Aj/Aj−1 →Ai+j/Ai+j−1, induced by the map Ai ⊗ Aj → Ai+j.

Definition 2.8. A graded module over a graded ring is a module M that may be writtenas M = ⊕

i∈ZM i with M i = 0 for i 0 such that AiM j ⊂ M j+i. We may alternatively write

this is Ai ⊗M j →M j+i

Definition 2.9. A filtered module over a filtered ring A is a module M with a filtration ofabelian groups · · ·Mi− 1 ⊆Mi ⊆ · · · such that the action acts as Ai ⊗Mj →Mi+j

Note that for M a filtered module over a filtered ring A, gr(M) is naturally a gradedmodule over gr(A). Consider the following lemma relating finite generation of the associatedgraded of a module to finite generation of the original module.

Lemma 2.10. Let M be a filtered module over a filtered ring A. Suppose that gr(M) isfinitely generated over gr(A). Then M is finitely generated over A.

Proof. Consider homogenous generators m1,m2, . . .mn, which are respectively ingri1(M), gri2(M), . . . grin(M). Let us consider lifts m1,m2, . . .mn ∈M .

Then consider the filtered module A[−ij] which is A as a module over itself shifted by ij.Thus A[−ij]ij contains 1 ∈ A0. Now let us consider a map of filtered modules

A[−i1]⊕ A[−i2]⊕ · · · ⊕ A[−in]→M

22


given by mapping 1 ∈ A[−ij]ij to mj ∈Mij . Then the associated map of gradeds

gr(A[−i1])⊕ gr(A[−i2])⊕ · · · ⊕ gr(A[−in])→ gr(M)

is surjective, so the original map is surjective, which means that the mj generated M , asdesired.

This lemma gives us a corollary which, along with considering finitely generated algebrasover Noetherian rings, is one of the best ways to create Noetherian rings. Thus, Prof.Gaitsgory says that it is “pretty damn useful”.

Corollary 2.11. Let A be a filtered ring such that gr(A) is Noetherian. Then A is Noethe-rian.

Proof. Consider some ideal I ⊂ A with the induced filtration. Then gr(I) ⊂ gr(A) is an idealso gr(I) is finitely generated over gr(A), so applying the above lemma I is finitely generatedover A, as desired.

———Let A be a filtered ring, and M be a finitely generated A module.

Definition 2.12. A filtration on M is good if gr(M) is finitely generated over gr(A).

Lemma 2.13. A good filtration exists.

Proof. Since M is finitely generated as and A module, there exists a surjective map of Amodules A⊕n M . Then if we take the induced filtration on M from this surjection, asnoted before we have gr(A)⊕n gr(M), as so the filtration is good, as desired.

Remark 2.14. One might think that we could take a more simple filtration like Mi = Mfor i ≥ 0 and Mi = 0 for i < 0¿ Then we have gri(M) = 0 for i 6= 0 and gri(M) = M fori = 0. But this does not work because of all the Ai only A0 sends gri(M) to itself, so we areessentially only acting on M with A0, so this should not be expected to be finitely generated.

Corollary 2.15. (Artin Rees) Let gr(A) be Noetherian. Let M be a module and Mi a goodfiltration. Let N ⊂M be a submodule. Then the induced filtration on N is good.

Proof. Note that we have an injection gr(N) → gr(M) because the filtration on N ⊂ M isthe induced one. Thus, since gr(M) is finitely generated and gr(A) is Noetherian, gr(N) isalso finitely generated, as desired.

Note that good filtrations are not unique, since when we proved the existence we couldhave chosen generators any way we wanted. We can, however, control the non-uniqueness ofof the filtrations, as the per the following lemma:

Theorem 2.16. Let Fil(M) and Fil′(M) be two good filtrations on M . Then there is k suchthat for any i

Fil′i−k(M) ⊆ Fili(M) ⊆ Fil′i+k(M)

23

2.2 Dimension of Modules over a Polynomial Algebra

Proof. First note that we only need to prove one side of the inequality, as the two sides aresymmetric, and once we have one k for each, the maximum of the two ks works.

Before we proceed, consider the following lemma:

Lemma 2.17. Let Fil be a good filtration. Then there exists n and i0 such that for alli ≥ 0,

Fili+n ⊆ Ai+i0Filn

Proof of Lemma. Let mα be homogenous generators of gr(M) over gr(A). Then choose nand i0 such that

n− i0 ≤ deg(mα) < n

where deg(x) is defined to be the smallest d such that x ∈ Fild, and for x ∈ gr(M) ahomogenous element, deg(x) is the label of the summand of gr(M) to which x belongs. Wewill show that the inclusion holds for this pair of n and i0. Let us induct on i. So supposethat for i we have Fili+n ⊆ Ai+i0Filn, and we want to show that Fili+n+1 ⊆ Ai+i0+1Filn+1.So consider some element m ∈ Fili+n+1. Then note that we may write m ∈ gri+1+n as

m =∑

mαaα

as an element of gr(m)i+1+n for some aα ∈ gr(A).Note that we may choose aα ∈ gr(A)i+1+n−deg(mα),because if we only consider this component of each aα, we get the same answer. Thusdeg(aα) = i+ n+ 1− deg(mα) ≤ i+ i0 + 1.

Thus m−∑mαaα = 0 in gri+n+1(M), so m−

∑mαaα ∈ Fili+nM .

The base case is clear.

To get to the theorem from the Lemma: Let n be as in the lemma, and choose m such thatFiln ⊆ Fil′n+m. Then we have

Fili+n ⊆ Ai+i0Filn ⊆ Ai+i0Fil′n+m ⊆ Fil′i+n+(m+i0)

as desired.


Consider the polynomial algebra A = k[x1, x2, . . . xn]. Let M be a finitely generate A module,and choose a good filtration on M . Then set Φ(i) = dimk(Mi) i.e., we are taking thedimension of Mi as a k vector space. Loosely speaking, we want our notion of dimension tomodel the rate at which Φ grows with i.

Theorem 2.18. For Φ as defined above,

1. There is a polynomial P with deg(P ) ≤ n such that Φ(i) = P (i) for i 0.

2. The degree of the polynomial and the leading coefficient are independent of choice ofgood filtration.

24


Before we proceed let us note this small lemma:

Lemma 2.19. Let φ be a function on the integers such that Φ(i + 1) − Φ(i) is eventuallya polynomial of degree less than or equal to m − 1. Then Φ is eventually a polynomial ofdegree less than or equal to m.

DG: “OK. You are an olympiad crowd. You know how to do this.”Proof Outline: If Φ(i + 1) − Φ(i) = P (i) for i ≥ N , then Φ(i) =

∑j≤i P (j) + C for i > N

and some constant C, which is essentially the sum of the differences between Φ and P fori < N . Thus we just need to show that

∑j≤i P (j) is eventually a polynomial of degree less

than or equal to m. Write P (x) = P0 + P1x+ · · ·+ Pm−1xm−1. Then we want that

∑j≤i

m−1∑α=1

Pαjα

is eventually a polynomial. It suffices to show this for each α,∑

j≤i jα is a polynomial in j.

But then we may induct on α, since if we set

X = 1α+1 + 2α+1 + · · ·+ iα+1,

thenX − iα+1 = 0 + 1α+1 + · · ·+ (i− 1)α+1,

so subtracting, we have

iα+1 = (1α+1) + (2α+1 − 1α+1) + · · ·+ (iα+1 − (i− 1)α+1)

and jα+1 − (j − 1)α+1 is a polynomial in j of degree α, say a0 + · · · + aαjα. Then the left

side is a polynomial in i, and on the rights side by the induction hypothesis we may rewriteall the terms of degree less than or equal to α − 1 as polynomials of i, so we are left withthe α term, which is now a polynomial of i.

Now let us prove the theorem:Proof of Theorem First note that as vector spaces dim(Mi/Mi−1) = dim(Mi) − dim(Mi−1),so we have that as vector spaces∑

j≤i

dim(gr(M)j) = dim(Mi).

We proceed by induction on n, the number of generators of A over k.Consider some submodule N of M with the induced filtration. Then we may consider

N [1] which is N , with the filtration shifted by 1, i.e. N [1]i = Ni+1. Then consider the map

N -xn

N [1]

given by multiplication by xn. Then let K be the kernel, and C the cokernel of this map, sowe have exact sequence

0 - K - N -xn

N [1] - C - 0

25


and taking the dimension of the ith line of the above we get

ΦN(i+ 1)− ΦN(i) = ΦK(i)− ΦC(i).

Now note that xn acts as 0 on both K and C, so we have that K and C are finitelygenerated modules over A[x1, · · ·xn−1], where the action is the same so we may now applythe induction hypothesis. Thus, the right side is eventually a polynomial, so we may applythe lemma.

The base case is clear: if A = k, then M is a finite dimensional vector space and dim(Mi)can never exceed dimk(M), so it eventually becomes constant.

So this gives us point 1 of the theorem.

Remark 2.20. Note that it may seem like the same proof holds if instead of saying “iseventually a polynomial”, we said “is a polynomial”. This is false, because though theinduction step still works, the base case does not.

Now to show point 2 of the theorem. Applying theorem 2.16, we see that for two filtra-tions Fil and Fil′,

Fil′i−k ⊆ Fili ⊆ Fil′i+k

so we haveΦ′(i− k) ≤ Φ(i) ≤ Φ′(i+ k)

so for i 0P ′(i− k) ≤ P (i) ≤ P ′(i+ k)

which means that P and P ′ must have the same degree.

Now, we can define the dimension of M , dimA(M) to be the degree of this polynomial.

Definition 2.21. Define the dimension of a finitely generated module M over a polynomialring A to be the the degree of the polynomial P described in theorem 2.18.

Due to point 2 of theorem 2.18, we see that this notion of dimension is well defined.Now let us derive some properties of the dimension.

2.2.1 Properties of Dimension of Modules over Polynomial Rings

Lemma 2.22. If M is a finitely generated module over A, a polynomial ring, and M1 is asubmodule, then dimA(M1) ≤ dimA(M). (Note that M1 is finitely generated because A isNoetherian, so dimA(M1) is well defined).

Proof. Since A is a polynomial algebra, it is Noetherian. This means that gr(A) is alsoNoetherian, so invoking lemma 2.15, we see that the induced filtration on M1 is good. Thuswe have (M i

1) ⊆ (M i) so dimk(Mi1) ≤ dimk(M

i)

Lemma 2.23. If M is a finitely generated module over A, a polynomial ring, and M M1,then dimA(M1) ≤ dimA(M).

Proof. This is the same thing.

26

2.3 For an arbitrary finitely generated algebra

Lemma 2.24. Suppose we have a short exact sequence of modules

0 - M1- M - M2

- 0

over a polynomial ring A. Then

dimA(M) = max(dimA(M1), dimA(M2)).

Proof. Note that the induced filtrations on M1 and M2 are good, as in the above lemmas.Then, taking the dimension of each level, we get

ΦM(i) = ΦM1(i) + ΦM2(i).

Thus,dimA(M) = max(dimA(M1), dimA(M2))

as desired.

Lemma 2.25. The dimension of the ring as a module over itself, dim(k[x1, . . . xn]) = n

Proof. So we consider the usual filtration by degree. There are(n+i−1n−1

)monomials of degree

i, so Φ(i)− Φ(i− 1) has degree n− 1, so Φ has degree n, as desired.

Lemma 2.26. Consider an injection of modules f : M → M . Then dim(coker(f)) <dim(M).

Proof. Consider the short exact sequence

0 - M1f- M - coker(f) - 0

where M1 = M , but we denote them differently here because they do not have the samefiltration, but rather we start with a filtration on M and then consider the induced filtrationon M1. Then, note that as we showed in one of the above lemmas,

ΦM(i) = ΦM1(i) + Φcoker(f)(i).

However although the filtrations are different we have shown that the leading coefficientand degree are independent of filtration, so this means that the degree of the polynomialassociated to Φcoker(f) is strictly smaller, as desired.


Up to now, we have only been looking at dimensions of modules over k[x1, . . . xn], the poly-nomial algebra over a field k. Now we consider arbitrary finitely generated algebras. A bean arbitrary finitely generated algebra over k, and let M be a finitely generated A module.

Then since A is finitely generated we may consider some k[x1, . . . xn] A. Then M isalso a finitely generated k[x1, . . . xn] module, and we should like to define:

dimA(M) = dimk[x1,...xn](M).

In order for this to make sense, we need to show that it is independent of generators, a factthat would result from the following theorem

27


Theorem 2.27. Consider k[x1, . . . xn, y1, . . . ym] and M a module over this algebra which isfinitely generated as a k[x1, . . . xn] module. Then

dimk[x1,...xn](M) = dimk[x1,...xn,y1,...ym](M)

Proof. It suffices to show that if M is a k[x1, . . . xn, y] module, and M is finitely generatedover k[x1, . . . xn] then its dimension over k[x1, . . . xn] is the same as that over k[x1, . . . xn, y].

Let A = k[x1, . . . xn], and B = A[y] = k[x1, . . . xn, y]. Let Ai be the ith filtration, i.e, thepolynomials of degree at most i. Define Bi similarly. We want to show dimA(M) = dimB(M).Since M is finitely generated over A, consider M0 ⊂ M which is the k span of the (finitelymany) generators of M over A. Then set Mi = AiM0, and note that M0 ⊂ M1 ⊂ · · · is anA-good filtration of M , which we may use to compute the Hilbert dimension of M as an Amodule.

Now let us define Mi = BiM0. This is clearly a B-good filtration. Note that Mi ⊃ Mi,so dimB(M) ≥ dimA(M).

We now want the other direction. Since M0 is a finite dimensional vector space, we mayconsider Mk such that yM0 ⊂Mk.Claim Mi ⊂Mik.Proof of Claim: Note that Bi = Ai ⊕ Ai−1y ⊕ · · · ⊕ A0y

i. So it suffices to show AjylM0 ⊂

Mj+lk, and for this it suffices to show

ylM0 ⊂Mlk.

For this, it suffices that yMt ⊂Mt+k, but this is true because

yMt = yAtM0 = AtMk ⊂Mt+k

as desired. So we have proven the claim.Now let us extract the theorem from the claim. Mi ⊂ Mil means that ΦB(i) ≤ ΦA(ik).

So if we consider the corresponding polynomials PB and PA, for sufficiently large x PB(x) ≤PA(kx) for some constant k, which means deg(PB) ≤ deg(PA), so dimB(M) ≤ dimA(M) asdesired.

Using this theorem it is easy to establish that the dimension is independent of choiceof generators. Namely, if we have generators x1, . . . xn and y1, . . . ym, then we may considerM as a module over k[x1, . . . xn] and k[y1, . . . ym], but both will have the same dimension asconsidered over k[x1, . . . xn, y1, . . . ym], by the above theorem, giving the desired independenceof generators.

k[x1, . . . xn] -- A

k[x1, . . . xn, y1, . . . ym]

66

--

--

k[y1, . . . ym]

66

Now that we have established independence of choice of generators, the suggested defi-nition makes sense:

28


Definition 2.28. Let A be a finitely generated k algebra and M a finitely generated Amodule. Then consider some polynomial algebra k[x1, . . . xn] with k[x1, . . . xn] A, anddefine

dimA(M) = dimk[x1,...xn](M).

Also definedim(A) = dimA(A).

2.3.1 Some Properties of Dimension

Lemma 2.29. dim(k[x1, . . . xn]) = n

Proof. This is manifest.

Lemma 2.30. If we have a map of algebras A → B such that B is finitely generated overA, then dimB(B) = dimA(B).

Proof. This follows from Theorem 2.27

Lemma 2.31. For M a finitely generated module over a finitely generated k algebra A,

dimA(M) ≤ dimA(A)

Proof. Note that we have A⊕n M , and the dimension of a quotient is less than or equalto the dimension of the original module. Also the dimension of a direct sum is the maximalof the dimensions of the direct summands, which gives us the desired inequality.

Lemma 2.32. For a map of algebras A → B such that B is finitely generated as an Amodule, then dim(B) ≤ dim(A).

Proof. Applying the above two lemmata dim(B) = dimB(B) = dimA(B) ≤ dim(A).

Lemma 2.33. For an injective map of finitely generated k algebras A → B, dim(A) ≤dim(B).

Proof. Note that if we have k[x1, . . . xn] A, and extend this to k[x1, . . . xn, y1, . . . ym] B,then ΦA(i) is the dimension of the image of k[x1, . . . xn]≤i in A, which is the same as thedimension of its image in B. ΦB(i) is the dimension of k[x1, . . . xn, y1, . . . ym]≤i in B, and theformer is less than or equal to the latter.

Lemma 2.34. Let A be a domain and I 6= 0 an ideal in A. Then dim(A) > dim(A/I).

Proof. Let 0 6= f ∈ I be an element. Then A/fA A/I, so dim(A/fA) ≥ dim(A/I).Thus, it is enough to show the statement for principal ideals. For this case, we may considerthe short exact sequence

0 - Af×- A - A/fA - 0

and then we may apply lemma 2.26 to see that the dimension must strictly drop.

29


For the interested there is actually a stronger statement associated with the proof of theabove lemma, namely:

Theorem 2.35. For a domain A, if V (fA) 6= 0 then dim(A/fA) = dim(A)− 1

The proof of this requires some more machinery that is developed in a later sectionof these notes. It is Theorem 7.8. Also, there is another proof that uses more high-techhomological algebra not covered in this course. This proof is in the Appendix.

Lemma 2.36. If we have a map of algebras A→ B and M a module over B such that Mis finitely generated as a module over A, then dimA(M) = dimB(M)

Proof. This follows from the definition of dimension and from theorem 2.27.

30

3 Dimension via Transcendency Degree

3.1 Definitions

Here we will introduce another notion of dimension, via the transcendency degree.Let us start with some definitions.

Definition 3.1. For K ⊃ k a field extension, we say that an element in K is integral over kif it satisfies a polynomial equation with coefficients from k. We say that K is algebraic overk if every element if K is integral. We say that K is transcendental if it is not algebraic.

Definition 3.2. We say that a field extension K ⊃ k is finitely generated if there are finitelymany elements y1, . . . yn ∈ K such that K does not contain a proper subfield containingy1, . . . yn.

(Note that this is not the same as being finitely generated as an algebra. For example,k(y1, . . . yn) is a finitely generated field extension of k, but not a finitely generated algebra.)

Definition 3.3. We say that a field extension K ⊃ k is finite if it K is a finite dimensionalvector space over k.

Lemma 3.4. If K ⊃ k is both algebraic and finitely generated then it is finite.

Proof. By induction, it suffices to show that if we adjoin one element x, i.e, we considerk ⊂ k(x) ⊂ K then the field got by adjoining one x is finite over k. But this is so because xis integral, so this field can be seen as k[t]/P (t) for some polynomial P .

Observation 3.5. Transitivity If k ⊂ K ⊂ L are field extensions such that K is algebraicover k and L is algebraic over K, then L is algebraic over k.

Definition 3.6. For K a field extension of k, a subset xα, indexed by a set A, are calledalgebraically independent over k if there does not exist a polynomial p ∈ k[t1, . . . tn] suchthat p(x1, . . . xn) = 0. Equivalently, one might say that the map k[tα] → K given bytα → xα extends to an injection k(tα)→ K.

Definition 3.7. A subset xα|α ∈ A is maximal if for any y ∈ K the collection xα|α ∈A ∪ y is not algebraically independent.

Because the above definition is somewhat ad-hoc, we also give this other equivalentdefinition:

Definition 3.8. A subset xα|α ∈ A is maximal if K is algebraic over the subfield of Kgenerated by xα|α ∈ A.

It is easy to see that the above definitions are equivalent. Here is a note on the existenceof finite maximal collections.

Lemma 3.9. K ⊃ k admits a finite maximal collection if and only if K is algebraic over afinitely generated sub-extension of k.

31

3.2 Agreement with Hilbert dimension

Proof. If y1, . . . yn is a finite maximal collection, then let K ′ ⊆ K be generated by it. Thecollection being maximal means that K is algebraic over K ′. For the other direction, if wehave K ′ finite generated, then let y1, . . . yn be its generators. Then y1, . . . yn form a maximalcollection.

Proposition 3.10. Let K be a field extension of k and let x1, . . . xn be algebraicallyindependent elements. Let y1, . . . ym be a maximal set of elements. Then m ≥ n.

Proof. This proof echoes the analogous theorem in linear algebra regarding spanning subsetsand linearly independent subsets of vector spaces. Suppose for the sake of contradiction thatn > m. Let us proceed with the proof of the proposition. Consider (x1, y1, . . . ym).

Then we claim that there is i such that (x1, y1, . . . , yi, . . . , ym) is maximal. To showthis consider a polynomial P ∈ k[t0, . . . tm] such that P (x1, y1, . . . ym) = 0, which existsbecause (y1, . . . ym) is maximal. Let i be such that the polynomial involves ti in a nontrivialway. Then we obtain that yi is algebraic over (x1, y1, . . . , yi, . . . , ym), so by transitivity, Kis also algebraic over (x1, y1, . . . , yi, . . . , ym). Let us re-index the yjs so that we have that(x1, y2, . . . ym) is maximal.

Now let us induct the above: if we have obtained that (x1, . . . xj, yj+1, . . . ym) is max-imal, then (x1, . . . xj, xj+1, yj+1, . . . ym) are algebraically dependent, and then we may con-sider a polynomial P that takes it to 0. Then P must involve the ys because the xsare algebraically independent, so then we may repeat the above process, and obtain that(x1, . . . xj+1, yj+2, . . . ym) is maximal. Repeating this until the end, we get that (x1, . . . xm)is maximal. However, (x1, . . . xn) are algebraically independent, so n ≤ m, as desired.

The above corollary inspires us to define the notion of transcendency degree, echoing thedefinition of dimension of a vector space, as follows:

Definition 3.11. Let k ⊂ K be an extension that admits a finite maximal collection.Let n be the maximal number of algebraically independent elements. Then n is called thetranscendency degree of K over k, which may be denoted tr.deg(K/k).

The above proposition shows that the definition makes sense because it gives us thefollowing corollary:

Corollary 3.12. For K an extension of k that admits a finite maximal collection andn = tr.deg(K/k) then

1. For any maximal collection y1, . . . ym, m ≥ n

2. For any algebraically independent y1, . . . ym, m ≤ n.


Now let us retrieve the notion of the dimension of an algebra from this:

Lemma 3.13. LetA be a finitely generated domain over k. Then dim(A) = tr.deg(Frac(A)/k)where Frac(A) denotes the field of fractions of A, i.e., the localization at the prime ideal(0) ⊂ A.

32


Proof. Let n = tr.deg(Frac(A)/k) and let x1, . . . xn be algebraically independent elements inFrac(A). Note that this means that (x1, . . . xn) is a maximal subset. We wish to show thatdim(A) = n.

First note from that in a problem set it was shown that in a domain, localizing at anelement does not affect the dimension (See the worked exercises section, Problem set 10,problem 9), so by localizing at product of the denominators of the xi, we may assume thatxi ∈ A. Now we have a map

k[x1, . . . xn] → A

where the map is injective because the xi are algebraically independent. Note that here,Frac(A) is finitely generated as a k(x1, . . . xn) extension because (x1, . . . xn) is a maximalsubset, but it is also algebraic, so it is a finite extension. Now applying another problem setproblem, we see that dim(A) = dim(k[x1, . . . xn]), as desired (see Problem Set 10, problem10).

33

4 Dimension via Chains of Primes

4.1 Definition via chains of primes and Noether Normalization

Now we have defined dimension of a finitely generated algebra A over a field k via Hilbertfunctions, and we have also defined transcendency degree and shown that for A a finitegenerated domain over k, dim(A) = tr.deg(Frac(A)/k). Here, we would like to define thedimension of a finitely generated domain over a field k to be the length of a maximal chainof primes

0 = p1 ( p1 ( · · · ( pn = A

where being maximal means that another prime cannot be added between any two of theabove primes. So we want to set dim(A) = n.

We will need to show that this is well defined, i.e., does not depend on the choice ofmaximal chain of primes, and that it is equivalent to our previous notions of dimension.

As a step in proving this, we will show the following theorem:

Theorem 4.1 (Noether Normalization Lemma). Let A B be a surjection of finitelygenerated algebras. Then there are polynomials rings k[x1, . . . xn] with an injection to A,and k[x1, . . . xm] with an injection to B such that

A -- B

k[x1, . . . xn]∪

6

⊃ k[x1, . . . xm]∪

6

commutes and A and B are finitely generated as modules over k[x1, . . . xn] and k[x1, . . . xm]respectively.

Proof. Actually our proof will only work when A is not finite, but this theorem still holdswhen A is finite.3

To start consider the following lemma

Lemma 4.2. Consider k[x1 . . . xn] Φ- A such that A is finitely generated as a moduleover k[x1, . . . xn]. Suppose Φ is not injective. Then there are y1, . . . yn−1 in the vectorspace spanned by x1, . . . xn such that k[y1, . . . yn−1] → A and A is finitely generated as ak[y1, . . . yn−1] module.

Proof of Lemma Consider some polynomial F which is in the kernel, so that F (x1, . . . xn) = 0in A. Let f be the top degree component of F . Then suppose xn enters into f . Then, becausethe field is not finite, there are some λ′1, . . . λ

′n ∈ k such that f(λ′1, . . . λ

′n) 6= 0, and λ′n 6= 0.

Then, renormalizing, we get f(λ1, . . . , λn−1, 1) 6= 0. Define

c := f(λ1, . . . λn−1, 1).

3see Eisenbud, Commutative Algebra with a View Towards Algebraic Geometry, theorems 13.2, 13.3

34


Let us consider yi = xi−λixn, and we claim that A is finitely generated over y1, . . . , yn−1.To show this, it suffices to show that xn satisfies a monic polynomial over k[y1, . . . yn−1] andindeed it does satisfy

g(t) = c−1F (y1 + λ1t, . . . yn−1 + λn−1t, t)

which can easily check to be monic.Now we prove the theorem from the lemma. Start with some k[β1, . . . , βs] → B which

is surjective, so in particular B is finite as a module. Then consider k[α1, . . . αt] → A, alsosurjective. Then we may consider

A -- B

k[α1, . . . αt, β1, . . . , βs]

66

⊃ k[β1, . . . , βs]

66

Now if the map k[β1, . . . , βs] → B is not injective, applying the above lemma we may re-choose a basis of the vector space spanned by β1, . . . βs and then throw out 1 variable. Thisdoes not affect the surjectivity of the left arrow, because the αs alone make this surjective.And the right arrow still has B finitely generated as a module. Then we may repeat thisprocess until the left arrow is surjective, and then do the same with the αs.

We may extract Hilbert’s Nullstellensatz as a corollary to the Noether NormalizationTheorem:

Corollary 4.3 (Nullstellensatz). A field extension of a field k which is finitely generated asan algebra is a finite field extension.

Proof. Now note that k[x1, . . . xn] → A is injective, so Spec(A) → Spec(k[x1, . . . xn]) hasdense image, but by the going up theorem it is closed, so it is surjective. But A was a fieldso Spec(A) is a point, so n = 0. Then we have that A is a finitely generated module over k,so it is finite, as desired.


Here are two theorems that we only give an outline of a proof of.

Theorem 4.4. Let A be a finitely generated domain over a field k. Let

p0 ⊂ p1 ⊂ · · · ⊂ pn ⊂ A

be a maximal chain of prime ideals. Then n = dim(A).

Theorem 4.5. Let A be a domain and let p ⊂ A be a sub-minimal prime ideal (i.e. p 6= 0but there are no primes between 0 and p). Then dim(A/p) = dim(A)− 1.

These two theorems are mostly proved in problem set 11, a problem set which could befixed by:

35


1. Modify statement 3, so that you may require A to be integrally closed.

2. For problem 5, note that you actually only need 3 for the integrally closed case. Alsoremove the phrase “whenever Noether Normalization holds”, because it always does.

3. To do part 3, consider the following theorem:

Theorem 4.6. Going Down Theorem Suppose that we have an injection of Noethe-rian domains, A → B with A is integrally closed and B is finitely generated as an Amodule. Then if p ∈ Spec(B) is a subminimal prime, then its image in Spec(A) is alsosubminimal.

Miaowrl?

36

5 Kahler Differentials

Let A be a k algebra and M be an A module. Then

Definition 5.1. A derivation ψ : A→M is a k-linear map such that

ψ(ab) = aψ(b) + bψ(a).

Let Derk(A,M) denote the set of derivations from A to M .

You can think of this as similar to a derivative, and the condition on derivations isessentially a product rule. To illustrate this, one may consider the following observation

Observation 5.2. For ψ : A → M a derivation, let P (x) ∈ k[x]. Then for any a ∈ A,ψ(P (a)) = P ′(a)ψ(a).

The above observation can be obtained by writing out the P (a), dividing it into mono-mials (using the face that ψ is k-linear), so in particular, additive, and then repeatedly usingthe above “product rule” on each monomial.

Proposition-Construction 5.3. For A an algebra over the field k, then there is a uniqueA module Ωk(A), called the module of Kahler differentials of A over k such that for any Amodule M ,

HomA(Ωk(A),M) ' Derk(A,M)

Proof. The uniqueness follows from Yoneda’s lemma, which is established in Problem set 6,problem 4. We will show existence by constructing the module explicitly.

Consider the free module generated by formal elements da and then modding the relationthat makes it a derivation.

Ωk(A) = SpanAda, a ∈ A/(adb+ bda− d(ab)).

It is easy to check that that mapping out of this is the same as derivations out of A.

Now let us establish some properties of the module of Kahler differentials.

Lemma 5.4. Consider the multiplication map mult : A⊗kA→ A, where we consider A⊗

kA

as an A module by acting A on the left side, and consider its kernel I:

0 - IA - A⊗kA mult- A - 0.

Then consider IA/I2A = IA ⊗

A⊗AA. We have IA/I

2A = Ωk(A).

37

Proof. First let us construct a map Ωk(A)→ IA/I2A. Such a map is by definition a derivation

ψ : A→ IA/I2A. Let this derivation be given by

ψ(a) = a⊗ 1− 1⊗ a.

We need to check that it is a derivation. Note that

ψ(ab)− aψ(b)− bψ(a) = ab⊗ 1− 1⊗ ab− a(b⊗ 1− 1⊗ b)− b(a⊗ 1− 1⊗ a)

= ab⊗ 1− 1⊗ ab− ab⊗ 1 + a⊗ b− ab⊗ 1 + b⊗ a= −1⊗ ab− a⊗ b− ab⊗ 1 + b⊗ a= (a⊗ 1− 1⊗ a)(1⊗ b− b⊗ a) ∈ I2

A

so it is indeed a derivation.In order to map in the other direction,

φ : IA/I2A → Ωk(A),

let us consider the explicit for Ωk(A), namely Ωk(A) = SpanAda, a ∈ A/(adb+bda−d(ab)).Then let our map be

φ(∑

ai ⊗ bi) =∑

bidai.

To show that this is well defined, we need that it vanishes on I2A. An element of I2

A is of theform (∑

i

ai ⊗ bi

)·

(∑j

cj ⊗ ej

)for∑

i ai ⊗ bi and∑

j cj ⊗ ej in IA. Then

φ

((∑i

ai ⊗ bi

)·

(∑j

cj ⊗ ej

))=∑i,j

biejd(aicj) =∑i,j

biejaid(cj) +∑i,j

biejcjd(ai) = 0

where the last equality is because∑

i aibi = 0 and∑

j cjej = 0 because∑

i ai ⊗ bi and∑j cj ⊗ ej are in IA.It is easy to check that these maps are mutually inverse.

Now let us work out an example.

Lemma 5.5. For A = k[x1, . . . xn], Ωk(A) is freely generated by dx1, . . . dxn.

Proof. The assertion is equivalent to saying that a derivation ψ : A→ M is the same as ann tuple of elements of M , m1, . . .mn. Given a derivation, we may consider mi = ψ(xi).

Given some mi, we may consider ψ(P ) =∑

∂P∂ximi, and check that it is a derivation. It

is clear that these two maps are inverses to each-other.

Now let us introduce a lemma that will provide a good way to show exactness of sequencesinvolving Ωk(A) using the universal property:

38

Lemma 5.6. A sequence of modules

M α- Nβ- L→ 0

is exact if and only if for every module M1 the sequence

0→ HomA(L,M1)β- HomA(N,M1) α- HomA(M,M1)

is exact

Proof. The proof of this is a simple diagram chase. As an example, let us show that if thebottom line is exact at HomA(N,M1), then the top is exact at N . The exactness of thebottom at HomA(N,M1) happens if and only if φ vanishes on the image of α if and only ifit factors through β, i.e. if and only if it vanishes on the kernel of β. So is suffices to showthat if we have submodules N1 and N2 of N , and for any M1 and φ ∈ Hom(N,M1) vanisheson N1 if and only if it vanishes in N2 then N1 = N2. In the spirit of Yoneda’s lemma, it iseasy to show this by checking on M1 = N/N1 and M1 = N/N2.

Lemma 5.7. Consider a map of k algebras A→ B, and the kernel I

0 - I - A - B - 0.

Then the there is an exact sequence

I/I2 - Ωk(A) - Ωk(B) - 0

where the first map is given by a 7→ da.

Proof.

For a map of rings A0 → A, define ΩA0(A), analogously to Ωk(A). Note that inProposition-Construction 5.3, we did not really use the field structure, so this makes sense.Then if A0 and A are k algebras, then note that A0 derivations are also k derivations, so wehave a natural inclusion

0→ DerA0(A,M)→ Derk(A,M)

so that we have0→ HomA(ΩA0(A),M)→ HomA(Ωk(A),M).

This is a natural transformation, so by Yoneda’s it uniquely defines a map

Ωk(A) - ΩA0(A).

Lemma 5.8. There is an exact sequence

A ⊗A0

Ωk(A0) - Ωk(A) - ΩA0(A) - 0.

Proof. This is a special case of problem set 12 problem 1.

39

Definition 5.9. Let A be a finitely generated algebra over k, with k algebraically closed.Then let fm ∈ Specm(A), and we have a short exact sequence

0 - m - Aξ- k - 0.

We define the cotangent space to A at m to be

T ∗m(Spec(A)) = m/m2 ' m⊗Ak

where A acts on k via ξ.We define the tangent space to be Tm(Spec(A)) = (T ∗m(Spec(A)))∗.

Lemma 5.10. T ∗mSpec(A) = Ωk(A)⊗Ak, where A acts on k via ξ.

Proof. This is Problem Set 12 Problem 5, see the end.

Now let us study the dimension of the contangent space to Spec(A) at m as a k vectorspace. In particular, note the following theorem.

Theorem 5.11. Let k algebraically closed and A a domain over k. Then

1. For any m ∈ Spec(A), dimk(Ωk(A)⊗Ak) ≥ dim(A).

2. There is an open set U ⊂ Spec(A) such that for any m ∈ U , dimk(Ωk(A)⊗Ak) = dim(A).

Points which are such that the dimensions are equal will informally be called “smoothpoints”. I will not write down the proof of the theorem as it is on the 221 course website.Here, we give some examples, and counterexamples if conditions are dropped.

1. Let A be a local Artinian ring, such as k[x]/x2. Then note that dim(A) = 0, becausethe ring is Artinian. However, for m the maximal idea, m⊗

Ak is not zero dimensional.

2. Consider A = k[y, z]/y3 − z2. Then m = (y, z) and m/m2 = Span(y, z) which is twodimensional. Note that the dimension of A is 1, by lemma 2.34. The dimension ofm/m2 is two, so this is an example of a point which is not a smooth point.

3. Let us consider further in depth the case when the field is not algebraically closed.Recall that a finite field extension k′ ⊃ k is called separable if k′ ⊗

kk is reduced as an

algebra over k, where k is the algebraic closure of k.

Lemma 5.12. If k1 ⊃ k2 ⊃ k then k2 is separable over k if and only if k1 is separableover k and k2 is separable in k.

MEOW?!

Lemma 5.13. Set k′ = k[x]/P (x). Then k′ is separable if and only if P has no multipleroots in the algebraic closure.

40

Proof. Note that k′ ⊗kk = k[x]/P (x) isn’t reduced if and only if there is some (x −

a)2|P (x), which is exactly there being multiple roots.

Observation 5.14. P has distinct roots in k if and only if P is coprime with P ′.However, P is irreducible, so (P, P ′) 6= 1 if and only if P ′ = 0. This may seem slightlystrange, because it may seem like this can never happen, but it isn’t so strange if weare in positive characteristic. For example xp− a for a not a pth root is an irreduciblepolynomial which has zero derivative. In the algebraic closure, it is (x− b)p for bp = a.

So we see that all field extensions of fields of zero characteristic are separable, butseparability is a non-trivial question when we are in fields of positive characteristic.

Now suppose A = k′ ⊃ k is a finite non-separable extension. Then dim(k′) = 0 butΩk(k

′) is zero if and only if k′ is separable over k, by Problem Set 12, problem 6,showing how this theorem could not work out if k is not algebraically closed.

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6 Completions

6.1 Introduction to inverse limits and completions

First let us define a notion of “inverse limit”.

Definition 6.1. Given a sequence of modules equipped with maps

M0φ01

M1φ12

M2 · · ·

we may define the inverse limit as

lim←−Mi = (mi) ∈∏

Mi

∣∣φi,i+1(mi+1) = mi.

Let the map lim←−Mi → Mi be the called the natural projection. We call the Mis equippedwith maps an inverse system.

The following lemma gives an alternate definition by universal property:

Lemma 6.2. For any N and any inverse system (Mi, φi,i+1),

Hom(N, lim←−Mi) = (τi) ∈∏

Hom(N,Mi)∣∣ψi,i+1 τi+1 = τi.

Proof. To go from τ on the left side to the τi, just compose with the projections, and thisworks by definition. For the other direction, use the τi to get a map from N to

∏Mi and

then the condition on the τs exactly says that you land in the inverse limit, which is a subsetof∏Mi.

Note that it is clear that inverse limits are functorial, where maps of inverse systems(M,φM)→ (N, φN) are maps Mi → Ni for each i that commute with the φs. We will showthat this functor is left exact, i.e:

Lemma 6.3. If we have a short exact sequence of inverse systems: i.e, if we have a diagram

...

0 - Mi

αi - Ni

βi - Li - 0

0 - Mi+1

6

αi+1- Ni+1

6

βi+1- Li+1

6

- 0

...

with exact rows then the sequence of inverse limits

0 - lim←−Miα- lim←−Ni

β- lim←−Li

is exact.

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Proof. First let us show the injectivity of α. Well if some (mi) maps to zero then eachcomponent must map to zero, but the αi are injective, so that means the mi are zero, asdesired.

Now we will show it is exact at lim←−Ni. Suppose that β(ni) = (0). Then for each i thereis mi such that αi(mi) = ni, by the exactness of the rows. So we want to show that (mi) isin the inverse limit. Note that

αi(φi,i+1(mi+1)) = φi,i+1(αi+1(mi+1)) = αi(mi)

but αi is injective, so φi,i+1(mi+1) = mi, as desired.

It is not true that the map lim←−Niβ- lim←−Li is surjective, but let us try to prove this

anyway: Consider some (li) ∈ lim←−Li and since βi are surjective, let us consider pre-imagesni ∈ Ni for each li. Obviously, we can’t just try to prove that (ni) is in the inverse limit,because we made a choice in selecting them.

So, we want to correct our choice. Suppose φ(n1) = n′0. Then n′0 and n0 have the sameimage in L0, so n0 − n′0 = α(m0) for some m0. We want to correct our choice of n1 tomatch n0, so we want m1 such that φ(n1 + α(m1)) = n′0 + α(m0), but this means we needφ(α(m1)) = α(m0). This isn’t always possible, but would be possible if we had also that theφ corresponding to the system Mi are surjective, giving us the Mittag Leffler theorem:

Theorem 6.4. Mittag-Leffler Given a diagram as in the previous lemma, if the maps inthe inverse system Mi are surjective, then

lim←−Niβ- lim←−Li

- 0

is also exact.

The proof follows pretty much the same reasoning as the pre-amble to this, where wecorrected for n1; one may continue correcting n2, etc.

Now let introduce the notion of completions. Consider M an abelian group, consider achain of subgroups

M ⊃M1 ⊃M2 ⊃ · · ·and let Mi = M/M i. Then we have projections Mi+1 →Mi

0 = M0 M1

M2 · · · .

We will use the M i to define a topology on M .

Definition 6.5. Let U ⊂M be open if for any m ∈ U there is i such that m+M i ⊂ U .

It is easy to see that this satisfies the conditions of a topology. Furthermore, this topologyis by definition countable. Let us consider whether it is Hausdorff.

Lemma 6.6. The topology as defined above is Hausdorff if and only if ∩M i = 0.

Proof. If ∩M i 6= 0 then consider some m in all M i. Then m ∈ ∩M i then every neigh-bourhood of m contains 0, so the topology is not Hausdorff.

If ∩M i = 0 then for any m1 and m2, choose M i such that m1−m2 6∈M i, and consideropen sets m1 +M i and m2 +M i.

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Now let us talk about completions. Note that the usual topological notion of completionsfrom you analysis class requires more than just a topology; they also require a metric.

Definition 6.7. For a sequence (mn) we say it converges to an element m

limmn = m

if for any i there is N such that for any n ≥ N , m−mn ∈M i.

Definition 6.8. We say that a sequence (mn) is Cauchy if for any i there is some N suchthat for any n, n′ ≥ N , mn−mn′ ∈M i. We define two Cauchy sequences (mn) and (m′n) to

be equivalent if (mn −m′n) converges to 0. Let M , the completion of M be the equivalenceclasses of Cauchy sequences.

The following theorem gives us a different way to view the completion:

Theorem 6.9. M = lim←−M/M i.

Proof. Let us first construct a map M → lim←−M/M i. For a Cauchy sequence (mn), for eachi consider let N be as in the definition of a Cauchy sequence. Then pick α > N and map(mn) to mα ∈M/M i, and let this be the term in the completion.

For the other direction, for (mi) compatible elements of∏M/M i, just choose any mi ∈M

with πi(mi) = mi. Then these form a Cauchy sequence.

Remark 6.10. M i ⊂M is a closed set, because it is clear that its complement is open.

Note that this new definition of completions as the inverse limit of M/M i gives us someinteresting properties, such as:

Lemma 6.11. Suppose we have an exact sequence of modules

0 - M ′ - Mp- M ′′ - 0

and suppose M is given a filtration

M ⊃M1 ⊃M2 ⊃ · · · .

Then let us consider the induced filtrations on M ′ and M ′′, namely

(M ′)i = M i ∩M ′ and (M ′′)i = p(M i)

Then there is short exact sequence

0 - M ′ - M - M ′′ - 0

Proof. One can easily check that

0 - M ′/(M i ∩M ′) - M/M i p- M ′′/p(M i) - 0

and then this becomes an application of Mittag-Leffler.

44


Note that if M and N are modules equipped with filtration and there is a map M → Nwhich is compatible with filtration, then the induced map of completions does not need tobe injective or surjective. For example, if we take the filtration on M to be M ⊃ 0 ⊃ 0 · · ·and that on N to be N ⊃ N ⊃ N ⊃ · · · then the completion of the right side is 0, but thecompletion of the left side is not zero, so the map cannot be injective, even if M → N is anisomorphism.

However, if the topology on M is induced from N then we do have this.Note that the above lemma means that ifM ′ ⊂M and the filtration onM ′ is induced from

the filtration on M , then the map M ′ →M is also injective. Applying this to M ′ = M i ⊂M ,we get the following corollary:

Corollary 6.12.M = M

Proof. First apply Lemma 6.11 to M ′ = M i and we get that

M/M i = M/M i

but with the filtration induced, M/M i has the discrete topology, so M/M i = M/M i. Now,taking inverse limits on both sides gives us the desired.

Let A be a commutative ring, a ⊂ A an ideal, and M an A module.

Definition 6.13. Consider the filtration M i = aiM . This is called the a-adic filtration ofM .

Let us consider some examples

1. A = k[t], m = tk[t], and M = k[t]. Then

A = k[[x]] = pi ∈ k[x]/tik[x]

2. Set A = Z, a = pZ and M = A. Then

A = Zp

Note that if we set M = A, and consider the a-adic topology, A is still a ring. Further-more, since there is a natural map A → A, it is also naturally and A-algebra. Also M isnaturally an A module. Let us further consider the structure of these modules.

Lemma 6.14. Let a be maximal. Then with the a-adic filtration, A is local, with maximalideal a.

Proof. To show this we need to show that for any a ∈ A, a is invertible if and only if it isnot in a.

If a 6∈ a then we need b ∈ A such that ba = 1. Take bi ∈ A/aiA such that aibi = 1 inA/aiA...

This material is all in Atiyah-Macdonald, Chapter 10, in pretty much thesame way, so the rest of the notes for this section will be very terse. I might fillin later if I have the time.

45

6.2 Artin Rees Pattern


For a ⊂ A, and M an A-module, we may consider the filtration on M

M ⊃ aM ⊃ a2M ⊃ · · ·

and then for a submodule M ′ ⊂M , we may consider two filtrations on it:

M ′ ⊃ aM ′ ⊃ a2M ′ ⊃ · · ·

and the induced filtration on it as a sub-module of M . The following theorem relates thesetwo filtrations:

Theorem 6.15. Artin-Rees For M finitely generated over Noetherian ring A, the abovetwo topologies on M ′ ⊂M coincide.

Definition 6.16. Given an A module M with a filtration M ⊃ M1 ⊃ M2 · · · , we saythat the filtration is compatible with a if aM i ⊂ M i+1. We say a filtration is a-good if forsufficiently large i, aM i = M i+1.

Observation 6.17. It is easy to see that if Fili(M) and Fili(M) are two good filtrations

then for some k

Fili+k

(M) ⊂ Fili(M) ⊂ Fili−k

(M).

This observation has the following consequence:

Corollary 6.18. Any two good filtrations define the same topology.

Note that the Artin Rees theorem can be reformulated in these terms:

Theorem 6.19 (Equivalent to Artin-Rees). The filtration M ′ ∩ aiM is a-good.

We will prove get back to this theorem.

Definition 6.20. For a ring A with ideal a, we may consider a graded ring

A∗ = A⊕ a⊕ a2 ⊕ · · ·

a graded ring. This ring is called the Artin ring.

Lemma 6.21. For A Noetherian, A∗ is also Noetherian.

Proof. Consider t1, . . . tn the generators of a as an A module. Then A[t1, . . . tn] A∗, so A∗

is also Noetherian.

Note also that if we have M ⊃M1 ⊃ · · · compatible with filtration then

M∗ = M ⊕M1 ⊕ · · ·

is a module over A∗.

Theorem 6.22. Let A be Noetherian. Then the following are equivalent

46


1. M is finitely generated and the filtration is a-good.

2. M∗ is finitely generated.

Proof. (2) ⇒ (1) It is finitely generated because it is a quotient of M∗. We want to show itif a good. Say the generators are in the first i gradings. Then for something in Mk we canget there by multiplying asome power.(1) ⇒ (2) Let i0 be such that for i ≥ i0 aM i = aM i+1. Then M ⊕M1 ⊕ · · · generates M∗

over A∗.

Corollary 6.23 (This will give us Artin-Rees). Let A be Noetherian and M be a finitelygenerated A module. Let M ′ ⊂ M and M ⊃ M1 ⊃ M2 ⊃ · · · be a good. Let M ′ ⊂ M .Then M ′i = M ′ ∩M i is an a good filtration on M ′.

Corollary 6.24. Let A be Noetherian and M2 finitely generated in the following short exactsequence of A modules:

0 - M1- M2

- M3- 0.

Then the sequence of completions with respect to the a-adic filtrations

0 - M1- M2

- M3- 0

is also exact.

Proof. Well, we already knew this was true for induced filtrations, and by Artin-Rees theseare the same as the a-adic filtrations.

Corollary 6.25. Let A be local Noetherian and M finitely generated over A. Then M → M .

Proof. Let N be the kernel. Then we have commutative diagram

N ⊂ - M

N

?⊂ - M

?

N → M → M is 0 by definition So N → N → M is 0. However N → M , so N → N is 0,but this is not possible by Nakayama, since we have

N → N → N/aN

for maximal ideal a.

Corollary 6.26. Let A be Noetherian and M finitely generated. Then

A⊗AM ' M

47


Proof. Note that M = lim←−M/aiM and A/aiA acts on this M/aim, so M is a A module.Consider An M and M ′ the kernel. Then we have

0 - M ′ - An - M - 0

and note that we have

A⊗AAn - An

A⊗AM

?- M

?

where the top map is an isomorphism because tensor products and completions commutewith finite direct sums. The arrow on the right is surjective because of surjectivity of inverselimits. The bottom map is surjective because the top map is an isomorphism and the rightmap is surjective. Then we have the following diagram:

A⊗AM ′ - A⊗

AAn - A⊗

AM - 0

M ′

??- An

∼?

- M

??- 0

and now we get the desired by applying a diagram chase (or the five lemma does it immedi-ately).

The following two corollaries are immediate from the above one.

Corollary 6.27. A is A flat if A is Noetherian. This is because it is enough to check flat-nesson finitely generated modules, and we have shown that completions are exact in Corollary6.24.

Corollary 6.28. Let A be Noetherian such that A ' A let M be a finitely generated Amodule. Then M = M .

Theorem 6.29. If A is Noetherian, then A is also Noetherian.

Consider M ⊃ M1 ⊃ M2 ⊃ · · · and gr(M) = ⊕M i/M i+1 a graded abelian group. Wealso consider gr(A) = ⊕ai/ai+1.

Lemma 6.30. suppose we have a map φ : M → N compatible with filtrations andgr(φ) : gr(M) → gr(N) is injective (resp. surjective). Then φ : M → N is injective(resp. surjective).

Proof. We are in the situation of Mittag Leffler, so it is enough to show M/M i → N/N i isinjective (resp. surjective), which reduces to the case of increasing filtrations.

48


Theorem 6.31. Let A be a commutative ring and a ⊂ A such that A ' A. Let M be an Amodule and

M ⊃M1 ⊃M2 ⊃ · · ·

be a a compatible filtration such that M is Hausdorff. Assume that gr(M) is finitely gener-ated over gr(A).

1. M is finitely generated over A

2. M is complete.

Proof. Let m1,m2, . . .mn be the homogenous generators of gr(M) and lift them to elementsm1, . . .mn. Then the corresponding map

⊕gr(A)[−di] - gr(M)

is surjective. Now apply the surjection case of Lemma 6.30.

Observation 6.32. If A is Noetherian then gr(A) is Noetherian. This is because

gr(A) = A/a⊕ a/a2 ⊕ · · ·

so gr(A) is a quotient of A∗, so it is the quotient of a Noetherian ring.

Theorem 6.33. Let A ' A and gr(A) Noetherian. Then A is Noetherian.

Proof. Let I ⊂ A. We want to show that I is finitely generated. Consider I∩ai as a filtrationof I. It is a compatible. Then gr(I) is an ideal of gr(A) so it is finitely generated, but thetopology on I is Hausdorff because it is induced by a Hausdorff topology. Thus, applyingthe above theorem we see that I is finitely generated, as desired.

49

7 Local Rings and Other Notions of Dimension

7.1 Dimension theory for Local Noetherian Rings

We will introduce three notions of dimension and show that they are the same.First let us define the Hilbert Dimension.For A ⊇ m a local Noetherian ring, and M a finitely generated A module, we may define

dimHilb(M) analogously to how we did it for finitely generated algebras:Choose any m-good filtration on M and set Φ(i) = len(M/Mi). Note that this is finite

because this does not depend on choice of filtration and we have M/miM ⊇ mM/miM ⊇ · · · ,and then each successive quotient has finite length because it is a module over A/m, whichis Artinian.

Then by a similar argument to what we did with Hilbert dimension earlier, we can showthat Φ is eventually a polynomial, with leading coefficient and degree independent of thegood filtration. So we can set dimHilb(M) = deg(P ).

Also define dimHilb(A) = dimHilb(M)Recall that the Artin-Rees lemma says that for M ′ ⊂M , a good filtration on M ′ induces

a good filtration on M . Therefore, by the same argument as in Lemmas 2.24 and 2.26 weget:

Lemma 7.1. 1. For exact sequence

0 - M ′ - M - M ′′ - 0

we havedimHilb(M) = max dimHilb(M

′), dimHilb(M′′)

2. For φ : M →M we have

dimHilb(M/im(φ)) ≤ dimHilb(M)− 1

Observation 7.2. Note that if we have

mk ⊂ a ⊂ m

then a and m determine the same dimension, because

M/mikM M/aiM M/miM.

Thus, we could have calculated with any ideal a such that rad(a) = m.

Also, define dimHilb(A) = dimHilb(A) for A as a module over itself. dimHilb(A/I) for A/Iover itself is the same as for A/I over A? (Note A/I is also a local ring (its Spec is a closedsubset), so this question makes sense).

Also we define Krull Dimension, dimKrull(A) to be the maximum length of a chain ofprime ideals

p0 ⊂ p1 ⊂ · · · ⊂ pn = m.

And we define the Generator Dimension To be the minimal number d such that thereare x1, · · ·xd ∈ m such that rad((x1, . . . xd)) = m, where rad denotes the radical of the ideal.

50


Theorem 7.3. dimhilb = dimkrull = dimgen.

Proof. First note that dimhilb ≥ dimkrull can be done by a similar argument to problems 1and 2 of pset 11.

Also dimgen ≥ dimhilb because we may set a = (x1, . . . xd), and because the Hilbertpolynomial has degree at most the number of generators of a.

It remains to show dimKrull ≥ dimgen.

Definition 7.4. For p ⊂ A a prime, define the height of p, to be height(p) = the maximallength of a chain of prime ideals p0 ⊂ p1 ⊂ · · · ⊂ pn = p. Equivalently, since primes of thelocalization Spec(Ap) are primes in Spec(A) which are contained in p, one may say

height(p) = dimkrull(Ap)

For every i = 1, . . . d, where d is the Krull dimension, we are going to construct someelements x1, . . . xi ∈ m such that for any prime p ⊃ (x1, . . . xi), height(p) ≥ i. Indeed if wedid this, this would mean for any p ⊃ (x1, . . . xd), height(p) ≥ d. But this would mean thatp = m because d is the Krull dimension. So m is the only prime ideal in A/(x1, . . . xd) whichmeans rad(x1, . . . xd) = m, so the generator dimension would be at most d, as desired.

So we now want to construct the xs. Assume, inductively, that x1, . . . xi−1 have alreadybeen constructed. Then let pα be the minimal primes containing x1, . . . xi−1 of heightexactly i−1. Note pα 6= m. Note that if a prime ideal is not contained in any of the pα, thenit is not contained in their union. So m 6⊂ ∪pα, so there is some xi ∈ m such that xi 6∈ pαfor any α. Then consider (x1, . . . xi) and consider any prime p ⊃ (x1, . . . xi). Then p mustcontain some minimal prime q which contains x1, . . . xi−1. Then consider two cases:

1. If q = pα for some α, then p ) pα, because qα does not contain xi. So dim(p) >dim(pα) = i− 1.

2. If q is not one of them, then q has height at least i, so p which contains it, does also.

Corollary 7.5. If A is a local Noetherian ring then dim(A) ≤ dimA/m(m/m2).

Proof. Consider a basis of m/m2 and let xi ∈ m be their lifts to m. Then by Nakayama’slemma the xi generate m, so the radical of the ideal the generate is still m, so applying thegenerator definition of dimension, the statement becomes obvious.

Here is a useful theorem that gives the dimensions of some localizations:

Theorem 7.6. Krull’s Hauptidealsatz (a.k.a Krull’s Principal Ideal Theorem). Let Abe a Noetherian ring and x ∈ A a non-unit, non zero-divisor. Then for any minimal primep ⊃ (x), height(p) = 1.

Proof. We need to show that dim(Ap) = 1. Note that dim(Ap) ≤ 1. In fact, if we considerx ∈ pp ⊂ Ap, then rad(x) = pp, because pp is simultaneously maximal and the minimal primecontaining x.

However, if dim(Ap) = 0, then Ap is local Artinian, which means that p is nilpotent inAp, but x ∈ p, so xn = 0, which contradicts x not being a zero-divisor.

51


Now we consider the dimension of a localization at a maximal ideal:

Lemma 7.7. Let A be a finitely generated k algebra and m a maximal ideal. Thendim(Am) = max(dim(Ai)), where Spec(Ai) are the irreducible components of Spec(A) whichcontain m.

Proof. Note that the right hand side is the maximal length of prime ideals terminating inm, and these are in bijection with chains of prime ideals in the localization, by the Krulldimension definition.

Now let us get back to proving the theorem that was mentioned in the section on Hilbertdimensions:

Theorem 7.8. For a finitely generated domain A over a field k, if V (fA) 6= ∅ thendim(A/fA) = dim(A)− 1.

Proof. We will in fact prove that every irreducible component Yi of Spec(A/fA) satisfies

dim(Yi) = dim(A)− 1.

Note that Yi = V (pi) = Spec(A/pi) where pi is a minimal prime which contains f , soapplying Krull’s Hauptidealsatz, we see that height(pi) = 1, so dim(A/pi) = dim(A) − 1,using the definition of dimension of a finitely generated domain via chains of primes.

As mentioned earlier, another proof of this using higher-tech homological algebra is givenin the appendix.

52

8 Smoothness and Regularity

8.1 Regular Local Rings

Note that when we were considering the dimension of a local Noetherian ring A, we definedthe Hilbert dimension via the function

φ(i) = lenA/mi =i−1∑j=0

len(mj/mj−1) =i−1∑j=0

dimA/m(mj/mj+1)

where mj/mj+1 can naturally be viewed as a vector space over A/m because the action of Aon mj/mj+1 can be factored through A/m. But note that

∞⊕i=0

mi/mi+1 = gr(A)

with respect to the filtration A ⊃ m ⊃ m2 ⊃ · · · so

dim(A) = dim(gr(A))

where on the right we are considering the dimension of gr(A) as an A/m algebra. It isa finitely generated A/m algebra, so we take the Hilbert dimension for finitely generatedalgebras for it.

Notice that gr(A) is generated by m/m2 i.e, SymA/m(m/m2) gr(A).

Definition 8.1. A local ring is called regular if the above map is an isomorphism.

Recall that Corollary 7.5 states that for any ring dim(A) ≤ dimA/m(m/m2). The followinglemma gives another equivalent definition of regularity of a local ring:

Lemma 8.2. A local ring is regular if and only if the inequality

dim(A) ≤ dimA/m(m/m2)

is an equality.

Proof. If SymA/m(m/m2) ' gr(A), then SymA/m(m/m2) has the same Hilbert polynomial asgr(A), but the SymA/m(m/m2) is just a polynomial algebra in dimA/m(m/m2) variables sothe degree of its Hilbert polynomial is dimA/m(m/m2), so dim(gr(A)) = dimA/m(m/m2), butrecall dim(A) = dim(gr(A)), so dim(A) = dimA/m(m/m2).

For the other direction, if dimA/m(m/m2) = dim(A), then dim(A) = dim(gr(A)), andagain applying the face that SymA/m(m/m2) is a polynomial algebra in dimA/m(m/m2) vari-ables, so it has dimension dimA/m(m/m2), we see that

dim(SymA/m(m/m2)) = dim(gr(A)).

But note that SymA/m(m/m2) is a domain, so if the surjection is not an isomorphism,then the dimension must strictly drop by Lemma 2.34 so the map must be an isomorphism,as desired.

53

8.1 Regular Local Rings

Remark 8.3. Let A be a local ring and A its completion. Then dim(A) = dim(A), becauseA/mn = A/mn, so the Hilbert functions are the same. Similarly, gr(A) = gr(A). However,by 6.14 A is also a local ring. So applying the above lemma, we see A is regular if and onlyif A is regular.

Lemma 8.4. A regular local ring is a domain.

Proof. Let a, b 6= 0. Note that ∩mn = 0, so there are k1 and k2 such that a 6= 0 ∈ mk1/mk1+1,and (b) 6= 0 ∈ mk2/mk2+1, which is to say that a ∈ mk1 , a 6∈ mk1+1. But then ab 6= 0 ∈ gr(A),because gr(A) = Sym(m/m2) is a domain. So ab 6= 0, as desired.

Lemma 8.5. If A is a regular local ring, and f ∈ m is such that f 6= 0 ∈ m/m2. ThenA′ = A/fA is also regular of dimension dim(A)− 1.

Proof. First let us show the dimension part of the statement. We know from 7.1 that thedimension has to drop. On the other hand if the images of rad(f1, . . . fn) = m/(f) in A/fA,then rad(f, f1, . . . fn) = m in A, giving us the other direction.

Now we want to show that it is regular. This amounts to showing that dimA′/mA′(mA′/m

2A′) =

dim(A′), or in other words that

dimA′/mA′(mA′/m

2A′) ≤ dimA/mA(mA/m

2A)− 1.

But note that we have a surjection

A A′ = A/fA

and mA′ = mA/f , so we also have a surjection

mA mA′

and since we havem2A → m2

A′

the map factors asα : mA/m

2A mA′/m

2A′ .

However, all of these maps are maps of A modules, so this last map is a map of A/mA vectorspaces (note that A/mA = A′/mA′ because f is in mA).

Also note that f 6= 0 ∈ mA/m2A, but α maps the image of f to 0, so α is not injective as a

map of vector spaces. It is however, surjective, so the dimension must drop, as desired.

Corollary 8.6. Consider elements f1, . . . fm in m such that f1, . . . fm ∈ m/m2 are indepen-dent. Then A/(f1, . . . fm) is regular with dim(A/(f1, . . . fm)) = dim(A)−m

Proof. This follows from Lemma 8.5 by induction. One just needs to check that in A1 =A/(f1), m1 = m/(f1), we have that f2, . . . fm are still linearly independent in m1/m

21. This

is easy to check.

54

8.2 Smoothness/Regularity for algebraically closed fields

Theorem 8.7. Let A0 be a regular local ring of dimension n, and consider a short exactsequence

0 - I - A0- A - 0

where A is a ring. Note that Spec(A) is a closed subset of Spec(A0) so A is also local. Thenthe following are equivalent

1. A is regular

2. There are elements f1, . . . fm ∈ I such that f1, . . . fm are linearly independent inmA0/m

2A0

where m = n− dim(A) such that (f1, . . . fm) = I.

Proof. (2) ⇒ (1) This is exactly the statement of Corollary 8.6(1) ⇒ (2) Just as in the proof of Lemma 8.5, i.e, using Problems 2a and 5a in Problem Set12, we see that there is exact sequence

I ⊗A0

A0/mA0- mA0/m

2A0

- mA/m2A

- 0.

When we proved this in Lemma 8.5, we used f ∈ mA, but that still works because I ⊂ mA0

because A0 is local with maximal ideal mA0 so all ideals are in mA0 .By assumption A0 and A are regular local, and we know A0/mA0 = A/mA, because

A = A0/I, mA = mA0/I so

dimA0/mA0(mA0/m

2A0

) = dim(A0) = n

anddimA0/mA0

(mA/m2A) = dim(A)

so the image of I ⊗A0

A0/mA0 in mA0/m2A0

has dimension m = n− dim(A). Let f1, . . . fm be a

set of linearly independent generators of mA0/m2A0

, and let f1, . . . fm be liftings to I.Let I ′ be the ideal generated by f1, . . . fm and consider A′ = A0/I

′. Then by Corollary8.6, we know that A′ is a regular local ring with dimension n−m = dim(A). Also I ′ ⊂ I sowe have

0 - J - A′ - A - 0

But Lemma 8.4, this means that it is a domain with the same dimension as A. But we knowfrom Lemma 7.1 that if we mod a domain out by a nonzero ideal, the dimension strictlydrops, so since the dimensions are the same, J = 0, so I is generated by f1, . . . fm, as desired.


For all of this section k be an algebraically closed field and A a finitely generatedalgebra over k. Let m ∈ Specm(A).

Definition 8.8. We say that A is smooth/regular at m if Am is a regular local ring.

55


Note that this is consistent with the definition given in Theorem 5.11, because Ωk(A)⊗Ak =

m/m2, by Problem Set 12, Problem 5a, and dim(A) = dim(Am) by Lemma 7.7.Let us denote x = x1, . . . xn. Then since A any finitely generated algebra over k, we

may consider a short exact sequence

0 - I - k[x] - A - 0.

Note that for c ∈ kn = Specm(k[x]), c ∈ Specm(A) if and only if c ∈ V (I).

Theorem 8.9. For c ∈ Specm(A), A is smooth at c if and only if there are f1, . . . fm′ ∈ I,where m′ = n − dimAm such that f1, . . . , ¯fm′ ∈ mk[x]/m

2k[x] are linearly independent, and

(f1, . . . fm′)m∼- Im.

Proof. Let mk[x] and mA denote the maximal ideals corresponding to c in k[x] and A respec-tively. We apply Lemma 7.7 to show that k[x]c is a regular local ring of dimension n: thedimension of the ring results from the lemma, and to calculate the dimension of mk[x]/m

2k[x]

it is clear that all points look the same, so we may as well take mk[x] = (x1, . . . xn), and forthis we know the dimension is n.

Now we have0 - Imk[x]

- k[x]mk[x]- Amk[x]

- 0

where k[x]mk[x] is regular local of dimension n, so the theorem follows from 8.7.

Remark 8.10. The above theorem is analogous to the Implicit Function Theorem. Wemay think of fis as elements of mA0/m

2A0' Ωk(A0) ⊗

A0

k. Now, we may think of Ωk(A0) =

A0dx1 ⊕ · · · ⊕ A0dxn, since it is the free module on n generators. In this language,

dfi ∈ mA0/m2A0

= Ωk(A0) ⊗A0

k =

(∂fi∂x1

(c), · · · ∂fi∂xn

(c)

)and we are saying that if the dfi are linearly independent and locally generate I then thealgebra A0/I is smooth at m.

In more geometric terms, this is saying that if we consider the following diagram

Spec(A) ⊂ - Spec(k[x1, . . . xn])

pt?⊂ - Spec(k[y1, . . . yn′ ])

Φ

?

Where pt maps to the origin of Spec(k[y1, . . . yn′ ]), i.e. to the maximal ideal (y1, . . . yn′),and Φ corresponds to the map φ : k[y1, . . . yn′ ]→ k[x1, . . . xn] given by yi 7→ fi, for f1, . . . fn′ .

The above diagram is saying that for the map

Spec(k[x1, . . . xn]) - Spec(k[y1, . . . yn′ ])

56


Spec(A) is the pre-image of a point, which is true because Φ(V (I)) = V (φ−1(I)) = (y1, . . . yn′),by Problem Set 4, Problem 1d.

Note that by Problem Set 4 Problem 7, Φ : kn → kn′

is given by (f1, . . . fn′), which leadsus to the Implicit Function Theorem for differential manifolds which states that if we have

U ⊂ Rn

V?

⊂ Rn′

f = (f1, . . . fn′)

?

the pre-image of a point is a manifold if df is surjective on the pre-image, which in this casemeans if the partials are linearly independent.

Proposition 8.11. Specm(A) is smooth at m ∈ Specm(A) if and only if Ωk(A) is free ona neighbourhood of m of rank dim(Am), or equivalently if Ωk(Am) is a free Am module ofrank dim(Am). (Note that when we showed that flatness is equivalent to local free-ness, weshowed that if Mp is a free Ap module, then there is a neighbourhood Uf around p on whichM is free of the same rank).

Proof. If Ωk(Am) is a locally free Am module of rank dim(Am) then

m/m2 = Ωk(A)⊗Ak = Ωk(A)⊗

AAm ⊗

Am

Am/mm

is a vector space of dimension dim(Am), which means that Am is regular, as desired.For the other direction, if Am is regular, we will show that Ωk(A) is free on a neighbour-

hood of m of rank dim(Am). By Theorem 8.9, for

0 - I - (A0 = k[x]) - A - 0

and there are f1, . . . fn′ such that Im = (f1, . . . , fn′)m and such that (df1(c), . . . dfn′(c)) arelinearly independent, where we are saying dfi =

∑ ∂fi∂xjdxj.

Note, by the way, that by Lemma 8.13 (which is shown using nothing from this section),we have that df1, . . . dfn′ are linearly independent in a Zariski neighbourhood of m. Although,this is not actually used in the proof.

Recall from Problem Set 12, problem 2a, that we have an exact sequence

I/I2 - A ⊗A0

Ωk(A0) - Ωk(A) - 0,

so localized at m, we have

Im/I2m

- Am ⊗A0

Ωk(A0) - Ωk(Am) - 0,

where the latter is an exact sequence of Am modules. However, because Im = (f1, . . . fn′)m,the above exact sequence may be written as

An′

m- Am ⊗

A0

Ωk(A0) - Ωk(Am) - 0,

57


where the generators of An′

m map to f1, . . . fn′ . Note that f1, . . . fn′ still satisfy that theirimages in k ⊗

Am

Am ⊗A0

Ωk(A0) are linearly independent.

Thus for the map An′

m- Am ⊗

A0

Ωk(A0), we have that

An′

m ⊗Am

k - (Am ⊗A0

Ωk(A0)) ⊗Am

k

is injective.The proposition now follows from the second point of Lemma 8.12, (shown below using

nothing from this section), which gives us that Ωk(Am) is locally free of the correct rank, asAn′

m is free of rank n′ and Ωk(A0) is free over A0 of rank n, so Am ⊗A0

Ωk(A0) is free over Am

of rank n, and n− n′ = dim(Am) by definition.

Lemma 8.12. Let A be a local Noetherian ring and let Φ : P1 → P0 be a map of locally freefinitely generated modules, such that for k = A/m, we have Φ : P1⊗

Ak → P0⊗

Ak is injective.

Then

1. Φ is injective

2. coker(Φ) is locally free of rank rank(P1)− rank(P0).

(Note that for a finitely generated module over a Noetherian ring, to being locally free isthe same as being projective, and over a local Noetherian ring, being projective is the sameas being free, which is the same as being flat, so throughout this lemma, we may replace thewords “locally free” with “projective”, “free”, or “flat”, as we like.)

Proof. Consider Q to be the image of P1, in P0. Let T be the kernel of the map, and R thecokernel. Then we have short exact sequences

0 - T - P1- Q - 0

and0 - Q - P0

- R - 0

For a module M over A, denote M = M ⊗Ak.

Because tensoring is right exact, we have P1 Q. However, we are given P1 → P0, andthe map factors as P1 → Q→ P0, so we must have

Q → P0.

But note that short exact sequence

0 - Q - P0- R - 0

induces a long exact sequence of Tors

TorA1 (P0, k) - TorA1 (R, k) - Q - P0- R - 0

58


but P0 is locally free and therefore flat, so TorA1 (P0, k) = 0. This along with Q - P0 beinginjective means that

TorA1 (R, k) = 0.

From here, we will show that R is free. Take a basis of R and lift the elements to R. Thenby Nakayama’s lemma, we have that these elements generate R. Then consider R′ the kernelof the map An → R given by these generators and we have a short exact sequence

0 - R′ - An - R - 0.

Now, considering the long exact sequence or Tors, and since TorA1 (R, k) = 0, this gives

0 - R′ - An - R - 0

but then An - R is an isomorphism, so R′ = 0, which, by Nakayama’s, means that R′ = 0,

so R is free.Also, our proof also gives that its rank is the same as the dimension of R, which is the

dimension of the cokernel of P1 → P0, but the latter map is injective, so the dimension ofthe cokernel is the difference of the dimensions. However, the dimensions of P1 and P0 areby a similar reasoning also equal to the ranks of P1 and P0 respectively, which gives us thesecond point of the proposition.

The first point of the proposition is equivalent to T = 0. Note that since R is projective,the short exact sequence

0 - Q - P0- R - 0

splits so thatP0 = Q⊕R

but this means that Q is a direct summand of a free module, so it is also projective, whichmeans the short exact sequence

0 - T - P1- Q - 0

also splits. So we haveP1 = T ⊕Q

so now our map P1 → P0 becomes

Q⊕ T ' P1 → P0 ' Q⊕R

and it factors asQ⊕ T Q → Q⊕R

But now, applying our assumption P1 → P0, we have that the composed map

Q⊕ T Q → Q⊕R

is injective, which means T = 0, so by Nakayama’s T = 0, as desired.

Lemma 8.13. If df1, . . . dfn′ ∈ Ωk(A0) are independent at c then are also independent in aZariski neighbourhood of it.

59

8.3 A taste of smoothness and regularity for non-algebraically closed fields

Proof. Let us write out what is going on. So we have df1, df2, . . . df′n which are

df1 =

(∂f1

∂x1

, . . .∂f1

∂xn

)...

dfn′ =

(∂fn′

∂x1

, . . .∂fn′

∂xn

)when we identify Ωk(A0) with An0 , where we view ∂fi

∂xjas an element of A0.

Then our assumption as that or some c, (the point in Specm(k[x1, . . . xn]) which cor-responds to m), df1, . . . dfn′ evaluated at c, (which are now elements of kn) are linearlyindependent.

Let us choose some n − n′ more vectors in kn, and then lift them to An0 , and call themxn′+1, . . . xn. Then df1, . . . dfn′ , xn′+1, . . . xn are linearly independent when evaluated at c.But this means the determinant at c is nonzero. But the determinant, which is a polynomialkn

2 → k can actually be seen as a map (A0)n2 → A0, and it factors through evaluation at c.

So the determinant of df1, . . . dfn′ , xn′+1, . . . xn, when we see this as a matrix directly in An2

0

is an element of A0, which is such that evaluated at c it is non-zero.So it is non-zero as an element of A0. Call it s. Then note that evaluated at any maximal

ideal in Us, it is non-zero. So we have produced a Zariski open, such that df1, . . . dfn′ arelinearly independent on it, as desired.

Remark 8.14. A condition which is such that if it happens as a point m ∈ Specm(A), thenit happens in a neighbourhood of m is called a “Zariski open condition”.

Theorem 8.15. Let A be a finitely generated algebra over an algebraically closed field kwith no nilpotent elements. Then there is a nonempty open subset U ⊂ Spec(A) such thatfor any m ∈ U , A is smooth at m.

The proof of this theorem is on the course website, labelled Differentials and Dimension.

8.3 A taste of smoothness and regularity for non-algebraicallyclosed fields

Definition 8.16. Let A be a finitely generated algebra over a field k. We say that m ∈Specm(A) is a regular point of A if Am is a regular local ring. We say that m is a smooth pointof A if Ωk(A)m is locally free. Note that we have shown that these two notions are equivalentif k is algebraically closed.

Note that in our proof of the equivalence of regularity and smoothness for algebraicallyclosed fields, we did not use the algebraic closed-ness in the direction of smooth impliesregular, so we have the following. So we have that if A is smooth at m, then it is alsoregular. The converse, however, is not always true.

60

8.3 A taste of smoothness and regularity for non-algebraically closed fields

Theorem 8.17. Consider a short exact sequence

0 - I - k[x1, . . . xn] - A - 0

and let n′ = n − dim(A). Then A is regular at m if and only if there are f1, . . . fn′ suchthat Im = (f1, . . . fn′)m and the images f1, . . . fn′ in mA/m

2A are linearly independent. A is

smooth at m if and only if there are f1, . . . fn′ such that Im = (f1, . . . fn′)m and df1, . . . dfn′ ∈Ωk(A0) ⊗

A0

k′ are linearly independent, where k′ = A/mA = A0/m

Note that there is always a map mA/m2A → Ωk(A0)⊗

A0

k′, so smooth-ness implies regularity.

However, this map may be neither injective nor surjective, and can in fact be very badlybehaved if the field is not separable.

We will now work only in the case when the field is separable. Recall from the ramblingsaround Theorem 5.11 that a finite field extension k′ ⊃ k is called separable if k′⊗

kk is reduced

as an algebra over k, where k is the algebraic closure of k, and that an equivalent definitionis that k′ = k[x]/P (x) is separable if and only if P has no multiple roots in the algebraicclosure.

MIAAOOWW!

61

9 Appendix

This is a collection of awesome pieces of math from office hours which are not incrediblyintegral to the course, but quite fun.

Theorem 9.1. For a domain A, if V (fA) 6= 0 then dim(A/fA) = dim(A) − 1. (This wasproven earlier as Theorem 7.8, which was also Theorem 2.35, but here we give a morehigh-tech proof).

Proof. Let A0 = k[x1, . . . xn] → A. Then note that n − dim(A) is the smallest k such thatExtkA0

(A,A0), but note that we have a short exact sequence

0→ A→ A→ A/fA→ 0

which induces a long exact sequence of Exts

ExtiA0(A/fA,A0)→ ExtiA0

(A,A0)→ ExtiA0(A,A0)→ Exti+1

A0(A/fA,A0)

and consider the term with i = n− dim(A). We know that the dimension must drop, so thefirst term is 0. If the dimension dropped by more than 1, then the last term would also be0, so we would have that the map is an isomorphism. But the map

ExtkA0(A,A0)→ ExtkA0

(A,A0)

is the action of f , so it suffices to show that if

Mf- M

is an isomorphism then f is invertible. Well, if f is not invertible, then choose some maximalideal m such that f ∈ m. Then we would have

M/mMf- M/mM

is still a surjection, but f ∈ m, a so it is actually 0.

Theorem 9.2. Cayley-Hamilton For a square matrix T (over some ring), consider thepolynomial P (λ) = detT − λId. Then P (T ) = 0.

Proof. First note that the desired statement is clearly true if T is diagonalizable. We willextract the general case from the diagonalizable case. We will use the Grothendieck Spec,that is for an algebra A over a field k, consider Spec(A) to be a functor from k-algebras tosets, given by

Spec(A)(R) = Hom(A,R)

What we are attempting to show is a polynomial identity, so to show that it holds overany ring, it suffices to show that it holds over C. Let k = C. Let Matn×n(R) denote thespace of n×n matrices over R for a ring R. Note that Matn×n(R) = Spec(k[x1, . . . xn2 ])(R),and we are given two maps

Matn×n-- Matn×n

62

namely Φ, which is given by for a matrix T , considering Det(T−λ) and applying it to T , andthe other being 0. These are both natural transformations, and it suffices to show that theyagree. Let A = B = Spec(k[x1, . . . xn2 ]). Then we have these two natural transformations

Spec(B) -- Spec(A)

and we want to show that they agree. Note that by Yoneda’s lemma they correspond tomaps

A -- B

and it suffices to show that the latter coincide.Consider some f ∈ B which is not a zero divisor (we will choose one later). Then note

that to show that two maps A -- B coincide it suffices to show the composed maps

A -- B - Bf

coincide. Because since f is not a zero divisor, the map B → Bf is injective.So this means it suffices to show that the maps

Spec(Bf )-- Spec(A)

agree, but since A, B are polynomial algebras and f is not a zero divisor, this means thatBf is reduced, which means that it suffices to show that the two maps

Spec(Bf )(k) - Spec(A)(k)

agree (this is by Problem Set 6 Problem 3)So now let us consider what we should choose for f . In particular, note that we would

be done if we could produce f a non-zero divisor, such that if T ∈ Spec(Bf )(k) then P (λ) =det(T − λ) has distinct roots, because then we could apply the fact that we already knowthis problem works for diagonalizable matrices, so we would know that the maps agree onSpec(Bf )(k) ⊂ Spec(B)(k).

To produce the f consider the map Spec(B) → Poln, where Poln = kn+1 = k[x1, . . . xn]sends a ring to the set of monic polynomials of degree n with coefficients in the ring, andthe map takes a matrix to its characteristic polynomial. This is a natural transformation,so it corresponds by Yoneda’s to a map k[x1, . . . xn]→ B

Now consider the discriminant (which was shown to exist in Problem Set 6 Problem 9),and is a polynomial g in k[x1, . . . xn] such that if g(a0, . . . an−1) 6= 0, then the polynomialxn + an−1x

n−1 + · · · + a0 has no multiple roots. It is easy to see that the image of g in Bsatisfies the desired.

Theorem 9.3. On Mapping Into Direct Limits. For a module M over a ring A thefollowing are equivalent

1. For any directed Pi the natural map

lim−→(Hom(M,Pi))α- Hom(M, lim−→(Pi))

(given by noting that a map out of a direct limit is a compatible system of maps outof each) is an isomorphism.

63

2. M is finitely presented. This means that M can be written as the cokernel of a mapA⊕n → A⊕m. (Note that if A is Noetherian, then this is the same as being finitelygenerated).

Proof. 2 =⇒ 1. Let νij : Ni → Nj be the maps of the directed system.Assume that M is finitely presented. We will show that α is an isomorphism. First, let

us show injectivity. Suppose there is some element φi ∈ lim−→(Hom(M,Pj)) that maps to 0(i.e, we are taking φi ∈ Hom(M,Pi) to represent the element of the element of the directlimit). Then note that by definition

α(φi)(m) = νiφi(m)

for νi : Ni → N the universal maps. so we have that for any m, φi(m) = 0 ∈ lim−→(Nj), so forany fixed m there is some j such that for I > j, νiI(φi(m)) = 0.

But M is finitely presented, so in particular it is finitely generated, so say it is generatedby m1, . . .mk. Consider the corresponding j1, . . . jk, and let j be the largest of them. Then,for any m, νiIφi(m) = 0, for I > j, so we get that νiI φi = 0, which means that as anelement of lim−→Hom(M,Ni), φi = 0, as desired.

So using only that M is finitely generated we have established injectivity.Now we will show surjectivity. Consider a short exact sequence

0 - M ′ - A⊕n - M - 0

where M ′ is finitely generated. This is possible because M is finitely presented. Thenconsider the diagram

0 - lim−→Hom(M ′, Ni) - lim−→Hom(An, Ni) - lim−→Hom(M,Ni)

0 - Hom(M ′, lim−→Ni)

α′

?- Hom(An, lim−→Ni)

αfr

?- Hom(M, lim−→Ni)

α

?

The rows are exact because Hom(−, N) is a left exact functor for modules N , and directlimits are exact in the category of modules.

We have shown that α and α′ are injective, since M and M ′ are finitely generated (recallwe only used finite generation and not the full force of finite presentation to show injectivity).It is easy to check that αfr is an isomorphism, because

lim−→Hom(An, Ni) = lim−→N⊕ni = (lim−→Ni)⊕n = Hom(An, lim−→Ni)

since direct limits commute with direct sums.So we have by the “Four Lemma” (a part of the “Five Lemma”; we could alternately

just do a diagram chase), that α is a surjection, completing this part of the proof.1 =⇒ 2. We are given that for any direct system Ni the map α is an isomorphism, and

we want to show that M is finitely presented.

64

First we will show that it is finitely generated, and from there we will show that it isfinitely presented. Since M is the direct limit of its finitely generated submodules, with themaps of the directed system being inclusions, we have

M = lim−→Mi

for Mi finitely generated. Let µi : Mi → M be the canonical map. We will apply the givenisomorphism to the system Mi. We have

lim−→Hom(M,Mi)∼- Hom(M, lim−→Mi) = Hom(M,M)

and we may now consider the element of the left side corresponding to the identity on theright. Let φi ∈ Hom(M,Mi) be a representative of it. Then note that by definition of α

α(φi)(m) = µiφi(m)

where µi is the inclusion from Mi to M . But this means that we have Mi = M , becauseeverything in M needs to be in the image of µi : Mi → M . This concludes the proof thatM is finitely generated.

Now we will show that it is finitely presented. The proof is very similar to the above.Note that any finitely generated module may be written as a direct limit of finitely

presented modules. To do this consider An → M and let M ′ be the kernel. Let us considerits finitely generated submodules M ′

i , with the maps being inclusions. This is clearly adirected system with limit M ′. Now consider Mi so that

0 - M ′i

- An - Mi- 0.

Here, since M ′i ⊂ M ′, we have that M is a quotient of Mi. Since M ′ is the direct limit of

M ′i and direct limits are exact in the category of modules, we may conclude that M is the

direct limit of the Mi. But the Mi are finitely presented, so we have that M is the directlimit of finitely presented modules where the universal maps to M are surjective.

Retracing our earlier steps that we employed to prove that M is finitely generated, wehave that there is some φi : M → Mi such that µi φi = IdM where µi : M → Mi is theuniversal map (which is also the canonical map to a quotient module). But this means thatif we let K be the kernel of the projection Mi →M so that we have exact sequence

0 - K - Mi- M

this short exact sequence splits. So

Mi = M ⊕K

But Mi is finitely presented, so to show that M is finitely presented it suffices to show thatK is finitely generated, because

M = Mi/K = An/(K +M ′i).

But K is finitely generated because it is a quotient of Mi, which is finitely generated, sowe are done.

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math 221 (commutative algebra) notes, 2nd halfpeople.math.harvard.edu/~gaitsgde/comalg_2008/... ·...

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