elementary commutative algebra

ELEMENTARYCOMMUTATIVE ALGEBRA

LECTURE NOTES

H.A. NIELSEN

DEPARTMENT OF MATHEMATICAL SCIENCESUNIVERSITY OF AARHUS

2005

Elementary Commutative Algebra

H.A. Nielsen

Contents

Prerequisites 7

1. A dictionary on rings and ideals 91.1. Rings 91.2. Ideals 111.3. Prime ideals 131.4. Chinese remainders 141.5. Unique factorization 151.6. Polynomials 161.7. Roots 181.8. Fields 191.9. Power series 20

2. Modules 212.1. Modules and homomorphisms 212.2. Submodules and factor modules 232.3. Kernel and cokernel 252.4. Sum and product 282.5. Homomorphism modules 302.6. Tensor product modules 342.7. Change of rings 37

3. Exact sequences of modules 413.1. Exact sequences 413.2. The snake lemma 453.3. Exactness of Hom 503.4. Exactness of tensor 533.5. Projective modules 553.6. Injective modules 563.7. Flat modules 60

4. Fraction constructions 634.1. Rings of fractions 634.2. Modules of fractions 654.3. Exactness of fractions 674.4. Tensor modules of fractions 694.5. Homomorphism modules of fractions 704.6. The polynomial ring is factorial 71

5. Localization 735.1. Prime ideals 735.2. Localization of rings 75

5

6 CONTENTS

5.3. Localization of modules 775.4. The local-global principle 795.5. Flat ring homomorphisms 815.6. Faithfully flat ring homomorphisms 83

6. Finite modules 876.1. Finite modules 876.2. Free modules 896.3. Cayley-Hamilton’s theorem 916.4. Nakayama’s lemma 936.5. Finite presented modules 946.6. Finite ring homomorphisms 99

7. Modules of finite length 1017.1. Simple modules 1017.2. The length 1027.3. Artinian modules 1057.4. Artinian rings 1077.5. Localization 1097.6. Local artinian ring 110

8. Noetherian rings 1138.1. Noetherian modules 1138.2. Noetherian rings 1158.3. Finite type rings 1178.4. Power series rings 1188.5. Localization of noetherian rings 1208.6. Prime filtrations of modules 121

9. Primary decomposition 1239.1. Zariski topology 1239.2. Support of modules 1269.3. Ass of modules 1289.4. Primary modules 1329.5. Decomposition of modules 1349.6. Decomposition of ideals 136

10. Dedekind rings 13910.1. Principal ideal domains 13910.2. Discrete valuation rings 14110.3. Dedekind domains 143

Bibliography 145

Index 147

Prerequisites

The basic notions from algebra, such as groups, rings, fields and their homomor-phisms together with some linear algebra, bilinear forms, matrices and determi-nants.Linear algebra: Fraleigh & Beauregard, Linear algebra, New York 1995.Algebra: Niels Lauritzen, Concrete abstract algebra, Cambridge 2003.Also recommended: Jens Carsten Jantzen, Algebra 2, Aarhus 2004.The propositions are stated complete and precise, while the proofs are quite short.No specific references to the literature are given. But lacking details may all befound at appropriate places in the books listed in the bibliography.A proposition being important when working in commutative algebra or a propo-sition containing a final result is named “Theorem”.

Nielsen, University of Aarhus, Spring 2005

7

1

A dictionary on rings and ideals

1.1. Rings

1.1.1. Definition. An abelian group is a setAwith an additionA×A→ A, (a, b) 7→a+ b and a zero 0 ∈ A satisfying

(1) associative: (a+ b) + c = a+ (b+ c)(2) zero: a+ 0 = a = 0 + a(3) negative: a+ (−a) = 0(4) commutative: a+ b = b+ a

for all a, b, c ∈ A. A subset B ⊂ A is a subgroup if 0 ∈ B and a − b ∈ B forall a, b ∈ B. The factor group A/B is the abelian group whose elements are thecosets a+B = {a+ b|b ∈ B} with addition (a+B) + (b+B) = (a+ b) +B. Ahomomorphism of groups φ : A→ C respects addition φ(a+ b) = φ(a) + φ(b).The projection π : A→ A/B, a 7→ a+B is a homomorphism. If φ(b) = 0 for allb ∈ B, then there is a unique homomorphism φ′ : A/B → C such that φ = φ′ ◦ π.

1.1.2. Definition. A ring is an abelian group R, addition (a, b) 7→ a+ b and zero0, together with a multiplication R × R → R, (a, b) 7→ ab and an identity 1 ∈ Rsatisfying

(1) associative: (ab)c = a(bc)(2) distributive: a(b+ c) = ab+ ac, (a+ b)c = ac+ bc(3) identity : 1a = a = a1(4) commutative : ab = ba

for all a, b, c ∈ R. If (4) is not satisfied then R is a noncommutative ring. Asubring R′ ⊂ R is an additive subgroup such that 1 ∈ R′ and ab ∈ R′ for alla, b ∈ R′. The inclusion R′ ⊂ R is a ring extension. A homomorphism of ringsφ : R → S is an additive group homomorphism respecting multiplication andidentity

φ(a+ b) = φ(a) + φ(b), φ(ab) = φ(a)φ(b), φ(1) = 1

An isomorphism is a homomorphism φ : R → S having an inverse map φ−1 :S → R which is also a homomorphism. The identity isomorphism is denoted1R : R→ R.

1.1.3. Remark. (1) A bijective ring homomorphism is an isomorphism.(2) Recall the usual formulas: a+ (−b) = a− b, 0a = 0, (−1)a = −a.(3) The identity 1 is unique.(4) A ring R is nonzero if and only if the elements 0 6= 1.(5) If φ : R → S is a ring homomorphism, then φ(0) = 0 and φ(R) is a subring

of S.(6) The unique additive group homomorphism Z → R, 1 7→ 1 is a ring homo-

morphism.

9

10 1. A DICTIONARY ON RINGS AND IDEALS

1.1.4. Proposition. Let R1, R2 be rings. The product ring is the product of addi-tive groupsR1×R2, ((a1, a2), (b1, b2)) 7→ (a1+b1, a2+b2), with coordinate multi-plication ((a1, a2), (b1, b2)) 7→ (a1b1, a2b2). The element (1, 1) is the identity. Theprojections R1×R2 → R1, (a1, a2) 7→ a1 and R1×R2 → R2, (a1, a2) 7→ a2 arering homomorphisms.

1.1.5. Lemma. In a ring R the binomial formula is true

(a+ b)n =n∑

k=0

(n

k

)an−kbk

a, b ∈ R and n a positive integer.

Proof. The multiplication is commutative, so the usual proof for numbers works.Use the binomial identity (

n

k − 1

)+

(n

k

)=

(n+ 1k

)together with induction on n.

1.1.6. Definition. a ∈ R is a nonzero divisor if ab 6= 0 for all b 6= 0 otherwise azero divisor. a is a unit if there is a b such that ab = 1.

1.1.7. Remark. (1) A unit is a nonzero divisor.(2) If ab = 1 then b is uniquely determined by a and denoted b = a−1.

1.1.8. Definition. A nonzero ring R is a domain if every nonzero element is anonzero divisor and a field if every nonzero element is a unit. Clearly a field is adomain.

1.1.9. Example. The integers Z is a domain. The units in Z are {±1}. The rationalnumbers Q, the real numbers R and the complex numbers C are fields. Thenatural numbers N is not a ring.

1.1.10. Example. The set of n× n-matrices with entries from a commutative ringis an important normally noncommutative ring.

1.1.11. Exercise. (1) Show that the product of two domains is never a domain.(2) Let R be a ring. Show that the set of matrices

U2 ={(

a b0 a

) ∣∣∣a, b ∈ R}with matrix addition and matrix multiplication is a ring.

(3) Show that the set of matrices with real number entries{(a −bb a

) ∣∣∣a, b ∈ R}

with matrix addition and multiplication is a field isomorphic to C.(4) Show that the composition of two ring homomorphisms is again a ring homomor-

phism.(5) Show the claim 1.1.3 that a bijective ring homomorphism is a ring isomorphism.(6) Let φ : 0→ R be a ring homomorphism from the zero ring. Show that R is itself the

zero ring.

1.2. IDEALS 11

1.2. Ideals

1.2.1. Definition. Let R be a ring. An ideal I is an additive subgroup of R suchthat ab ∈ I for all a ∈ R, b ∈ I . A proper ideal is an ideal I 6= R.

1.2.2. Lemma. Let {Iα} be a family of ideals in R. Then the additive subgroups∑α Iα and

⋂α Iα are ideals.

Proof. The claim for the intersection is clear. Use the formulas∑bα +

∑cα =∑

(bα + cα) and a∑bα =

∑abα to conclude the claim for the sum.

1.2.3. Definition. The intersection, 1.2.2, of all ideals containing a subset B ⊂ Ris the ideal generated by B and denoted (B) = RB = BR. It is the smallest idealcontaining B. A principal ideal (b) = Rb is an ideal generated by one element. Afinite ideal (b1, . . . , bn) is an ideal generated by finitely many elements. The zeroideal is (0) = {0}. The ring itself is a principal ideal, (1) = R. The ideal productof two ideals I, J is denoted IJ and is the ideal generated by all ab, a ∈ I, b ∈ J .This generalizes to the product of finitely many ideals. The power of an ideal isdenoted In. The colon ideal I : J is the ideal of elements a ∈ R such that aJ ⊂ I .

1.2.4. Example. (1) Every ideal in Z is principal.(2) In a field (0), (1) are the only ideals.(3) A subring is normally not an ideal.(4) Let K be a field. In K ×K there are four ideals (0), (1), ((1, 0)), ((0, 1)).

1.2.5. Proposition. Let R be a ring and B a subset, then RB =∑

b∈B Rb

RB = {a1b1 + · · ·+ anbn|ai ∈ R, bi ∈ B}A principal ideal is

(b) = Rb = {ab|a ∈ R}A finite ideal is

(b1, . . . , bn) = Rb1 + · · ·+Rbn

Proof. The righthand side is contained in the ideal RB. Moreover the righthandside is an ideal containing B, so equality.

1.2.6. Definition. Let φ : R → S be a ring homomorphism. For an ideal J ⊂ Sthe contracted ideal is φ−1(J) ⊂ R and denoted J ∩ R. The kernel is the idealKerφ = φ−1(0). For an ideal I ⊂ R the extended ideal is the ideal φ(I)S ⊂ Sand denoted IS. Note that (J ∩R)S ⊂ J and I ⊂ (IS) ∩R.

1.2.7. Lemma. Let I ⊂ R be an ideal and let R/I be the additive factor group.The multiplication

R/I ×R/I → R/I, (a+ I, b+ I) 7→ ab+ I

is well defined. Together with the addition the conditions of 1.1.2 are satisfied.

Proof. If a + I = a′ + I and b + I = b′ + I then a − a′, b − b′ ∈ I and soab − a′b′ = a(b − b′) + b′(a − a′) ∈ I . Therefore ab + I = a′b′ + I and themultiplication is well defined. Clearly 1.1.2 are satisfied.

1.2.8. Definition. LetR be a ring and I an ideal, then the factor ring is the additivefactor group R/I with addition (a + I, b + I) 7→ (a + b) + I and multiplication,1.2.7, (a+ I, b+ I) 7→ ab+ I . The projection π : R→ R/I, a 7→ a+ I is a ringhomomorphism.


1.2.9. Theorem. Let φ : R→ S be a ring homomorphism.(1) Let I ⊂ Kerφ be an ideal. Then there is a unique ring homomorphism

φ′ : R/I → S such that φ = φ′ ◦ π.

R

π��

φ // S

R/Iφ′

==

(2) The homomorphism φ′ : R/Kerφ → S is a ring isomorphism onto thesubring φ(R) of S.

R

π

��

φ // φ(R)

R/Kerφφ′

::

(3) For any ideal J ⊂ S, I = φ−1(J) ⊂ R is an ideal and the map φ′ : R/I →S/J is an injective ring homomorphism.

Proof. The statements are clear for the addition and the factor map φ′(a + I) =φ(a) is clearly a ring homomorphism.

1.2.10. Corollary. Let π : R → R/I be the projection. The map I ′ 7→ J =π−1(I ′) gives a bijective correspondence between ideals I ′ in R/I and ideals J inR containing I . Also I ′ = π(J) = J/I . This correspondence preserves inclusions,sums and intersections of ideals.

1.2.11. Corollary. Let I ⊂ J ⊂ R be ideals. Then there is a canonical isomor-phism

R/J → (R/I)/(J/I)

Proof. The kernel of the surjective east-south composite

R

��

// R/I

��R/J // (R/I)/(J/I)

is J . By 1.2.9 the horizontal lower factor map gives the isomorphism.

1.2.12. Example. For any integer n the ideals in the factor ring Z/(n) correspondto ideals (m) ⊂ Z where m divides n.

1.2.13. Definition. Let R be a ring. The additive kernel of the unique ring homo-morphism Z→ R is a principal ideal generated by a natural number char(R), thecharacteristic of R. Z/(char(R)) is isomorphic to the smallest subring of R.

1.2.14. Proposition. If the characteristic char(R) = p is a prime number, thenthe Frobenius homomorphism R→ R, a 7→ ap is a ring homomorphism.

Proof. By the binomial formula 1.1.4

(a+ b)p =p∑

k=0

(p

k

)ap−kbk = ap + bp

1.3. PRIME IDEALS 13

since a prime number p divides(pk

), 0 < k < p. Clearly (ab)p = apbp.

1.2.15. Exercise. (1) Let I, J be ideals in R. Show that the ideal product

IJ = {a1b1 + · · ·+ anbn|ai ∈ I, bi ∈ J}

(2) Let I ⊂ R be an ideal. Show that I = I : R.(3) Show that a ∈ R is a unit if and only if (a) = R.(4) Show that a ring is a field if and only if (0) 6= (1) are the only two ideals.(5) Show that a nonzero ring K is a field if and only if any nonzero ring homomorphism

φ : K → R is injective.(6) Let m,n be natural numbers. Determine the ideals in Z

(m,n), (m) + (n), (m) ∩ (n), (m)(n)

as principal ideals.(7) Show that a additive cyclic group has a unique ring structure.(8) Let p be a prime number. What is the Frobenius homomorphism on the ring Z/(p)?

1.3. Prime ideals

1.3.1. Definition. Let R be a ring and P 6= R a proper ideal.(1) P is a prime ideal if for any product ab ∈ P either a ∈ P or b ∈ P . This

amounts to: if a, b /∈ P then ab /∈ P .(2) P is a maximal ideal if no proper ideal 6= P contains P .

1.3.2. Proposition. LetP be a prime ideal and I1, . . . In ideals such that I1 . . . In ⊂P , then some Ik ⊂ P .

Proof. If there exist ak ∈ Ik\P for all k, then since P is prime a1 . . . an ∈I1 . . . In\P contradicting the inclusion I1 . . . In ⊂ P .

1.3.3. Proposition. Let R be a ring and P an ideal.(1) P is a prime ideal if and only if R/P is a domain.(2) P is a maximal ideal if and only if R/P is a field.

Proof. Remark P 6= R ⇔ R/P 6= 0. (1) Assume a + P, b + P are nonzero inR/P . Then a, b /∈ P . So if P is prime then by 1.3.1 ab /∈ P and ab+P is nonzeroin R/P . It follows that R/P is a domain. The converse is similar.(2) Assume R/P is a field and a /∈ P . Then a + P is a unit in R/P and there isb such that ab − 1 ∈ P . It follows that the ideal (a) + P = R and therefore P ismaximal. The converse is similar.

1.3.4. Corollary. (1) A maximal ideal is a prime ideal.(2) A ring is an domain if and only if the zero ideal is a prime ideal.(3) A ring a field if and only if the zero ideal is a maximal ideal.

1.3.5. Corollary. (1) If φ : R→ S is a ring homomorphism and Q ⊂ S a primeideal then φ−1(Q) is a prime ideal of R.

(2) Let I ⊂ R be an ideal. An ideal I ⊂ P is a prime ideal in R if and only ifP/I is a prime ideal in R/I .

(3) Let I ⊂ R be an ideal. An ideal I ⊂ P is a maximal ideal in R if and only ifP/I is a maximal ideal in R/I .

Proof. (1) By 1.2.9 R/φ−1(Q) is a subring of the domain S/Q. (2) (3) By 1.2.11R/P ' (R/I)/(P/I).


1.3.6. Example. An ideal in Z is a prime ideal if it is generated by 0 or a primenumber. Any nonzero prime ideal is a maximal ideal.

1.3.7. Definition. For an ideal I in a ring R the radical is√I = {a ∈ R|an ∈ I for some n}

a is nilpotent is an = 0 for some positive integer n. A ring is reduced if thenilradical

√0 = 0, that is if 0 is the only nilpotent element.

1.3.8. Proposition. (1) The radical of an ideal is an ideal.(2) The nilradical is contained in any prime ideal.(3) A domain is reduced.

Proof. (1) If am, bn ∈ I then by the binomial formula

(a+ b)m+n =m+n∑k=0

(m+ n

k

)am+n−kbk ∈ I

and the radical is an ideal. (2) (3) Clearly a nilpotent element is contained in anyprime ideal.

1.3.9. Exercise. (1) Show that the characteristic of a domain is either 0 or a primenumber.

(2) Let m,n be a natural numbers. Show that n+ (m) ∈ Z/(m) is a unit if and only ifm,n are relative prime.

(3) Let m be a natural number. Show that Z/(m) is reduced if m is square free.(4) Show that a product of reduced rings is reduced.(5) Let a be nilpotent. Show that 1− a is a unit.(6) Let I, J be ideals. Show that

√IJ =

√I ∩ J =

√I ∩√J .

(7) Assume an ideal I is contained in a prime ideal P . Show that√I ⊂ P .

1.4. Chinese remainders

1.4.1. Definition. Ideals I, J ⊂ R are comaximal ideals if I + J = R.

1.4.2. Theorem (Chinese remainder theorem). Let I1, . . . , Ik be pairwise comax-imal ideals in a ring R. Then

(1) For a1, . . . , ak ∈ R there is a a ∈ R, such that a−am ∈ Im form = 1, . . . , k(2)

I1 · · · Ik = I1 ∩ · · · ∩ Ik(3) The product of projections

R/I1 · · · Ik → R/I1 × · · · ×R/Ikis an isomorphism.

Proof. (1) For each m

R =∏n6=m

(Im + In) = Im +∏n6=m

In

So choose um ∈ Im and vm ∈∏

n6=m In with um +vm = 1. Put a = a1v1 + · · ·+akvk. Then a−am = · · ·+amum + · · · ∈ Im. (2) For a in the intersection assumeby induction that a ∈ I2 · · · Ik. From the proof of (1) a = u1a + av1 ∈ I1 · · · Ik.(3) Surjectivity follows from (1). The kernel is the intersection which by (2) is theproduct. 1.2.9 gives the isomorphism.

1.5. UNIQUE FACTORIZATION 15

1.4.3. Corollary. Let P1, . . . , Pk be pairwise different maximal ideals in a ring R.Then

Pn11 · · ·P

nkk = Pn1

1 ∩ · · · ∩ Pnkk

andR/Pn1

1 · · ·Pnkk → R/Pn1

1 × · · · ×R/Pnkk

is an isomorphism.

1.4.4. Definition. An element e in a ring R is idempotent if e = e2. A nontrivialidempotent is an idempotent e 6= 0, 1.

1.4.5. Proposition. A ring R is a product of two nonzero rings if and only if itcontains a nontrivial idempotent e.

Proof. Use that the ideals Re and R(1− e) are proper and comaximal.

1.4.6. Exercise. (1) Show that for a prime number p the rings Z/(p2) and Z/(p) ×Z/(p) are not isomorphic.

(2) Let n = pn11 . . . pnk

k be a factorization into different primes. Show that

Z/(pn11 · · · p

nk

k )→ Z/(pn11 )× · · · × Z/(pnk

k )

is an isomorphism.(3) Let elements e1 + e2 = 1 with e1e2 = 0 be given in a ring R. Show that R '

R/(e1)×R/(e2).(4) Let I, J be ideals such that

√I,√J are comaximal. Show that I, J are comaximal.

1.5. Unique factorization

1.5.1. Lemma. Let R be a domain and (a) = (b) principal ideals. Then there is aunit u ∈ R such that b = ua.

Proof. b = ua and a = vb giving b = uvb. If b 6= 0 then uv = 1 showing that u isa unit.

1.5.2. Definition. Let R be a domain and P the set of principal ideals differentfrom (0) and R. By 1.5.1 multiplication of generators gives a well defined multi-plication of principal ideals onP . Inclusion inP , (a) ⊂ (b) is division and writtenb|a. An element in P maximal for inclusion is an irreducible principal ideal. Agenerator of an irreducible element is an irreducible element in R. A principalprime ideal is a prime divisor. A generator of a prime divisor is a prime elementin R. Clearly a prime divisor is irreducible.

1.5.3. Definition. A domain R is a unique factorization domain if(1) Every irreducible ideal in P is a prime divisor.(2) Every element in P is a product of irreducible elements.

1.5.4. Proposition. In a unique factorization domain the factorization of elementsin P into prime divisors is unique up to order.

Proof. Proceed by induction on the shortest factorization of an element in P . Let(a1) . . . (am) = (b1) . . . (bn) be factorizations into irreducibles. (a1) is a primeideal, so by 1.3.2 and reordering (b1) = (a1). By cancellation m− 1 = n− 1 and(ai) = (bi) after a reordering.

1.5.5. Definition. A domain R is a principal ideal domain if every ideal is princi-pal.


1.5.6. Theorem. A principal ideal domain R is a unique factorization domain.

Proof. Let (a) be irreducible and x, y /∈ (a). Then (a, x), (a, y) are principal idealsproperly containing (a) giving (a, x) = (a, y) = R. Let ba + cx = da + ey = 1and look at (ba+cx)(da+ey) = 1 to see that xy /∈ (a). It follows that (a) is prime,part (1) of 1.5.3. If (b) is not irreducible, then (b) = (a1)(b1) for some irreducible(a1) and (b) ⊂ (b1). Continue this process to get (b) = (a1) . . . (an)(bn) for someirreducibles (a1), . . . (an). The chain of ideals (b) ⊂ (b1) ⊂ · · · ⊂ (bn) has aunion

⋃n(bn) which is a principal ideal. The generator of the union must be in

some (bn). Therefore (bn) = (bn+1) giving that (bn) is irreducible. This gives afactorization required in 1.5.3 part (2).

1.5.7. Example. The integers Z is a principal ideal domain and therefore a uniquefactorization domain.

1.5.8. Definition. The supremum in the set of all principal ideals of a set of ele-ments in P is the greatest common divisor and an infimum is the least commonmultiple.

1.5.9. Corollary. In a unique factorization domain the greatest common divisorand the least common multiple of a finite set of elements exist.If (a) = (p1)m1 . . . (pk)mk and (b) = (p1)n1 . . . (pk)nk withmi, ni ≥ 0 then great-est common divisor is (p1)min(m1,n1) . . . (pk)min(mk,nk) and least common multipleis (p1)max(m1,n1) . . . (pk)max(mk,nk).

1.5.10. Exercise. (1) Show that an irreducible element in a principal ideal domain gen-erates a maximal ideal.

(2) Show that there are infinitely many prime numbers.(3) Let Z[

√−1] be the smallest subring of C containing

√−1. Show that Z[

√−1] is a

principal ideal domain.(4) Let Z[

√−5] be the smallest subring of C containing

√−5. Show that Z[

√−5] is not

a unique factorization domain.

1.6. Polynomials

1.6.1. Definition. Let R be a ring. The polynomial ring R[X] is the additivegroup given by the direct sum

⊕nRX

n, n = 0, 1, 2, . . . consisting of all finitesums f = a0 + a1X + . . . amX

m, a polynomial with an ∈ R being the n′thcoefficient. Multiplication is given by XiXj = Xi+j extended by linearity. Ifg = b0 + b1X + . . . bnX

n is an other polynomial, then

f + g = (a0 + b0) + (a1 + b1)X + · · ·+ (ak + bk)Xk + . . .

fg = a0b0 + (a0b1 + a1b0)X + · · ·+ (a0bk + a1bk−1 + · · ·+ akb0)Xk + . . .

A monomial is polynomial of form aXn. The construction may be repeated togive the polynomial ring in n-variables R[X1, . . . , Xn] or even in infinitely manyvariables.

1.6.2. Definition. The degree, deg(f), of a polynomial 0 6= f ∈ R[X] is theindex of the highest nonzero coefficient, the leading coefficient. A polynomialwith leading coefficient the identity is a monic polynomial.

1.6.3. Remark. (1) R is identified with the subring of constants in the polyno-mial ring R[X1, . . . , Xn].

(2) The nonzero constants are the polynomials of degree 0.

1.6. POLYNOMIALS 17

(3) The constant polynomial 1 is the unique monic polynomial of degree 0 andthe identity in the polynomial ring.

1.6.4. Proposition. Let 0 6= f, g ∈ R[X].(1) If fg 6= 0 then deg(fg) ≤ deg(f) + deg(g).(2) If the leading coefficient of f or g is a nonzero divisor in R, then fg 6= 0 and

deg(fg) = deg(f) + deg(g)

Proof. (1) This is clear. (2)Clearly the leading coefficient of the product is theproduct of the leading coefficients.

1.6.5. Corollary. Let R be a domain.

(1) The polynomial ring R[X] is a domain.(2) The units in R[X] are the constants, which are units in R.

1.6.6. Theorem. Let R be a domain, 0 6= f, d ∈ R[X] polynomials with d monic.Then there are a unique q, r ∈ R[X] such that

f = qd+ r, r = 0 or deg(r) < deg(d)

Proof. Induction on deg(f). If deg(f) < deg(d) then q = 0, r = f . Otherwiseif a is the leading coefficient of f , then f − adXdeg(f)−deg(d) has degree less thandeg(f). By induction f − adXdeg(f)−deg(d) = qd+ r giving the claim.

1.6.7. Proposition. Let φ : R → S be a ring homomorphism. For any elementb ∈ S there is a unique ring homomorphism R[X]→ S extending φ and mappingX 7→ b.

Proof.

a0 + a1X + . . . amXm 7→ φ(a0) + φ(a1)b+ . . . φ(am)bm

is clearly the one and only choice.

1.6.8. Definition. The homomorphism in 1.6.7 is the evaluation map at b in S.The image of a polynomial f ∈ R[X] is denoted f(b) ∈ S.

1.6.9. Proposition. Let I ⊂ R be an ideal. Then there is a canonical isomorphism.

(R/I)[X] ' R[X]/IR[X]

Proof. There is an obvious pair of inverse homomorphisms constructed by 1.2.9and 1.6.7.

1.6.10. Corollary. If P ⊂ R is a prime ideal, then PR[X] ⊂ R[X] is a primeideal.

1.6.11. Definition. Let φ : R→ S be a ring homomorphism and B ⊂ S a subset.The ring generated over R by B is

R[B] = φ(R)[B] ⊂ S

the smallest subring of S containing φ(R) ∪ B. If there is a finite subset B suchthat R[B] = S then S is a finite type ring or a finitely generated ring over R.

1.6.12. Corollary. Let φ : R→ S be a ring homomorphism.


(1) If bα ∈ S and Xα is a family of variables, then there is a surjective ringhomomorphism

R[Xα]→ R[bα], Xα 7→ bα

making R[bα] a factor ring of the polynomial ring R[Xα].(2) If S is a finite type ring over R, then S is a factor ring of a polynomial ring

in finitely many variables over R.

1.6.13. Exercise. (1) LetK be a field. Show that there are infinitely many prime idealsin K[X].

(2) What are the units in the ring Z[X]/(1− 2X)?(3) Determine the prime ideals in Q[X]/(X −X2).(4) Show that the ring Z[X] is not a principal ideal domain.(5) Show that the ring Q[X,Y ] is not a principal ideal domain.

1.7. Roots

1.7.1. Definition. Let φ : R → S be a ring homomorphism and f ∈ R[X] apolynomial. An element b ∈ S is a root of f (in S) if the evaluation f(b) = 0.

1.7.2. Proposition. Let R be a domain. An element a ∈ R is a root of the polyno-mial f ∈ R[X] if and only if there is a q ∈ R[X] such that

f = q(X − a)

Proof. By 1.6.6 f = q(X−a)+r. It follows that a is a root if and only if r = 0.

1.7.3. Corollary. Let R is a domain. There are at most deg(f) roots in a nonzeropolynomial f ∈ R[X].

1.7.4. Definition. The multiplicity of a root a of a nonzero polynomial f ∈ R[X]is highest m such that

f = q(X − a)m

A root of multiplicity m = 1 is a simple root.

1.7.5. Corollary. Let R is a domain. If m1, . . . ,mk are the multiplicities of theroots of a nonzero polynomial f ∈ R[X], then m1 + · · ·+mk ≤ deg(f).

1.7.6. Definition. The derivative of a polynomial f =∑anX

n ∈ R[X] is

f ′ =∑

nanXn−1

1.7.7. Lemma. The derivative satisfies(1) (f + g)′ = f ′ + g′.(2) (fg)′ = f ′g + fg′

(3) If f is constant, then f ′ = 0.

1.7.8. Proposition. Let R is a domain. An element a ∈ R is a root of multiplicitym > 1 of a nonzero f ∈ R[X] if and only if a is a root of f and f ′.

Proof. By 1.6.6 f = q(X−a)2 + cX+d and by 1.7.7 f ′ = q′(X−a)2 +2q(X−a) + c. I follows that a is a root of multiplicity m > 1 if and only if c = d = 0.

1.7.9. Exercise. (1) Let a1, . . . ak be roots with multiplicities m1, . . . ,mk in a poly-nomial f . Show that m1 + · · ·+mk ≤ deg(f).

(2) LetK be a field and let a1, . . . , an ∈ K. Show that the ideal (X1−a1, . . . , Xn−an)is maximal in K[X1, . . . , Xn].

(3) Let the characteristic char(R) = n > 0. What is (Xn)′ in R[X].

1.8. FIELDS 19

1.8. Fields

1.8.1. Definition. Let p ∈ Z be a prime number. The factor ring Fp = Z/(p) is afield with p elements. Together with Q they constitute the prime fields.

1.8.2. Theorem. Let K be a field then the polynomial ring K[X] is a principalideal domain.

Proof. Let d 6= 0 be a polynomial of lowest degree in an ideal I . Given f ∈ I thenby 1.6.6 f = qd + r with r = f − qd ∈ I . By degree considerations r = 0 andI = (d).

1.8.3. Corollary. Let K be a field then the polynomial ring K[X] is a uniquefactorization domain.

Proof. Follows from 1.5.6.

1.8.4. Definition. A subfield is a subring, which is a field. A field extension is theinclusion of a subfield K ⊂ L in a field. A finite field extension K ⊂ L is anextension, where L is finite dimensional as a vector space over K.

1.8.5. Example. (1) Let K be a field and f an irreducible polynomial in K[X].Then K ⊂ K[X]/(f) is a finite field extension.

(2) Let K ⊂ R ⊂ L be a subring in a finite field extension. Then R is a field.Namely multiplication on R with a nonzero element of R is a K-linear mapon the finite dimensional K-vector space R and therefore an isomorphism.

1.8.6. Proposition. (1) LetK be a field and f a polynomial inK[X]. Then thereis a finite field extension K ⊂ L such that f factors in linear factors in L[X].

(2) If K ⊂ L1 and K ⊂ L2 are finite field extensions then there is a finite fieldextension K ⊂ L such that L1 ∪ L2 ⊂ L.

Proof. (1) Assume f irreducible. In L = K[X]/(f) the class X + (f) is a root off . In general proceed adjoining roots of irreducible factors of f . (2) An element xin a finite field extension K ⊂ K ′ is the root of the irreducible monic polynomialf generating the kernel of the evaluation homomorphism K[X] → K ′, X 7→ x.Now proceed by (1) adjoining elements in L2 to L1.

1.8.7. Proposition. Let p ∈ Z be a prime number. For any power q = pn there isa field Fq with q elements, unique up to isomorphism.

Proof. Let Fp ⊂ K be a field extension, where Xq − X factors into linear fac-tors, 1.8.6 (1). The subset of roots is the set of elements fixed under n-times theFrobenius and therefore a subring being a subfield by 1.8.5 (2). The derivative(Xq − X)′ = −1 so by 1.7.8 there are q elements in this subfield. Uniquenessfollows from 1.8.6 (2).

1.8.8. Exercise. (1) Show that the ring R[X]/(X2 + 1) is isomorphic to the field ofcomplex numbers.

(2) Show that the ring F2[X]/(X2 +X + 1) is a field with 4 elements.(3) Show that the ring F2[X]/(X3 +X + 1) is a field with 8 elements.(4) Let K ⊂ L be a field extension of fields of characteristic 0 and let a ∈ L be a root of

an irreducible polynomial f ∈ K[X]. Show that a is a simple root.(5) Let p be a prime number. Show that Fp is the only ring with p elements.(6) Let p be a prime number. Show that a ring with p2 elements is isomorphic to one of

four non isomorphic Z/(p2),Fp × Fp,Fp[X]/(X2),Fp2 .


1.9. Power series

1.9.1. Definition. Let R be a ring. The power series ring R[[X]] is the additivegroup

∏nRX

n, n = 0, 1, 2, . . . of all power series∑anX

n with n′th coefficientan ∈ R. Multiplication is given by XiXj = Xi+j extended by linearity. Foranother power series

∑bnX

n the sum and product are∑anX

n +∑

bnXn =

∑(an + bn)Xn∑

anXn ·

∑bnX

n =∑

(∑

k

an−kbk)Xn

The construction may be repeated to give the power series ring in n-variablesR[[X1, . . . , Xn]] or even in infinitely many variables.

1.9.2. Remark. The polynomial ring is identified as a subring R[X] ⊂ R[[X]] ofpower series with only finitely many nonzero terms.

1.9.3. Definition. The order, o(f), of a power series 0 6= f ∈ R[[X]] is the indexof the least nonzero coefficient.

1.9.4. Proposition. If R is a domain, then R[[X]] is a domain and for 0 6= f, g ∈R[X]

o(fg) = o(f) + o(g)

Proof. Clearly the lowest nonzero coefficient in the product is the product of thetwo lowest nonzero coefficients.

1.9.5. Proposition. A power series f =∑anX

n is a unit if and only if a0 is aunit.

Proof. It suffices to look at a power series f = 1 − gX . Then the power series1/f = 1 + gX + g2X2 + · · ·+ gnXn + . . . is well defined and f · 1/f = 1.

1.9.6. Theorem. If K is a field, then K[[X]] is a principal ideal domain. and (X)is the only nonzero prime ideal.

Proof. If the lowest order of an element in an ideal I is n. Then clearly I =(Xn).

1.9.7. Corollary. If K is a field, then K[[X]] is a unique factorization domain.

1.9.8. Proposition. Let I ⊂ R be an ideal. Then there is a canonical surjectivehomomorphism

R[[X]]/IR[[X]]→ R/I[[X]]

1.9.9. Corollary. If Q ⊂ R[[X]] is a maximal ideal, then P = Q ∩ R ⊂ R is amaximal ideal and Q = (P,X).

Proof. X ∈ Q so R/Q ∩R ' R[[X]]/Q.

1.9.10. Exercise. (1) Show that the ring Z[[X]] is not a principal ideal domain.(2) Show that the ring Q[[X,Y ]] is not a principal ideal domain.(3) Let K be a field. Show that (X1, . . . , Xn) is the unique maximal ideal in the power

series ring K[[X1, . . . , Xn]].(4) Let a ∈ R be nilpotent. Show that the ring R[[X]]/(X − a) is isomorphic to R.(5) What is R[[X]]/(X − a) if a ∈ R is a unit?(6) Let I ⊂ R be an ideal. Show that IR[[X]] ⊂ R[[X]] is not a maximal ideal.

2

Modules

2.1. Modules and homomorphisms

2.1.1. Definition. Let R be a ring. A module (R-module) is an abelian groupM , addition (x, y) 7→ x + y and zero 0, together with a scalar multiplicationR×M →M, (a, x) 7→ ax satisfying

(1) associative : (ab)x = a(bx)(2) bilinear : a(x+ y) = ax+ ay, (a+ b)x = ax+ bx(3) identity: 1x = x

for all a, b ∈ R, x, y ∈ M . A submodule M ′ ⊂ M is an additive subgroup suchthat ax ∈ M ′ for all a ∈ R, x ∈ M ′. A homomorphism is an additive grouphomomorphism f : M → N respecting scalar multiplication

f(x+ y) = f(x) + f(y), f(ax) = af(y)

for all a ∈ R, x, y ∈ M . An isomorphism is a homomorphism f : M → Nhaving an inverse map f−1 : N → M which is also a homomorphism. Theidentity isomorphism is denoted 1M : M →M .

2.1.2. Lemma. Let R be a ring and M a module.(1) a0 = 0 = 0x(2) (−1)x = −x(3) (−a)x = −(ax) = a(−x)

for all a ∈ R, x ∈M .

Proof. (1) Calculate a0 = a(0+0) = a0+a0 and cancel to get a0 = 0. Similarly0x = 0. (2) By (1) 0 = 0x = (1 + (−1))x = x + (−1)x, so conclude −x =(−1)x. (3) Calculate (−a)x = ((−1)a)x = (−1)(ax) = −(ax). Similarlya(−x) = −(ax).

2.1.3. Lemma. Let R be a ring and f : M → N a homomorphism of modules.(1) f(0) = 0.(2) f(ax+ by) = af(x) + bf(y).(3) f(−x) = −f(x) for all x ∈M .

for all a, b ∈ R, x, y ∈M .

Proof. (1) Calculate f(0) = f(0 + 0) = f(0) + f(0) and conclude f(0) = 0.(2) Calculate f(ax + by) = f(ax) + f(by) = af(x) + bf(y). (3) By 2.1.2f(−x) = f((−1)x) = (−1)f(x) = −f(x).

2.1.4. Example. (1) The zero group is the zero module.(2) Over the zero ring the zero module is the only module.(3) The zero subgroup of a module is the zero submodule.(4) The ring R is a module under multiplication. An ideal is a submodule.

21

22 2. MODULES

(5) If R is a field, a module is a vector space and a homomorphism is a linearmap.

(6) A module over Z is an abelian group and an additive map of abelian groupsis a homomorphism.

2.1.5. Proposition. A bijective homomorphism is an isomorphism.

Proof. Let f : M → N be a bijective homomorphism of R-modules and let g :N → M be the inverse map. For x, y ∈ N write x = f(g(x)), y = f(g(y)) andget additivity of g, g(x + y) = g(f(g(x)) + f(g(y))) = g(f(g(x) + g(y))) =g(x) + g(y). Similarly for a ∈ R g(ax) = g(af(g(x))) = g(f(ag(x))) = ag(x),so g respects scalar multiplication and is a homomorphism.

2.1.6. Lemma. Let a ∈ R and M be a module. The map M → M,x 7→ ax is ahomomorphism.

Proof. Let f(x) = ax and calculate f(x+y) = a(x+y) = ax+ay = f(x)+f(y)and f(bx) = a(bx) = (ab)x = (ba)x = b(ax) = bf(x) to get that f is ahomomorphism. Remark that the last calculation uses that R is commutative 1.1.2(4).

2.1.7. Definition. Let a ∈ R and M be a module.(1) The scalar multiplication with a is the homomorphism, 2.1.6,

aM : M →M,x 7→ ax

(2) a ∈ R is a nonzero divisor on M if scalar multiplication aM is injective, i.e.ax 6= 0 for all 0 6= x ∈M . Otherwise a is a zero divisor.

2.1.8. Remark. The two notions of nonzero divisor on R : 1.1.6 as element in thering and 2.1.4 as scalar multiplication on the ring coincide.

2.1.9. Example. IfR is a field, then scalar multiplication on a vector space is eitherzero or an isomorphism.

2.1.10. Lemma. Let φ : R → S be a ring homomorphism and N an S-module.The map

R×N → N, (a, x) 7→ φ(a)xis an R-scalar multiplication on N , 2.1.1.

Proof. Let a, b ∈ R, x, y ∈ N and µ(a, x) = φ(a)x. Calculate µ(a + b, x) =φ(a + b)x = φ(a)x + φ(b)x = µ(a, x) + µ(b, x), µ(a, x + y) = φ(a)(x + y) =φ(a)x + φ(a)y = µ(a, x) + µ(a, y), µ(1, x) = φ(1)x = 1x = x and µ(ab, x) =φ(ab)x = φ(a)φ(b)x = µ(a, µ(bx)) showing the conditions 2.1.1.

2.1.11. Definition. Let φ : R → S be a ring homomorphism. The restriction ofscalars of an S-module N is the same additive group N viewed as an R-modulethrough φ. The scalar multiplication is 2.1.10

R×N → N, (a, x) 7→ ax = φ(a)x

An S-module homomorphism g : N → N ′ is also an R-module homomorphism.

2.1.12. Example. (1) The scalar multiplication with a Restriction of scalars forthe unique ring homomorphism Z→ R give just the underlying abelian groupof a module, 2.1.4 (6).

2.2. SUBMODULES AND FACTOR MODULES 23

(2) Let I ⊂ R be an ideal. Restriction of scalars along the projection R → R/Igives any R/I-module as an R-module.

2.1.13. Proposition. Let R be a ring. There is a dictionary:(1) To anR[X]-moduleN associate the pair (N, f) consisting ofN asR-module

through restriction of scalars and f = XN : N → N, f(y) = Xy scalarmultiplication withX as anR-module homomorphism. AnR[X]-homomorphismg : N → N ′ gives an R-homomorphism such that g ◦ f = f ′ ◦ g.

(2) To a pair (N, f) of anR-module and a homomorphism f : N → N associatetheR[X]-module with abelian groupN and scalar multiplication determinedby Xy = f(y) for y ∈ N . Note

(∑

anXn)y =

∑anf

◦n(y)

An R-homomorphism g : N → N ′ such that g ◦ f = f ′ ◦ g is an R[X]-homomorphism.

Proof. The statement is an algorithm to follow.

2.1.14. Proposition. Let R be a ring and M a module. The abelian group R⊕Mwith multiplication

(a+ x)(b+ y) = ab+ (ay + bx)is a ring. R is a subring and M is an ideal.

Proof. Simple calculations show that the conditions 1.1.2 are satisfied.

2.1.15. Exercise. (1) Show that a composition of homomorphisms is a homomorphism.(2) Show that composition of scalar multiplications with a, b ∈ R on a module M is a

scalar multiplication with the product, aM ◦ bM = (ab)M .(3) Let φ : R → S be a ring homomorphism. Show that φ is an R-module homomor-

phism, when S is viewed as R-module through restriction of scalars 2.1.11.(4) Fill out the dictionary 2.1.13.

2.2. Submodules and factor modules

2.2.1. Lemma. Let R be a ring and M a module. Let Nα be a family of submod-ules. Then the additive subgroups

∑αNα and

⋂αNα are submodules.

Proof. Use the formulas∑xα +

∑yα =

∑(xα + yα) and a

∑xα =

∑axα to

conclude that∑Nα is a submodule. If x, y ∈ Nα for all α, then x + y, ax ∈ Nα

for all α, so⋂Nα is a submodule.

2.2.2. Definition. LetR be a ring andM a module. The intersection of all submod-ules containing a subset Y ⊂ M is the submodule generated by Y and denotedRY . This is the smallest submodule, 2.2.1, of M containing Y . The module M isgenerated by Y if RY = M . Let I be an ideal. The submodule generated by allproducts ax, a ∈ I, x ∈M is denoted IM .

2.2.3. Proposition. Let R be a ring and M a module. If Y ⊂ M , then RY =∑y∈Y Ry,

RY = {a1y1 + · · ·+ anyn|ai ∈ R, yi ∈ Y }

Proof. The righthand side is contained in the submodule RY . Moreover the right-hand side is a submodule containing Y , so equality.

2.2.4. Corollary. Let I ⊂ R be an ideal and M a module.

24 2. MODULES

(1)IM = {a1y1 + · · ·+ anyn|ai ∈ I, yi ∈M}

(2) If I = (a) is principal, then

aM = (a)M = {ay|y ∈M}

Proof. (1) is clear. (2) By (1) an element in aM is∑biayi = a

∑biyi = ay for

y =∑biyi ∈M as claimed.

2.2.5. Lemma. LetR be a ring,M a module andN ⊂M a submodule. LetM/Nbe the abelian factor group, then the map

R×M/N →M/N, (a, x+N) 7→ ax+N

is well defined and a scalar multiplication, 2.1.1.

Proof. If x+N = y+N then x−y ∈ N and so a(x−y) = ax−ay ∈ N . Thereforeax + N = ay + N and the multiplication is well defined. Since representativesmay be chosen such that (x+N) + (y+N) = x+ y+N, a(x+N) = ax+N ,the laws for scalar multiplication are satisfied.

2.2.6. Definition. LetR be a ring,M a module andN ⊂M a submodule, then thefactor module is the additive factor group M/N with, 2.2.5, scalar multiplicationa(x+N) = ax+N . The projection p : M → M/N, x 7→ x+N is a surjectivehomomorphism.

2.2.7. Lemma. Let R be a ring, N ⊂ M a submodule and p : M → M/Nthe projection. p is surjective and if Y ⊂ M generates M , then p(Y ) ⊂ M/Ngenerates the factor module.

Proof. Clearly if RY = M then Rp(Y ) = p(RY ) = M/N .

2.2.8. Example. (1) A submodule of R is the same as an ideal.(2) Both an ideal I ⊂ R and a factor ring R/I are modules.(3) The module structure on R/I as a factor module and the structure by restric-

tion of scalars through the projection R→ R/I are identical.

2.2.9. Proposition. LetR = R1×R2 be the product ring 1.1.4. There is a bijective(up to natural isomorphism) correspondence.

(1) If M1 is an R1-module and M2 is an R2-module, then M = M1 ×M2 is anR-module with coordinate scalar multiplication. A pair of homomorphismsinduce a homomorphism on the product.

(2) If M is an R-module then M1 = (1, 0)M is an R1-module and M2 =(0, 1)M is an R2-module. A homomorphism induces a pair of homomor-phisms.

2.2.10. Remark. The correspondence 2.2.9 indicates that the structure of modulesand homomorphisms over a product ring is identified with the structure of pairs ofmodules and homomorphisms over each component ring in the product.

2.2.11. Exercise. (1) Give an example of two submodules N,L ⊂ M such that theunion N ∪ L is not a submodule.

(2) Let R be a ring and a ∈ R. Show that the R-module R[X]/(X − a) is isomorphicto R.

(3) Show that the projection p 2.2.6 is a homomorphism.(4) Fill in the details in the dictionary 2.2.9

2.3. KERNEL AND COKERNEL 25

2.3. Kernel and cokernel

2.3.1. Lemma. Let R be a ring and f : M → N a homomorphism of modules.Given submodules M ′ ⊂ M,N ′ ⊂ N , then f−1(N ′) ⊂ M and f(M ′) ⊂ N aresubmodules.

Proof. If x, y ∈ f−1(N ′) then f(x + y) = f(x) + f(y) ∈ N ′ and for a ∈ Rf(ax) = af(x) ∈ N ′ so x+y, ax ∈ f−1(N ′) proving f−1(N ′) to be a submodule.The same equations prove that f(M ′) is a submodule.

2.3.2. Definition. Let f : M → N be a homomorphism of modules. Then thereare submodules, 2.3.1.

(1) The kernel Ker f = f−1(0).(2) The image Im f = f(M).(3) The cokernel Cok f = N/ Im f .

2.3.3. Proposition. Let f : M → N be a homomorphism of modules.(1) f is injective if and only if Ker f = 0.(2) f is surjective if and only if Cok f = 0.(3) f is an isomorphism if and only if Ker f = 0 and Cok f = 0.

Proof. (1) If f is injective and x ∈ Ker f then f(x) = 0 = f(0) so x = 0.Conversely if Ker f = 0 and f(x) = f(y) then f(x − y) = 0 so x = y. (2) Thefactor module N/ Im f = 0 if and only if Im f = N . (3) This follows from (1)and (2).

2.3.4. Example. Let a ∈ R give scalar multiplication aM : M →M,x 7→ ax.(1) Im aM = aM = {ax ∈M |x ∈M}.(2) Ker aM = {x ∈M |ax = 0}.(3) Cok aM = M/aM .

2.3.5. Theorem. Let f : M → N be a homomorphism of modules.(1) Let L ⊂ Ker f be a submodule. Then there is a unique homomorphism f ′ :

M/L→ N such that f = f ′ ◦ p.

M

p

��

f // N

M/Lf ′

<<

(2) The homomorphism f ′ : M/Ker f → N is a module isomorphism onto thesubmodule Im f of N .

M

p

��

f // Im f

M/Ker ff ′

99

Proof. (1) If x+L = x′+L then x−x′ ∈ L giving f(x) = f(x′). The factor mapf ′ : M/L → N,x + L 7→ f(x) is therefore well defined and f = f ′ ◦ p. Sincef, p are homomorphisms and p is surjective it follows that f ′ is a homomorphism.(2) The kernel of f ′ is Ker f/Ker f = 0 so by 2.3.3 it is an isomorphism.

26 2. MODULES

2.3.6. Corollary. Let p : M →M/N be the projection onto a factor module. Themap L′ 7→ L = p−1(L′) gives a bijective correspondence between submodulesin M/N and submodules in M containing N . Also L′ = p(L) = L/N . Thiscorrespondence preserves inclusions, additions and intersections of submodules.

Proof. If L′ is a submodule of M/N then clearly p(p−1(L′)) = L′. If N ⊂ Lis a submodule of M then clearly L ⊂ p−1(p(L)). Moreover if x ∈ p−1(p(L))then p(x) = p(y) for some y ∈ L and therefore y − x ∈ N . It follows thatL = p−1(p(L)) and the correspondence is bijective. Inclusions are easily seen tobe preserved. Also easily p(L1 + L2) = p(L1) + p(L2) and p−1(L′1 ∩ L′2) =p−1(L′1) ∩ p−1(L′2) hold. The two resting equalities are consequences of this andbijectivity of the correspondence.

2.3.7. Corollary. Let L ⊂ N ⊂ M be submodules. Then there is a canonicalisomorphism

M/N → (M/L)/(N/L)

Proof. The kernel of the surjective east-south composite

M

��

// M/L

��M/N // (M/L)/(N/L)

is N . By 2.3.5 the horizontal lower factor map gives the isomorphism.

2.3.8. Corollary. Let L,N ⊂ M be submodules. Then there is a canonical iso-morphism

N/N ∩ L→ N + L/L

given by x+N ∩ L 7→ x+ L.

Proof. The kernel of the east-south composite

N

��

// N + L

��N/N ∩ L // N + L/L

is N ∩ L. Since x + y + L = x + L for x ∈ N, y ∈ L this composite is alsosurjective. By 2.3.5 the horizontal lower factor map gives the isomorphism.

2.3.9. Proposition. Let f : M → N and g : N → L be homomorphisms such thatIm f ⊂ Ker g. Then there is a unique homomorphism g′ : Cok f → L such thatg = g′ ◦ p.

Mf // N

p

""FFFFFFFF

g // L

Cok f

g′

OO

Proof. This follows from 2.3.5.

2.3.10. Lemma. Let R be a ring and M a module. For x ∈ M the map R →M, a 7→ ax is the unique homomorphism such that 1 7→ x.

2.3. KERNEL AND COKERNEL 27

Proof. Let f(a) = ax. f(ab) = (ab)x = a(bx) = af(b) shows that f is ahomomorphism. This argument does not use the commutativity 1.1.2 of R.

2.3.11. Definition. Denote the homomorphism 2.3.10

1x : R→M,a 7→ ax

The annihilator of x is the ideal

Ann(x) = Ker 1x = {a ∈ R|ax = 0}For a subset Y ⊂ M the annihilator Ann(Y ) =

⋂y∈Y Ann(y) is the ideal of

elementsAnn(Y ) = {a ∈ R|ay = 0, for all y ∈ Y }

2.3.12. Proposition. Let R be a ring and Y a subset of a module M .(1) Ann(Y ) = Ann(RY ).(2) a ∈ Ann(M) if and only if aM = 0.(3) If modules M 'M ′ then Ann(M) = Ann(M ′).(4) The induced homomorphism

1′x : R/Ann(x)→ Rx, a+ Ann(x) 7→ ax

is an isomorphism.

Proof. (1) If a ∈ Ann(Y ) then a∑biyi =

∑biayi = 0 giving the not so obvious

Ann(Y ) ⊂ Ann(RY ). (2) Clear since aM (x) = ax, 2.1.7. (3) Let f : M → M ′

be an isomorphism and a ∈ R. Then af(x) = f(ax) expresses that f ◦ aM =aM ′ ◦ f . Since f is bijective, aM = 0 if and only if aM ′ = 0. By (2) Ann(M) =Ann(M ′). (4) This follows from 2.3.5.

2.3.13. Corollary. Let I, J ⊂ R be ideals.(1) Ann(R/I) = I(2) If R/I ' R/J then I = J .

2.3.14. Lemma. Let I ⊂ R be an ideal and M an R-module. If I ⊂ Ann(M)then M is an R/I-module with the scalar multiplication

R/I ×M →M, (a+ I, x) 7→ ax

That is, M is an R/Ann(M)-module.

2.3.15. Example. Let a ∈ R give scalar multiplication aM : M →M,x 7→ ax.(1) a ∈ Ann(Ker aM ).(2) a ∈ Ann(Cok aM ).(3) Ker aM and Cok aM are modules over the factor ring R/(a), 2.3.14.

2.3.16. Definition. Let R be a ring and L,N ⊂ M submodules. The colon idealN : L is the ideal of elements a ∈ R such that aL ⊂ N .

2.3.17. Proposition. The colon ideal of submodules L,N ⊂M is

N : L = Ann(N + L/N) = Ann(N/N ∩ L)

Proof. If a ∈ N : L then aL ⊂ N . Therefore a(N + L) ⊂ N and a ∈ Ann(N +L/N). Conversely if a ∈ Ann(N +L/N) then aL ⊂ N and therefore a ∈ N : L.The last equality follows from 2.3.8.

2.3.18. Exercise. (1) Give an example of a homomorphism f : M → N submodulesM1,M2 ⊂M such that f(M1 ∩M2) 6= f(M1) ∩ f(M2).

28 2. MODULES

(2) Give an example of a homomorphism f : M → N submodules N1, N2 ⊂ N suchthat f−1(N1 +N2) 6= f−1(M1) + f−1(M2).

(3) Let R be a ring and M a module. Show that M may be regarded as an R/Ann(M)-module in a natural way.

(4) Let L,N ⊂M be submodules. Show that Ann(L+N) = Ann(L) ∩Ann(N).(5) Let f : M → N be a surjective homomorphism. Show that Ann(M) ⊂ Ann(N).(6) Let f : M → N be an injective homomorphism. Show that Ann(N) ⊂ Ann(M).

2.4. Sum and product

2.4.1. Lemma. Let R be a ring and (Mα) a family of modules. The product∏αMα is the abelian group of all families (xα), xα ∈ Mα with term wise ad-

dition. The settingr(xα) = (rxα)

is a scalar multiplication on∏

αMα. The direct sum⊕

αMα is the subgroupof

∏αMα consisting of families with only finitely many nonzero terms. This is a

submodule.

Proof. The laws in 2.1.1 are true for each factor and therefore trivially verified forthe product and sum.

2.4.2. Definition. Let R be a ring and Mα a family of modules. By 2.4.1 there aremodules and homomorphisms

(1) The direct product is∏

αMα.(2) The projections pβ :

∏αMα → Mβ are the homomorphisms pβ((xα)) =

xβ .(3) The direct sum is

⊕αMα. Elements in

⊕αMα are written as finite sums∑

xα.(4) The injections iβ : Mβ →

⊕αMα are the homomorphisms given by iβ(x) =

(xα), where xβ = x and xα = 0, α 6= β.

2.4.3. Theorem. Let R be a ring and Mα a family of modules.

(1) Given a family of homomorphisms fα : L → Mα, then there exists a uniquehomomorphism f : L→

∏Mα such that fα = pα ◦ f .

L

fα @@@

@@@@

@f //

∏αMα

pα{{vvvvv

vvvv

Mα

(2) Given a family of homomorphisms gα : Mα → L, then there exists a uniquehomomorphism g :

⊕Mα → L such that gα = g ◦ iα.⊕

αMαg // L

Mα

iα

ddHHHHHHHHH gα

>>~~~~~~~~

Proof. (1) f(y) = (fα(y)) is the unique homomorphism. (2) g(∑xα) =

∑gα(xα)

is well defined since only finitely many xα 6= 0 and a homomorphism.

2.4. SUM AND PRODUCT 29

2.4.4. Definition. A family of submodules Mα ⊂ M constitutes a direct sum ifany element x ∈

∑αMα has a unique finite representation

x =∑α

xα, xα ∈Mα

2.4.5. Proposition. The following conditions are equivalent:(1) The family Mα ⊂M constitutes a direct sum.(2) The natural homomorphism⊕

α

Mα →∑α

Mα

is an isomorphism.(3) For all β

Mβ ∩∑α 6=β

Mα = 0

Proof. (1)⇔ (2): This is clear. (1)⇒ (3): If xβ =∑

α 6=β xα ∈Mβ ∩∑

α 6=β Mα,then xβ−

∑α 6=β xα = 0. Therefore by uniqueness xβ = 0. (3)⇒ (1): If

∑α xα =

0, then xβ = −∑

α 6=β xα ∈ Mβ ∩∑

α 6=β Mα = 0. for any β. This showsuniqueness.

2.4.6. Definition. Let R be a ring. A module isomorphic to a direct sum⊕

αR ofcopies of the ring R is a free module.A basis of a module, is a subset Y such that any element admits a unique finiterepresentation

∑α aαyα, where aα ∈ R, yα ∈ Y .

The standard basis of⊕

αR consists of the elements eα, where each is a familywith 0 for β 6= α and exactly 1 at index α.

2.4.7. Proposition. A module is free if and only if it admits a basis. If yα is a basisof F then F = Ryα as direct sum and there is an isomorphism

f :⊕

α

R→ F

where f(∑aα) =

∑aαyα.

Proof. Given a free module f :⊕

αR → F then yα = f(eα) is a basis. Con-versely given a basis yα ∈ F then 1yα : R → F 2.3.11 is a family of homomor-phisms giving a homomorphism f :

⊕αR→ F by 2.4.3. As f(

∑aα) =

∑aαyα

it follows that f is bijective and therefore by 2.1.5 an isomorphism.

2.4.8. Remark. The polynomial ring R[X1, . . . , Xn] is free as R-module.

2.4.9. Example. (1) A nonzero ideal is a free module if and only if it has a basisconsisting of a nonzero divisor. Namely if x1 6= x2 where in a basis then theproduct x1x2 has two different representations.

(2) Let I ⊂ R be an ideal. The moduleR/I is free if and only if I = 0 or I = R.

2.4.10. Proposition. Any module over a field is free. Conversely if any moduleover a nonzero ring is free, then the ring is a field

Proof. By Zorn’s lemma any vector space admits a basis. If I ⊂ R is an ideal andR/I is free, then I = Ann(R/I) is either 0 or R. So R is a field.

30 2. MODULES

2.4.11. Proposition. Let F be a free module with basis yα. For a module M anda family of elements xα ∈ M there is a unique homomorphism g : F → M suchthat g(yα) = xα given by g(

∑aαyα) =

∑aαxα.

Proof. The basis yα ∈ F gives the isomorphism 2.4.7 f :⊕

αR → F . Thefamily 1xα : R → M gives a homomorphism g′ :

⊕αR → M by 2.4.3. Then

g = g′ ◦ f−1.

2.4.12. Corollary. Let M be an R-module and⊕

M R the free module with basisex indexed by x ∈M . The homomorphism⊕

M

R→M,∑

axex 7→∑

ax x

is surjective identifying M as a factor module of a free module in a natural way.

2.4.13. Definition. A module is indecomposable if it is not isomorphic to a directsum of two nonzero submodules, otherwise decomposable.

2.4.14. Example. Q is an indecomposable Z-module. Namely if m1n1, m2

n2are

nonzero numbers in two submodules, then n1m2m1n1

= n2m1m2n2

is a nonzero num-ber in the intersection.

2.4.15. Exercise. (1) Show that if a ring is decomposable as a module, then it is theproduct of two nonzero rings.

(2) Let Mα be a finite family of modules. Show that⊕Mα =

∏Mα,

(3) Let Nα ⊂Mα be a family of submodules modules. Show that⊕Mα/

⊕Nα '

⊕Mα/Nα

and that ∏Mα/

∏Nα '

∏Mα/Nα

2.5. Homomorphism modules

2.5.1. Lemma. Let R be a ring and f, g : M → N homomorphisms.(1) (f + g)(x) = f(x) + g(x) is a homomorphism.(2) If a ∈ R, then (af)(x) = af(x) is a homomorphism.

Proof. Calculate according to 2.1.1. (1) (f + g)(x+ y) = f(x+ y) + g(x+ y) =f(x) + f(y) + g(x) + g(y) = (f + g)(x) + (f + g)(y) and (f + g)(ax) =f(ax)+ g(ax) = a(f(x)+ g(x)) = a(f + g)(x). (2) (af)(x+y) = af(x+y) =af(x) + af(y) = (af)(x) + (af)(y) and (af)(bx) = af(bx) = abf(x) =baf(x) = b(af)(x). The last calculation uses that R is commutative 1.1.2 (4).

2.5.2. Definition. Let R be a ring and M,N modules. By 2.5.1, the homomor-phism module HomR(M,N) is the additive group of all homomorphism withscalar multiplication

R×HomR(M,N)→ HomR(M,N), (a, f) 7→ af = [x 7→ af(x)]

2.5.3. Definition. Let a ∈ R be a ring and f : M → M ′, g : N → N ′, h, k :M ′ → N homomorphisms of modules. By 2.5.1(1) (h+ k) ◦ f = h ◦ f + k ◦ f .(2) (ah) ◦ f = a(h ◦ f).(3) g ◦ (h+ k) = g ◦ h+ g ◦ k.(4) g ◦ (ah) = a(g ◦ h).

2.5. HOMOMORPHISM MODULES 31

There is induced a homomorphism

Hom(f, g) : HomR(M ′, N)→ HomR(M,N ′)

(h : M ′ → N) 7→ (g ◦ h ◦ f : M → N ′)of R-modules.

2.5.4. Definition. Let R,S be rings. A functor is a construction T , which to anR-module M associates S-modules T (M)

M 7→ T (M)

and for two R-modules M,N there is an additive homomorphism

HomR(M,N)→ HomS(T (M), T (N))

f 7→ T (f)

such that(1) T (1M ) = 1T (M)

(2) T (g ◦ f) = T (g) ◦ T (f)In case the homomorphism changes domain and target

HomR(M,N)→ HomS(T (N), T (M))

f 7→ T (f)

and(1) T (1M ) = 1T (M)

(2) T (g ◦ f) = T (f) ◦ T (g)the functor is contravariant.Clearly (contravariant) functors transform isomorphisms into isomorphisms.Given functors T, T ′ a natural homomorphism is a family νM : T (M)→ T ′(M)of homomorphisms, such that for each f : M → N the following diagram com-mutes

T (M)

T (f)

��

νM // T ′(M)

T ′(f)��

T (N)νN // T ′(N)

In the contravariant case the diagram is

T (M)νM // T ′(M)

T (N)

T (f)

OO

νN // T ′(N)

T ′(f)

OO

A natural isomorphism is a natural homomorphism such that each νM is an iso-morphism.

2.5.5. Proposition. Let R be a ring.(1) The construction

N 7→ HomR(M,N)

g 7→ Hom(1M , g)

32 2. MODULES

is a functor and

HomR(N,N ′)→ HomR(HomR(M,N),HomR(M,N ′))

is an R-homorphism.(2) The construction

M 7→ HomR(M,N)

f 7→ Hom(f, 1N )

is a contravariant functor and

HomR(M,M ′)→ HomR(HomR(M ′, N),HomR(M,N))

is an R-homorphism.

Proof. By 2.5.3 the construction on homomorphisms are homomorphisms. Givenalso homomorphisms f ′ : M ′ →M ′′ and g′ : N ′ → N ′′. Then

Hom(1M , g′ ◦ g) = Hom(1M , g

′) ◦Hom(1M , g)

andHom(f ′ ◦ f, 1N ) = Hom(f, 1N ) ◦Hom(f ′, 1N )

showing the conditions on compositions. Clearly also

Hom(1, g + g′) = Hom(1, g) + Hom(1, g′)

Hom(1, ag) = aHom(1, g)and similarly for the resting properties.

2.5.6. Corollary. Let a ∈ R give scalar multiplications aM , aN .(1) The three constructions

Hom(aM , 1N ) = Hom(1M , aN ) = aHomR(M,N)

are the same homomorphism

HomR(M,N)→ HomR(M,N) f 7→ af

(2) The map R→ HomR(M,M), a 7→ aM is a homomorphism.

2.5.7. Example. Let R = R1×R2 be the product ring. The constructions in 2.2.9is: (1) A functor which to a pair of an R1-module and an R2-module associates anR-module.(2) A functor which to an R-module associates a pair of an R1-moduleand an R2-module.

2.5.8. Proposition. Let R be a ring and Mα a family of modules. For any moduleN there are natural isomorphisms

(1) HomR(⊕

αMα, N) '∏

α HomR(Mα, N)(2) HomR(N,

∏αMα) '

∏α HomR(N,Mα)

Proof. This is 2.4.3 reformulated. (1) A homomorphism g :⊕

αMα → N isuniquely determined by the family gα = g ◦ iα : Mα → N . (2) A homomorphismf : N →

∏αMα is uniquely determined by the family fα = pα ◦f : N →Mα.

2.5.9. Lemma. Let R be a ring and M,N modules. For x ∈ M there is a homo-morphism HomR(M,N)→ N, f 7→ f(x).

Proof. Calculate according to 2.1.1 (f + g) 7→ (f + g)(x) = f(x) + g(x) and(af) 7→ (af)(x) = af(x).

2.5. HOMOMORPHISM MODULES 33

2.5.10. Definition. The natural homomorphism 2.5.9

evx : HomR(M,N)→ N, f 7→ f(x)

is the evaluation at x.

2.5.11. Lemma. There is a natural homomorphism

M → HomR(HomR(M,N), N), x 7→ evx

Proof. Calculate according to 2.1.1 evx+y(f) = f(x + y) = f(x) + f(y) =evx(f) + evy(f). and evax(f) = f(ax) = af(x) = aevx(f) to see that the mapis a homomorphism.

2.5.12. Proposition. Let R be a ring and M a module. The evaluation

ev1 : HomR(R,M) 'M, f 7→ f(1)

is a natural isomorphism. x 7→ 1x 2.3.11 is the inverse.

Proof. Calculate the composite ev1(x 7→ 1x) = 1x(1) = x and 1f(1)(a) =af(1) = f(a) proving the claims.

2.5.13. Definition. Let R be a ring and M a module. The dual module is

M∨ = HomR(M,R)

If f : M → N is a homomorphism, then the dual homomorphism is

f∨ = Hom(f, 1R) : N∨ →M∨

This construction is a contravariant functor.

2.5.14. Lemma. There is a natural homomorphism

M →M∨∨ = HomR(HomR(M,R), R), x 7→ evx

where evx(f) = f(x) 2.5.10.

Proof. This is a special case of 2.5.11

2.5.15. Definition. A moduleM is a reflexive module if the homomorphism 2.1.14,

M →M∨∨

is an isomorphism.

2.5.16. Example. Let R be a ring.

(1) The module R is reflexive.(2) If (a) 6= R, (0) in a domain, then R/(a) is not reflexive.

2.5.17. Exercise. (1) Show that HomZ(Q,Z) = 0.(2) Calculate HomZ(Z/(m),Z/(n)) = 0 for integers m,n.(3) Let I ⊂ R be an ideal andM a module. Show that HomR(R/I,M) = {x ∈M |I ⊂

Ann(x)}.(4) Let R be a ring. Show that a free module with a finite basis is a reflexive module.(5) If (n) ⊂ (m) ⊂ Z, then show that (m)/(n) is a reflexive Z/(n)-module.

34 2. MODULES

2.6. Tensor product modules

2.6.1. Definition. Let R be a ring and M,N modules. The tensor product module

M ⊗R N = F/F ′

is the factor module F/F ′, where F = ⊕M×NR is the free module with basis(x, y) = e(x,y) 2.4.6 and F ′ is the submodule generated by all elements of form

(x1 + x2, y)− (x1, y)− (x2, y), (x, y1 + y2)− (x, y1)− (x, y2)

(ax, y)− a(x, y), (x, ay)− a(x, y)The projection of the basis element (x, y) onto M ⊗R N is x⊗ y = (x, y) + F ′.

2.6.2. Remark. The relations are interpreted.(1) There are identities in M ⊗R N

(x1 + x2)⊗ y = x1 ⊗ y + x2 ⊗ y, x⊗ (y1 + y2) = x⊗ y1 + x⊗ y2

ax⊗ y = a(x⊗ y) = x⊗ ay(2) The map

⊗ : M ×N →M ⊗R N, (x, y) 7→ x⊗ yhas partial maps x 7→ x⊗y : M →M⊗RN and y 7→ x⊗y : N →M⊗RNthat are all homomorphisms.

(3) The formation of partial homomorphism are again homomorphisms.

N → HomR(M,M ⊗R N)), y 7→ (x 7→ x⊗ y)and

M → HomR(N,M ⊗R N)), x 7→ (y 7→ x⊗ y)2.6.3. Theorem. Given a map µ : M × N → L such that the partial maps x 7→µ(x, y) : M → L and y 7→ µ(x, y) : N → L are homomorphisms. Then thereexists a unique homomorphism u : M ⊗R N → L such that u(x⊗ y) = µ(x, y).

M ×N

µ&&MMMMMMMMMMMM

⊗ // M ⊗R N

u

��L

On a general element u satisfies

u(∑

xα ⊗ yα) =∑

u(xα ⊗ yα) =∑

µ(xα, yα)

Proof. By 2.6.1 M ⊗R N = F/F ′. The homomorphism 2.4.11 F → L, (x, y) 7→µ(x, y) has F ′ in the kernel. 2.3.5 gives the homomorphism u.

2.6.4. Remark. Assume xα generate M and yβ generate N .(1) xα ⊗ yβ generate M ⊗R N .(2) Two homomorphisms u, v : M ⊗R N → L are equal if u(xα ⊗ yβ) =

v(xα ⊗ yβ).

2.6.5. Proposition. Let R be a ring and f : M → M ′, g : N → N ′ homomor-phisms of modules. Then there is induced a homomorphism

M ⊗R N →M ′ ⊗R N′, x⊗ y 7→ f(x)⊗ g(y)

2.6. TENSOR PRODUCT MODULES 35

Proof. The map south-east

M ×N

f×g

��

⊗ // M ⊗R N

��M ′ ×N ′ ⊗ // M ′ ⊗R N

′

satisfies the assumptions in 2.6.3 to induce the right vertical map satisfying x⊗y 7→f(x)⊗ g(y).

2.6.6. Definition. f ⊗ g : M ⊗R N →M ′ ⊗R N′ is the induced homomorphism

2.6.5. On a general element

f ⊗ g(∑

xα ⊗ yα) =∑

f(xα)⊗ g(yα)

2.6.7. Proposition. Let R be a ring. The constructions

(1)M 7→M ⊗R N, f 7→ f ⊗ 1N

(2)N 7→M ⊗R N, g 7→ 1M ⊗ g

are functors and

HomR(M,M ′)→ HomR(M ⊗R N,M′ ⊗R N)

HomR(N,N ′)→ HomR(M ⊗R N,M ⊗R N′)

are R-homomorphism.

Proof. Given also homomorphisms f ′ : M ′ → M ′′ and g′ : N ′ → N ′′. Then by2.6.4

f ′ ◦ f ⊗ g′ ◦ g = f ′ ⊗ g′ ◦ f ⊗ gThe rest follows directly from 2.6.5. E.g.

(f + g)⊗ 1 = f ⊗ 1 + g ⊗ 1, (af)⊗ 1 = a(f ⊗ 1)

and similarly.

2.6.8. Corollary. Let a ∈ R give scalar multiplications aM , aN . The homomor-phisms

aM ⊗ 1N = 1M ⊗ aN = aM⊗RN

are the same homomorphism

M ⊗R N →M ⊗R N, x⊗ y 7→ a(x⊗ y)2.6.9. Example. Let R be a ring.

(1) Then there is an isomorphism

R⊗R R ' R, a⊗ b 7→ ab

(2) Let M,N,L be modules. Composition of maps gives a homomorphism

HomR(N,L)⊗R HomR(M,N)→ HomR(M,L), g ⊗ f 7→ g ◦ f

by 2.5.3. This is a natural homomorphism in each variable.

36 2. MODULES

(3) For a module M composition

HomR(M,M)⊗R HomR(M,M)→ HomR(M,M), g ⊗ f 7→ g ◦ f

gives HomR(M,M) a structure of a normally noncommutative ring. Themap R→ HomR(M,M), a 7→ aM is a ring homomorphism.

2.6.10. Proposition. LetR be a ring andM,N,Lmodules. Then there are naturalisomorphisms(1) (unit:) M ⊗R R 'M, x⊗ a 7→ ax(2) (commutative:) (M ⊗R N) ' N ⊗R M, x⊗ y 7→ y ⊗ x(3) (associative:) (M⊗RN)⊗RL 'M⊗R(N⊗RL), (x⊗y)⊗z 7→ x⊗(y⊗z)

Proof. (1) M × R → M, (x, a) 7→ ax induces the homomorphism M ⊗R R →M, x⊗ a 7→ ax by 2.6.3. The map M → R⊗R M, x 7→ 1⊗ x is the inverse. (2)M ×N → N ⊗R M, (x, y) 7→ y ⊗ x induces the homomorphism (M ⊗R N) 'N ⊗R M, x ⊗ y 7→ y ⊗ x by 2.6.3. The inverse is constructed similarly and thecomposites are the identities by the uniqueness statement in 2.6.3. (3) For a fixedz ∈ L the map M × N → M ⊗R (N ⊗R L), (x, y) 7→ x ⊗ (y ⊗ z) induces thehomomorphism µz : M ⊗RN →M ⊗R (N ⊗RL), x⊗y 7→ x⊗ (y⊗ z) by 2.6.3.Finally the map (M⊗RN)×L→M⊗R(N⊗RL), (x⊗y, z) 7→ µz(x⊗y). inducesthe homomorphism (M⊗RN)⊗RL 'M⊗R(N⊗RL), (x⊗y)⊗z 7→ x⊗(y⊗z)by 2.6.3. The inverse is constructed similarly and the composites are the identitiesby the uniqueness statement in 2.6.3.

2.6.11. Proposition. LetR be a ring andMα a family of modules. For any moduleN there is a natural isomorphism

(⊕

α

Mα)⊗R N '⊕

α

(Mα ⊗R N)

giving the identification (∑xα)⊗ y =

∑(xα ⊗ y).

Proof. 2.4.3 and 2.6.3 give a pair of inverse homomorphisms. Fix y ∈ N . Thefamily gα : Mα →

⊕(Mα ⊗R N), xα 7→ xα ⊗ y induces by 2.4.3 a homomor-

phism gy :⊕Mα →

⊕(Mα ⊗R N). The map (

⊕Mα) × N →

⊕(Mα ⊗R

N), (∑xα, y) 7→ gy(

∑xα) induces by 2.6.3 the homomorphism (

⊕Mα) ⊗R

N →⊕

(Mα⊗RN), (∑xα)⊗y 7→

∑xα⊗y. The family iα⊗1N : Mα⊗RN →

(⊕Mα)⊗R N induces by 2.4.3 the inverse.

2.6.12. Example. Let R be a ring, F a free module with basis yα and G a freemodule with basis zβ . Then F ⊗R G is a free module with basis yα ⊗ zβ .

2.6.13. Theorem. Let R be a ring and M,N,L modules. Then there is a naturalisomorphism

HomR(M ⊗R N,L) ' HomR(M,HomR(N,L))

f 7→ (x 7→ [y 7→ f(x⊗ y)])(x⊗ y 7→ g(x)(y))←[ g

Proof. A given f : M ⊗R N → L is mapped to the composite homomorphismM → HomR(N,M ⊗RN)→ HomR(N,L), 2.5.4 and 2.6.2. This is a homomor-phism as map of f by 2.5.4. Given g : M → HomR(N,L) the map M × N →L, (x, y) 7→ g(x)(y) induces a homomorphism M ⊗R N → L, x⊗ y 7→ g(x)(y)

2.7. CHANGE OF RINGS 37

by 2.6.3. Clearly the maps are inverse to each other and therefore giving an iso-morphism by 2.1.5.

2.6.14. Exercise. (1) Show that Q⊗Z Q/Z ' 0.(2) Show that Z/(m)⊗Z Z/(n) = 0 if (m,n) = Z.(3) Let P,Q ⊂ R be different maximal ideals and M a module. Show that M/PM ⊗R

M/QM = 0.

2.7. Change of rings

2.7.1. Proposition. (1) Let φ : R → S be a ring homomorphism and N an Smodule. The restriction scalars 2.1.11 viewing N as an R-module throughφ, R × N → N, (a, x) 7→ ax = φ(a)x, is a functor from S-modules toR-modules.

(2) Let I ⊂ R be an ideal. Restriction of scalars along R → R/I identifiesR/I-modules M with R-modules such that I ⊂ Ann(M). For R/I-modulesM,N there is a natural isomorphism

HomR(M,N) ' HomR/I(M,N)

Proof. This is a restatement of 2.1.11 using 2.5.4.

2.7.2. Lemma. Let R → S be a ring homomorphism, M an R-module and N anS-module. Then

S ×M ⊗R N →M ⊗R N, (b, x⊗ y) 7→ x⊗ by

is an S-scalar multiplication.

Proof. For fixed b ∈ S the map M × N → M ⊗R N, (x, y) 7→ x ⊗ by inducesthe homomorphism µb : M ⊗R N → M ⊗R N, x ⊗ y 7→ x ⊗ by by 2.6.3. Thisgives a well defined scalar multiplication S×M ⊗RN →M ⊗RN, (b, x⊗y) 7→µb(x⊗ y).

2.7.3. Definition. Let R→ S be a ring homomorphism and M an R-module. Thechange of ring S-module is M ⊗R S with S-scalar multiplication 2.7.2

S ×M ⊗R S →M ⊗R S, (b, x⊗ c) 7→ x⊗ bc2.7.4. Proposition. The construction

M 7→M ⊗R S

andf : M →M ′ 7→ f ⊗ 1S : M ⊗R S →M ′ ⊗R S

is a functor from R-modules to S-modules.

HomR(M,M ′)→ HomS(M ⊗R S,M′ ⊗R S)

is an R-homomorphism.

Proof. By 2.6.7 this is a functor to R-modules and by 2.7.2 f ⊗ 1S is an S-homomorphism.

2.7.5. Proposition. Let R → S be a ring homomorphism, M,M ′ two R-modulesand N an S-modules.

38 2. MODULES

(1) There is a natural isomorphism of S-modules.

M ⊗R S ⊗S N 'M ⊗R N

x⊗ b⊗ y 7→ x⊗ by

(2) Then there is a natural isomorphism of S-modules.

(M ⊗R S)⊗S (M ′ ⊗R S) ' (M ⊗R M′)⊗R S

x⊗ b⊗ y ⊗ c 7→ x⊗ y ⊗ bc

Proof. (1) The homomorphism v : S ⊗S N → N, b⊗ y 7→ by is an isomorphism,2.6.10. The homomorphism 1M ⊗ v : M ⊗R S⊗S N 'M ⊗R N is an R-moduleisomorphism, 2.6.7. The identity x ⊗ bc ⊗ y = x ⊗ b ⊗ cy proves this to be anS-module homomorphism. (2) Use (1) on N = M ′ ⊗R S and conclude by thecanonical isomorphisms 2.6.10.

2.7.6. Theorem. Let R → S be a ring homomorphism, M,M ′ two R-modulesand N an S-modules.

(1) For any R-homomorphism f : M → N , then there exists a unique S-homomorphism f ′ : M ⊗R S → N,x ⊗ b 7→ bf(x) such that f(x) =f ′(x⊗ 1).

M

$$IIIIIIIII

f // N

M ⊗R Sf ′

::

(2) There is a natural isomorphism

HomR(M,N) ' HomS(M ⊗R S,N)

f 7→ (x⊗ b 7→ bf(x))

(3) There is a natural homomorphism

HomR(M,N)⊗R S → HomS(M ⊗R S,M′ ⊗R S)

f ⊗ c 7→ (x⊗ b 7→ f(x)⊗ bc)

Proof. (1) From 2.6.3 f ′ is the unique R-homomorphism and by 2.7.3 this is anS-homomorphism. (2) This is a restatement of (1). A given f is mapped to thecomposite M ⊗R S → N ⊗R S → N which is an S-homomorphism. Given ahomomorphism g : M ⊗R S → N then the composite M →M ⊗R S → N is anR-homomorphism and an inverse to the first given map. (3) This follows from (1)applied to HomR(M,N)→ HomS(M ⊗R S,M

′ ⊗R S).

2.7.7. Lemma. Let R → S be a ring homomorphism, M an R-module and N anS-module. Then

S ×HomR(N,M)→ HomR(N,M)

(b, f : N →M) 7→ (y 7→ f(by))

is an S-scalar multiplication.

Proof. The map (b, f) 7→ f ◦ bN satisfies the laws 2.1.1.

2.7. CHANGE OF RINGS 39

2.7.8. Definition. Let R→ S be a ring homomorphism and M an R-module. Theinduced module is the S-module HomR(S,M) with S-scalar multiplication 2.7.7

S ×HomR(S,M)→ HomR(S,M)

(b, f : S →M) 7→ (c 7→ f(bc))

2.7.9. Proposition. The induced module

M 7→ HomR(S,M)

and

f : M →M ′ 7→ Hom(1S , f) : HomR(S,M)→ HomR(S,M ′)

is a functor from R-modules to S-modules.

HomR(M,M ′)→ HomS(HomR(S,M),HomR(S,M ′))

is an R-homomorphism.

Proof. By 2.5.5 this is a functor to R-modules and by 2.7.7 Hom(1, f) is an S-homomorphism.

2.7.10. Theorem. Let R→ S be a ring homomorphism and M an R-module andN an S-modules. Then there is a natural isomorphism

HomR(N,M) ' HomS(N,HomR(S,M))

f 7→ (y 7→ [b 7→ f(by)])

Proof. g 7→ (y 7→ g(y)(1)) is an inverse.

2.7.11. Example. Let I ⊂ R be an ideal and R→ R/I the projection.

(1) The change of ring functor maps anR-moduleM to theR/I-moduleM/IM .The natural isomorphism 2.7.6 is

HomR(M,N) ' HomR/I(M/IM,N)

for any R/I-module N .(2) The induced module functor maps an R-module M to the R/I-module {x ∈

M |I ⊂ Ann(x)}. The natural isomorphism 2.7.10 is

HomR(N,M) ' HomR/I(N,HomR(R/I,M))

for any R/I-module N .

2.7.12. Definition. Let R → S, S′ be ring homomorphisms. The tensor productring over R is S ⊗R S′ with multiplication given by (b ⊗ b′)(c ⊗ c′) = bc ⊗ b′c′extended by linearity. R → S ⊗R S′, r 7→ r ⊗ 1 = 1 ⊗ r is the natural ringhomomorphism.

2.7.13. Proposition. Let φ, φ′ : R → S, S′ and ψ,ψ′ : S, S′ → T give a commu-tative diagram of ring homomorphisms, ψ ◦φ = ψ′ ◦φ′. Then b⊗ b′ 7→ ψ(b)ψ′(b′)

40 2. MODULES

is the unique homomorphism making the following diagram commutative.

R

��

// S′

��

��555

5555

5555

5555

5

S //

))SSSSSSSSSSSSSSSSSSSS S ⊗R S′

##T

Proof. This is clear by 2.6.3.

2.7.14. Example. Let R→ S be a ring homomorphism. Then

R[X]⊗R S ' S[X]

is an isomorphism.

2.7.15. Exercise. (1) Show that the change of rings of a free R-module is a free S-module.

(2) Let φ : R → S be a ring homomorphism. Show that the change of rings of ascalar multiplication a : M → M on an R-module is a scalar multiplication φ(a) :M ⊗R S →M ⊗R S.

(3) Show that the change of rings of the composition of two homomorphisms is thecomposition of the change of rings of each homomorphism.

(4) Show the isomorphism

R[X]⊗R R[Y ] ' R[X,Y ]

3

Exact sequences of modules

3.1. Exact sequences

3.1.1. Definition. Let f : M → N and g : N → L be homomorphisms ofmodules. The sequence

Mf // N

g // L

of homomorphisms is a

(1) 0-sequence: g ◦ f = 0 or equivalently Im f ⊂ Ker g(2) exact sequence: Im f = Ker g

For a sequence of more homomorphisms the conditions should be satisfied forevery consecutive composition. E.g. The sequence

Mf // N

g // Lh // K

is a 0-sequence if g ◦ f = 0 and h ◦ g = 0. The sequence is exact if Im f = Ker gand Im g = Kerh.

3.1.2. Remark. An interpretation of 2.3.3 gives:

(1) The sequence

0 // Mf // N

is exact if and only if f is injective.(2) The sequence

Mf // N // 0

is exact if and only if f is surjective.(3) The sequence

0 // Mf // N // 0

is exact if and only if f is an isomorphism.

3.1.3. Proposition. (1) For a homomorphism f : M → N the sequence

0 // Ker f // Mf // N // Cok f // 0

is exact.(2) For scalar multiplication with a ∈ R on M the sequence

0 // Ker aM// M

aM // M // M/aM // 0

is exact.

41

42 3. EXACT SEQUENCES OF MODULES

3.1.4. Proposition. Given a 0-sequence

0 // Mf // N

g // L

then the following statements are equivalent:

(1) The sequence is exact.(2) f is an isomorphism onto Ker g.(3) Given a homomorphism h : K → N such that g ◦ h = 0 then there is a

unique h′ : K →M such that h = f ◦ h′.

0 // Mf // N

g // L

K

h′

OOh

>>||||||||

Proof. (1) ⇔ (2): This is clear. (1) ⇒ (3): Imh ⊂ Ker g = Im f, so by (2) puth′ = f−1 ◦ h. (3)⇒ (2): Apply it to Ker g →M to see that (2) is satisfied.

3.1.5. Proposition. Given a 0-sequence

Mf // N

g // L // 0

then the following statements are equivalent:

(1) The sequence is exact.(2) The factor homomorphism 2.3.9 g′ : Cok f → L induced by g is an isomor-

phism.(3) Given a homomorphism k : N → K such that k ◦ f = 0 then there is a

unique k′ : L→ K such that k = k′ ◦ g.

Mf // N

g //

k

AAA

AAAA

A L

k′

��

// 0

K

Proof. (1) ⇔ (2): This follows from 2.3.9. (1) ⇒ (3): By 2.3.5 there is k′′ :Cok f → K such that k′′ ◦ p = k. By (1) put k′ = k′′ ◦ g−1. (3)⇒ (2) satisfiedand apply it to N → Cok f to see that (2) is satisfied.

3.1.6. Proposition. Let

Mα// Nα

// Lα

be a family of exact sequences. Then there are exact sequences:

(1) The sum ⊕Mα

//⊕Nα

//⊕Lα

(2) The product ∏Mα

//∏Nα

//∏Lα

Proof. Clear since kernel and image are calculated componentwise.

3.1. EXACT SEQUENCES 43

3.1.7. Definition. An exact sequence

0 // Mf // N

g // L // 0

is a short exact sequence. That is f is injective, Im f = Ker g and g is surjective.

3.1.8. Proposition. (1) Let I ⊂ R be an ideal, then there is a short exact se-quence

0 // I // R // R/I // 0

(2) Let M ⊂ N be a submodule, then there is a short exact sequence

0 // M // N // N/M // 0

(3) For scalar multiplication with nonzero divisor a ∈ R on M the sequence

0 // MaM // M // M/aM // 0

is a short exact sequence.(4) Given a homomorphism f : M → N there are associated two short exact

sequences.

0 // Ker f // Mf // Im f // 0

and

0 // Im f // N // Cok f // 0

(5) For scalar multiplication with any a ∈ R onM there are associated two shortexact sequences.

0 // Ker aM// M

aM // aM // 0

and

0 // aM // M // M/aM // 0

3.1.9. Definition. Let f : M → N be a homomorphism.

(1) f has a retraction if there is a homomorphism u : N →M such that u ◦ f =1M .

(2) f has a section if there is a homomorphism v : N →M such that f ◦v = 1N .

3.1.10. Proposition. Let f : M → N be a homomorphism.

(1) If f has a retraction u : N →M then f is injective, u is surjective and

N = Im f ⊕Keru

(2) If f has a section v : N →M then f is surjective, v is injective and

M = Ker f ⊕ Im v

Proof. (1) u(f(x)) = x so f is injective and u is surjective. If y ∈ N theny = f(u(y)) + (y− f(u(y)) and u(y− f(u(y))) = 0, so N = Im f + Keru. Lety ∈ Im f ∩ Keru. Then y = f(x) gives x = u(f(x)) = u(y) = 0, so y = 0.Conclude by 2.4.5 that the sum is direct. (2) y = f(v(y)) so f is a retraction of v.Finish by (1).


3.1.11. Lemma. For a short exact sequence

0 // Mf // N

g // L // 0

the following statements are equivalent:(1) f has a retraction.(2) g has a section.

For any retraction u there is a unique section v and wise-verse such that

1N = f ◦ u+ v ◦ g

Proof. If u is a retraction of f , then Ker g = Im f ⊂ Ker(1N − f ◦ u). By 3.1.5there is a homomorphism v : L → N such that v ◦ g = 1N − f ◦ u. This is asection of g. Conversely if v is a section of g then Im(1N − v ◦ g) ⊂ Ker g, sothere is a homomorphism u : N → M such that f ◦ u = 1N − v ◦ g, 3.1.4. u is aretraction of f . The equation is clearly satisfied.

3.1.12. Definition. LetR be a ring and f : M → N, g : N → L homomorphisms.A short exact sequence

0 // Mf // N

g // L // 0

is a split exact sequence if equivalently 3.1.11 f has a retraction or g has a section.

3.1.13. Theorem. A sequence

0 // Mf // N

g // L // 0

is a split exact sequence if and only if there are homomorphism u : N → M,v :L→ N satisfying

u ◦ f = 1M , g ◦ v = 1L, f ◦ u+ v ◦ g = 1N

If the sequence is split exact then

0 // Lv // N

u // M // 0

is split exact and (x, y) 7→ f(x)+v(y) and z 7→ u(z)+g(z) gives the isomorphism

M ⊕ L ' N

Proof. The sequence is a 0-sequence f is injective and g is surjective. From f ◦u+ v ◦ g = 1N follows that z ∈ Ker g ⊂ Im f , so the sequence is short exact. Therest is contained in 3.1.10.

3.1.14. Corollary. A (contravariant) functor preserves split exact sequences. If

0 // Mf // N

g // L // 0

is split exact and T a functor, then

0 // T (M)T (f) // T (N)

T (g) // T (L) // 0

is split exact. If T is contravariant, then

0 // T (L)T (g) // T (N)

T (f) // T (M) // 0

is split exact.

3.2. THE SNAKE LEMMA 45

Proof. By 3.1.13 a split exact sequence is characterized by a set of equations.These are preserved by the functor, 2.5.4.

3.1.15. Example. A short exact sequence

0 // Mf // N

g // F // 0

where F is a free module is a split exact sequence. Namely let xα ∈ F be abasis and choose yα ∈ N with g(yα) = xα. The define v : L → N by settingv(xα) = yα, 2.4.11.

3.1.16. Example. Let Zi be the family of modules each a copy of Z indexed bythe natural numbers. Then the short exact sequence

0 //⊕

i Zi//∏

i Zi//∏

i Zi/⊕

i Zi// 0

is not split exact.The element f = (1, 2, 22, . . . , 2n, . . . ) +

⊕i Zi is divisible by 2k for all k. If

fk = (0, . . . , 0, 2n−k, . . . ) +⊕

i Zi for n ≥ k, then 2kfk = f in∏

i Zi/⊕

i Zi.But in

∏i Zi the only element divisible with all 2k is 0, so no section exists.

3.1.17. Exercise. (1) Show that the sequence

0 // Z // Q // Q/Z // 0

is short exact, but not split exact.(2) Show that the sequence

0 // Zn // Z // Z/(n) // 0

is exact, but not split exact for n 6= 0, 1.(3) Show that the sequence

0 // Z/(2) 1 7→2 // Z/(4) // Z/(2) // 0

is exact, but not split exact.(4) Show that the sequence

0 // Z/(2) 1 7→3 // Z/(6) // Z/(3) // 0

is split exact.

3.2. The snake lemma

3.2.1. Example. Given a commutative diagram of homomorphisms

M

u

��

f // N

v

��M ′ f ′ // N ′

there is induced a commutative diagram

0 // Ker f

u

��

// M

u

��

f // N

v

��

// Cok f

v

��

// 0

0 // Ker f ′ // M ′ f ′ // N ′ // Cok f ′ // 0

where the rows are exact sequences.


The diagram splits into two diagrams

0 // Ker f

u

��

// M

u

��

f // Im f

v

��

// 0

0 // Ker f ′ // M ′ f ′ // Im f ′ // 0

and

0 // Im f

v

��

// N

v

��

// Cok f

v

��

// 0

0 // Im f ′ // N ′ // Cok f ′ // 0

where the rows are short exact sequences.

3.2.2. Lemma. Given a commutative diagram of homomorphisms

M

u

��

f // N

v

��

g // L

w

��

// 0

0 // M ′ f ′ // N ′ g′ // L′

where the rows exact sequences. The snake homomorphism δ : Kerw → Cokuis well defined by: For z ∈ Kerw choose y ∈ N such that g(y) = z. Theelement v(y) ∈ Ker g′ so there is x′ ∈ M ′ such that f ′(x′) = v(y). Then δ(z) =x′ + Imu ∈ Coku.

Proof. Assume g(y′) = z and f ′(x′′) = v(y′). There is x ∈M with f(x) = y−y′.Now f ′(u(x)) = v(f(x)) = v(y − y′) = f ′(x′ − x′′) so u(x) = x′ − x′′ since f ′

is injective. Then x′ + Imu = x′′ + Imu as wanted. The choices made respectaddition and scalar multiplication showing that δ is a homomorphism.

3.2.3. Remark. The snake is

Keru

��

f // Ker v

��

g // Kerw

��jmkl

on_________________________________

hi!!!!

..^

M

u

��

f // N

v

��

g // L

w

��

// 0

0 // M ′

��

f ′ // N ′

��

g′ // L′

��Coku

f ′ // Cok vg′ // Cokw


The construction of δ is schematically

Kerw

��N

v

��

g // L

M ′

��

f ′ // N ′

Coku

z_

��y_

��

� // z

x′_

��

� // v(y)

δ(z)

3.2.4. Theorem (snake lemma). Given a commutative diagram of homomorphisms

M

u

��

f // N

v

��

g // L

w

��

// 0

0 // M ′ f ′ // N ′ g′ // L′

where the rows exact sequences. There is induced a six term long exact sequence

Keruf // Ker v

g // Kerw*+/. δ��

Cokuf ′ // Cok v

g′ // Cokw

Proof. By construction of δ it is clear that the sequence is a 0-sequence: If y ∈Ker v then to calculate δ(g(y)) the choice v(y) = 0 gives δ ◦ g = 0. Alsof ′(δ(z)) = v(y) + Im v shows that f ′ ◦ δ = 0. Exactness at Ker v and Cok vare clear. Given z ∈ Kerw such that δ(z) = 0. By 3.2.2 choose y, g(y) = zand x′, f ′(x′) = v(y). Then δ(z) = x′ + Imu = 0, so choose x, u(x) = x′.Now v(f(x)) = f ′(u(x)) = v(y) so y − f(x) ∈ Ker v and g(y − f(x)) =g(y) = z. Therefore exactness at Kerw. Given x′ + Imu ∈ Coku such thatf ′(x′) + Im v = 0 ∈ Cok v. Choose y, v(y) = f ′(x′) and put z = g(y). Thenw(g(y)) = g′(v(y)) = g′(f ′(x′)) = 0. Now z ∈ Kerw and δ(z) = x′ + Imu.Therefore exactness at Coku.

3.2.5. Corollary. If f is injective then the f : Keru → Ker v is injective and thelong exact sequence is

0 // Keruf // Ker v

g // Kerw*+/. δ��

Cokuf ′ // Cok v

g′ // Cokw

If g′ is surjective then g′ : Cok v → Cokw is surjective and the long exact se-quence is

Keruf // Ker v

g // Kerw*+/. δ��

Cokuf ′ // Cok v

g′ // Cokw // 0


3.2.6. Corollary. (1) If v is injective and u is surjective, then w is injective.(2) If v is surjective and w is injective, than u is surjective.(3) If v is an isomorphism, then w is injective if and only if u is surjective.

3.2.7. Proposition. Given submodules N,L ⊂ M , then there is a short exactsequence

0 // M/N ∩ Lx 7→(x,x)// M/N ⊕M/L

(x,y) 7→x−y// M/N + L // 0

Proof. The commutative diagram

0 // N ∩ L

��

// N ⊕ L

��

// N + L

��

// 0

0 // Mx 7→(x,x)// M ⊕M

(x,y) 7→x−y // M // 0

where the rows are short exact sequences, gives by 3.2.4 a five term long exactsequence

0 // M/N ∩ L // M/N ⊕M/L // M/N + L // 0

3.2.8. Proposition (five lemma). Given a commutative diagram of homomorphisms

M1

u1

��

// M2

u2

��

// M3

u3

��

// M4

u4

��

// M5

u5

��M ′

1// M ′

2// M ′

3// M ′

4// M ′

5

where the rows are exact sequences. If u1 is surjective, u2, u4 are isomorphismand u5 is injective, then u3 is an isomorphism.

Proof. Let fi : Mi →Mi+1, f′i : M ′

i →M ′i+1. Split the given diagram in three as

follows

M1

u1

��

// M2

u2

��

// Cok f1

u′3��

// 0

0 // Ker f ′2 // M ′2

// Im f ′2// 0

0 // Cok f2

u′′3��

// M4

u4

��

// Im f4

u5

��

// 0

0 // Ker f ′4 // M ′4

// M ′5

0 // Im f2

u′3��

// M3

u3

��

// Cok f2

u′′3��

// 0

0 // Im f ′2// M ′

3// Cok f ′2 // 0

Note that Cok f1 ' Im f2 and Cok f ′2 ' Ker f ′4. Now use 2.3.3 and the snakelemma to conclude that Keru3 = 0 and Coku3 = 0 and u3 is therefore an isomor-phism.


3.2.9. Corollary. (1) Given a commutative diagram of homomorphisms

M

u

��

f // N

v

��M ′ f ′ // N ′

If u, v are isomorphism, then the induced homomorphisms u : Ker f →Ker f ′ and v : Cok f → Cok f ′ are isomorphisms.

(2) Given a commutative diagram of homomorphisms

M

u

��

// N

v

��

// L

w

��M ′ // N ′ // L′

If u, v, w are isomorphism, then the upper row is exact if and only if the lowerrow is exact.

3.2.10. Proposition (windmill lemma). Given homomorphism Mf // N

g // L .There is induced an eight term long exact sequence

0 // Ker f // Ker g ◦ ff // Ker g*+

/.��

Cok fg // Cok g ◦ f // Cok g // 0

Proof. Look at the two diagrams

0

��

// M

f

��

1 // M

g◦f��

// 0

0 // Ker g // Ng // L

M

g◦f��

f // N

g

��

// Cok f

��

// 0

0 // L1 // L // 0

By the snake lemma the sequences

0 // Ker f // Ker g ◦ f*+/. δ=f��

Ker g // Cok f // Cok g ◦ f

Ker g ◦ f // Ker g // Cok f*+/. δ=g��

Cok g ◦ f // Cok g // 0

are exact and overlap to give the windmill sequence.


3.2.11. Remark. The windmill is

Ker g ◦ f //

$$IIIIIIIII

Ker g

%%LLLLLLLLLL

zzvvvvv

vvvv

v

Ker f //

99ssssssssssM

f //

g◦f ��@@@

@@@@

@ N //

g��

��

��Cok f

��

��

��

��

�

0

OO

L

))RRRRRRRRRRRRRRR

uullllllllllllllll

Cok g

eeLLLLLLLLLLL

Cok g ◦ foo

3.2.12. Exercise. (1) Given a short exact sequence

0 // M // N // L // 0

Show that Ann(N) ⊂ Ann(M) ∩Ann(L).(2) Give a short exact sequence

0 // M // N // L // 0

such that Ann(N) 6= Ann(M) ∩Ann(L).(3) Given ideals I, J ⊂ R. Show that there is a short exact sequence.

0 // R/I ∩ J // R/I ⊕R/J // R/I + J // 0

3.3. Exactness of Hom

3.3.1. Definition. A functor T is a left exact functor if given an exact sequence

0 // Mf // N

g // L

the following sequence is exact

0 // T (M) // T (N) // T (L)

T is a right exact functor if given an exact sequence

Mf // N

g // L // 0


T (M) // T (N) // T (L) // 0

A functor is an exact functor if it is both left and right exact.A contravariant functor T is a left exact contravariant functor if given an exactsequence

0 // Mf // N

g // L


T (L) // T (N) // T (M) // 0

T is a right exact contravariant functor if given an exact sequence

Mf // N

g // L // 0

3.3. EXACTNESS OF HOM 51


0 // T (L) // T (N) // T (M)

A contravariant functor is an exact functor if it is both left and right exact.

3.3.2. Proposition. Let T be a functor. The following conditions are equivalent:(1) T is a exact.(2) Given an exact sequence

Mf // N

g // L

then the following sequence is exact

T (M) // T (N) // T (L)

(3) Given a short exact sequence

0 // Mf // N

g // L // 0

Then the following sequence is short exact

0 // T (M) // T (N) // T (L) // 0

Let T be a contravariant functor. The following conditions are equivalent:(1) T is a exact.(2) Given an exact sequence

Mf // N

g // L


T (L) // T (N) // T (M)


0 // Mf // N

g // L // 0


0 // T (L) // T (N) // T (M) // 0

Proof. Let T be a functor. (1)⇒ (2): The exact sequence M → Im f → 0 givesT (M) → T (Im f) → 0 exact and the exact sequence 0 → Im f → N gives0 → T (Im f) → T (N) exact. It follows that ImT (f) ' T (Im f). The sequence0 → Im f → N → L is exact so 0 → T (Im f) → T (N) → T (L) is exact. Nowconclude that ImT (f) = KerT (g). (2) ⇒ (3): This is clear. (3) ⇒ (1): Given

0 → Mf // N

g // L exact, then the sequences 0 → T (M) → T (N) →T (Im g)→ 0 and 0→ T (Im g)→ T (L) are exact, so ImT (f) = kerT (g) and Tis left exact. Similarly T is rigth exact.

3.3.3. Corollary. Let T be an exact functor. For a homomorphism f : M → Nthere are natural isomorphisms.(1) KerT (f) ' T (Ker f).(2) ImT (f) ' T (Im f).(3) CokT (f) ' T (Cok f).


Let T be a contravariant functor. For a homomorphism f : M → N there arenatural isomorphisms.

(1) KerT (f) ' T (Cok f).(2) ImT (f) ' T (Im f).(3) CokT (f) ' T (Ker f).

Proof. Represent the statements using short exact sequences. (1) The kernel isdetermined by the exact sequence 0→ Ker f →M → N , 3.1.4. (3) The cokernelis determined by the exact sequence M → N → Cok f → 0, 3.1.5. (2) Theimage is determined by the exact sequence 0→ Ker f →M → Im f → 0, 2.3.5,3.1.4.

3.3.4. Corollary. Let T be an exact functor. For submodules N,L ⊂M there arenatural identifications

(1) T (M/N) = T (M)/T (N).(2) T (N + L) = T (N) + T (L).(3) T (N ∩ L) = T (N) ∩ T (L).

Proof. Represent the statements using short exact sequences. (1) 0→ N →M →M/N → 0 is short exact giving 0 → T (N) → T (M) → T (M/N) → 0. Thewanted isomorphism follows form 3.1.5. (2) N + L is the image of N ⊕ L→M .(3) N ∩ L is the kernel of N ⊕ L→M .

3.3.5. Proposition. The functor HomR(K,−) is left exact. Given an exact se-quence

0 // Mf // N

g // L

and an R-module K. Then the following sequence is exact

0 // HomR(K,M) // HomR(K,N) // HomR(K,L)

Proof. Given h : K → M such that f ◦ h = 0 then h = 0 since f is injective.Given k : K → N such that g ◦ k = 0 then by 3.1.4 there is h : K →M such thatf ◦ h = k. So the sequence is exact.

3.3.6. Proposition. The contravariant functor HomR(−,K) is right exact. Givenan exact sequence

Mf // N

g // L // 0and a module K. Then the following sequence is exact

0 // HomR(L,K) // HomR(N,K) // HomR(M,K)

Proof. Given h : L → K such that h ◦ g = 0 then h = 0 since g is surjective.Given k : N → K such that k ◦ f = 0 then by 3.1.5 there is h : L→ K such thath ◦ g = k. So the sequence is exact.

3.3.7. Proposition. Given a sequence

Mf // N

g // L // 0

such that for any module K, the following sequence is exact

0 // HomR(L,K) // HomR(N,K) // HomR(M,K)

Then the original sequence is exact.

3.4. EXACTNESS OF TENSOR 53

Proof. ForK = L the identity 1L is mapped to 1L ◦g ◦f = 0 so it is a 0-sequence.Take K = Cok g then pg : L → Cok g has pg ◦ g = 0, but by exactness 0 is theunique homomorphism satisfying this, so pg = 0. Therefore Cok g = 0 and g issurjective. Take K = Cok f , p : N → Cok f the projection. p ◦ f = 0 so byexactness there exists a unique q : L→ Cok f such that q ◦ g = p. It follows thatKer g ⊂ Ker p = Im f . All together the original sequence is exact.

3.3.8. Proposition. Given a sequence

0 // Mf // N

g // L // 0

The following conditions are equivalent:(1) The sequence is split exact.(2) For any K the following sequence is exact

0 // HomR(K,M) // HomR(K,N) // HomR(K,L) // 0

(3) For any K the following sequence is exact

0 // HomR(L,K) // HomR(N,K) // HomR(M,K) // 0

If the conditions are true, then the sequences (2) and (3) are split exact.

Proof. (1)⇒ (2), (1)⇒ (3) are clear by 3.1.14 giving that the sequences (2), (3)are split exact. (2)⇒ (1): Let K = L, then there is a section to g. By 3.1.11 theoriginal sequence is split exact. (3)⇒ (1): Let K = M , then there is a retractionto f . By 3.3.7and 3.1.11 the original sequence is split exact.

3.3.9. Exercise. (1) Show that the sequence

0 // HomZ(Q/Z,Z) // HomZ(Q/Z,Q) // HomZ(Q/Z,Q/Z)

is exact, but the rightmost map is not surjective.(2) Show that the sequence

0 // HomZ(Z/(n),Z) n // HomZ(Z/(n),Z) // HomZ(Z/(n),Z/(n))

is exact, but the rightmost map is not surjective.

3.4. Exactness of tensor

3.4.1. Theorem. The functor −⊗R K is right exact. Given an exact sequence

Mf // N

g // L // 0

and an R-module K. Then the following sequence is exact

M ⊗R K // N ⊗R K // L⊗R K // 0

Proof. Let K ′ be any module. By 3.3.7 it is enough to see that the sequence

0 // HomR(L⊗R K,K′) // HomR(N ⊗R K,K

′)

��HomR(M ⊗R K,K

′)


is exact. By 2.6.13 it amounts to see that the sequence

0 // HomR(L,HomR(K,K ′)) // HomR(M,HomR(K,K ′))

��HomR(N,HomR(K,K ′))

is exact. This follows from 3.3.6

3.4.2. Proposition. Given a split exact sequence

0 // Mf // N

g // L // 0

and a module K. Then the following sequence is split exact

0 // K ⊗R M // K ⊗R N // K ⊗R L // 0

Proof. This follows from the functor properties 3.1.14.

3.4.3. Proposition. Let I ⊂ R be an ideal. For any moduleM , the homomorphism

M ⊗R R/I →M/IM, x⊗ a+ I 7→ ax+ IM

is an isomorphism.

Proof. There is a commutative diagrm

M ⊗R I

��

// M ⊗R R

��

// M ⊗R (R/I)

��

// 0

0 // IM // M // M/IM // 0

The conclusion follows from the snake lemma, 3.2.4.

3.4.4. Corollary. Let I, J ⊂ R be ideals. Then

R/I ⊗R R/J → R/(I + J), a+ I ⊗ b+ J 7→ ab+ I + J

is an isomorphism.

3.4.5. Proposition. Let I1, . . . , Ik ⊂ R be pairwise comaximal ideals and M amodule. Then the product of projections

M/I1 · · · IkM →M/I1M × · · · ×M/IkM

is an isomorphism.

Proof. By Chinese remainders 1.4.2

R/I1 · · · Ik → R/I1 × · · · ×R/Ikis an isomorphism. Tensor with M and use 3.4.3 to get the isomorphism.

3.4.6. Exercise. (1) Calculate Z/(m)⊗Z Z/(n) for all integers m,n.(2) Let I ⊂ R be an ideal. Show that R/I ⊗R R/I ' R/I .(3) Let I ⊂ R be an ideal. Show that I ⊗R R/I ' I/I2.(4) Let 2Z be scalar multiplication. Show that 2Z⊗1Z/(2) : Z⊗Z Z/(2)→ Z⊗Z Z/(2)

is not injective.

3.5. PROJECTIVE MODULES 55

3.5. Projective modules

3.5.1. Definition. An R-module F is a projective module if for any surjecticehomomorphism N → L the homomorphism

HomR(F,N)→ HomR(F,L)

is surjective.

3.5.2. Proposition. Let F be an R-module. The following conditions are equiva-lent:

(1) F is projective.(2) The functor HomR(F,−) is exact.(3) Given an exact sequence

M // N // L


HomR(F,M) // HomR(F,N) // HomR(F,L)


0 // Mf // N

g // L // 0


0 // HomR(F,M) // HomR(F,N) // HomR(F,L) // 0

Proof. This is clear from 3.3.2. (1) ⇔ (2): This follows since HomR(F,−) isallways left exact. (1)⇔ (3)⇔ (4): This is true for any exact functor.

3.5.3. Proposition. A module F is projective if and only if any surjective homo-morphism M → F → 0 has a section.

Proof. Assume F projective and g : M → F surjective. Then HomR(F,M) →HomR(F, F ) → 0 is exact. So there exists a v : F → M such that g ◦ v = 1F .v is then a section. Conversely given g : N → L surjective and h : F → L. LetM = KerN⊕F → L, (y, z) 7→ g(y)−h(z) and pN : M → N, pF : M → F theprojections, then g ◦pN = h◦pF . Now pF is surjective since g is. Let v : F →Mbe a section of pF , then h′ = pN ◦ v satisfies h = g ◦ h′.

Ng // L // 0

M

pN

OO

pF

// F

h′``

vtth

OO

3.5.4. Corollary. A short exact sequence

0 // Mf // N

g // F // 0

where F is a projective module is a split exact sequence.

3.5.5. Corollary. A free module is projective. Over a field every module is projec-tive.

3.5.6. Example. Let I ⊂ R be an ideal. If R/I is projective, then there is a ringdecomposition R/I ×R′ ' R.


3.5.7. Proposition. A direct summand in a projective module is projective.

Proof. Let F ⊕ F ′ be projective and g : M → F surjective. By 3.5.3 there is asection v′ : F ⊕ F ′ →M ⊕ F ′ to (g, 1F ′). Then v(y) = pM ◦ v′(y, 0) is a sectionto g and F is projective.

3.5.8. Proposition. A module is projective if and only if it is a direct summand ina free module.

Proof. By 2.4.12 any module is a factor module of a free module. By 3.5.3 aprojective factor module has a section, and is therefore by 3.1.10 a direct summand.

3.5.9. Proposition. Let Fα be a family of projective modules, then the direct sum⊕α Fα is a projective module.

Proof. Let N → L be surjective. Then by 2.5.8

HomR(⊕

Fα, N)→ HomR(⊕

Fα, L)

is the product ∏HomR(Fα, N)→

∏HomR(Fα, L)

which is surjective by 3.1.6. So⊕Fα is projective.

3.5.10. Proposition. Let F, F ′ be projective modules. Then F ⊗R F′ is projective.

Proof. F ⊗R F′ is clearly a direct summand in a free module.

3.5.11. Proposition. Let R → S be a ring homomorphism and F a projectivemodule. The change of ring module F ⊗R S is a projective S-module.

Proof. A direct summand of a free R-module is changed to a direct summand of afree S-module.

3.5.12. Proposition. Any module M admits an exact sequence

F →M → 0

where F is a projective module. That is, any module is a factor module of a pro-jective module.

Proof. Take F free, 2.4.12.

3.5.13. Exercise. (1) Let R = R1 ×R2. Show that R1, R2 are projective ideals in R.(2) Show that the ideal (2)/(6) in the ring Z/(6) is projective.(3) Show that the ideal (2)/(4) in the ring Z/(4) is not projective.

3.6. Injective modules

3.6.1. Definition. An R-module E is an injective module if for any injective ho-momorphism M → N the homomorphism

HomR(N,E)→ HomR(M,E)

is surjective.

3.6.2. Proposition. Let E be an R-module. The following conditions are equiva-lent:

(1) E is injective.

3.6. INJECTIVE MODULES 57

(2) The contravariant functor HomR(−, E) is exact.(3) Given an exact sequence

M // N // L


HomR(L,E) // HomR(N,E) // HomR(M,E)


0 // Mf // N

g // L // 0


0 // HomR(L,E) // HomR(N,E) // HomR(M,E) // 0

Proof. This is clear from 3.3.2. (1) ⇔ (2): This follows since the contravariantfunctor HomR(−, E) is allways right exact. (1)⇔ (3)⇔ (4): This is true for anyexact contravariant functor.

3.6.3. Proposition. A module E is an injective module if and only if any injectivehomomorphism 0→ E → L has a retraction.

Proof. Assume E injective and f : E → L injective. Then HomR(L,E) →HomR(E,E) → 0 is exact. So there exists a u : L → E such that u ◦ f = 1E .Then u is a retraction. Conversely given f : M → N injective and h : M → E.Let L = CokM → E ⊕N, x 7→ h(x)− f(x) and iE : E → L, iN : N → L theinjections, then iN ◦ f = iE ◦ h. Now iE is injective since f is. Let u : L→ E bea retraction of iE , then h′ = u ◦ iN satisfies h = h′ ◦ f .

0 // M

h��

f // N

iN��

h′

~~E

iE // Lu

ii

3.6.4. Corollary. A short exact sequence

0 // Ef // N

g // L // 0

where E is an injective module is a split exact sequence.

3.6.5. Example. Let I ⊂ R be an ideal. If I is injective, then there is a ringdecomposition R/I ×R′ ' R.

3.6.6. Proposition. A direct summand in an injective module is injective.

Proof. Let E ⊕ E′ be an injective module and f : E → L an injective homomor-phism. By 3.6.3 there is a retraction u′ : L ⊕ E′ → E ⊕ E′ to (f, 1E′). Thenu(y) = pE ◦ u′(y, 0) is a retraction to f and E is injective.

3.6.7. Proposition. LetEα be a family of injective modules, then the direct product∏αEα is an injective module


Proof. Let M → N be injective. Then by 2.5.8

HomR(N,∏

Eα)→ HomR(M,∏

Eα)

is the product ∏HomR(N,Eα)→

∏HomR(M,Eα)

which is surjective by 3.1.6. So∏Eα is injective.

3.6.8. Proposition. A module E is injective, if for any ideal I ⊂ RHomR(R,E)→ HomR(I, E)→ 0

is exact.

Proof. Let f : E → L be an injective homomorphism. The set of submodulesf(E) ⊂ L′ ⊂ L and retractions u′ : L′ → E is nonempty and inductively ordered.By Zorn’s lemma a maximal L′, u′ exists. If L/L′ 6= 0 choose a y ∈ L\L′. Thehomomorphism Ann(y + L′) → E, a 7→ u′(ay) extends to u′′ : R → E byhypothesis. The setting L′ +Ry → E, x+ ay 7→ u′(x) + u′′(a) is a well definedretraction. This contradicts maximality.

3.6.9. Definition. Let R be a domain. M is a divisible module if scalar multipli-cation with a nonzero a ∈ R is surjective.

3.6.10. Proposition. (1) An injective module over a domain is divisible.(2) A divisible module over a principal ideal domain is injective.(3) Over a field any module is injective.

Proof. (1) Let E be an injective module and a 6= 0. Choose x ∈ E and lookat 1x : R → E. Scalar multiplication aR is injective, so there is an extension1y : R → E such that 1x = 1y ◦ aR. Then ay = x. (2) Let E be divisible. Toextend f : (a)→ E, choose x ∈ E such that ax = f(a), then 1x : R→ E, 1→ xextends f . (3) This is clear.

3.6.11. Proposition. Let R → S be a ring homomorphism and E an injectiveR-module. The induced module HomR(S,E) is an injective S-module.

Proof. Let M → N be an injective homomorphism of S-modules. Then by2.7.10 HomS(N,HomR(S,E)) → HomS(M,HomR(S,E)) is HomR(N,E) →HomR(M,E) being surjective since E is an injective R-module.

3.6.12. Lemma. (1) Q and Q/Z are divisible and therefore injective Z modules.(2) The homomorphism 2.5.11

x 7→ evx : M → HomZ(HomZ(M,Q/Z),Q/Z)

is injective.

Proof. (1) Clear by 3.6.10. (2) Let x 6= 0 and choose h : Z/Ann(x) → Q/Znonzero. Now Z/Ann(x) ' Zx ⊂ M so extend h to h′ : M → Q/Z. Thenevx(h′) = h(1) 6= 0.

3.6.13. Theorem. Any module M admits an exact sequence

0→M → E

where E is an injective module. That is, any module is a submodule of an injectivemodule.

3.6. INJECTIVE MODULES 59

Proof. By 2.4.12 choose a surjection F → HomZ(M,Q/Z) where F is a freeR-module. Then

0→M → HomZ(F,Q/Z)is exact by 3.6.12. Since F '

⊕αR the module

HomZ(F,Q/Z) '∏α

HomZ(R,Q/Z)

is injective, 3.6.7 and 3.6.11.

3.6.14. Proposition. A homomorphism f : M → N is injective if and only if

HomR(N,E)→ HomR(M,E)→ 0

is surjective for any injective module E.

Proof. Assume that Ker f 6= 0 and choose 0→ Ker f → E, 3.6.13. The sequence

HomR(N,E)→ HomR(M,E)→ HomR(Ker f,E)→ 0

is exact. So HomR(N,E)→ HomR(M,E) is not surjective.

3.6.15. Lemma. Given an injective homomorphism 0 // Ni // M , then the fol-

lowing conditions are equivalent:(1) Any nonzero submodule L ⊂M has nonzero retraction i−1(L) 6= 0.(2) Given a homomorphism f : M → K such that the restriction f ◦i is injective,

then f is injective.

3.6.16. Definition. An injective homomorphism 0 // Ni // M satisfying the

equivalent conditions 3.6.15 is an essential extension. An essential extension

0 // Mi // E is an injective envelope of M .

3.6.17. Proposition. Any module M has an injective envelope. If M ⊂ E,E′ aretwo injective envelopes, then there is an isomorphism f : E → E′ fixing M .

Proof. By 3.6.13 choose M ⊂ E′ with E′ injective. By Zorn’s lemma chooseM ⊂ E ⊂ E′ maximal among the essential extensions of M . If E 6= E′ then theset of modules 0 6= N ′ ⊂ E′ such that E ∩ N ′ is nonempty by maximality of E.Let N be maximal among these by Zorn’s lemma. Since E′ is injective there is acommutative diagram

E

��

// E′ // E′/N

vvmmmmmmmmmmmmmmm

E′

The composite f : E′ → E′/N → E′ has E ⊂ Im f . If Im f 6= E then bymaximality of E there is a submodule 0 6= L ⊂ Im f with E ∩ L = 0. NowN f−1(L) and E ∩ f−1(L) = E ∩ L = 0 gives a contradiction. ThereforeIm f = E and f is a retraction making E a direct summand in E′. By 3.6.6 E isan injective module. Now uniqueness: given two envelopes, let f : E → E′ beany homomorphism fixing M . Then f is injective, since M ⊂ E is essential. If fis not surjective, then E′ ' Im f ⊕ E′′ contradicting that M ⊂ E′ is essential.

3.6.18. Exercise. (1) Let R be a domain. Show that the fraction field is an injectivemodule.


(2) Let R be a domain. Show that the torsion free divisible module injective.(3) Show that for a ring that all modules are projective if and only all modules are injec-

tive.

3.7. Flat modules

3.7.1. Definition. An R-module F is a flat module if for any injective homomor-phism M → N the homomorphism

M ⊗R F → N ⊗R F

is injective.


(1) F is a flat.(2) The functor −⊗R F is exact.(3) Given an exact sequence of R-modules

M // N // L

Then the sequence of change of rings modules is exact

M ⊗R F // N ⊗R F // L⊗R F


0 // Mf // N

g // L // 0

Then the following sequence is exact

0 // M ⊗R F // N ⊗R F // L⊗R F // 0

Proof. This is follows from 3.3.2. (1)⇔ (2): This follows since −⊗R is allwaysright exact. (1)⇔ (3)⇔ (4): This is true for any exact functor.

3.7.3. Proposition. A direct summand in a flat module is flat.

Proof. Let F ⊕ F ′ be flat and M → N injective. Then M ⊗R (F ⊕ F ′) →N ⊗R (F ⊕ F ′) is injective. Conclusion by 2.6.11.

3.7.4. Proposition. Let Fα be family of flat modules, then the direct sum⊕

α Fα

is a flat module.

Proof. Let M → N be injective. Then M ⊗R (⊕Fα) → N ⊗R (

⊕Fα) is

injective by 2.6.11 and 3.1.6, so the product is flat.

3.7.5. Corollary. A free module is flat.

Proof. R is flat since M ⊗R R 'M , so⊕

αR is flat.

3.7.6. Corollary. A projective module is flat.

Proof. By 3.5.8 a projective module is a direct summand in a free. Conclusion by3.7.4.

3.7.7. Proposition. Let F, F ′ be flat modules. Then F ⊗R F′ is flat.

Proof. Let M → N be injective. Then by 2.6.10 M ⊗R (F ⊗R F ′) → N ⊗R

(F ⊗R F′) is (M ⊗R F )⊗R F

′ → (N ⊗R F )⊗R F′ being injective by using the

definition twice.

3.7. FLAT MODULES 61

3.7.8. Proposition. Let R → S be a ring homomorphism and F a flat R-module.The change of ring module F ⊗R S is a flat S-module.

Proof. Let M → N be an injective homomorphism of S-modules. Then by 2.7.5M ⊗S (F ⊗R S)→ N ⊗S (F ⊗R S) is M ⊗R F → N ⊗R F being injective sinceF is a flat R-module.

3.7.9. Theorem. Let R be a ring and F a module. The following conditions are :(1) F is a flat module.(2) HomR(F,E) is an injective module for any injective module E.

Proof. LetM → N be injective. By 3.6.14M⊗RF → N⊗RF is injective if andonly if HomR(N ⊗R F,E)→ HomR(M ⊗R F,E) is surjective for any injectivemoduleE. By 2.6.13 this is HomR(N,HomR(F,E))→ HomR(M,HomR(F,E)).

3.7.10. Corollary. Given a short exact sequence

0 // M // N // F // 0

where F is a flat module. Then M is flat if and only if N is flat.

Proof. Let E be an injective module. By 3.7.9 and 3.6.4 the sequence

0 // HomR(F,E) // HomR(N,E) // HomR(M,E) // 0

is split exact. By 3.6.6 and 3.6.7 HomR(N,E) is injective if and only if HomR(M,E)is so. Conclusion by 3.7.9.


0 // M // N // F // 0

where F is a flat module. For any module L there is a short exact sequence

0 // L⊗R M // L⊗R N // L⊗R F // 0

Proof. Let E be an injective module. By 3.7.9 and 3.6.4 the sequence

0 // HomR(F,E) // HomR(N,E) // HomR(M,E) // 0

is split exact. So also the sequence

0 // HomR(L,HomR(F,E)) // HomR(L,HomR(N,E))

��HomR(L,HomR(M,E)) // 0

is split exact. By 2.6.13 this is natural isomorphic to the sequence

0 // HomR(L⊗R F,E) // HomR(L⊗R N,E)

��HomR(L⊗R M,E) // 0

Conclusion by 3.6.14.


3.7.12. Theorem. A module F is flat, if for any ideal I ⊂ R0→ I ⊗R F → R⊗R F

is exact.

Proof. By 3.7.9 it suffices to see that HomR(F,E) is injective for any injective E.By 3.6.8 this amounts to HomR(R,HomR(F,E))→ HomR(I,HomR(F,E)) be-ing surjective. By 2.6.13 this homomorphism is HomR(R⊗RF,E)→ HomR(I⊗R

F,E) which is surjective since E is injective.

3.7.13. Corollary. A module F is flat, if for any finite ideal J ⊂ R0→ J ⊗R F → R⊗R F

is exact.

Proof. Let I be any ideal. Given∑bi ⊗ xi ∈ I ⊗R F such that

∑bixi = 0, then∑

bi ⊗ xi = 0 ∈ (bi)⊗R M and therefore also in I ⊗R M .

3.7.14. Exercise. (1) Show that if a ∈ R is a nonzero divisor and M is a flat modules,then aM is injective and a is a nonzero divisor on M .

(2) Show that Z/(n) is not a flat Z-module for n 6= 0, 1.(3) Show that Z/(2) is a flat Z/(6)-module.(4) Show that Z/(2) is not a flat Z/(4)-module.

4

Fraction constructions

4.1. Rings of fractions

4.1.1. Lemma. Let R be a ring and U ⊂ R such that 1 ∈ U and for u, v ∈ U theproduct uv ∈ U . On U ×R is defined a relation

(u, a) ∼ (u′, a′)⇔ there is v ∈ U such that vu′a = vua′

(1) The relation is an equivalence relation.(2) If (u, a) ∼ (u′, a′) and (v, b) ∼ (v′, b′) then (uv, va + ub) ∼ (u′v′, v′a′ +

u′b′).(3) If (u, a) ∼ (u′, a′) and (v, b) ∼ (v′, b′) then (uv, ab) ∼ (u′v′, a′b′).

Proof. The claims are proved by simple calculations. (1) Symmetry is clear. Re-flexive follows as 1 ∈ U . Transitive: if (u, a) ∼ (u′, a′), (u′, a′) ∼ (u′′, a′′) thenvu′a = vua′, v′u′′a′ = v′u′a′′. Since multiplication is commutative (v′vu′)u′′a =v′u′′vua′ = (v′vu′)ua′′ give (u, a) ∼ (u′′, a′′). (2) From wu′a = wua′, w′v′b =w′vb′ follow that w′vv′wu′a = w′vv′wua′, wuu′w′v′b = wuu′w′vb′. So nowww′(u′v′)(va+ ub) = ww′(uv)(v′a′ + u′b′) as needed. (3) This is similar.

4.1.2. Definition. Let R be a ring. U ⊂ R is a multiplicative subset if 1 ∈ Uand for u, v ∈ U the product uv ∈ U . The ring of fractions U−1R is given byequivalence classes

a

u= {(u′, v′)|(u′, v′) ∼ (u, v)}

on U ×R under the relation 4.1.1

(u, a) ∼ (u′, a′)⇔ there is v ∈ U such that vu′a = vua′

The addition isa

u+b

v=va+ ub

uv

and the multiplication isa

u· bv

=ab

uv

This is well defined and U−1R is a ring, verified analog to the rational numbers Qbeing a ring. The canonical ring homomorphism is

ι : R→ U−1R, a 7→ a

14.1.3. Theorem. Let φ : R → S be a ring homomorphism and U ⊂ R a multi-plicative subset. If all elements in φ(U) ⊂ S are units, then there exists a unique

63

64 4. FRACTION CONSTRUCTIONS

ring homomorphism φ′ : U−1R→ S such that φ = φ′ ◦ ι.

R

ι ""FFF

FFFF

FFφ // S

U−1R

φ′

<<

Proof. φ′( au) = φ(a)φ(u)−1 is the well defined unique ring homomorphism. Ob-

serve that the elements of form bv−1 satisfy the rules for fractions.

4.1.4. Proposition. Let U ⊂ R be a multiplicative subset.(1) a

1 6= 0 in U−1R if and only if Ann(a) ∩ U = ∅.(2) The canonical ring homomorphism R → U−1R is injective if and only if U

consists only of nonzero divisors.

Proof. a1 = 0 if and only if va = 0 for some v ∈ U .

4.1.5. Corollary. IfR is a domain and U is the nonzero elements thenK = U−1Ris a field and the canonical homomorphism identifies R as a subring.

4.1.6. Definition. The field K in 4.1.5 is the fraction field of R.

4.1.7. Corollary. If R is a domain with fraction field K and φ : R → L is aninjective ring homomorphism into a field, then there is a unique homomorphismK → L extending φ.

4.1.8. Definition. The total ring of fractions of R is U−1R, where U is the set ofnonzero divisors of R.

4.1.9. Corollary. A ring is a subring of its total ring of fractions.

4.1.10. Proposition. Let U = {un} the powers of an element u ∈ R. There is anisomorphism

R[X]/(uX − 1)→ U−1R, X 7→ 1u

The ring of fractions with only one denominator is of finite type.

Proof. A pair of inverse homomorphism are constructed by 1.6.7 and 4.1.3.

4.1.11. Definition. Let U = {un} the powers of an element u ∈ R. The ringU−1R, 4.1.10, is denoted

R[1u

]

4.1.12. Proposition. If U ⊂ V ⊂ R are multiplicative, then there is a canonicalidentification

V −1R ' (V/1)−1U−1R,a

v7→ a/1

v/14.1.13. Proposition. Let U ⊂ R1 × R2 be a multiplicative subset of a product ofrings and let Ui ⊂ Ri be the projections of U . Then there is an isomorphism

U−1(R1 ×R2) ' U−11 R1 × U−1

2 R2

(a1, a2)(u1, u2)

7→(a1

u1,a2

u2

)

4.2. MODULES OF FRACTIONS 65

Proof. Injective: If a1u1

= 0 and a2u2

= 0 then there is (v1, v2), (w1, w2) ∈ U

such that v1a1 = 0 and w2a2 = 0. It follows that (v1, v2)(w1, w2)(a1, a2) = 0.Surjective: Given

(a1u1, a2

u2

)then there is (u1, v2), (v1, u2) ∈ U . It follows that

(a1v1,a2v2)(u1v1,u2v2) 7→

(a1u1, a2

u2

).

4.1.14. Exercise. (1) Show that if U contains a nilpotent element, then U−1R = 0.(2) Show that

KerR→ U−1R = {a ∈ R|ua = 0, for some u ∈ U}

(3) Let U = {u1, . . . , um} and u = u1 · · ·um. Then show that

U−1R = {un}−1R

(4) Let R1, R2 be domains with fraction fields K1,K2. Show that the total ring of frac-tions of R1 ×R2 is K1 ×K2.

(5) Show that the ring

{ a2n∈ Q|a ∈ Z, n ∈ N}

is of finite type over Z.

4.2. Modules of fractions

4.2.1. Lemma. Let R be a ring and U ⊂ R multiplicative. On U ×M is defineda relation

(u, x) ∼ (u′, x′)⇔ there is v ∈ U such that vu′x = vux′

(1) The relation is an equivalence relation.(2) If (u, x) ∼ (u′, x′) and (v, y) ∼ (v′, y′) then (uv, vx + uy) ∼ (u′v′, v′x′ +

u′y′).(3) If (u, a) ∼ (u′, a′) in U × R 4.1.1 and (v, x) ∼ (v′, x′) then (uv, ax) ∼

(u′v′, a′x′).

Proof. The claims are proved by simple calculations. See the proof 4.1.1.

4.2.2. Definition. Let R be a ring and U a multiplicative subset. The module offractions U−1M is given by equivalence classes x

u on U ×M under the relation4.2.1

(u, x) ∼ (u′, x′)⇔ there is v ∈ U such that vu′x = vux′

The addition isx

u+y

v=vx+ uy

uv

and the U−1R-scalar multiplication isa

u· yv

=ay

uv

The canonical homomorphism is

M → U−1M, x 7→ x

1


4.2.3. Theorem. Let U ⊂ R be a multiplicative subset. Given an R-module M , aU−1R-module N and a R-homomorphism f : M → N , then there exists a uniqueU−1R-homomorphism f ′ : U−1M → N, x

u 7→1uf(x) such that f = f ′ ◦ i.

M

i ##GGG

GGGG

GGf // N

U−1M

f ′

;;

Proof. The claims are proved by simple calculations. If (u, x) ∼ (u′, x′) thenvu′x = vux′ and therefore vu′f(x) = vuf(x′) showing 1

uf(x) = 1u′ f(x′). So the

map is well defined. The rest is similar.

4.2.4. Corollary. Let f : M → N a homomorphism of R-modules. Then there isa homomorphism U−1f : U−1M → U−1N, x

u 7→f(x)

u of U−1R-modules.

4.2.5. Proposition. The construction 4.2.2

M 7→ U−1M

and 4.2.4

f : M → N 7→ U−1f : U−1M → U−1N

is a functor from R-modules to U−1R-modules.

Proof. Follows from the definitions by simple calculations as in the proof of 4.2.3.For example, U−1(f + g) = U−1f + U−1g, follows from f(x)+g(x)

u = f(x)u +

g(x)u .

4.2.6. Remark. The induced homomorphism relates to the canonical homomor-phism such that the diagram is commutative.

M

��

f // N

��U−1M

U−1f // U−1N

That is, the canonical homomorphism is a natural homomorphism.

4.2.7. Proposition. Let U ⊂ R be a multiplicative subset and M a module.

(1) x1 6= 0 in U−1M if and only if Ann(x) ∩ U = ∅.

(2) The canonical homomorphism M → U−1M is injective if and only if Uconsists only of nonzero divisors on M .

Proof. x1 = 0 if and only if vx = 0 for some v ∈ U .

4.2.8. Proposition. If Mα is a family of modules, then the homomorphism

U−1(⊕

α

Mα)→⊕

α

U−1Mα

is a natural isomorphism of U−1R-modules.

4.3. EXACTNESS OF FRACTIONS 67

Proof. By 2.4.3 there is an R-homomorphism⊕

αMα →⊕

α U−1Mα which

gives the U−1R-homomorphism in in question by 4.2.3. An inverse is given by2.4.3. This is also the method of common denominators in a finite sum.∑

i

xi

ui=

1Πiui

∑i

(Πj 6=iuj)xi

4.2.9. Definition. Let U = {un} the powers of an element u ∈ R. The R[ 1u ]-

module, 4.1.10, U−1M is denoted

M [1u

]

4.2.10. Proposition. If U ⊂ V ⊂ R are multiplicative, then there is a canonicalidentification

V −1M ' (V/1)−1U−1M,x

v7→ x/1

v/14.2.11. Exercise. (1) Show that if U contains a nilpotent element, then U−1M = 0.

(2) Show that

KerM → U−1M = {x ∈M |ux = 0, for some u ∈ U}(3) Let U = {u1, . . . , um} and u = u1 · · ·um. Then show that

U−1M = {un}−1M

(4) Show that U−1M = 0 if and only if U ∩Ann(x) 6= ∅ for all x ∈M .(5) Show that the fraction homomorphism of a composition is the composition of the

respective fraction homomorphisms.(6) Let M be a free R-module. Show that U−1M is a free U−1R-module(7) Show that the homomorphism

(∏N

Z)[12]→

∏N

Z[12]

is not surjective.

4.3. Exactness of fractions

4.3.1. Theorem. Let R be a ring and U a multiplicative subset. The functorU−1(−) is exact. Given an exact sequence of R-modules

Mf // N

g // L


U−1M // U−1N // U−1L

Proof. If yu ∈ U

−1N maps to g(y)u = 0 then there is v ∈ U such that 0 = vg(y) =

g(vy). Choose x ∈ M such that f(x) = vy. Then xvu maps to f(x)

vu = yu proving

exactness.


0 // Mf // N

g // L // 0


0 // U−1M // U−1N // U−1L // 0


If the first sequence is split exact, also the second sequence is split exact.

4.3.3. Corollary. For a homomorphism f : M → N there are natural isomor-phisms of U−1R-modules.

(1) U−1 Ker f ' KerU−1f .(2) U−1 Im f ' ImU−1f .(3) U−1 Cok f ' CokU−1f .

Proof. Represent the statements using short exact sequences. (1) The kernel isdetermined by the exact sequence 0→ Ker f →M → N , 3.1.4. (3) The cokernelis determined by the exact sequence M → N → Cok f → 0, 3.1.5. (2) Theimage is determined by the exact sequence 0→ Ker f →M → Im f → 0, 2.3.5,3.1.4.

4.3.4. Corollary. For submodules N,L ⊂ M there are natural identifications ofU−1R-submodules and factor modules.

(1) U−1(M/N) = U−1M/U−1N .(2) U−1(N + L) = U−1N + U−1L.(3) U−1(N ∩ L) = U−1N ∩ U−1L.

Proof. Represent the statements using short exact sequences. (1) 0→ N →M →M/N → 0 is short exact giving 0→ U−1N → U−1M → U−1(M/N)→ 0. Thewanted isomorphism follows form 3.1.5. (2) N + L is the image of N ⊕ L→ Mso conclude by 4.3.3. (3) N ∩ L is the kernel of N ⊕ L → M so conclude by4.3.3.

4.3.5. Corollary. For ideals I, J ⊂ R there are natural identifications in U−1R.

(1) U−1(R/I) = U−1R/U−1I .(2) U−1(I + J) = U−1I + U−1J .(3) U−1(I ∩ J) = U−1I ∩ U−1J .(4) U−1(IJ) = U−1IU−1J .

Proof. (1) (2) (3) These are special cases of 4.3.4. (4) Both sides have the samegenerators ab

u , a ∈ I, b ∈ J, u ∈ U .

4.3.6. Proposition. Let R→ U−1R be the canonical homomorphism.

(1) For an ideal I ⊂ R the extended ideal

IU−1R = U−1I

(2) For an ideal J ⊂ U−1R the extended contracted ideal

U−1(J ∩R) = J

(3) For an ideal I ⊂ R the contracted extended ideal

I ⊂ IU−1R ∩R

Proof. (1) This is clear. (2) U−1(J ∩ R) ⊂ J is true for any ring homomorphism1.2.6. If b

u ∈ J then b1 ∈ J giving b ∈ J ∩R and 1

ub = bu ∈ U

−1(J ∩R). (3) Thisis true for any ring homomorphism 1.2.6.

4.4. TENSOR MODULES OF FRACTIONS 69

4.3.7. Proposition. Let R → U−1R be the canonical homomorphism. For anideal I ⊂ R the contracted extended ideal

I = IU−1R ∩Rif and only if each u ∈ U is a nonzero divisor on R/I .

Proof. Apply the snake lemma 3.2.4 to

0 // I //

��

R //

��

R/I //

��

0

0 // 0 // U−1R/U−1I U−1R/U−1I // 0

and get the exact sequence

0→ I → IU−1R ∩R→ Ker(R/I → U−1(R/I))→ 0

Conclusion from 4.2.7.

4.3.8. Corollary. Let P ⊂ R be a prime ideal. Then U−1P ⊂ U−1R is either aprime ideal or the whole ring.

Proof. By 4.1.5 and 4.3.5 U−1R/U−1P = U−1(R/P ) is either 0 or a domain.

4.3.9. Corollary. Let R be a principal ideal domain. Then U−1R is a principalideal domain.

Proof. A restricted ideal is principal by hypothesis and the extension of a principalideal is principal. Conclude by 4.3.6.

4.3.10. Theorem. LetR be a unique factorization domain. ThenU−1R is a uniquefactorization domain.

Proof. By 4.3.7 the extension of an irreducible element is either a unit or an irre-ducible element. Since a principal ideal ( a

u) = (a1 ), a factorization into irreducibles

in R gives a factorization in U−1R. Now if (a) = (p1) . . . (pn) is a factorizationin R and (a

1 ) is irreducible in U−1R. Then all but one pi

1 is a unit in U−1R so(a1 ) = (pi

1 ) is a prime ideal 4.3.9. The conditions 1.5.3 are satisfied.

4.3.11. Exercise. (1) Let U ⊂ R be multiplicative. Show that

U−1(R[X]) = (U−1R)[X]

(2) Let φ : R → S be a ring homomorphism and U ⊂ R a multiplicative subset. Showthat

U−1S = φ(U)−1S

4.4. Tensor modules of fractions

4.4.1. Theorem. Let R be a ring and U a multiplicative subset. For any moduleM , the homomorphism

M ⊗R U−1R→ U−1M, x⊗ a

u7→ ax

u

is a natural isomorphism of U−1R-modules.

Proof. By 2.6.3 there is an R-module homomorphism x⊗ au →

axu . By definition

bv (x ⊗ a

u) = x ⊗ bavu 7→

baxvu = b

vaxu , so the this is a U−1R-homomorphism. The

map U−1M →M ⊗R U−1R, x

u 7→ x⊗ 1u is an inverse.


4.4.2. Corollary. The two constructions, module change of ring to a fraction ringand fraction module are natural isomorphic functors from modules to modules overthe fraction ring.

4.4.3. Corollary. Let R be a ring and U a multiplicative subset. Then U−1R is aflat R-module.

Proof. This follows from 4.4.1 and 4.3.1.

4.4.4. Corollary. Let R be a ring and U a multiplicative subset and M,N mod-ules. Then there is a natural isomorphism

U−1(M ⊗R N) ' U−1M ⊗U−1R U−1N


4.4.5. Corollary. Let I ⊂ R be an ideal and U a multiplicative subset For amodule M module

U−1(IM) ' U−1IU−1M

Proof. Use that IM = Im(I ⊗R M →M) and 4.4.3,4.4.4.

4.4.6. Proposition. Let U ⊂ R be a multiplicative subset and M an R-module.Let i : M → U−1M be the canonical homomorphism and N ⊂ U−1M anyU−1R-submodule. Then N is extended

U−1(i−1(N)) ' N

Proof. The inclusion U−1(i−1(N)) ⊂ N is clear. If xu ∈ N then x

1 ∈ N givingx ∈ i−1(N) and 1

ux = xu ∈ U

−1(i−1(N)).

4.4.7. Exercise. (1) Let F be a flat R-module. Show that U−1F is a flat U−1R-module.

(2) Let M be a projective R-module. Show that U−1M is a projective U−1R-module.(3) Show that the homomorphism

(∏N

Z)⊗Z Z[12]→

∏N

Z⊗Z Z[12]

is not surjective.

4.5. Homomorphism modules of fractions

4.5.1. Proposition. LetR be a ring andU a multiplicative subset. For any modulesM,N there is a natural homomorphism

U−1 HomR(M,N)→ HomU−1R(U−1M,U−1N)

of U−1R-modules.

Proof. Given f : M → N and u ∈ U the setting xv 7→

f(x)uv is a U−1R-

homomorphism U−1M → U−1N .

4.5.2. Proposition. Let R be a ring and U a multiplicative subset. For any R-module M and any U−1R-modules N there is a natural isomorphism

HomR(M,N)→ HomU−1R(U−1M,N)

of U−1R-modules.

4.6. THE POLYNOMIAL RING IS FACTORIAL 71

Proof. This is the change of rings isomorphism 2.7.6 interpreted according to4.4.2. Also this is a reinterpretation of 4.2.3.

4.5.3. Example. Let R be a ring and U a multiplicative subset. R → U−1R thecanonical homomorphism. The induced module functor maps an R-module M tothe U−1R-module HomR(U−1R,M). The natural isomorphism 2.7.10 is

HomR(N,M) ' HomU−1R(N,HomR(U−1R,M))

for any U−1R-module N .

4.5.4. Example. The homomorphism 4.5.1 is in general neither injective nor sur-jective.

(1) Not surjective:

0 = HomZ(Q,Z)→ HomQ(Q,Q) = Q

(2) Not injective:

0 6= HomZ(Q,Q/Z)⊗Z Q→ HomQ(Q,Q/Z⊗Z Q) = 0

4.5.5. Exercise. (1) Let M be a Z-module and N a Q-module. Show that there is anisomorphism HomZ(M,N) ' HomQ(M ⊗Z Q, N).

4.6. The polynomial ring is factorial

4.6.1. Definition. Let R be a unique factorization domain and let f = anXn +

· · · + a0 be a polynomial over R. Then the content of polynomial f , c(f), is thegreatest common divisor of the coefficients a0, . . . , an.

4.6.2. Lemma (Gauss’ lemma). LetR be a unique factorization domain. For poly-nomials f, g ∈ R[X]

c(fg) = c(f)c(g)

Proof. Assume by cancellation that c(f), c(g) are units in R. For any irreduciblep ∈ R the projections of f, g in R/(p)[X] are nonzero. Since R has unique factor-ization the ideal (p) is a prime ideal. It follows that the projection of the productfg in R/(p)[X] is also nonzero and therefore p is not a common divisor of thecoefficients of the product fg.

4.6.3. Theorem. Let R be a unique factorization domain. Then the ring of poly-nomials R[X] is a unique factorization domain.

Proof. Let K be the fraction field of R, then the polynomial ring K[X] is a prin-cipal ideal domain. Let f ∈ R[X] and use unique factorization in K[X] to get0 6= a ∈ R and p1, . . . , pn ∈ R[X], irreducible in K[X], such that

af = p1 . . . pn

Assume by 4.6.2 that a = 1 and c(p1), . . . , c(pn) are units in R. Apply 4.6.2 and1.6.5 to see that p1, . . . , pn are irreducible inR[X]. An irreducible p ∈ R generatesa prime ideal (p) ⊂ R[X]. A non constant irreducible p ∈ R[X] generates a primeideal (p) ⊂ K[X] and therefore also a prime ideal (p) ⊂ R[X]. So conditions1.5.3 are satisfied.

4.6.4. Corollary. Let K be a field. Then the polynomial ring K[X1, . . . , Xn] is aunique factorization domain.

Proof. Follows by induction from 4.6.3.


4.6.5. Exercise. (1) Let f ∈ Z[X] be monic and assume f = gh where g, h ∈ Q[X]are monic. Show that g, h ∈ Z[X].

(2) Let f ∈ Z[X] be monic and irreducible in Z/(n)[X]. Show that f is irreducibleQ[X].

(3) Let K be a field. Show that the polynomial ring K[X1, X2, . . . ] in countable manyvariables is a unique factorization domain

5

Localization

5.1. Prime ideals

5.1.1. Theorem (Krull). A nonzero ring contains a maximal ideal.

Proof. The nonempty set of ideals different from R is ordered by inclusion. Givenan increasing chain Iα then ∪Iα is an ideal different from R which is a maximumfor the chain. Conclusion by Zorn’s lemma.

5.1.2. Corollary. Any proper ideal in a ring is contained in a maximal ideal.

Proof. If I 6= R then the factor ring R/I 6= 0 and contains a maximal ideal.Conclusion by 1.2.10.

5.1.3. Theorem. Let P1, . . . , Pn ⊂ R be ideals with at most 2 not being primeideals. If an ideal

I ⊂ P1 ∪ · · · ∪ Pn

then I ⊆ Pi for some i.

Proof. Assume n > 1 and I is not contained in any sub union. Moreover assumethe numbering such that P3, . . . , Pn are prime ideals. Then for each i there is

ai ∈ (I ∩ Pi)\ ∪j 6=i Pj

The elementan + a1 . . . an−1

is in I but not in any Pi, giving a contradiction. So n = 1.

5.1.4. Proposition. Let P1, . . . , Pn ⊂ R be prime ideals and I any ideal. If forsome a ∈ R

a+ I ⊂ P1 ∪ · · · ∪ Pn

then I ⊆ Pi for some i.

Proof. If a ∈ ∩iPi then conclusion by 5.1.3. On the contrary after renumberingthere exists j with 1 ≤ j < n such that

a ∈ P1 ∩ · · · ∩ Pj\Pj+1 ∪ · · · ∪ Pn

Assume no inclusions between the prime ideals and choose by 5.1.3

b ∈ I ∩ Pj+1 ∩ · · · ∩ Pn\P1 ∪ · · · ∪ Pj

Then a+ b /∈ ∪iPi contradicts the hypothesis.

5.1.5. Theorem. LetR→ U−1R be the canonical homomorphism. Extension andcontraction gives a bijective correspondence between prime ideals in R disjointfrom U and all prime ideals in U−1R.

(1) For a prime ideal P ⊂ R\U the extended ideal PU−1R is a prime ideal inU−1R and the contracted PU−1R ∩R = P .

73

74 5. LOCALIZATION

(2) For a prime ideal Q ⊂ U−1R the contracted ideal Q ∩ R is a prime idealand the extended (Q ∩R)U−1R = Q

R→ U−1R

P 7→ PU−1R

Q ∩R←[ Q

Proof. Conclusion by 4.3.6, 4.3.7, 4.3.8.

5.1.6. Corollary. Let R be a ring and U a multiplicative subset.(1) An ideal P ⊂ R is maximal among the ideals disjoint from U if and only if

PU−1R is a maximal ideal in U−1R.(2) An ideal maximal among the ideals disjoint from U is a prime ideal.(3) Any ideal disjoint from U is contained in a prime ideal disjoint form U .

Proof. The prime ideals disjoint from U are the prime ideals in U−1R.

5.1.7. Theorem. The nilradical of a ring R is the intersection of all prime idealsP . √

0 =⋂P

P

Proof. By 1.3.8 the nilradical is contained in any prime ideal. Suppose u ∈ R isnot nilpotent. Then {un}−1R is nonzero. Then contraction of a maximal ideal,5.1.1, is a prime ideal in R not containing u.

5.1.8. Corollary. Let R be a ring.(1) The radical of an ideal I is the intersection of all prime idealsP containing I

√I =

⋂I⊂P

P

(2) For ideals I, J ⊂ R,√I ∩ J =

√I ∩√J .

(3) If U is a multiplicative subset, then U−1√

0 =√

0 in U−1R.If R is reduced, then U−1R is reduced.

Proof. (1) Use 5.1.7 on the factor ring R/I . (2) This follows from (1). (3) ClearlyU−1√

0 ⊂√

0. If au is nilpotent, then van = 0 for some v ∈ U . So (va)n = 0 and

au = va

vu ∈ U−1√

0.

5.1.9. Definition. A prime ideal minimal for inclusion among prime ideals is aminimal prime ideal.

5.1.10. Theorem. Any prime ideal of Q ⊂ R contains a minimal prime idealP ⊂ Q.

Proof. The set of prime ideals in R is ordered by inclusion. Given a decreasingchain Pα then ∩Pα is a prime ideal. Conclusion by Zorn’s lemma.

5.1.11. Corollary. Let R ⊂ S be a subring and P ⊂ R a minimal prime ideal.Then there is a minimal prime ideal Q ⊂ S contracting to P = Q ∩R.

5.1.12. Proposition. Let R be a domain. The following conditions are equivalent:(1) R is a unique factorization domain.(2) Any nonzero prime ideal contains a nonzero principal prime ideal.

5.2. LOCALIZATION OF RINGS 75

Proof. (1)⇒ (2): A prime ideal P 6= 0 contains an irreducible element generatinga prime ideal. (2)⇒ (1): LetU be the multiplicative subset generated by generatorsof principal prime ideals. Given a ∈ R nonzero and not a unit. If a /∈ U then by5.1.6 there is a prime ideal (a) ⊂ P such that P ∩ U = ∅. Such a P contains noprincipal prime ideal, so a ∈ U and 1.5.3 are satisfied.

5.1.13. Exercise. (1) Let K ⊂ R be an infinite subfield and I, P1, . . . , Pn any ideals.Show that if I ⊂ P1 ∪ · · · ∪ Pn then I ⊂ Pi for some i.

(2) Let P, P1, P2 be proper ideals. Show that if P is a maximal ideal and Pn ⊂ P1 ∪P2

then P = P1 of P = P2.

5.2. Localization of rings

5.2.1. Definition. A ring R which contains precisely one maximal ideal P is alocal ring and denoted (R,P ). The residue field of R is R/P denoted by k(P ).A ring homomorphism φ : R → S of local rings (R,P ), (S,Q) is a local ringhomomorphism if φ(P ) ⊂ Q.

5.2.2. Proposition. Let R be a ring.

(1) If (R,P ) is a local ring, then R\P is the set of units of R.(2) If the subset of non units in R is an ideal P , then (R,P ) is a local ring.

Proof. (1) If u /∈ P then by 5.1.1 (u) = R and u is a unit. (2) Any ideal I 6= Rcontains only non units, so I ⊂ P .

5.2.3. Proposition. A ring homomorphism φ : R→ S of local rings (R,P ), (S,Q)is a local ring homomorphism if the extended ideal PS ⊂ Q or the contracted idealQ ∩R = P . The residue homomorphism k(P )→ k(Q) is a field extension.

Proof. The contraction Q ∩R is a prime ideal containing P . The rest is clear.

5.2.4. Lemma. Let R be a ring and P a prime ideal. Then U = R\P is a multi-plicative subset. The ring of fractions U−1R is a local ring. The maximal ideal isthe extended ideal PU−1R. The residue field is U−1R/PU−1R which is canoni-cal isomorphic to the fraction field of R/P .

5.2.5. Definition. Let R be a ring, P a prime ideal and U = R\P . The localizedring at P is the local ring RP = U−1R. The residue field is denoted k(P ) =RP /PRP .

R

��

// RP

��R/P // k(P )

Note that

k(P ) = RP /PRP = (R/P )P = (R/P )(0)5.2.6. Proposition. Given a ring homomorphism R → S and a prime ideal Q ⊂S. Then P = Q ∩ R is a prime ideal and there is a local ring homomorphism

76 5. LOCALIZATION

(RP , PRP )→ (SQ, QSQ) fitting into the following commutative diagram.

S

��

// SQ

��

R

��

//

55kkkkkkkkkkkkkkkkkkkkRP

��

55kkkkkkkkkkkkkkkkkk

S/Q // k(Q)

R/P //

55lllllllllllllllllk(P )

55lllllllllllllllll

Proof. This is clear from the constructions.

5.2.7. Example. Let the ring be Z.(1) The local ring at (0) is the fraction field Q = Z(0).(2) The local ring Z(p) for a prime number p is identified with a subring of Q

Z(p) = {mn|p not dividing n}

The residue field Fp = Z(p)/(p).(3) Any nonzero ideal in Z(p) is principal of the form (pn) for some n.

5.2.8. Proposition. Let (R,P ) be a local ring. One of the following conditions issatisfied:(1) The characteristic char(R) = 0. P ∩ Z = (0) and Q ⊂ R is a subfield.

Q→ k(P ) is a field extension.(2) The characteristic char(R) = 0. P ∩ Z = (p), p a prime number. Z(p) ⊂ R

is a local subring. Fp → k(P ) is a field extension.(3) The characteristic char(R) = pn, a power of a prime number. Z/(pn) ⊂ R

is a local subring. Fp → k(P ) is a field extension.

Proof. (1) (2) are clear by 5.2.3 and 5.2.7. (3) If the characteristic is nonzero thenany prime ideal contracts Q ∩ Z = (p). So a prime number q 6= p gives a unit inR. There is a local homomorphism Z(p) → R, 4.1.3. The nontrivial kernel is (pn)by 5.2.7.

5.2.9. Example. (1) A field K is a local ring with maximal ideal (0). The powerseries ringK[[X]] is a local ring with maximal ideal (X) and residue fieldK.

(2) Let (R,P ) be a local ring. The power series ring R[[X]] is a local ring withmaximal ideal (P,X) and residue field k(P,X) = k(P ).

5.2.10. Proposition. Let R be a domain.(1) The local ring at (0) is the fraction field K = R(0).(2) Any local ring RP is identified with a subring of the fraction field K.(3) The intersection

R =⋂P

RP , P a maximal ideal

5.2.11. Proposition. Let R× S be a product of rings.

5.3. LOCALIZATION OF MODULES 77

(1) A prime ideal is of the form P × S or R ×Q for uniquely determined primeideals P ⊂ R or Q ⊂ S.

(2) The local ring at P ×S is identified withRP through the projectionR×S →R.

(3) The local ring atR×Q is identified with SQ through the projectionR×S →S.

5.2.12. Theorem. Let P be a prime ideal and R → RP the canonical homomor-phism. Extension and contraction gives a bijective correspondence between primeideals in R contained in P and all prime ideals in RP .

(1) For a prime ideal Q ⊂ P the extended ideal QRP is a prime ideal in RP andthe contracted QRP ∩R = Q.

(2) For a prime ideal Q′ ⊂ RP the contracted ideal Q′ ∩R ⊂ P is a prime idealand the extended (Q′ ∩R)RP = Q′

R→ RP

P → PRP

Q→ QRP

Q′ ∩R←[ Q′

Proof. This is a special case of 5.1.5.

5.2.13. Proposition. Let U ⊂ R be a multilicative subset and P ∩ U = ∅ a primeideal. Then PU−1R is a prime ideal in U−1R and canonically

RP = (U−1R)PU−1R

Proof. This follows from 4.1.12 and reflects the fraction rule au/

wv = av

uw .

5.2.14. Corollary. Let Q ⊂ P ⊂ R be prime ideals. Then QRP is a prime idealin RP and canonically

RQ = (RP )QRP

5.2.15. Definition. The intersection of all maximal ideals in a ring is the Jacobsonradical.

5.2.16. Remark. The Jacobson radical contains the nilradical. In a local ring theJacobson radical is the maximal ideal.

5.2.17. Exercise. (1) Show that a local ring is never a product of two nonzero rings.(2) Show that a ∈ R is in the Jacobson radical if and only if 1+ab is a unit for all b ∈ R.(3) Let p be a prime number. Describe the prime ideals in the ring Z(p).(4) Let P be a prime ideal. Show that k(P ) is the fraction field of R/P .(5) Let (R,P ) be a local ring. Show that (R[[X]], (P,X) is a local ring and the canonical

homomorphism R→ R[[X]] is a local homomorphism.

5.3. Localization of modules

5.3.1. Definition. Let R be a ring, P a prime ideal and U = R\P . For a moduleM , the localized module at P is the module MP = U−1M over the local ring RP .For a homomorphism f : M → N the localized homomorphism is fP : MP →NP and the residue homomorphism is f(P ) : M ⊗R k(P )→ N ⊗R k(P ).

78 5. LOCALIZATION

The constructions are functors, 2.7.4, 4.2.5. If f : M → N is a homomorphism,then the following diagram is commutative.

N

��

// NP

��

M

��

//

44iiiiiiiiiiiiiiiiiiiiiiiMP

��

44iiiiiiiiiiiiiiiiiiiii

N/PN // N(P )

M/PM //

44jjjjjjjjjjjjjjjjjjM(P )

44jjjjjjjjjjjjjjjjjjj

5.3.2. Proposition. Let R be a ring, P a prime ideal and M a module.(1) MP 'M ⊗R RP are natural isomorphic exact functors.(2) MP /PRPMP 'M ⊗R k(P ) are natural isomorphic functors.

Proof. See 4.4.1.

5.3.3. Corollary. For a homomorphism f : M → N

(1) (Ker f)P ' Ker fP .(2) (Im f)P ' Im fP .(3) (Cok f)P ' Cok fP .

Proof. See 4.3.3.

5.3.4. Corollary. LetR be a ring and P a prime ideal. For submodulesN,L ⊂M(1) (M/N)P 'MP /NP .(2) (N + L)P ' NP + LP .(3) (N ∩ L)P ' NP ∩ LP .

Proof. See 4.3.4.

5.3.5. Proposition. Let R be a ring and P a prime ideal. If Mα is a family ofmodules, then the homomorphism

(⊕

α

Mα)P →⊕

α

(Mα)P

is an isomorphism of RP -modules.

Proof. See 4.2.8.

5.3.6. Proposition. Let R be a ring, P a prime ideal.(1) For an R module M and an RP -module N there is a natural isomorphism

M ⊗R RP ⊗RPN 'M ⊗R N

(2) For an R module M,L there is a natural isomorphism

(M ⊗R L)P 'MP ⊗RPLP

(3) For an R module M,L there is a natural isomorphism

(M ⊗R L)(P ) 'M(P )⊗k(P ) L(P )

Proof. See 4.4.4 and 2.7.4.

5.4. THE LOCAL-GLOBAL PRINCIPLE 79

5.3.7. Proposition. Let U ⊂ R be a multilicative subset and P ∩ U = ∅ a primeideal. Then PU−1R is a prime ideal in U−1R and canonically for any R-moduleM

MP = (U−1M)PU−1R

Proof. This follows from 4.2.10 extending the set of denominators.

5.3.8. Corollary. Let Q ⊂ P ⊂ R be prime ideals and M an R-module ThenQRP is a prime ideal in RP and canonically

MQ = (MP )QRP

5.3.9. Definition. Let R be a ring. F is a locally free module is FP is a freeRP -module for all prime ideals P .

5.3.10. Proposition. Let R be a ring. F is a locally free module if FQ is a freeRQ-module for all maximal ideals Q.

Proof. A prime ideal P ⊂ Q is contained in a maximal ideal. By 5.3.8 FP '(FQ)PQ

is free.

5.3.11. Example. A free module is a locally free module.

5.3.12. Exercise. (1) Let P ⊂ R be a prime ideal and R → S a ring homomorphism.Show that RP → SP is a ring homomorphism.

(2) Let Q ⊂ S be a prime ideal and R→ S a ring homomorphism. Show that RQ∩R →SQ is a local ring homomorphism.

(3) Let R = K × L be a product of fields. Show that ideal K × {0} is locally free butnot free.

5.4. The local-global principle

5.4.1. Theorem. Let R be a ring and M a module. The following conditions areequivalent:(1) M = 0.(2) MP = 0 for all prime ideals P .(3) MP = 0 for all maximal ideals P .

Proof. (1) ⇒ (2) ⇒ (3) is clear. (3) ⇒ (1): Let 0 6= x ∈ M be given. ThenAnn(x) ⊂ P is contained in a maximal ideal, 5.1.2. Clearly 0 6= x

1 ∈ MP

contradicts (3).

5.4.2. Corollary. Let R be a ring and f : M → N a homomorphism. The follow-ing conditions are equivalent:(1) f is injective.(2) fP is injective for all prime ideals P .(3) fP is injective for all maximal ideals P .

Proof. Use 5.4.1 on Ker f .

5.4.3. Corollary. Let R be a ring and f : M → N a homomorphism. The follow-ing conditions are equivalent:(1) f is surjective.(2) fP is surjective for all prime ideals P .(3) fP is surjective for all maximal ideals P .

80 5. LOCALIZATION

Proof. Use 5.4.1 on Cok f .

5.4.4. Corollary. Let R be a ring and f : M → N a homomorphism. The follow-ing conditions are equivalent:(1) f is an isomorphism.(2) fP is an isomorphism for all prime ideals P .(3) fP is an isomorphism for all maximal ideals P .

5.4.5. Corollary. Let R be a ring and

Mf // N

g // L

a sequence of homomorphisms. The following conditions are equivalent:(1) The sequence is exact.(2) The sequence

MPfP // NP

gP // LP

is exact for all prime ideals P .(3) The sequence

MPfP // NP

gP // LP

is exact for all maximal ideals P .


0 // Mf // N

g // L // 0

a sequence of homomorphisms. The following conditions are equivalent:(1) The sequence is short exact.(2) The sequence

0 // MPfP // NP

gP // LP// 0

is short exact for all prime ideals P .(3) The sequence

0 // MPfP // NP

gP // LP// 0

is short exact for all maximal ideals P .

5.4.7. Corollary. Let R be a ring and F a module. The following conditions areequivalent:

(1) F is flat.(2) FP is flat for all prime ideals P .(3) FP is flat for all maximal ideals P .

Proof. Let 0→M → N . Use 5.3.6 and 5.4.2 on M ⊗R F → N ⊗R F .

5.4.8. Proposition. Let R be a ring and M a module. Then there is an exactsequence

0→M →∏

P maximal

MP

Proof. Let 0 6= x ∈ M be given. Then Ann(x) ⊂ P is contained in a maximalideal, 5.1.2. Clearly 0 6= x

1 ∈MP .

5.5. FLAT RING HOMOMORPHISMS 81

5.4.9. Corollary. Let R be a ring. Then there is an injective ring homomorphism

R→∏

P maximal

RP

5.4.10. Corollary. Let R be a ring. The following conditions are equivalent:(1) R is reduced.(2) RP is reduced for all prime ideals P .(3) RP is reduced for all maximal ideals P .

Proof. Use 5.1.8 and 5.4.9.

5.4.11. Exercise. (1) Let R be a ring and

0 // Mf // N

g // L // 0

a split exact sequence. Show that the localized sequence is split exact for all primeideals P .

5.5. Flat ring homomorphisms

5.5.1. Definition. A ring homomorphism R → S is a flat ring homomorphism ifS is a flat R-module, 3.7.1.

5.5.2. Proposition. Let R → S be a ring homomorphism. The following condi-tions are equivalent:

(1) R→ S is a flat ring homomorphism.(2) If M → N is injective, then the change of rings homomorphism M ⊗R S →

N ⊗R S is injective.(3) The change of rings functor −⊗R S is an exact functor.(4) Given an exact sequence of R-modules

M // N // L

Then the sequence of change of rings modules is exact

M ⊗R S // N ⊗R S // L⊗R S


0 // Mf // N

g // L // 0


0 // M ⊗R S // N ⊗R S // L⊗R S // 0

Proof. This follows from the definition 3.7.1 and 3.7.2.

5.5.3. Corollary. (1) Let U ⊂ R be a multiplicative subset. Then the canonicalring homomorphism R→ U−1R is flat.

(2) Let P ⊂ R be a prime ideal. Then the canonical ring homomorphism R →RP is flat.

5.5.4. Corollary. Let R→ S be a flat ring homomorphism. For a homomorphismf : M → N of R-modules there are natural isomorphisms of S-modules.

(1) Ker f ⊗R S ' Ker f ⊗ 1.(2) Im f ⊗R S ' Im f ⊗ 1.(3) Cok f ⊗R S ' Cok f ⊗ 1.

82 5. LOCALIZATION

Proof. See 3.3.3. Represent the statements using short exact sequences. (1) Thekernel is determined by the exact sequence 0→ Ker f →M → N , 3.1.4. (3) Thecokernel is determined by the exact sequence M → N → Cok f → 0, 3.1.5. (2)The image is determined by the exact sequence 0 → Ker f → M → Im f → 0,2.3.5, 3.1.4.

5.5.5. Corollary. Let R → S be a flat ring homomorphism. For submodulesN,L ⊂M there are natural identifications of S-submodules and factor modules.

(1) (M/N)⊗R S = M ⊗R S/N ⊗R S.(2) (N + L)⊗R S = N ⊗R S + L⊗R S.(3) (N ∩ L)⊗R S = N ⊗R S ∩ L⊗R S.

Proof. See 3.3.4. Represent the statements using short exact sequences. (1) 0 →N → M → M/N → 0 is short exact giving 0 → N ⊗R S → M ⊗R S →(M/N)⊗R S → 0. The wanted isomorphism follows form 3.1.5. (2) N +L is theimage of N ⊕ L→M . (3) N ∩ L is the kernel of N ⊕ L→M .

5.5.6. Corollary. Let R → S be a flat ring homomorphism. For ideals I, J ⊂ Rthere are natural identifications in S.

(1) I ⊗R S = IS.(2) (I ∩ J)S = IS ∩ JS.

Proof. (1) The sequence 0 → I ⊗R S → R ⊗R S → (R/I)⊗R S → 0. is exact.(2) This follows from (1) and 5.5.4.

5.5.7. Proposition. Let φ : R→ S be a flat ring homomorphism.(1) If F is a flat S-module, then the restriction of scalars makes F a flat R-

module.(2) If S → T is a flat ring homomorphism, then the composite R → T is a flat

ring homomorphism.

Proof. (1) There is a natural isomorphismM⊗RF ' (M⊗RS)⊗SF and clearlythe composition of two exact functors is exact. (2) This follows from (1).

5.5.8. Proposition. Let φ : R → S be a ring homomorphism. The followingconditions are equivalent:

(1) φ is flat.(2) φP : RP → SP is flat for all prime ideals P ⊂ R.(3) φP : RP → SP is flat for all maximal ideals P ⊂ R.

Proof. Use 5.4.7.


(1) R→ S is flat.(2) RP → SQ is flat for all prime ideals Q ⊂ S and P = Q ∩R.(3) RP → SQ is flat for all maximal ideals Q ⊂ S and P = Q ∩R.

Proof. RP → SQ is the composite RP → SP → SP QSP= SQ, 5.3.8. (1) ⇒

(2): The composite RP → SP → SQ is flat by 5.5.8 and 5.5.3. (3) ⇒ (1): If0 → M → N is exact, then 0 → M ⊗R RP → N ⊗R RP is exact, 5.5.3. By thehypothesis 0→M ⊗RRP ⊗RP

SQ → N ⊗RRP ⊗RPSQ is exact, so cancelation

and canonical isomorphism, 2.7.5 and 5.3.2, give 0→ (M ⊗RS)Q → (N ⊗RS)Q

5.6. FAITHFULLY FLAT RING HOMOMORPHISMS 83

exact for all maximal ideals Q. By 5.4.2 0 → M ⊗R S → N ⊗R S exact andR→ S is flat.

5.5.10. Exercise. (1) Let R → S and S → T be flat homomorphisms. Show that thecomposite R→ S is flat.

(2) Show that Q is a flat but not faithfully flat Z-module.(3) Let R be a ring and I =

√0 the nilradical. Show that IR[X] is the nilradical of

R[X].

5.6. Faithfully flat ring homomorphisms

5.6.1. Definition. A flat R-module F is faithfully flat if for any R-module M 6= 0the tensor product M ⊗R F 6= 0.

5.6.2. Proposition. (1) A nonzero free module is faithfully flat.(2) Let F, F ′ be faithfully flat modules, then F ⊗R F

′ is faihfully flat.(3) Let R → S be a ring homomorphism and F a faithfully flat R-module. The

change of ring module F ⊗R S is a faithfully flat S-module.

Proof. (1) This is clear from properties of tensor product. (2) F ⊗R F ′ is flat by3.7.7 and M 6= 0 gives M ⊗R F 6= 0 so M ⊗R F ⊗RF ′ 6= 0. (3) F ⊗R S is flatby 3.5.8 and if N 6= 0 is an S-module (3) F ⊗R S is flat by 3.5.8 and if N 6= 0 isan S-module then N ⊗S (F ⊗R S) ' N ⊗R F 6= 0 by 2.7.5.


(1) F is faithfully flat.(2) A homomorphism M → N is injective if and only if M ⊗R F → N ⊗R F is

injective.(3) A sequence of R-modules

Mf // N

g // L

is exact if and only if the sequence

M ⊗R F // N ⊗R F // L⊗R F

is exact.(4) A sequence

0 // Mf // N

g // L // 0is short exact if and only if the sequence

0 // M ⊗R F // N ⊗R F // L⊗R F // 0

is short exact.

Proof. (1)⇔ (2): f : M → N is injective if and only if Kerf = 0 if and only ifKer f ⊗ 1F = (Ker f)⊗R F = 0. (1)⇔ (3)⇔ (4): M → N → L is exact if andonly if (Im f + ker g)/Ker g = 0 = (Im f + ker g)/ Im f .

5.6.4. Proposition. Let F be a flat R-module. The following conditions are equiv-alent:(1) F is faithfully flat.(2) PF 6= F for all maximal ideals P ⊂ R.(3) F ⊗ k(P ) 6= 0 for all maximal ideals P ⊂ R.

84 5. LOCALIZATION

Proof. (1)⇒ (2)⇔ (3): k(P ) ⊗R R ' F/PF 6= 0. (2)⇒ (1): For 0 6= x ∈ Mlet Ann(x) = I ⊂ P be a maximal ideal. Then R/I ⊗R F → R/P ⊗R F → 0and 0→ R/I ⊗R F →M ⊗R F → 0 are exact, so M ⊗R F 6= 0.

5.6.5. Proposition. Let R be a ring and F a module. The following conditions areequivalent:(1) F is faithfully flat.(2) FP is faithfully flat for all prime ideals P .(3) FP is faithfully flat for all maximal ideals P .

Proof. (1) ⇒ (2) ⇒ (3): 5.6.2. (3) ⇒ (1): F is flat by 5.4.7. If M 6= 0, thenMP 6= 0 for some P . Therefore (M ⊗R F )P 'MP ⊗RP

FP 6= 0.

5.6.6. Definition. A ring homomorphism R → S is a faithfully flat ring homo-morphism if S is a faithfully flat R-module.


(1) R→ S is faithfully flat.(2) A homomorphism M → N is injective if and only if the change of rings

homomorphism M ⊗R F → N ⊗R S is injective.(3) A sequence of R-modules

Mf // N

g // L

is exact if and only if the sequence

M ⊗R S // N ⊗R S // L⊗R S

is exact.(4) A sequence

0 // Mf // N

g // L // 0is short exact if and only if the sequence

0 // M ⊗R S // N ⊗R S // L⊗R S // 0

is short exact.

Proof. See 5.6.3.

5.6.8. Proposition. Let R → S be a faithfully flat ring homomorphism and F anR-module.

(1) If F ⊗R S is a flat S-module, then F is a flat R-module.(2) If F ⊗R S is a faithfully flat S-module, then F is a faithfully flat R-module.

Proof. (1) If 0 → M → N is exact, then 0 → M ⊗R S ⊗S F ⊗R S → N ⊗R

S ⊗S F ⊗R S is exact. This is natural isomorphic to 0 → M ⊗R F ⊗R S →N ⊗R F ⊗R S, so by 5.6.3 0→M ⊗R F → N ⊗R F is exact. (2) If M 6= 0 thenM ⊗R S ⊗S F ⊗R S 'M ⊗R F ⊗R S 6= 0 and therefore M ⊗R F 6= 0.

5.6.9. Proposition. Let R→ S be a faithfully flat ring homomorphism.(1) If F is a faithfully flat S-module, then the restriction of scalars makes F a

faithfully flat R-module.(2) If S → T is a faithfully flat ring homomorphism, then the composite R → T

is a faithfully flat ring homomorphism.

5.6. FAITHFULLY FLAT RING HOMOMORPHISMS 85

Proof. (1) F is a flat R-module by 5.5.7. If M 6= 0 is an R-module, then M ⊗R

F 'M ⊗R S ⊗S F 6= 0. (2) This is a special case of (1).

5.6.10. Proposition. Let R→ S be a ring homomorphism.(1) If F is a faithfully flat S-module such that the restriction of scalars makes F

a flat R-module. Then R→ S is a flat ring homomorphism.(2) If F is a faithfully flat S-module such that the restriction of scalars makes F a

faithfully flat R-module. Then R→ S is a faithfully flat ring homomorphism.(3) If S → T is a faithfully flat ring homomorphism and the composite R → T

is a faithfully flat ring homomorphism. Then R → S is a faithfully flat ringhomomorphism.

(4) If S → T is a faithfully flat ring homomorphism and the composite R → Tis a flat ring homomorphism. Then R→ S is a flat ring homomorphism.

Proof. LetM → N be anR-homomorphism. ThenM⊗RS⊗SF → N⊗RS⊗SFis natural isomorphic to M ⊗R F → N ⊗R F . (1) Since F is flat as R-moduleand faithfully flat as S-module, it follows by 5.6.3 that if M → N is injectivethen M ⊗R S → N ⊗R S is injective. (2) Since F is faithfully flat both as R-module and S-module, it follows by 5.6.3 that M → N is injective if and only ifM ⊗R S → N ⊗R S is injective. (3),(4) These are special cases of (1),(2).

5.6.11. Proposition. Let R → S be a faithfully flat ring homomorphism. for anyideal I ⊂ R the extended contracted returns I , i.e.

I = IS ∩R

Proof. Tensor the homomorphismR/I → R/IS∩Rwith S. The inducedR/I⊗R

S → R/IS ∩ R ⊗R S is canonically isomorphic to the identity S/IS → S/IS.By 5.6.3 R/I → R/IS ∩R is injective giving I = IS ∩R.

5.6.12. Corollary. A faithfully flat ring homomorphism R→ S is injective.

5.6.13. Proposition. A flat local homomorphism (R,P ) → (S,Q) is faithfullyflat.

Proof. PS ⊂ Q 6= S so S is faithfully flat by 5.6.4.

5.6.14. Theorem (going-down). Let φ : R→ S be a flat ring homomorphism andQ ⊂ S a prime ideal. For any prime ideal P ′ ⊂ P = Q∩R there is a prime idealQ′ ⊂ Q contracting to P ′ = Q′ ∩R.

S

R

nnnnnnn

Q

P

oooooo

Q′

P ′

Proof. The local homomorphism RP → SQ is faithfully flat. The ring k(P ′) ⊗R

SQ is nonzero and therefor contains a maximal ideal Q′′. The contraction to S,Q′ = Q′′ ∩ S contracts to P ′ = R ∩Q′.

86 5. LOCALIZATION

5.6.15. Proposition. Let R → S be a flat homomorphism. The following condi-tions are equivalent:(1) R→ S is faithfully flat.(2) Any prime ideal P ⊂ R is the contraction P = Q ∩ R of a prime ideal

Q ⊂ S.S

R

oooooo

Q

P

Proof. (1)⇒ (2): The ring SP /PSP ' k(P ) ⊗R S 6= 0, so let Q′ be a maximalideal. The contraction Q = Q′ ∩ S is a prime ideal such that P = Q ∩ R. (2)⇒(1): Let M 6= 0 be an R-module. Then MP 6= 0 for some P . Let P = Q∩R, then(M ⊗R S)Q 'MP ⊗RP

SQ 6= 0 as RP → SQ is faithfully flat.

5.6.16. Proposition. The inclusion R → R[X1, . . . , Xn] is a faithfully flat homo-morphism

Proof. The R-module R[X1, . . . , Xn] is free.

5.6.17. Exercise. (1) Show that a free module is faithfully flat.(2) Let F be a faithfully flat module and G a flat module. Show that F ⊕G is faithfully

flat.(3) Show that Q is a flat but not faithfully flat Z-module.(4) Let R be a ring and I =

√0 the nilradical. Show that IR[X] is the nilradical of

R[X].(5) Show that

⊕RP over all maximal ideals P is a faithfully flat R-module.

6

Finite modules

6.1. Finite modules

6.1.1. Definition. Let R be a ring. A finite module is generated by finitely manyelements. The finite free module with standard basis e1, . . . , en is denoted Rn.

6.1.2. Lemma. Let R be a ring and M a module. The following conditions areequivalent:(1) M is generated by n elements x1, . . . , xn.(2) There is a surjective homomorphism Rn →M → 0, ei 7→ xi.

Proof. See 2.4.12.

6.1.3. Proposition. Let R → S be a ring homomorphism. If an R-module M isgenerated by n elements x1, . . . , xn. Then the change of rings S-module M ⊗R Sis generated by x1 ⊗ 1, . . . , xn ⊗ 1 over S.

Proof. Follows from 6.1.2 and 3.4.1

6.1.4. Corollary. Let R be a ring and U a multiplicative subset. If M is a finiteR-module, then U−1M is a finite U−1R-module.

6.1.5. Proposition. For a short exact sequence

0 // Mf // N

g // L // 0

the following hold(1) If N is finite, then L is finite.(2) If M,L are finite, then N is finite.

Proof. (1) If y1, . . . , yn generatesN , then g(y1), . . . , g(yn) generatesL. (2) Chooseu : Rn → M → 0 and v : Rm → L → 0 exact. By 3.5.5 there is w : Rm → Nsuch that g ◦ w = v. There is a diagram

0 // Rn //

u

��

Rn ⊕Rm //

f◦u+w

��

Rm //

v

��

0

0 // Mf // N

g // L // 0Conclusion by the snake lemma 3.2.4.

6.1.6. Corollary. Let

0 // Mf // N

g // L // 0

be a split exact sequence. Then N is finite if and only if M,L are finite.

Proof. Let u be a retraction of f . By 6.1.5 Imu = M is finite. The rest is containedin 6.1.5.

87

88 6. FINITE MODULES

6.1.7. Corollary. Let f : M → N be a homomorphism.(1) If M is finite, then Im f is finite.(2) If Ker f, Im f are finite, then M is finite.(3) If N is finite, then Cok f is finite.(4) If Im f,Cok f are finite, then N is finite.

Proof. Use 6.1.5 on the exact sequences 3.1.8.

6.1.8. Corollary. Let M,N be modules. Then M ⊕ N is finite if and only if Mand N are finite.

Proof. Use 6.1.6.

6.1.9. Proposition. Let R be a ring and M,N finite modules. Then M ⊗R N isfinite.

Proof. Use 6.1.2 and 6.1.5. Let Rm → M and Rn → N be surjective. ThenRm ⊗R R

n →M ⊗R N is surjective, 3.4.1.

6.1.10. Proposition. Let R be a ring and F a module. The following conditionsare equivalent:(1) F is finite and projective.(2) F is a direct summand in a finite free module.

Proof. Use 6.1.2 and 6.1.8.

6.1.11. Proposition. Let R be a ring and F a finite projective module.(1) If N is finite, then HomR(F,N) is finite.(2) If N is projective, then HomR(F,N) is projective.

Proof. Use 6.1.8.

6.1.12. Proposition. Let F be a finite projective module and E an injective mod-ule.

(1) F ⊗R E is injective.(2) HomR(F,E) is injective.

Proof. (1) (2) Both modules become summands in injective modules.

6.1.13. Proposition. Let R be a ring and U a multiplicative subset. For a finitemodule M the following hold:(1) U−1M = 0 if and only if there is a u ∈ U such that uM = 0.(2) Ann(U−1M) = U−1 Ann(M) = Ann(M)U−1R in the ring U−1R.

Proof. Let x1, . . . , xn generate M . (1) U−1M = 0 if and only if u1x1 = · · · =unxn = 0. Put u = u1 . . . un. (2) Ann(M) = Ann(x1)∩· · ·∩Ann(xn). The exactsequence 0 → Ann(x1) → R → Rx1 → 0 localizes to 0 → U−1 Ann(x1) →U−1R → U−1Rx1 → 0 giving that U−1 Ann(x1) = Ann(U−1Rx1). Concludeby 4.3.4.

6.1.14. Proposition. Let R be a ring and Mα a family of modules. For any finitemodule N there is a natural isomorphism

HomR(N,⊕

α

Mα) '⊕

α

HomR(N,Mα)

6.2. FREE MODULES 89

Proof. A homomorphism f : N →⊕Mα has an image in a finite sum.

6.1.15. Exercise. (1) Show that⊕

n Z/(n) is not a finite Z-module.(2) Let K be a field and R = K[X1, X2, . . . ] the polynomial ring in countable many

variables. Show that R is a finite module, but the ideal (X1, X2, . . . ) is not a finitemodule.

6.2. Free modules

6.2.1. Definition. Let R be a ring and let Rn be the free module with standardbasis e1, . . . , en.(1) Let A = (aij) be a m × n-matrix with m rows and n columns, where the

entry aij ∈ R. Identify Rn with n-columns. Then matrix multiplication

x = (xj) 7→ y = Ax, yi =∑

j

aijxj

defines a homomorphism Rn → Rm.(2) Let f : Rn → Rm be a homomorphism. Then define a m × n-matrix A =

(aij) by

f(ej) =∑

i

aijei

6.2.2. Proposition. (1) The dictionary defined in 6.2.1 gives a canonical isomor-phism between the module of m× n-matrices and HomR(Rn, Rm).

(2) Matrix multiplication corresponds to composition of homomorphisms and theidentity matrix corresponds to the identity homomorphism.

(3) Invertible matrices correspond to isomorphisms.

Proof. Do linear algebra homework.

6.2.3. Definition. Let R be a ring.(1) Let A = (aij) be a m × n-matrix. The (m − 1) × (n − 1) matrix derived

from A be deleting i-row and j-column is Aij .(2) For a square matrix A the determinant is defined by row expansion and in-

duction:det(a11) = a11

detA =∑

j

(−1)1+ja1j detA1j

(3) The determinant of a k × k-matrix derived from A by choosing entries fromk rows and columns is a k-minor of A.

(4) If A is a n× n-matrix, then the cofactor matrix A′ = (a′ij) has entries

a′ij = (−1)i+j detAji

given by (n− 1)-minors.

6.2.4. Proposition. (1) For a fixed a ∈ R, the scaled determinant adet is theunique map from n× n-matrices to R, which satisfies(a) multilinear: For n− 1 fixed columns a•1, . . . , a•j−1, a•j+1, . . . , a•n, the

map

Rn → R, x 7→ adet(a•1, . . . , a•j−1, x, a•j+1, . . . , a•n)


is a homomorphism:

adet(..bx+ cy..) = badet(..x..) + cadet(..y..)

(b) alternating: For two equal columns

adet(..x..x..) = 0

(c) normed:adet(1n) = a

(2) The determinant is calculated by any expansion

detA =∑

i

(−1)i+jaij detAij =∑

j

(−1)i+jaij detAij

(3) If A,B are n× n-matrices then the product rule holds

detAB = detAdetB

(4) Let A be an n×n-matrix with cofactor matrix A′. Then the matrix Cramer’srule holds

AA′ = A′A = detA(1n)Where (1n) is the n× n identity matrix.

(5) A square matrix A is invertible if and only if detA is a unit in R.

Proof. More linear algebra homework. (1) First prove that any map satisfying (a)-(c) must satisfy the rule for interchanging columns and the rule for a modificationof a column. Then continue proof by induction. (2) Use the uniqueness from (1).(3) To get the product rule detAB = detAdetB, you fix A. Then both sidessatisfy (1) with norm detA. (4) Writeout (2) and use alternating from (1).

6.2.5. Theorem. Let f : Rn → Rn be a homomorphism represented by an n× n-matrix A.

(1) f is a surjective if and only if detA is a unit.(2) f is injective if and only if detA is a nonzero divisor.

Proof. (1) If f is surjective, then a section is represented by a matrix B such thatAB = (1n). Then detA is a unit. Conversely by 6.2.4. (2) If detA is a nonzerodivisor, then by 6.2.4 A′A = detA(1n) gives an injective homomorphism, so f isinjective. If detA is a zero divisor, then there is number k < n such that Ann(1−minors) = · · · = Ann(k −minors) = 0 and 0 6= b ∈ Ann((k + 1)−minors).Assume that the k−minor from first k rows and columns ck+1 has ck+1b 6= 0. Letcj(−1)k+1+j be the k-minor from first k rows and first k + 1 columns excludingnumber j and put cj = 0, j > k + 1. Then A(cj) is a column with entries being(k + 1)−minors so f((bcj)) = A(bcj) = bA(cj) = 0 and f is not injective.

6.2.6. Proposition. Let R be a ring and f : Rn → Rm homomorphism.(1) If f is surjective, then n ≥ m.(2) If f is injective, then n ≤ m.(3) f is an isomorphism if and only if m = n and f is surjective.

Proof. (1) If n < m let p : Rm → Rn be the projection onto first n coordinates.Then f◦p is surjective and represented by anm×m-matrixAwithm-column zero.A section to f ◦p is represented by anm×m-matrixB such thatBA = (1m). Butthe product AB must have a zero m-column, so the contradiction gives n ≥ m.(2) If n > m let i : Rm → Rn be injection into first m coordinates. Then i ◦ f is

6.3. CAYLEY-HAMILTON’S THEOREM 91

injective and represented by an n× n-matrix A with n-row zero. Then detA = 0contradicting 6.2.5. So n ≤ m. (3) If f is an isomorphism then m = n by (1), (2).If f is surjective, then f is an isomorphism by 6.2.5 (1) and 6.2.4 (5).

6.2.7. Theorem. (1) A finite free module has a finite basis.(2) The number of elements in a basis for a finite free module is independent of

the basis.

Proof. Let F be finite free generated by n elements. If y1, . . . , ym is part of abasis, then by projection F → Rm there is a surjective homomorphismRn → Rm.Conclusion by 6.2.6.

6.2.8. Definition. The number of elements 6.2.7 in a basis for a finite free moduleF is the rank, rankR F .

6.2.9. Proposition. If x1, . . . , xn generates a free module F of rank n, then theyconstitutes a basis.

Proof. Choose a basis and an isomorphism f : Rn → F . The homomorphismg : Rn → F, ei 7→ xi is surjective. The composite f−1 ◦ g : Rn → Rn issurjective and therefore by 6.2.6 an isomorphism. Then g is an isomorphism andxi a basis.

6.2.10. Proposition. Let F, F ′ be finite free modules. Then(1) F ⊕ F ′ is free and rankR F ⊕ F ′ = rankR F + rankR F

′.(2) F ⊗R F

′ is free and rankR F ⊗R F′ = rankR F · rankR F

′.(3) HomR(F, F ′) is free and rankR HomR(F, F ′) = rankR F · rankR F

′.

6.2.11. Exercise. (1) Let Rn → Rn be a surjective homomorphism. Show that it is anisomorphism.

6.3. Cayley-Hamilton’s theorem

6.3.1. Remark. Let R be a ring and f : M → M a homomorphism. By 2.1.13view M as an R[X]-module, where Xx = f(x) for x ∈ M . The homomorphism1.6.7, 2.6.9, R[X] → HomR(M,M), a 7→ aM , X 7→ f is a ring homomorphism.The image is R[f ] the smallest subring containing 1M , f . M is naturally a R[f ]-module and the R[X]-module above is the restriction of scalars.

6.3.2. Proposition. Let A be an n × n-matrix and I the ideal generated by theentries aij . The polynomial

det(X1n −A) = a0 + a1X + . . . an−1Xn−1 +Xn

has a0, . . . , an−1 ∈ I and gives the relation

a0(1n) + a1A+ . . . an−1An−1 +An = 0

as n× n-matrix.

Proof. View Rn as a module over the ring R[X], 6.3.1, with scalar multiplication

Xx = Ax, x ∈ Rn

Let p : R[X]n → Rn be the R[X]-homomorphism determined by p(ej) = ejand denote U = X(1n) − A the n × n-matrix with entries from R[X]. Thenp(Uej) = p(Xej − Aej) = 0 in Rn, so p(Uy) = 0 for all y ∈ R[X]n. Let Uhave cofactor matrix U ′ over R[X]. Then p(UU ′ej) = 0. By Cramer’s rule 6.2.4,


UU ′ej = detU ej in R[X]n. Therefore p(detU ej) = detU p(ej) = detU ej =0 in Rn. By calculation

detU = a0 + a1X + · · ·+ an−1Xn−1 +Xn

in R[X], with ai ∈ I and for all x ∈ Rn

detU x = (a0(1n) + a1A+ . . . an−1An−1 +An)x = 0

6.3.3. Theorem. Let I ⊂ R be an ideal and f : M → M a homomorphism ona finite module generated by n elements. Suppose Im f ⊂ IM , then there exista0, . . . , an−1 ∈ I such that

a01M + a1f + . . . an−1fn−1 + fn = 0

in HomR(M,M).

Proof. Let x1, . . . , xn generate M and write

f(xj) =∑

i

aijxi

for an n× n-matrix A with entries aij ∈ I . View Rn, x 7→ Ax and M,x 7→ f(x)as modules over the ring R[X], 6.3.1. Then the R-homomorphism p : Rn →M, ej 7→ xj is an R[X]-homomorphism. By 6.3.2

det(X1n −A) = a0 + a1X + . . . an−1Xn−1 +Xn

gives 0 = p(det(X(1n)−A) ej) = det(X(1n)−A)xj . It follows that

(a01M + a1f + . . . an−1fn−1 + fn)xj = 0

for all j.

6.3.4. Proposition. Let I ⊂ R be an ideal and M a finite module. If IM = Mthen I + Ann(M) = R. That is there is a ∈ I such that (1 + a)M = 0.

Proof. By 6.3.3

a01M + · · ·+ 1M = (a0 + · · ·+ an−1)1M + 1M = 0

Put a = a0 + · · ·+ an−1.

6.3.5. Corollary. Let I ⊂ R be an ideal and M a finite module. If IM = M andall elements 1 + I are nonzero divisors on M , then M = 0.

6.3.6. Corollary. Let I ⊂ R be an ideal andN ⊂M a submodule. SupposeM/Nis a finite module and M = N + IM . Then I + (N : M) = R.

6.3.7. Proposition. Let R be a ring and M a finite module. If a homomorphismf : M →M is surjective, then it is an isomorphism.

Proof. Regard M,f as a module over R[X] 6.3.1. Then (X)M = M , so by 6.3.4there is p ∈ R[X] such that 1 + pX ∈ Ann(M). For any u ∈ Ker f , calculateu = u+ p(f) ◦ f(u) = (1 + pX)u = 0. So f is an isomorphism.

6.3.8. Exercise. (1) Let√

0M = M . Show that M = 0.

6.4. NAKAYAMA’S LEMMA 93

6.4. Nakayama’s lemma

6.4.1. Theorem (Nakayama). Let (R,P ) be a local ring and M a finite module.The following conditions are equivalent:

(1) M = 0.(2) PM = M .(3) M ⊗R k(P ) = 0.

Proof. (1)⇒ (2)⇔ (3) is clear. (2)⇒ (1): Elements in 1 + P are units in R, soby 6.3.5 M = 0.

6.4.2. Corollary. Let (R,P ) be a local ring and N ⊂ M a submodule. SupposeM/N is a finite module and M = N + PM . Then N = M .

6.4.3. Corollary. Let (R,P ) be a local ring and M,N finite modules. If M ⊗R

N = 0, then M = 0 or N = 0.

Proof. If M,N 6= 0 then M ⊗R k(P ), N ⊗R k(P ) 6= 0 are vector spaces overk(P ). Now M ⊗R N ⊗R k(P ) 'M ⊗R k(P )⊗k(P ), N ⊗R k(P ) 6= 0, giving thestatement.

6.4.4. Corollary. LetR be a ring andM a finite module. The following conditionsare equivalent:

(1) M = 0.(2) PMP = MP for all prime ideals P .(3) PMP = MP for all maximal ideals P .(4) M ⊗R k(P ) = 0 for all prime ideals P .(5) M ⊗R k(P ) = 0 for all maximal ideals P .

Proof. Combine 6.4.1 with 5.4.1.

6.4.5. Corollary. Let (R,P ) be a local ring and f : M → N a homomorphism.Assume N is finite. The following conditions are equivalent:(1) f is surjective.(2) f(P ) is surjective.

Proof. f is surjective if and only if Cok f = 0. Cok f is finite, so it is zero if andonly if Cok f ⊗R k(P ) = Cok(f(P )) = 0, 6.4.1.

6.4.6. Corollary. Let R be a ring and f : M → N a homomorphism. Assume Nis finite. The following conditions are equivalent:(1) f is surjective.(2) f(P ) is surjective for all prime ideals P .(3) f(P ) is surjective for all maximal ideals P .

6.4.7. Corollary. Let (R,P ) be a local ring andM a finite module. Let x1, . . . , xn ∈M . The following conditions are equivalent:

(1) x1, . . . , xn generate M .(2) x1 ⊗ 1, . . . , xn ⊗ 1 generate M ⊗R k(P ).

The minimal number of generators of M is rankk(P )M ⊗R k(P ).

6.4.8. Corollary. Let R be a ring and M a finite module. Let x1, . . . , xn ∈ M .The following conditions are equivalent:

(1) x1, . . . , xn generate M .


(2) x1 ⊗ 1, . . . , xn ⊗ 1 generate M ⊗R k(P ) for all prime ideals P .(3) x1 ⊗ 1, . . . , xn ⊗ 1 generate M ⊗R k(P ) for all maximal ideals P .

6.4.9. Proposition. Let f : Rn → Rm be a homomorphism represented by anm× n-matrix A. The following statements are equivalent:(1) f is surjective.(2) n ≥ m and the ideal (m−minors) = R.

Proof. (1)⇒ (2): If f is surjective, then for any maximal ideal f(P ) is surjectivelinear map. So n ≥ m and some m-minor is nonzero in k(P ). Therefore (m −minors) is not contained in P , so (m − minors) = R. (2) ⇒ (1): f(P ) issurjective for all maximal ideals, so f surjective by 6.4.6.

6.4.10. Proposition. Let f : Rn → Rm be a homomorphism represented by anm× n-matrix A. The following statements are equivalent:(1) f is injective.(2) n ≤ m and the ideal Ann(n−minors) = 0.

Proof. (1)⇒ (2): See the proof 6.2.5 (2) and note that the construction only usesn ≤ m. (2) ⇒ (1): Assume f(x) = Ax = 0 and choose n rows in A to give an × n-matrix B. Then Bx = 0 and by Cramer’s rule detB x = 0. Since detBrun through all n-minors it follows that the coordinates of x belong to Ann(n −minors) = 0.

6.4.11. Exercise. (1) Let R be a domain and f : Rn → Rm a homomorphism repre-sented by an m× n-matrix. Show that f is injective if and only if n ≤ m and somen−minor 6= 0.

6.5. Finite presented modules

6.5.1. Definition. Let R be a ring. A finite presented module is a module Mhaving an exact sequence

Rn → Rm →M → 0

or equivalently there is a short exact sequence

0→ N → Rm →M → 0

with N finite.

6.5.2. Proposition. (1) Let R → S be a ring homomorphism and M a finitepresented R-module Then the change of rings S-module M ⊗R S is finitepresented.

(2) Let U ⊂ R be a multiplicative subset and M a finite presented R-module,then U−1M is a finite presented U−1R-module.

6.5.3. Proposition. For a short exact sequence

0 // Mf // N

g // L // 0

the following hold:(1) If M,L are finite presented, then N is finite presented.(2) If L is finite presented and N is finite , then M is finite.(3) If N is finite presented and M is finite , then L is finite presented.

6.5. FINITE PRESENTED MODULES 95

Proof. (1) Choose u : Rn → M → 0 and v : Rm → L → 0 exact with finitekernels. By 3.5.5 there is w : Rm → N such that g ◦ w = v. There is a diagram

0 // Rn //

u

��

Rn ⊕Rm //

f◦u+w

��

Rm //

v

��

0

0 // Mf // N

g // L // 0

By the snake lemma 3.2.4 the sequence 0→ Keru→ Ker f ◦u+w → Ker v → 0is exact. By 6.1.5 Ker f ◦ u+w is finite. (2) Choose v : Rm → L→ 0 exact withfinite kernel and w : Rm → N such that g ◦ w = v. There is a diagram

0 // 0 //

��

Rm

w

��

Rm //

v

��

0

0 // Mf // N

g // L // 0

By the snake lemma 3.2.4 the sequence 0 → Kerw → Ker v → M → Cokw →0 is exact. By 6.1.5 M is finite. (3) Choose w : Rm → N → 0 exact with finitekernel. Then v = g ◦ w : Rm → L→ 0 is exact and there is a diagram

0 // 0 //

��

Rm

w

��

Rm //

v

��

0

0 // Mf // N

g // L // 0

By the snake lemma 3.2.4 the sequence 0→ Kerw → Ker v → M → 0 is exact.By 6.1.5 Ker v is finite and L is finite presented.

6.5.4. Corollary. Let

0 // Mf // N

g // L // 0

be a split exact sequence. Then N is finite presented if and only if M,L are finitepresented.

Proof. By 3.1.13 there is a split exact sequence

0 // Lv // N

u // M // 0

so the statement follows from 6.5.3.

6.5.5. Corollary. Let f : M → N be a homomorphism.

(1) If M is finite and Im f finite presented, then Ker f is finite.(2) If Ker f, Im f are finite presented, then M is finite presented.(3) If N is finite presented and Im f finite, then Cok f is finite presented.(4) If Im f,Cok f are finite presented, then N is finite presented.

Proof. Use the sequences 3.1.8.

6.5.6. Proposition. Let R be a ring and M,N finite presented modules.

(1) M ⊕N is finite presented.(2) M ⊗R N is finite presented.


Proof. (1) This is clear from 6.5.4. (2) IfM = Rn thenM⊗RN is finite presentedby (1). In general chose u : Rn → M → 0 exact with finite kernel. The sequenceKeru ⊗R N → Rn ⊗R N → M ⊗R N → 0 is exact. So Keru ⊗ 1N is finite.Conclusion by 6.5.5.

6.5.7. Proposition. Given submodules N,L ⊂M . Then(1) If M/N,M/L are finite and M/N + L is finite presented, then M/N ∩ L is

finite.(2) If M/N,M/L are finite presented and M/N ∩ L is finite, then M/N + L is

finite presented.

Proof. Use the sequence 3.2.7

0 // M/N ∩ L // M/N ⊕M/L // M/N + L // 0

together with 6.5.3.

6.5.8. Theorem. Let M be a finite presented module and Nα a family of modules.Then there is a natural isomorphism

M ⊗R

∏α

Nα →∏α

(M ⊗R Nα)

Proof. If M = R this is an isomorphism. Then this is also an isomorphism forM = Rn since both functors respect finite direct sums. In general choose Rn →Rm →M → 0 exact. There is a diagram

Rn ⊗R∏Nα

//

��

Rm ⊗R∏Nα

//

��

M ⊗R∏Nα

��

// 0

∏Rn ⊗R Nα

//∏Rm ⊗R Nα

//∏M ⊗R Nα

// 0

giving isomorphism by the five lemma 3.2.8.

6.5.9. Theorem. Let R→ S be a flat ring homomorphism.(1) For a finite module M and any module N the natural homomorphism

HomR(M,N)⊗R S → HomS(M ⊗R S,N ⊗R S)

is injective.(2) For a finite presented module M and any module N the homomorphism

HomR(M,N)⊗R S → HomS(M ⊗R S,N ⊗R S)

is a natural isomorphism.

Proof. (1) If M = R this is an isomorphism. Then this is also an isomorphismfor M = Rn since both functors respect finite direct sums. In general choose0→ K → Rn →M → 0 exact. There is a diagram

0 // HomR(M, N)⊗R S //

��

HomR(Rn, N)⊗R S //

��

HomR(K, N)⊗R S

��0 // HomS(M ⊗R S, N ⊗R S) // HomS(Rn ⊗R S, N ⊗R S) // HomS(K ⊗R S, N ⊗R S)

giving injectivity. (2) In (1) K is finite, so the last vertical map is injective. Con-clusion by the five lemma 3.2.8.

6.5.10. Corollary. Let R be a ring and U a multiplicative subset.

6.5. FINITE PRESENTED MODULES 97

(1) For a finite module M and any module N the natural homomorphism


is injective.(2) For a finite presented module M and any module N the homomorphism


is a natural isomorphism.

Proof. Use that R→ U−1R is flat.

6.5.11. Corollary. Let P ⊂ R be a prime ideal and M a finite presented module.For any module N there is a natural isomorphism

HomR(M,N)P ' HomRP(MP , NP )


Mf // N

g // L

a sequence of homomorphisms with L a finite presented module. The followingconditions are equivalent:(1) The sequence is split exact.(2) The sequence

MPfP // NP

gP // LP

is split exact for all prime ideals P .(3) The sequence

MPfP // NP

gP // LP

is split exact for all maximal ideals P .

Proof. The statement for exactness is true in general 5.4.5. The sequence splits ifHomR(L,N)→ homR(N,N) is surjective. By 5.4.3 this holds if HomR(L,N)P →homR(N,N)P is surjective for all maximal ideals. Now L is finite presented, sothe localized is HomRP

(LP , NP )→ HomRP(NP , NP ) giving the conclusion.

6.5.13. Theorem. Let R be a ring and F a module. The following conditions areequivalent:

(1) F is flat.(2) For any module N and a relation 0 =

∑i yi ⊗ xi ∈ N ⊗R F , there exist

zj ∈ F and aij ∈ R such that 0 =∑

i aijyi ∈ N and xi =∑

j aijzj ∈ F .(3) For any relation 0 =

∑i bixi ∈ F , there exist zj ∈ F and aij ∈ R such that

0 =∑

i aijbi ∈ R and xi =∑

j aijzj ∈ F .

Proof. (1)⇒ (2): Let f : Rn → N, ei 7→ yj , then 0 → Ker f ⊗R F → Fn →N ⊗R F is exact. By assumption (xi) ∈ Ker f ⊗R F , so (xi) =

∑j aij ⊗ zj

with (aij) ∈ Ker f . (2) ⇒ (3) is clear. (3) ⇒ (1): Let I ⊂ R be an ideal and∑bi ⊗ xi ∈ Ker(I ⊗R F → F ). Then 0 =

∑i aijbi and xi =

∑j aijzj . Now

calculate∑bi ⊗ xi =

∑j

∑i aijbi ⊗ xi = 0. By 3.7.12 F is flat.

6.5.14. Theorem. Let (R,P ) be a local ring and F a finite presented module. Thefollowing conditions are equivalent:(1) F is free.


(2) F is projective.(3) F is flat.(4) P ⊗R F → F is injective.

Proof. (1) ⇒ (2) ⇒ (3) ⇒ (4) are clear. (4) ⇒ (1): Choose xi ∈ F such thatxi ⊗ 1 give a basis for F ⊗R k(P ). The homomorphism f : Rn → F, ei 7→ xi

is surjective by 6.4.5 and 0 → K → Rn → F → 0 is exact with K finite, 6.5.5.There is a diagram

P ⊗R K //

��

P ⊗R Rn //

��

P ⊗R F //

��

0

0 // K //

��

Rn //

��

F //

��

0

k(P )⊗R K // k(P )⊗R Rn ∼ // k(P )⊗R F // 0

By the snake lemma 3.2.4 k(P )⊗R K = 0 and therefore K = 0 by 6.4.1, so F isfree.

6.5.15. Corollary. Let (R,P ) be a local ring and f : F → F ′ a homomorphismof finite free modules. The following conditions are equivalent:

(1) f has a retraction u : F ′ → F .(2) f is injective and Cok f is free.(3) f(P ) is injective.

Proof. (1)⇒ (2): Cok f is projective, so free by 6.5.14. (2)⇒ (3) is clear. (3)⇒(1): Let f∨ : F ′∨ → F∨ be the dual homomorphism. f∨(P ) is surjective, so f∨ issurjective, 6.4.5. A section v of f∨ gives a retraction u = v∨.

6.5.16. Corollary. Let R be a ring and F a finite presented module. The followingconditions are equivalent:

(1) F is projective.(2) F is flat.(3) P ⊗R F → F is injective for all maximal ideals P .(4) FP is free for all maximal ideals P .

Proof. (1) ⇒ (2) ⇒ (3) are clear by 6.5.14. (3) ⇒ (4): PRP ⊗RPFP → FP is

(P ⊗R F )P → FP and therefore injective. Conclusion by 6.5.14. (4)⇒ (1): LetN → L → 0 be exact. By hypothesis HomRP

(FP , NP ) → HomRP(FP , LP ) →

0 is exact for all maximal ideals. By 6.5.11 HomR(F,N)P → HomR(F,L)P → 0is exact for all maximal ideals. By 5.4.3 HomR(F,N) → HomR(F,L) → 0 isexact and F is projective.

6.5.17. Exercise. (1) Show that a finite projective module is finite presented.(2) Let I ⊂ R be an ideal. Show that R/I is a finite presented R-module if and only if

I is a finite ideal.(3) Show that Q is a flat, but not projective Z-module.

6.6. FINITE RING HOMOMORPHISMS 99

6.6. Finite ring homomorphisms

6.6.1. Definition. A ring homomorphism φ : R → S is a finite ring homomor-phism if S is a finite R-module. If R ⊂ S is a subring, then a finite ring homomor-phism is a finite ring extension.

6.6.2. Proposition. Let R be a ring.(1) Let f ∈ R[X] be a monic polynomial. Then the homomorphism R →

R[X]/(f) is finite.(2) Let f : M → M be a homomorphism of a finite R-module. Then the homo-

morphism 6.3.1, R→ R[f ] is finite.

Proof. (2) Follows from (1) and 6.3.3.

6.6.3. Lemma. Let φ : R→ S be a finite ring homomorphism.(1) IfN is a finite S-module, then by restriction along φ theR-moduleN is finite.(2) Let U ⊂ R be a multiplicative subset, then U−1R → U−1S is a finite ring

homomorphism.

6.6.4. Proposition. Let R ⊂ S be a finite ring extension of domains. Then R is afield if and only if S is a field.

Proof. Let R be a field, a minimal equation 6.3.3 for scalar multiplication by anonzero b ∈ S, bS as R-module homomorphism

bn + · · ·+ a0 = 0

givesb−1 = −a−1

0 (an−1bn−2 + · · ·+ a1) ∈ S

Let S be a field and 0 6= a ∈ R. An equation 6.3.3 for scalar multiplication a−1S as

R-homomorphisma−n + · · ·+ a0 = 0

givesa−1 = −(a0a

n−1 + · · ·+ an−1) ∈ R

6.6.5. Corollary. LetR→ S be a finite ring homomorphism. A prime idealQ ⊂ Sis maximal if and only if the contraction Q ∩R is maximal.

Proof. R/Q ∩R ⊂ S/Q is a finite extension of domains.

6.6.6. Theorem. Let R ⊂ S be a finite ring extension and P ⊂ R a prime ideal.Then there is a prime ideal Q ⊂ S contracting P = Q ∩R.

S

R

oooooo

Q

P

Proof. RP ⊂ SP is a finite ring extension. Since SP 6= 0 there is a maximalideal in SP contracting to PRP by 6.6.5. The corresponding prime ideal Q ⊂ Scontracts to P .


6.6.7. Corollary (going-up). Let R → S be a finite ring homomorphism, Q ⊂ Sa prime ideal and P = Q ∩ R the contraction. For any prime ideal P ⊂ P ′ in Rthere is a prime ideal Q ⊂ Q′ in S contracting P ′ = Q′ ∩R.

S

R

nnnnnnn

Q′

P ′

Q

P

oooooo

Proof. R/P ⊂ S/Q is a finite ring extension. So there is a prime ideal in S/Qcontracting to P ′/P , 6.6.6. The corresponding prime ideal Q ⊂ Q′ ⊂ S contractsto P ′.

6.6.8. Theorem. Let R ⊂ S be a finite ring extension and E an R-module. Thefollowing conditions are equivalent:(1) E is an injective R-module.(2) HomR(S,E) is an injective S-module.

Proof. (1)⇒ (2): This is 3.6.11. (2)⇒ (1): Let E ⊂ E′ be an injective envelope,3.6.17. HomR(S,E) → HomR(S,E′) is injective and identifies HomR(S,E) asan essential submodule. For this, let f : S → E′ be nonzero and S = Rb1 +· · · + Rbn. Then use E ⊂ E′ essential to construct with renumbering a maximalsequence a1, . . . , am ∈ R such that

0 6= a1f(b1) ∈ E, . . . , 0 6= a1 · · · amf(bm) ∈ EThen a1 · · · amf : S → E is nonzero, so the extension of Hom’s is essential. Next,by (1) ⇒ (2) HomR(S,E′) is an injective S module, so the extension of Hom’sabove is trivial, 3.6.17. It follows that

HomR(S,E)

��

HomR(S,E′)

��HomR(R,E) // HomR(R,E′)

E // E′

with the right down map being surjective. So E = E′ and E is injective.

6.6.9. Exercise. (1) Show that Z→ Z[√−5] is finite.

(2) Let p be a prime number. Show that Z→ Z(p) is not finite.

7

Modules of finite length

7.1. Simple modules

7.1.1. Definition. A nonzero module M is a simple module if 0 and M are theonly submodules.

7.1.2. Lemma. Let f : M →M ′ be a homomorphism(1) If M is simple, then f is either zero or injective.(2) If M ′ is simple, then f is either zero or surjective.(3) If M,M ′ are both simple, then f is either zero or an isomorphism.

Proof. This follows from 2.3.3. (1) Ker f ⊂M is either 0 or M . (2) Im f ⊂ N iseither 0 or N .

7.1.3. Theorem. Let M be a simple R-module.(1) M is isomorphic to the factor module R/P , where P = Ann(M) is a maxi-

mal ideal.(2) M = Rx is finite and generated by any one nonzero element x ∈M .

Proof. A nonzero f : R→ M must be surjective. If Ker f = P then and R/P 'M is simple exactly when P is maximal. Clearly P = Ann(M).

7.1.4. Corollary. Any nonzero finite module has a simple factor module.

Proof. If M 6= 0 then MP 6= 0 for some maximal ideal 5.4.1. By 6.4.1 M ⊗R

k(P ) = M/PM 6= 0. Choose a nonzero linear projection M/PM → k(P ),giving M → k(P )→ 0 exact.

7.1.5. Corollary. Let (R,P ) be a local ring and M a finite module. The followingconditions are equivalent:

(1) M = 0.(2) HomR(M,k(P )) = 0.

7.1.6. Proposition. Let M,M ′ be simple modules. The following conditions areequivalent:

(1) M 'M ′.(2) M ⊗R M

′ 6= 0.(3) HomR(M,M ′) 6= 0.

Moreover if M 'M ′ there are non natural isomorphisms

M ⊗R M′ 'M ' HomR(M,M ′)

Proof. This follows from 7.1.3 with M = R/P .

7.1.7. Proposition. Let I ⊂ R be an ideal. A simple R/I-module M is a simpleR-module such that I ⊂ Ann(M).

101

102 7. MODULES OF FINITE LENGTH

Proof. Maximal ideals in R/I are of the form P/I and (R/I)/(P/I) ' R/P .

7.1.8. Proposition. Let U ⊂ R be a multiplicative subset and M is a simple R-module with Ann(M) = P .

(1) If U ∩ P = ∅, then is U−1M = M is a simple U−1R-module.(2) If U ∩ P 6= ∅, then U−1M = 0.

Proof. Let P ⊂ R be a maximal ideal, then the fractions U−1(R/P ) is either 0 orR/P , 5.1.5.

7.1.9. Proposition. Let R = R1 × R2 be a product of rings. A simple R-moduleis of the form M1 × 0 or 0×M2, where Mi is a simple Ri-module.

Proof. Clear by 2.2.9.

7.1.10. Example. (1) If K is a field, a one dimensional vector space is simple.(2) If R is a principal ideal domain, then R/(p) is a simple module for all irre-

ducible (p).(3) Z/(p) are simple for all prime numbers p.(4) Let K be a field. K[X]/(X − a) is a simple K[X]-module.

7.1.11. Exercise. (1) Show that Z does not contain any simple modules.(2) Show that Q does not have any simple factor Z-module.(3) Let L,L′ ⊂M be simple submodules. Show that either L ∩ L′ = 0 of L = L′.(4) Let L 6= L′ ⊂M be simple submodules. Show that L⊕ L′ = L+ L′.

7.2. The length

7.2.1. Definition. Let R be a ring. A module M has finite length if it admits acomposition series by submodules

0 = M0 ⊂M1 ⊂ · · · ⊂Mn−1 ⊂Mn = M

such that each factor Mi/Mi−1 is a simple module.

7.2.2. Lemma. Any two finite composition series have the same number of sub-modules.

Proof. Let l(M) be the least length of a composition series. M is simple if andonly if l(M) = 1. Let 0 = M0 ⊂ M1 ⊂ · · · ⊂ Mn = M be any compositionseries. For a submodule N ⊂ M there is a filtration 0 = N ∩M0 ⊂ N ∩M1 ⊂· · · ⊂ N ∩Mn = N with factors N ∩Mi/N ∩Mi−1 ⊂ Mi/Mi−1 being eithersimple or 0. It follows that l(N) ≤ l(M). If l(N) = l(M) then each factor isnonzero, so N ∩Mi/N ∩Mi−1 = Mi/Mi−1. This gives Mi ⊂ N and finallyN = M . Applying this to the composition series of M gives l(M0) < l(M1) <· · · < l(Mn) so n ≤ l(M) as needed.

7.2.3. Definition. The number of nontrivial submodules in a filtration as abovewill be denoted `R(M) and is the length of M . Remark that a module of finitelength is simple if and only if the length is 1.

7.2.4. Theorem. Given an exact sequence

0 // Mf // N

g // L // 0

then the following statements are equivalent:(1) N has finite length.

7.2. THE LENGTH 103

(2) M and L have finite length.If N has finite length, then

`R(N) = `R(M) + `R(L)

Proof. (1)⇒ (2): A finite filtrationNi with simple factors inN induces a filtrationf−1(Ni) in M and a filtration g(Ni) in L with factors being simple or 0. (2) ⇒(1): A finite composition series Mi in M and Lj in L may be spliced together

· · · ⊂ f(M) = g−1(0) ⊂ g−1(L1) ⊂ . . .to give a composition series of N . The length formula follows from 7.2.2.

7.2.5. Corollary. Let f : M → N be a homomorphism.(1) M has finite length if and only if Ker f, Im f have finite length. If finite length

`R(M) = `R(Ker f) + `R(Im f)

(2) N has finite length if and only if Im f,Cok f have finite length. If finite length

`R(N) = `R(Im f) + `R(Cok f)


7.2.6. Corollary. Let f : M → M be a homomorphism on a module of finitelength. Then

`R(Ker f) = `R(Cok f)and the following conditions are equivalent:(1) f is injective.(2) f is surjective.(3) f is an isomorphism.


7.2.7. Corollary. Let N ⊂M be a submodule and suppose M has finite length.(1) `R(N) ≤ `R(M).(2) `R(N) = `R(M) if and only if N = M .

7.2.8. Corollary. Let N,L ⊂ M be submodules and suppose N,L has finitelength. Then N + L,N ∩ L have finite length and

`R(N + L) + `R(N ∩ L) = `R(N) + `R(L)

Proof. By 2.3.8 N/N ∩ L ' N + L/L.

7.2.9. Proposition. Given submodules N,L ⊂ M . The following statements areequivalent:

(1) M/N,M/L have finite length.(2) M/N ∩ L has finite length.

If finite length

`R(M/N + L) + `R(M/N ∩ L) = `R(M/N) + `R(M/L)

Proof. Use the exact sequence 3.2.7

0 // M/N ∩ L // M/N ⊕M/L // M/N + L // 0



7.2.10. Proposition. A R-module M of finite length is finite and generated by`R(M) or less elements.

Proof. There is an exact sequence 0 // Nf // M

g // L // 0 withL simple, 7.1.4.If xi generate N and g(y) generate L, then f(xi), y generate M . Conclusion byinduction.

7.2.11. Proposition. Let I ⊂ R be an ideal. Suppose anR/I-moduleM has finitelength. Then M has finite length as R-module and

`R/I(M) = `R(M)

Proof. From 7.1.7 follows that a composition series as R/I-module is also a com-position series as R-module. and 7.2.2.

7.2.12. Proposition. Let U ⊂ R be a multiplicative subset. Suppose an R-moduleM has finite length. Then U−1M has finite length as U−1R-module and

`U−1R(U−1M) ≤ `R(M)

Proof. This follows from 7.1.8 and 7.2.2. The fraction construction of a composi-tion series may be refined to a composition series.

7.2.13. Proposition. Let (R,P ) be a local ring and M a module of finite length.The following conditions are equivalent:

(1) M = 0.(2) HomR(k(P ),M) = 0.(3) HomR(M,k(P )) = 0.

Proof. Clear by a composition series as k(P ) is the only simple module.

7.2.14. Proposition. Let M,M ′ be modules of finite length.(1) M ⊕M ′ has finite length and `R(M ⊕M ′) = `R(M) + `R(M ′).(2) M ⊗R M

′ has finite length and `R(M ⊗R M′) ≤ `R(M) · `R(M ′).

(3) HomR(M,M ′) has finite length and `R(HomR(M,M ′)) ≤ `R(M)·`R(M ′).

Proof. (1) This is clear from 7.2.4. (2) If M = R/P is simple then M ⊗R M′ '

M ′/PM ′ is a factor module, so of finite length and the equality follows from 7.2.4.In general there is an exact sequence 0→ N →M → L→ 0 withL simple givingan exact sequence N ⊗R M ′ → M ⊗R M ′ → L ⊗R M ′ → 0. Conclusion byinduction and 7.2.4. (2) If M = R/P is simple then HomR(M,M ′) ⊂ M ′ is asubmodule, so of finite length and the equality follows from 7.2.4. In general thereis an exact sequence 0 → N → M → L → 0 with L simple giving an exactsequence 0 → HomR(L,M ′) → HomR(M,M ′) → HomR(N,M ′). Conclusionby induction and 7.2.4.

7.2.15. Proposition. Let R = R1 × R2 be a product of rings. An R-moduleM1 × M2 is of the finite length if and only if Mi is a finite length Ri-module.Moreover

`R(M1 ×M2) = `R1(M1) + `R2(M2)


7.3. ARTINIAN MODULES 105

7.2.16. Example. Let K be a field. A module M is of finite length if and only if itis a finite dimensional vector space. Then

`K(M) = rankK(M)

7.2.17. Exercise. (1) Compute

`Z(Z/(pn11 . . . pnk

k )) = n1 + · · ·+ nk

(2) Compute

`K[X](K[X]/((X − a1)n1 . . . (X − ak)nk)) = n1 + · · ·+ nk

7.3. Artinian modules

7.3.1. Lemma. Let M be a module. The following conditions are equivalent:(1) Any descending chain · · · ⊂ Mi+1 ⊂ Mi of submodules is stationary: there

is n such that Mi+1 = Mi for i > n.(2) Any nonempty subset of submodules of M contains a minimal element.

Proof. (1) ⇒ (2): Suppose a nonempty subset of submodules do not contain aminimal element. Then choose a non stationary descending chain. (2) ⇒ (1): Adescending chain containing a minimal element is stationary.

7.3.2. Definition. A module M which satisfies the conditions of 7.3.1 is an ar-tinian module.

7.3.3. Example. (1) A module of finite length is artinian. Just note that a de-scending chain produces a decreasing sequence of lengths.

(2) A vector space is artinian if and only if it is finite.

7.3.4. Theorem. Let

0 // Mf // N

g // L // 0

be an exact sequence of modules over the ring R. The following statements areequivalent:

(1) N is artinian.(2) M and L are artinian.

Proof. (1) ⇒ (2): A chain Mi in M gives a chain f(Mi) in N which becomesstationary. So M is artinian. A chain in Li in L gives a chain in g−1(Li) in N ,which becomes stationary. Then also the original chain Li = g(g−1(Li)) becomesstationary and L is artinian. (2)⇒ (1): A chain Ni in N , induces chains f−1(Ni)in N and g(Ni) in L which become stationary. By the snake lemma 3.2.4

f−1(Ni)/f−1(Ni+1)→ Ni/Ni+1 → g(Ni)/g(Ni+1)

is exact and the original chain is stationary.

7.3.5. Corollary. (1) Let N ⊂ M be a submodule. Then M is artinian if andonly if both N and M/N are artinian.

(2) Let f : M → N be a homomorphism.(a) M is artinian if and only if Ker f, Im f are artinian.(b) N is artinian if and only if Im f,Cok f are artinian



7.3.6. Proposition. Let f : M → M be a homomorphism on an artinian module.The following conditions are equivalent:(1) f is injective(2) f is an isomorphism

Proof. There is a number n such that Im f◦n = Im f◦n+1. For x ∈ M there is ysuch that f◦n(x) = f◦n+1(y). Then f◦n(x − f(y)) = 0 so x = f(y) since f isinjective. It follows that f is surjective.

7.3.7. Proposition. A finite direct sum M1 ⊕ · · · ⊕Mn of artinian modules Mi isartinian.

Proof. Use the exact sequences

0→M1 ⊕ · · · ⊕Mn−1 →M1 ⊕ · · · ⊕Mn →Mn → 0

together with induction and 7.3.4.

7.3.8. Proposition. Given submodules N,L ⊂ M , then the following statementsare equivalent:(1) M/N,M/L are artinian(2) M/N ∩ L is artinian

Proof. Use the exact sequences 3.2.7

0 // M/N ∩ L // M/N ⊕M/L // M/N + L // 0


7.3.9. Proposition. Let M be a finite and N an artinian R-module.(1) M ⊗R N is artinian.(2) HomR(M,N) is artinian.

Proof. Choose Rn → M → 0 exact. (1) Nn ' Rn ⊗R N → M ⊗R N → 0 isexact, so conclusion by 7.3.4. (2) 0→ HomR(M,N)→ HomR(Rn, N) ' Nn isexact, so conclusion by 7.3.4.

7.3.10. Proposition. Let I ⊂ R be an ideal. Suppose an R/I-module M is ar-tinian. Then M is an artinian R-module.

Proof. This is clear since a sequence of R-submodules is a sequence of R/I-modules.

7.3.11. Proposition. Let U ⊂ R be a multiplicative subset. Suppose an R-moduleM is artinian. Then U−1M is an artinian U−1R-module.

Proof. Let i : M → U−1M . By 4.4.6 any U−1R-submodule N ⊂ U−1M isextended U−1(i−1(N)) ' N . So a chain is stationary.

7.3.12. Proposition. Let R = R1 × R2 be a product of rings. An R-moduleM1 ×M2 is artinian if and only if Mi is an artinian Ri-module.

Proof. This follows from the dictionary 2.2.9.

7.3.13. Exercise. (1) Show that a vector space is artinian if and only if it is finite di-mensional.

7.4. ARTINIAN RINGS 107

7.4. Artinian rings

7.4.1. Definition. A ring R is an artinian ring if R is an artinian module overitself.

7.4.2. Example. (1) A field is artinian.(2) A finite product of fields is artinian.

7.4.3. Proposition. (1) Let R be an artinian ring and I an ideal. Then the factorring R/I is artinian.

(2) A product ring R1 ×R2 is artinian if and only if each Ri is artinian.(3) Let R be an artinian ring and U a multiplicative subset. Then the fraction

ring U−1R is artinian.


7.4.4. Proposition. Given ideals I, J ⊂ R, then the following statements areequivalent:

(1) R/I,R/J are artinian(2) R/I ∩ J is artinian


7.4.5. Proposition. An artinian domain is a field.

Proof. By 7.3.6 scalar multiplication with a nonzero element is an isomorphism.

7.4.6. Theorem. LetR be an artinian ring. Then all prime ideals are maximal andthere are only finitely many such.

Proof. By 7.4.5 primes are maximal. If Pi are different maximal ideals, thenP1 · · ·Pn+1 ⊂ P1 · · ·Pn is a strictly decreasing chain. So there are only finitelymany maximal ideals.

7.4.7. Proposition. Let R be an artinian ring.(1) The factor ring R/

√0 is a finite product of fields.

(2) The nilradical√

0 is nilpotent,√

0k

= 0 for some k.

Proof. (1) Let P1, . . . , Pn be the prime and maximal ideals 7.4.6. The nilradical√0 = P1 ∩ · · · ∩ Pn by 5.1.7. Conclude by Chinese remainders 1.4.3. (2)

√0

k=√

0k+1

for some k. If√

0k 6= 0 let (a) be minimal among ideals I such that

I√

0k 6= 0. By minimality (a) = (a)

√0

k, so a = ab for some b ∈

√0

k. But b is

nilpotent 1.3.8, so a = 0 gives a contradiction. It follows, that√

0k

= 0.

7.4.8. Theorem. A ring R is artinian if and only if it has finite length.

Proof. The factor module√

0i/√

0i+1

is an artinian module over R/√

0 which by7.4.7 is a product of fields, so it has finite length. By the exact sequences

0→√

0i/√

0i+1 → R/

√0

i+1 → R/√

0i → 0

and induction follows that R/√

0k

is artinian for all k. Since√

0 is nilpotent 7.4.7it follows that R has finite length.

7.4.9. Corollary. Let R be an artinian ring and M a module. The following con-ditions are equivalent:


(1) M has finite length.(2) M is finite.(3) M is finite presented.

7.4.10. Corollary. Let R be artinian and R→ S a finite ring homomorphism.(1) S is artinian.(2) A finite length S-module N is by restriction of scalars a finite length R-

module.(3) A finite length R-module M gives by change of rings M ⊗R S as finite length

S-module.

7.4.11. Proposition. Let M be an R-module of finite length.(1) The ring R/Ann(M) is artinian.(2) There are only finitely many prime ideals Ann(M) ⊂ P .(3) Any prime ideal Ann(M) ⊂ P is maximal.

Proof. (1) Let x ∈ M , then R/Ann(x) ' Rx ⊂ M is artinian. If x1, . . . , xn

generate M , then Ann(M) = Ann(x1) ∩ · · · ∩ Ann(xn). By 7.4.4 R/Ann(M)is artinian. (2)(3) This follows by (1) and 7.4.6.

7.4.12. Proposition. Let M be a R-module. The following conditions are equiva-lent:

(1) M has finite length.(2) M is finite and artinian.

Proof. (1)⇒ (2): See 7.2.10 and 7.3.3. (2)⇒ (1): Let x ∈M , then R/Ann(x) 'Rx ⊂M is artinian. If x1, . . . , xn generate M , then Ann(M) = Ann(x1)∩ · · · ∩Ann(Xn). By 7.4.8 R/Ann(M) is artinian and a finite module over an artinianring has finite length.

7.4.13. Corollary. Let (R,P ) be a local ring and M an artinian module. Thefollowing conditions are equivalent:

(1) M = 0.(2) HomR(k(P ),M) = 0.

Proof. A finite submodule has finite length.

7.4.14. Example. Let K be a field. The ring R =∏

NK = {a : N→ K} is notartinian.(1) There are maximal ideals in R

Pi = {a : N→ K|a(i) = 0}(2) Corresponding simple types are

R/Pi ' Ji = {a : N→ K|a(j) = 0, i 6= j}(3) Ji are simple ideals and the sum is an ideal⊕

i

Ji = J 6= R

(4) J is an ideal which has no complement in R.

7.4.15. Exercise. (1) Show that a vector space is artinian if and only if it is finite di-mensional.

(2) Show that Z is not artinian.

7.5. LOCALIZATION 109

(3) Show that R[X] is not artinian for a nonzero ring R.(4) Show that Q[X]/(X2 −X) is artinian.(5) Show that a ring with finitely many ideals is artinian.(6) Let R be a principal ideal domain and a 6= 0. Show that R/(a) is artinian.

7.5. Localization

7.5.1. Theorem. Any artinian ring is the product of finitely many artinian localrings. Let R be artinian with maximal ideals P1, . . . , Pk. Then there is n such thatPn

1 . . . Pnk = 0 and

R ' R/Pn1 × · · · ×R/Pn

k

Each R/Pni is a local artinian ring.

Proof. This follows from 7.4.7, 7.4.8 and Chinese remainders 1.4.2.

7.5.2. Example. A reduced artinian ring is a finite product of fields.

7.5.3. Corollary. Let R be an artinian ring and P ⊂ R a prime ideal. Then thelocalization RP is an artinian ring.

7.5.4. Lemma. Let P ⊂ R be a maximal ideal and M a module. Assume u /∈ Pand n ∈ N.(1) (u) + Pn = R.(2) Scalar multiplication uM/P nM : M/PnM →M/PnM is an isomorphism.(3) The canonical map M/PnM → (M/PnM)P is an isomorphism.(4) If M is finite then M/PnM has finite length.

7.5.5. Theorem. LetM be anR-module of finite length and Ann(M) ⊂ P1, . . . , Pk

the maximal ideals. Then the sequence

0→ Pni M →M →MPi → 0

is exact and M/Pni M ' MPi is a finite length RPi-module. There are isomor-

phismsM '

⊕i

MPi '⊕

i

M/Pni M

and the length is`R(M) =

∑i

`RPi(MPi)

Proof. The ring R/Ann(M) ' R/Pn1 + Ann(M)× · · · ×R/Pn

k + Ann(M) by7.4.11 and 7.5.1. Therefore M 'M/Pn

1 M × · · · ×M/Pnk M and by localization

MPi 'M/Pni M .

7.5.6. Corollary. LetR be artinian with maximal ideals P1, . . . , Pk andM a finitemodule. Then the sequence


is exact and M/Pni M 'MPi is a finite RPi-module. There are isomorphisms

M '⊕

i

MPi '⊕

i

M/Pni M

and the length is`R(M) =

∑i

`RPi(MPi)


7.5.7. Proposition. Let (R,P ) → (S,Q) be a local ring homomorphism and as-sume that k(P )→ k(Q) is a finite extension. IfN is a finite length S-module, thenN is a finite length R-module and

`R(N) = rankk(P )(k(Q)) · `S(N)

Proof. Reduce to N = k(Q).

7.5.8. Proposition. Let (R,P ) → (S,Q) be a local ring homomorphism and as-sume that S/PS is a finite length S-module. Let M be a finite length R-module.(1) M ⊗R S is a finite length S-module.(2) In general

`S(M ⊗R S) ≤ `S(S/PS) · `R(M)(3) If R→ S is flat then

`S(M ⊗R S) = `S(S/PS) · `R(M)

Proof. The case M = k(P ) is clear. Conclude by induction.

7.5.9. Exercise. (1) Let K ⊂ L be a finite field extension and W a finite vector spaceover L. Show that

rankK(W ) = rankK(L) · rankL(W )

(2) Let K ⊂ L be a finite field extension and V a finite vector space over K. Show that

rankL(V ⊗K L) = rankK(V )

7.6. Local artinian ring

7.6.1. Lemma. Let (R,P ) be a local artinian ring and M any module. The fol-lowing conditions are equivalent:

(1) M = 0.(2) HomR(k(P ),M) = 0.

Proof. (2)⇒ (1): A nonzero submodule Rx ⊂M has finite length and contains asimple submodule k(P ) ' L ⊂M . So HomR(k(P ),M) 6= 0.

7.6.2. Proposition. Let (R,P ) be a local artinian ring and k(P ) ⊂ E an injectiveenvelope.(1) There are isomorphisms

k(P ) ' HomR(k(P ), k(P )) ' HomR(k(P ), E)

(2) For a finite module M , the module HomR(M,E) has finite length and

`R(HomR(M,E)) = `R(M)

Proof. (1) A nonzero homomorphism f : k(P ) → E has Im f = k(P ) sincethe extension is essential. (2) An exact sequence 0 → N → M → k(P ) → 07.1.4 gives an exact sequence with HomR(−, E). Conclude by (1) and inductionon `R(M).

7.6.3. Corollary. Let (R,P ) be a local artinian ring and k(P ) ⊂ E an injectiveenvelope. Then E has finite length

`R(E) = `R(R)

7.6. LOCAL ARTINIAN RING 111

7.6.4. Theorem. Let (R,P ) be a local artinian ring and k(P ) ⊂ E an injectiveenvelope. There is a natural isomorphism for any finite module M

x 7→ evx : M → HomR(HomR(M,E), E)

Proof. The case M = k(P ) is clear by 7.6.2. Let 0 → N → M → L → 0 be ashort exact sequence. Then the following diagram has exact rows.

0 // N //

��

M //

��

L //

��

0

0 // HomR(HomR(N, E), E) // HomR(HomR(M, E), E) // HomR(HomR(L, E), E) // 0

Conclusion by the five lemma 3.2.8 and induction on the length.

7.6.5. Corollary. Let (R,P ) be a local artinian ring and k(P ) ⊂ E an injectiveenvelope. There is an isomorphism

11E : R→ HomR(E,E)

7.6.6. Lemma. Let (R,P ) be a local artinian ring and k(P ) ⊂ E an injectiveenvelope. If E′ is injective and HomR(k(P ), E′) is finite, then E′ ' En, wheren = `R(HomR(k(P ), E′).

Proof. If n = 0 then E′ = 0 by 7.6.1. For n > 0 a nonzero k(P ) → E′ givesan injective extension 0 → E → E′. Since E is injective there is a splittingE′ ' E ⊕ E′′. Conclusion by induction on n.

7.6.7. Theorem. Let (R,P ) be a local artinian ring and k(P ) ⊂ E an injectiveenvelope. Let M → E′ be an injective envelope. Then(1) HomR(k(P ),M) ' HomR(k(P ), E′)(2) If M is artinian, then E′ ' En, where n = `R(HomR(k(P ),M).(3) An artinian M module has finite length and

`R(M) ≤ `R(HomR(k(P ),M)) `R(R)

Proof. (1) A homomorphism f : k(P ) → E′ has Im f ⊂ M since the extensionis essential. (2) E′ ' En by 7.6.6 and n is determined by (1) and 7.6.2. (3) Thisfollows from (2) and 7.6.3.

7.6.8. Corollary. Let R be an artinian ring and M an R-module. The followingconditions are equivalent:

(1) M has finite length.(2) M is finite.(3) M is artinian.

Proof. Reduce to the case where R is local by 7.5.1.

7.6.9. Theorem. Let (R,P ) be a local artinian ring and k(P ) ⊂ E an injectiveenvelope. The following conditions are equivalent:(1) R is injective.(2) R ' E.(3) HomR(k(P ), R) ' k(P ).

Proof. (1) ⇒ (2): By 7.6.6 R ' En and by 7.6.3 n = 1. (2) ⇒ (3): Immediatefrom 7.6.6. (3)⇒ (1): Let R ⊂ E′ be an injective envelope. By 7.6.7 R = E′.


7.6.10. Definition. A ring satisfying the conditions 7.6.9 is a local artinian Goren-stein ring.

7.6.11. Example. Let R be a principal ideal domain and p ∈ R an irreducibleelement. Then HomR(R/(p), R/(pn)) ' (pn−1)/(pn) ' R/P and R/(pn) is alocal artinian Gorenstein ring, 7.6.9.

7.6.12. Exercise. (1) Let p be a prime number. Show that Z/(pk) is a local artinianGorenstein ring.

8

Noetherian rings

8.1. Noetherian modules

8.1.1. Lemma. Let M be a module. The following conditions are equivalent:(1) Any increasing chain Mi ⊂ Mi+1 ⊂ . . . of submodules is stationary: there

is n such that Mi = Mi+1 for i > n.(2) Any nonempty subset of submodules of M contains a maximal element.(3) Any submodule of M is finite.

Proof. (1) ⇒ (2): Suppose a nonempty subset of submodules do not contain amaximal element. Then choose a non stationary ascending chain. (2)⇒ (1): Anascending chain containing a maximal element is stationary. (2)⇒ (3): Let N be asubmodule of M and choose a maximal element N ′ in the set of finite submodulesof N . For y ∈ N , the module N ′ +Ry is finite, so N ′ = N ′ +Ry gives N = N ′

finite. (3)⇒ (1): The union∪jMj is a submodule, generated by x1, . . . , xm ∈Mn,so Mi = ∪jMj , i > n.

8.1.2. Definition. A module M which satisfies the conditions of 8.1.1 is a noe-therian module.

8.1.3. Theorem. Let

0 // Mf // N

g // L // 0

be an exact sequence of modules over the ring R. The following statements areequivalent:

(1) N is noetherian.(2) M and L are noetherian.

Proof. (1) ⇒ (2): A chain Mi in M gives a chain f(Mi) in N which becomesstationary. So M is noetherian. A chain in Li in L gives a chain in g−1(Li)in N , which becomes stationary. Then also the original chain Li = g(g−1(Li))becomes stationary and L is noetherian. (2) ⇒ (1): A chain Ni in N , induceschains f−1(Ni) in N and g(Ni) in L which become stationary. By the snakelemma 3.2.4

f−1(Ni)/f−1(Ni−1)→ Ni/Ni−1 → g(Ni)/g(Ni−1)

is exact and the original chain becomes stationary.

8.1.4. Corollary. (1) Let M ⊂ N be a submodule. Then N is noetherian if andonly if both M and N/M are noetherian.

(2) Let f : M → N be a homomorphism.(a) M is noetherian if and only if Ker f, Im f are noetherian.(b) N is noetherian if and only if Im f,Cok f are noetherian

113

114 8. NOETHERIAN RINGS


8.1.5. Proposition. Let f : M →M be a homomorphism on a noetherian module.The following conditions are equivalent:

(1) f is surjective(2) f is an isomorphism

Proof. Analog of the proof of 7.3.4. There is a number n such that Ker f◦n+1 =Ker f◦n. For x ∈ Ker f there is y such that x = f◦n(y), since f is surjective.Then f◦n+1(y) = f(x) = 0, so y ∈ Ker f◦n+1 = Ker f◦n. Then x = f◦n(y) = 0and f is injective.

8.1.6. Proposition. A finite direct sum M1 ⊕ · · · ⊕Mn of noetherian modules Mi

is noetherian.

Proof. Use the exact sequences

0→M1 ⊕ · · · ⊕Mn−1 →M1 ⊕ · · · ⊕Mn →Mn → 0

together with induction and 8.1.3.

8.1.7. Proposition. Given submodules N,L ⊂ M , then the following statementsare equivalent:(1) M/N,M/L are noetherian(2) M/N ∩ L is noetherian

Proof. Use the exact sequence 3.2.7

0 // M/N ∩ L // M/N ⊕M/L // M/N + L // 0


8.1.8. Proposition. Let M be a finite and N a noetherian R-module.(1) M ⊗R N is noetherian.(2) HomR(M,N) is noetherian.

Proof. Choose Rn → M → 0 exact. (1) Nn ' Rn ⊗R N → M ⊗R N → 0 isexact, so conclusion by 8.1.3. (2) 0→ HomR(M,N)→ HomR(Rn, N) ' Nn isexact, so conclusion by 8.1.3.

8.1.9. Proposition. Let I ⊂ R be an ideal. Suppose an R/I-module M is noe-therian. Then M is a noetherian R-module.

Proof. This is clear since a chain of R-submodules is a chain of R/I-modules.

8.1.10. Proposition. Let U ⊂ R be a multiplicative subset. Suppose an R-moduleM is noetherian. Then U−1M is noetherian U−1R-module.

Proof. Let i : M → U−1M . By 4.4.6 any U−1R-submodule N ⊂ U−1M isextended U−1(i−1(N)) ' N . So a chain is stationary.

8.1.11. Proposition. Let R = R1 × R2 be a product of rings. An R-moduleM1 ×M2 is noetherian if and only if Mi is an noetherian Ri-module.


8.1.12. Proposition. Let M be a R-module. The following conditions are equiva-lent:

8.2. NOETHERIAN RINGS 115

(1) M has finite length.(2) M is noetherian and artinian.

Proof. (1)⇒ (2): A chain produces a monoton bounded sequence of lengths. (2)⇒ (1): M is finite and artinian, so conclusion by 7.4.12.

8.1.13. Example. (1) A vector space is noetherian if and only if it is finite.

8.1.14. Exercise. (1) Show that⊕

N Z is not noetherian.(2) Show that

∏N Z is not noetherian.

(3) Show that Q is not a noetherian Z-module.(4) Let p be a prime number. Show that the Z(p)-submodules of Q containing Z(p) is

of form p−nZ(p) and conclude that Q/Z(p) is an artinian but not noetherian Z(p)-module.

8.2. Noetherian rings

8.2.1. Definition. A ring R is a noetherian ring if R is a noetherian module.

8.2.2. Proposition. The following conditions are equivalent:(1) R is noetherian.(2) Any ideal is finite.(3) Any increasing sequence of ideals is stationary.(4) Any nonempty subset of ideals of contains an ideal maximal for inclusion.(5) Any ideal is noetherian.


8.2.3. Proposition. Let R be a noetherian ring.(1) The nilradical is nilpotent. For some n

√0

n= 0

(2) Some power of the radical of an ideal is contained in the ideal. For some n√I

n⊂ I

Proof. (1)√

0 = (b1, . . . , bm) such that bki = 0. Then (∑aibi)mk = 0. (2) Use

(1) on R/I .

8.2.4. Proposition. (1) Let I ⊂ R be an ideal in a noetherian ring. Then thefactor ring R/I is a noetherian ring.

(2) The product ring R1 ×R2 is noetherian if and only if each Ri is noetherian.(3) Let U ⊂ R be a multiplicative subset in a noetherian ring. Then the fraction

ring U−1R is a noetherian ring.

Proof. Use 8.1.3 and 8.1.10.

8.2.5. Proposition. Given ideals I, J ⊂ R, then the following statements areequivalent:

(1) R/I,R/J are noetherian(2) R/I ∩ J is noetherian

Proof. This is a special case of 8.1.7.

8.2.6. Theorem. LetR be a noetherian ring. Then any finiteR-module is noether-ian.


Proof. Let M be a finite R-module. There is a surjective homomorphism Rn →M → 0, 6.1.2. Conclusion by 8.1.3.

8.2.7. Corollary. Let R be a noetherian ring. Any finite R-module is finite pre-sented.

8.2.8. Proposition. Let M be an R-module. The following Statements are equiva-lent:(1) M is a noetherian.(2) R/Ann(M) is a noetherian ring and M is finite.

Proof. (1) ⇒ (2): Let x ∈ M , then R/Ann(x) ' Rx ⊂ M is noetherian. Ifx1, . . . , xn generate M , then Ann(M) = Ann(x1) ∩ · · · ∩ Ann(Xn). By 8.2.4R/Ann(M) is noetherian. (2)⇒ (1): By 8.2.6M is a finiteR/Ann(M)-module.Conclusion by 8.1.9.

8.2.9. Proposition. Let R be a noetherian ring and M,N noetherian modules.(1) M ⊗R N is noetherian.(2) HomR(M,N) is noetherian.

Proof. Conclusion by 6.1.8, 6.1.9, 8.2.6.

8.2.10. Theorem. Let R be a noetherian ring and U a multiplicative subset. Fora finite module M and any module N the homomorphism


is an isomorphism.

Proof. Conclusion by 6.5.10, 8.2.7.

8.2.11. Proposition. Let R be a noetherian ring. If E is an injective R-module,then U−1E is an injective U−1R-module.

Proof. Let I ⊂ R be an ideal. Then HomR(R,E) → HomR(I, E) → 0 is exact.By 8.2.10 HomU−1R(U−1R,U−1E) → HomU−1R(U−1I, U−1E) → 0 is exact.So U−1E is injective 3.6.8, using that any ideal is extended 4.3.6.

8.2.12. Theorem. For a ring R, the following statements are equivalent:(1) R is a noetherian ring.(2) For any family Eα of injective modules, the sum

⊕αEα is injective.

Proof. (1) ⇒ (2): Let I ⊂ R be an ideal. A homomorphism f : I →⊕Eα

has Im f contained in a finite sum, which is injective 3.6.7. So f extends to R →⊕Eα, giving injectivity. (2)⇒ (1): Let In ⊂ R be an increasing chain of ideals.

Put I = ∪In and choose an injective envelope I/In ⊂ En. The homomorphismf : I →

⊕En extends to f ′ : R →

⊕En. Since Im f is contained in a finite

sum, I/In = 0 for n >> 0.

8.2.13. Theorem. Let R be a noetherian ring and Fα a family of flat modules.Then the product

∏α Fα is flat.

Proof. An ideal I ⊂ R is finite presented. By 6.5.8 the homomorphsim I ⊗R∏Fα →

∏Fα is a product of injective homomorphisms

∏(I ⊗R Fα) →

∏Fα,

which is injective giving the conclusion.

8.3. FINITE TYPE RINGS 117

8.2.14. Theorem. If R is a ring such that every prime ideal is finite, then it isnoetherian.

Proof. If R is not noetherian, then the set of not finite ideals is nonempty and byZorn’s lemma it has a maximal element I . Since I is not prime there is a, b /∈ Iand ab ∈ I . The ideals I + (a) and I : (a) are both greater that I and thereforefinite. Let I + (a) = (a1, . . . , am, a), ai ∈ I and I : (a) = (b1, . . . , bn). Assumec = c1a1 + · · ·+ cmam +da ∈ I . Then d ∈ I : (a) and d = d1b1 + · · ·+dnbn. SoI = (a1, . . . , am, ab1, . . . , abn) is finite. It follows that R must be noetherian.

8.2.15. Exercise. (1) Show that a principal ideal domain is noetherian.(2) Let I ⊂ R be an ideal in a noetherian ring. Show that R/I is noetherian.(3) Let K be a field and R = K[X1, X2, . . . ] the polynomial ring in countable many

variables. Show that R is not noetherian.(4) Let K be a field. Show that the ring R =

∏NK is not noetherian.

8.3. Finite type rings

8.3.1. Theorem (Hilbert’s basis theorem). Let R be a noetherian ring. Then thering of polynomials R[X] is noetherian.

Proof. Assume I ⊂ R[X] to be a not finite ideal. Choose a sequence f1, f2, · · · ∈I such that

fi = aiXdi + terms of lower degree , ai 6= 0

and fi+1 has lowest degree in I\(f1, . . . , fi). The ideal of leading coefficients isfinitely generated by a1, . . . , an. Then an+1 = b1a1 + · · ·+ bnan and d1 ≤ · · · ≤dn+1 = d gives

fn+1 − b1Xd−d1f1 − · · · − bnXd−dnfn

in I\(f1 . . . , fn) of degree less than d. By contradiction the ideal I is finite.

8.3.2. Corollary. Let R be a noetherian ring.

(1) A polynomial ring R[X1, . . . , Xn] in finitely many variables is noetherian.(2) If R→ S a finite type ring over R, then S is noetherian.

8.3.3. Corollary. LetR be a noetherian ring andM a finite module. Then the ringR⊕M , 2.1.14, is noetherian.

Proof. Let x1, . . . , xn generate M . Then R[X1, . . . , Xn] → R ⊕M, Xi 7→ xi issurjective. So R→ R⊕M is of finite type.

8.3.4. Example. Let I ⊂ R be an ideal in a noetherian ring. Then there are noe-therian rings.

(1) GIR = ⊕n≥0In/In+1.

(2) BIR = ⊕n≥0In.

8.3.5. Theorem (Krull’s intersection theorem). Let I be an ideal in a noetherianring R. Then there is a ∈ I such that

1 + a ∈ Ann(⋂n

In)


Proof. Let I = (u1, . . . , um). If b ∈ In then b = fn(u1, . . . , um) where fn ∈R[X1, . . . , Xm] are homogeneous of degree n. By Hilbert’s basis theorem, 8.3.1there is N such that

fN+1 = f1g1 + · · ·+ fNgN

where gn is homogeneous of degree N − n + 1 > 0. By substitution b ∈ bI andI(∩In) = ∩In. Now ∩In ⊂ R is finite, so conclusion by 6.3.4.

8.3.6. Corollary. Let I be an ideal in a noetherian ring R such that the elements1 + a, a ∈ I are nonzero divisors. Then⋂

n

In = 0

8.3.7. Corollary. Let I be an ideal in a noetherian ring R and M a finite module.Then there is a ∈ I such that

1 + a ∈ Ann(⋂n

InM)

Proof. Use 8.3.5 on the ring R ⊕M and the ideal I ⊕M . Clearly (I + M)n =In + In−1M .

8.3.8. Corollary. Let I be an ideal in a noetherian ring R and M a finite modulesuch that the elements 1 + a, a ∈ I are nonzero divisors on M . Then⋂

n

InM = 0

8.3.9. Theorem. If R ⊂ S be a finite extension. Then R is noetherian if and onlyif S is noetherian.

Proof. Assume S is noetherian. Let Eα be a family of injective R-modules. Then⊕HomR(S,Eα) is an injective S-module. Since S is finite over R, there is an

isomorphism 6.1.14⊕

HomR(S,Eα) ' HomR(S,⊕Eα). By 6.6.8

⊕Eα is

injective and by 8.2.12 R is noetherian.

8.3.10. Example. Let K be a field and R = K[X1, X2, . . . ]/(X1 − x2X3, x2 −x3X4, . . . ). Put P = (X1, X2, . . . ).

(1) P is maximal(2) P 2 = P .(3) P (∩nP

n) = ∩nPn.

8.3.11. Exercise. (1) Show that if R[X] is noetherian, then R is noetherian.(2) Show that the subring Z[2X, 2X2, . . . ] ⊂ Z[X] is not noetherian. Conclude that the

extension is not finite.

8.4. Power series rings

8.4.1. Theorem. Let R be a noetherian ring. Then the power series ring R[[X]] isnoetherian.

Proof. Let P ⊂ R[[X]] be a prime ideal. Then P +(X)/(X) = (a1, . . . , an) ⊂ Ris a finite ideal. If X ∈ P then P = (a1, . . . , an, X) is finite. Suppose X /∈ Pand choose fi = ai + terms of positive degree ∈ P . If g ∈ P then write g =b10f1 + · · · + bn0fn + Xg1 for some bi0 ∈ R. Since P is prime, g1 ∈ P . Nowg1 = b11f1+ · · ·+bn1fn+Xg2 and so on. Put hi =

∑k bikX

k, then g =∑

i hifi.So P = (h1, . . . , hn) is finite and R[[X]] is noetherian by 8.2.14.

8.4. POWER SERIES RINGS 119

8.4.2. Corollary. Let R be a noetherian ring.

(1) A power series ring R[[X1, . . . , Xn]] in finitely many variables is noetherian.(2) Let I ⊂ R[[X1, . . . , Xn]] be an ideal, then the factor ringR[[X1, . . . , Xn]]/I

is noetherian.

8.4.3. Proposition. Let R be a principal ideal domain. Then the power series ringR[[X]] is a unique factorization domain.

Proof. By 5.1.12 it is enough to show that a nonzero prime contains a principalprime. Let P ⊂ R[[X]] be nonzero prime ideal. If X ∈ P then (X) ⊂ P isa principal prime. Suppose P 6= (X) and P + (X)/(X) = (a) ⊂ R. Choosef = a + terms of positive degree ∈ P . If g ∈ P then g = b0f + Xg1 forsome b0 ∈ R. Since P is prime, g1 ∈ P . Now g1 = b1f + Xg2 and so on. Puth =

∑k bkX

k, then g = hf . So P = (f) is a principal prime itself.

8.4.4. Proposition. Let I ⊂ R be an ideal in a noetherian ring. Then there is acanonical isomorphism

R[[X]]/IR[[X]] ' R/I[[X]]

Proof. The projection R[[X]]→ R[[X]]/IR[[X]] factors over R/I[[X]] since I isfinitely generated. Then there is an inverse to the homomorphism 1.9.8.

8.4.5. Corollary. If P ⊂ R is a prime ideal in a noetherian ring, then PR[[X]] ⊂R[[X]] is a prime ideal.

8.4.6. Proposition. Let R be a noetherian ring.

(1) The inclusion R[X] ⊂ R[[X]] is a flat homomorphism.(2) The inclusion R ⊂ R[[X]] is a faithfully flat homomorphism.

Proof. (1) Let I ⊂ R[X] be an ideal and let∑ai ⊗ fi ∈ K = Ker(I ⊗R[X]

R[[X]]→ R[[X]]. Then∑aifi = 0. Chase the diagram

I ⊗R[X] (Xn)

��I ⊗R[X] R[[X]]

��

// R[[X]]

��I ⊗R[X] R[[X]]/(Xn) // R[[X]]/(Xn)

Since R[X]/(Xn) = R[[X]]/(Xn) it follows that∑ai ⊗ fi =

∑aifi ⊗ 1 = 0 ∈

I⊗R[X]R[[X]]/(Xn). Therefore∑ai⊗fi ∈ Xn(I⊗R[X]R[[X]]). It follows that

K ⊂⋂

nXn(I⊗R[X]R[[X]]). I⊗R[X]R[[X]] is a finiteR[[X]]-module and 1+aX

is a unit, so conclusion by 8.3.8. (2) The homomorphism R → R[X] → R[[X]]is a composition of flat homomorphisms by (1). For any maximal ideal P ⊂ Rthe homomorphism RP → RP [[X]] is local. It follows that P = Q ∩ R for someQ ⊂ R[[X]], so faithfully flat by 5.6.15.

8.4.7. Exercise. (1) Show that if R[[X]] is noetherian, then R is noetherian.


8.5. Localization of noetherian rings

8.5.1. Theorem. Let R be a noetherian ring and P a prime ideal. Then the localring RP is a noetherian ring.

Proof. Any ideal is extended 4.3.6 or just 8.2.4.

8.5.2. Theorem. Let R be a noetherian ring and P a prime ideal. For a finitemodule M and any module N the homomorphism

HomR(M,N)P → HomRP(MP , NP )

is an isomorphism.

Proof. A special case of 8.2.10.

8.5.3. Proposition. Let R be a noetherian ring and M a finite module. Let P be aprime ideal. Then RP is a noetherian ring and MP is a finite RP -module.

8.5.4. Theorem (Krull’s intersection theorem). Let (R,P ) be a noetherian localring and M a finite module. Then ⋂

n

PnM = 0

Proof. From 8.3.7, (1 + a)⋂

n PnM = 0 for some a ∈ P . Now use that 1 + a is

a unit.

8.5.5. Corollary. Let (R,P ) be a noetherian local ring and I ⊂ P an ideal. Then⋂n

In = 0

8.5.6. Theorem. Let (R,P ) be a noetherian local ring and F a finite module. Thefollowing conditions are equivalent:

(1) F is free.(2) F is projective.(3) F is flat.(4) P ⊗R F → F is injective.

Proof. This follows from 6.5.14 as finite modules are finite presented.

8.5.7. Proposition. Let R be a noetherian ring and F a finite module. The follow-ing conditions are equivalent:

(1) F is projective.(2) F is flat.(3) P ⊗R F → F is injective for all maximal ideals P .(4) FP is free for all maximal ideals P .

Proof. This follows from 6.5.16 as finite modules are finite presented.

8.5.8. Exercise. (1) Is it true that if U−1R is noetherian, then R is noetherian?

8.6. PRIME FILTRATIONS OF MODULES 121

8.6. Prime filtrations of modules

8.6.1. Proposition. Let R be a ring and M 6= 0 a nonzero module. An idealAnn(x) maximal in the set of ideals {Ann(y)|0 6= y ∈M} is a prime ideal.

Proof. Let Ann(x) be a maximal annihilator. Suppose a, b ∈ R such that ab ∈Ann(x) and b /∈ Ann(x). Then

Ann(x) ⊆ Ann(bx) 6= R

Consequently Ann(x) = Ann(bx), in particular a ∈ Ann(x).

8.6.2. Corollary. Let R be a noetherian ring and M 6= 0 a nonzero module. Thenthere is x ∈M such that Ann(x) is a prime ideal.

8.6.3. Theorem. Let R be a noetherian ring and M 6= 0 a finite R-module. Thenthere exists a finite filtration of M by submodules

0 = M0 ⊂M1 ⊂ · · · ⊂Mn−1 ⊂Mn = M

such that Mi/Mi−1, i = 1, . . . , n is isomorphic to an R-module of the form R/Pi

where Pi is a prime ideal in R.

Proof. The set of submodules of M for which the theorem is true is nonempty by8.2.6. Let N ⊂ M be maximal in this set. Suppose N 6= M . By 8.2.6 applied toM/N there is a chainN ⊂ N ′ ⊂M such thatN ′/N is isomorphic to anR-moduleof the form R/P ′ where P ′ is a prime ideal. This contradicts the maximality of N .So N = M .

8.6.4. Corollary. Let R be a nonzero noetherian ring. Then there exists a finitefiltration of ideals

0 = I0 ⊂ I1 ⊂ · · · ⊂ In−1 ⊂ In = R

such that Ii/Ii−1, i = 1, . . . , n is isomorphic to an R-module of the form R/Pi

where Pi is a prime ideal in R.

8.6.5. Example. In Z there is a filtration

0 ⊂ (pn) ⊂ (pn−1) ⊂ · · · ⊂ (p) ⊂ Z

of any length with factors (pn) ' Z and (pi−1)/(pi) ' Z/(p) for any primenumber p.

8.6.6. Theorem. Let R be a ring. The following statements are equivalent:(1) R is artinian.(2) R is noetherian and all prime ideals are maximal.

Proof. (1)⇒ (2): By 7.4.8 an artinian ring has finite length and therefore noether-ian. Primes are maximal by 7.3.11. (2)⇒ (1): By 8.6.4 there is a finite compositionseries.

8.6.7. Corollary. Let R be a ring and M a module. The following statements areequivalent:(1) M has finite length.(2) R/Ann(M) is artinian and M is finite.

8.6.8. Theorem. If R ⊂ S be a finite extension. Then R is artinian if and only ifS is artinian.


Proof. Suppose S is artinian. By 8.3.9 R is noetherian. By 6.6.5 any prime idealin R is maximal. Conclusion by 8.6.6.

8.6.9. Proposition. Let R be a noetherian ring. The number of minimal primeideals is finite.

Proof. Choose a filtration 8.6.4, 0 = I0 ⊂ · · · ⊂ In = R with Ii/Ii−1 ' R/Pi,where Pi is a prime ideal. Let P be a minimal prime ideal inR. Then (Ii/Ii−1)P '(R/Pi)P 6= 0 if and only if Pi ⊂ P . Thus P = Pi for some i since RP 6= 0.

8.6.10. Proposition. Let R be a noetherian ring such that the local rings RP aredomains for all maximal ideals P . Then R is a finite product of domains.

Proof. Let P1, . . . , Pn be the minimal primes 8.6.9. The intersection is 0 since thering is reduced 5.4.10. They are comaximal since a domain has a unique minimalprime. Conclusion by Chinese remainders 1.4.2.

8.6.11. Exercise. (1) Compute a filtration 8.6.3 of the Z-module Z/(36).

9

Primary decomposition

9.1. Zariski topology

9.1.1. Definition. Let R be a ring.(1) The set of prime ideals is the spectrum and denoted X = Spec(R).(2) For a ring homomorphism φ : R→ S restriction defines the associated map

aφ : Spec(S)→ Spec(R)

Q 7→ φ−1(Q)

(3) For a subset B ⊂ RV (B) = {P ∈ Spec(R)|B ⊂ P}

is a subset of the spectrum.(4) For a subset B ⊂ R

XB = Spec(R)\V (B)

is the complement.

9.1.2. Lemma. Let R be a ring.(1) For a subset B ⊂ R

V (B) = V (RB) = V (√RB)

So V (B) depends only on the radical of the ideal generated by B.(2) For a subset B ⊂ R

V (B) =⋂b∈B

V (b)

(3) For a subset B ⊂ RXB =

⋃b∈B

Xb

9.1.3. Lemma. Let R be a ring.(1) Spec(R) = V (0), ∅ = V (1).(2) V (B1) ∪ V (B2) = V (B1B2).(3)

⋂α V (Bα) = V (

⋃αBα).

Proof. (2) Clearly V (Bi) ⊂ V (B1B2). If P /∈ V (Bi) then choose bi ∈ Bi\P .The product b1b2 /∈ P , so P /∈ V (B1B2).

9.1.4. Definition. Let R be a ring.(1) The subsets V (B), B ⊂ R are the closed sets in the Zariski topology on

Spec(R).(2) The subsetsXB ,B ⊂ R are the open sets in the Zariski topology on Spec(R).(3) Subsets Xb, b ∈ R are the principal open subsets.

123

124 9. PRIMARY DECOMPOSITION

9.1.5. Proposition. Let R be a ring.(1) Spec(R) = X1, ∅ = X0.(2) Xb1 ∩Xb2 = Xb1b2 .(3) Spec(R)\V (B) =

⋃b∈B Xb

(4) The subset Xb, b ∈ R is a basis for the open sets in the Zariski topology.

9.1.6. Lemma. Let R be a ring and I, J ideals. The following conditions areequivalent:

(1) V (I) ⊂ V (J).(2) J ⊂

√I .

(3)√J ⊂√I .

Proof. This is clear from√I =

⋂I⊂P P , 5.1.8.

9.1.7. Proposition. Let R be a ring and X = Spec(R).(1) V (I) = V (J) if and only if

√J =√I .

(2) V (I) = X if and only if I ⊂√

0.(3) V (I) = ∅ if and only if I = R.(4) Xb = X if and only if b is a unit.(5) Xb = ∅ if and only if b ∈

√0.


9.1.8. Proposition. Let φ : R→ S be a ring homomorphism.(1) The map aφ : Spec(S)→ Spec(R) is continuous.(2) If I ⊂ R is an ideal, then aφ−1(V (I)) = V (IS).(3) If b ∈ R and X = Spec(R) then aφ−1(Xb) = Xφ(b).(4) If J ⊂ S is an ideal, then the closure aφ(V (J)) = V (J ∩R).(5) If X = Spec(R), then the closure aφ(X) = V (Kerφ).

Proof. (1) This follows from (2). (2) Calculate aφ−1(V (I)) = {Q|aφ(Q) ∈V (I)} = {Q|I ⊂ Q ∩R} = V (IS).

9.1.9. Proposition. Let R be a ring and X = Spec(R).(1) Let I ⊂ R be an ideal. Then

Spec(R/I) ' V (I) ⊂ Spec(R)

is a homeomorphism onto the closed subset V (I).(2) Let b ∈ R. Then

Spec({bn}−1R) ' Xb ⊂ Spec(R)

is a homeomorphism onto the principal open subset Xb.

Proof. This is a restatement of 1.3.5 and 5.1.5.

9.1.10. Theorem. Let R be a ring. Then any principal open subset Xb is quasi-compact. That is, any open covering Xb =

⋃α Uα may be refined to an finite

covering Xb = Uα1 ∪ · · · ∪ Uαn .

Proof. Assume by 9.1.9 Xb = X . By 9.1.6 Uα = ∪Xbαβ. So V ({bαβ}) = ∅

and therefore some (bα1 , . . . , bαn) = R. Clearly X = Xbα1∪ · · · ∪ Xbαn

⊂Uα1 ∪ · · · ∪ Uαn .

9.1. ZARISKI TOPOLOGY 125

9.1.11. Theorem. Let R be a ring. Spec(R) is connected if and only if R is not aproduct of two nonzero rings.

Proof. IfR is a nontrivial product, then the are proper ideals I1+I2 = R, I1∩I2 =0. So Spec(R) = V (I1) ∪ V (I2) is not connected. Conversely if Spec(R) =V (I1)∪V (I2) is not connected, then the ideals are proper and I1+I2 = R, I1∩I2 ⊂√

0. Choose ai ∈ Ii such that a1 + a2 = 1 and n so big that an1a

n2 = 0. Now

V (an1 ) ∪ V (an

2 ) = V (a1) ∪ V (a2) = Spec(R) so for some bi, b1an1 + b2a

n2 = 1

and (b1an1 )(b2an

2 ) = 0. It follows that R ' R/(b1an1 ) × R/(b2an

2 ) is a product ofnonzero rings.

9.1.12. Lemma. Let X 6= ∅ be a topological space. The following conditions areequivalent:

(1) X is not a union of two proper closed subsets.(2) Any two nonempty open subsets intersects nonempty.(3) Any nonempty open subset is dense in X .

Proof. (1) ⇔ (2): This is clear. (2) ⇒ (3): If U is open and nonempty, thenX\U∩U = ∅ gives U = X . (3)⇒ (2): If two openU1∩U2 = ∅, then U1 ⊂ X\U2.So U1 and U2 cannot both be nonempty.

9.1.13. Definition. An irreducible space is a topological space satisfying the con-ditions in 9.1.10. A subset of a topological space is an irreducible subset if it isan irrecucible space in the induced topology. A maximal irreducible subset is anirreducible component.

9.1.14. Lemma. Let X 6= ∅ be a topological space.(1) If X is irreducible, then any nonempty open subset is irreducible.(2) If Y ⊂ X is an irreducible subset, then the closure Y is irreducible.(3) If X is irreducible, then X is connected.(4) IfX is irreducible and f : X → Y is continuous. Then f(X) is an irreducible

subset.(5) Any irreducible component in X is closed.(6) Any irreducible subset in X is contained in an irreducible component.(7) X is the union of irreducible components.

Proof. Do easy topology homework.

9.1.15. Theorem. Let R be a ring.(1) Spec(R) is irreducible if and only if the nilradical

√0 is a prime ideal.

(2) A closed subset V (I) is irreducible if and only if the radical√I is a prime

ideal.(3) An irreducible component is of the form V (P ) where P is a minimal prime

ideal.

Proof. LetX = Spec(R). (1) IfX is irreducible and a1, a2 /∈√

0, then V (a1a2) =V (a1) ∪ V (a2) 6= X . So a1a2 /∈

√0. Conversely, if

√0 is a prime, then replace

with R/√

0 and assume R is a domain. If V (I1I2) = V (I1) ∪ V (I2) = X thenI1I2 = 0. It follows that I1 = 0 or I2 = 0. (2) Use (1) on R/I . (3) From 5.1.10any prime ideal contains a minimal prime ideal.

9.1.16. Proposition. Let R be a ring. The following conditions are equivalent:


(1) Spec(R) is a discrete topological space.(2) R/

√0 is a finite product of fields.

Proof. (1) ⇒ (2): By 9.1.10 Spec(R) is finite and consists of maximal ideals.Conclusion by Chinese remainders 1.4.3.

9.1.17. Exercise. (1) LetK be a field. Show that Spec(K[X,Y ]/(XY )) is connected,but not irreducible.

9.2. Support of modules

9.2.1. Definition. Let R be a ring and M a module.(1) The support of M is

Supp(M) = {P ∈ Spec(R)|MP 6= 0}(2) A minimal prime ideal in Supp(M) is a minimal prime of M .

9.2.2. Proposition. Let I ⊂ R be an ideal. Then as R-module

Supp(R/I) = V (I)

Proof. This is a restatement of 1.3.5.

9.2.3. Proposition. Let0→M → N → L→ 0

be a short exact sequence of modules. Then

Supp(N) = Supp(M) ∪ Supp(L)

Proof. From 5.4.6 follows that 0→MP → NP → LP → 0 is exact.

9.2.4. Corollary. (1) Let N ⊂M be a submodule. Then

Supp(M) = Supp(N) ∪ Supp(M/N)

(2) Given submodules N,L ⊂M . Then(a) Supp(M/N ∩ L) = Supp(M/N) ∪ Supp(M/L) .(b) Supp(M/N + L) ⊂ Supp(M/N) ∩ Supp(M/L) .

Proof. (2) Use the exact sequence 3.2.7

0 // M/N ∩ L // M/N ⊕M/L // M/N + L // 0


9.2.5. Theorem. Let R be a ring and M a module.(1) M = 0 if and only if Supp(M) = ∅.(2) Let M be a module and P ∈ Supp(M), then V (P ) ⊂ Supp(M).(3) For any module

Supp(M) ⊂ V (Ann(M))(4) If M is finite, then the support is a closed subset in the Zariski topology

Supp(M) = V (Ann(M))

Proof. (1) See 5.4.1. (2) If P ⊂ Q, then MP = (MQ)PRQ. (3) If MP 6= 0 then

for u /∈ P there is x ∈M such that ux 6= 0. So Ann(M) ⊂ P . (4) Let x1, . . . , xn

generate M . If Ann(M) = ∩Ann(xi) ⊂ P then some Ann(xi) ⊂ P , so xi1 6= 0

in MP .

9.2. SUPPORT OF MODULES 127

9.2.6. Proposition. LetN1, . . . , Nk ⊂M be submodules such that Supp(M/Ni)∩Supp(M/Nj) = ∅, i 6= j. Then the homomorphism

M/ ∩i Ni →⊕

i

M/Ni

is an isomorphism.

Proof. By 3.2.7

0 // M/ ∩k1 Ni

// M/ ∩k−11 Ni ⊕M/Nk

// M/ ∩k−11 Ni +Nk

// 0

is exact. By 9.2.4 and induction on k

Supp(M/ ∩k−11 Ni +Nk) ⊂ (∪k−1

1 Supp(M/Ni)) ∩ Supp(M/Nk)

So the support of the cokernel is empty. Conclusion by 9.2.5.

9.2.7. Proposition. Let R be a ring and M,N modules.

(1)Supp(M ⊗R N) ⊂ Supp(M) ∩ Supp(N)

(2) If M,N are finite, then

Supp(M ⊗R N) = Supp(M) ∩ Supp(N)

(3) If M is finite, then

Supp(HomR(M,N)) ⊂ Supp(M) ∩ Supp(N)

Proof. There is an isomorphism (M ⊗R N)P ' MP ⊗RPNP . (1) This is clear.

(2) This follows from 6.4.3. (3) By 6.5.10 there is an injective homomorphism0→ HomR(M,N)P → HomRP

(MP , NP ).

9.2.8. Proposition. Let (R,P )→ (S,Q) be a local homomorphism andM a finiteR-module. If Supp(M) 6= ∅ then Supp(M ⊗R S) 6= ∅.

Proof. Calculate

M ⊗R S ⊗S k(Q) 'M ⊗R k(P )⊗k(P ) k(Q)

and conclude by Nakayama’s lemma 6.4.1.

9.2.9. Corollary. Let φ : R→ S be a ring homomorphism and M an R-module.

(1) For the change of rings S-module

Supp(M ⊗R S) ⊂ aφ−1(Supp(M))

(2) If M is finite, then

Supp(M ⊗R S) = aφ−1(Supp(M))

9.2.10. Corollary. LetR be a ring, I an ideal inR andM a finiteR-module. Thenaπ : Spec(R/I)→ Spec(R) defines a bijective correspondence

Supp(M/IM)→ Supp(M) ∩ V (I) = V (Ann(M) + I)Q 7→ P = Q ∩R


9.2.11. Corollary. LetR be a ring, U a multiplicative subset andM anR-module.Then ι : R→ U−1R, aι : Spec(U−1R)→ Spec(R) defines a bijective correspon-dence

Supp(U−1M)→ Supp(M) ∩ {P ∈ Spec(R)|P ∩ U = ∅}Q 7→ Q ∩R

Proof. Let Q ⊂ U−1R be a prime and P = Q ∩ R. Then U−1MQ ' MP givesthe result.

9.2.12. Proposition. Let M be a finite R-module and P ∈ Supp(M). Then thereis a nonzero homomorphism M → R/P , that is HomR(M,R/P ) 6= 0.

Proof. By 7.1.4 there is a surjective homomorphism f : MP → k(P ) → 0. Letx1, . . . , xn generate M and choose u ∈ R\P such that uf(xi) ∈ R/P . Thecomposite of M →MP with uf is a nonzero homomorphism M → R/P .

9.2.13. Proposition. Let R be a noetherian ring and M a finite module. M hasfinite length if and only if Supp(M) consists only of maximal ideals.

Proof. By 9.2.5 Supp(M) = V (Ann(M)), so conclusion by 8.6.7.

9.2.14. Definition. A ring R with only finitely many maximal ideals is a semi-local ring.

9.2.15. Proposition. Let R be a semi-local noetherian ring and F a finite module.If FP is free of rank m for all maximal ideals, then F is free of rank m.

Proof. Let P1, . . . , Pk be the maximal ideals. If R ' R/Pn1 × · · · × R/Pn

k isartinian, then by 7.5.5 F ' FP1×· · ·×FPk

' (R/Pn1 )m×· · ·× (R/Pn

k )m is freeof rank m. In general choose x1, . . . , xm ∈ F giving a basis for F/P1 · · ·PkF .as R/P1 · · ·Pk-module. The homomorphism Rm → F, ei 7→ xi is surjective byNakayama’s lemma 6.4.6 and the support of the kernel is empty, so the homomor-phism is injective by 9.2.5.

9.2.16. Exercise. (1) Let R = Z, M = Q and N = Z/(p). Show that Supp(M ⊗R

N) 6= Supp(M) ∩ Supp(N).

9.3. Ass of modules

9.3.1. Definition. Let M be an R-module. A prime ideal P ⊂ R is an associatedprime ideal of M if P = Ann(x) for some x ∈ M . The set of prime idealsassociated to M is Ass(M).

9.3.2. Proposition. Let R be a ring.(1) Let M be a module and P a prime. P ∈ Ass(M) if and only if there is an

injective homomorphism R/P →M .(2) For any module M

Ass(M) ⊂ Supp(M)(3) Let 0 6= N ⊂ R/P be a nonzero submodule, then

Ass(N) = {P}(4) Let I be an ideal and M an R/I-module. The inclusion Spec(R/I) →

Spec(R) identifies Ass(M) over the rings R/I and R.

9.3. ASS OF MODULES 129

Proof. This is clear from the definition. (1) R/P ' Rx ⊂ M . (2) 0 → R/P →M gives 0 → k(P ) → MP . (3) P = Ann(x) of any nonzero x ∈ R/P . (4)R/P = (R/I)/(P/I) gives by (1) identified associated primes.

9.3.3. Proposition. Let0→M → N → L→ 0

be a short exact sequence of modules. Then

Ass(M) ⊂ Ass(N) ⊂ Ass(M) ∪Ass(L)

Proof. The left inclusion is trivial. Assume M ⊂ N and L = N/M . Next letP ∈ Ass(N) such that P /∈ Ass(M). Choose a submodule K ⊂ N such thatK ' R/P . Then Ass(K ∩ M) ⊂ Ass(K) ∩ Ass(M). It follows 9.3.2 thatK ∩M = 0. Therefore K ' K/K ∩M ' K + M/M ⊂ N/M gives P ∈Ass(K) ⊂ Ass(L).

9.3.4. Corollary. Let 0 → M → N → L → 0 be a split exact sequence ofmodules.

Ass(N) = Ass(M) ∪Ass(L)

9.3.5. Corollary. (1) Let N ⊂M be a submodule. Then

Ass(N) ⊂ Ass(M) ⊂ Ass(N) ∪Ass(M/N)

(2) Let M,N be modules. Then

Ass(M ⊕N) = Ass(M) ∪Ass(N)

(3) Given submodules N,L ⊂M . Then

Ass(M/N ∩L) ⊂ Ass(M/N)∪Ass(M/L) ⊂ Ass(M/N ∩L)∪Ass(M/N +L)

Proof. (3) Use the exact sequence 3.2.7

0 // M/N ∩ L // M/N ⊕M/L // M/N + L // 0


9.3.6. Proposition. Let M be a module and P ⊂ Ass(M). Then there is a sub-module N ⊂M such that

Ass(N) = P , Ass(M/N) = Ass(M)\P

Proof. Choose by Zorn’s lemma N maximal in the set of submodules N ′ ⊂ Mfor which Ass(N ′) ⊂ P . Let Q ∈ Ass(M/N) and choose N ⊂ L ⊂ M withL/N ' R/Q. Then Ass(L) ⊂ P ∪ {Q}. By maximality of N follows thatAss(L) 6⊂ P . So Q /∈ P and Q ∈ Ass(L) ⊂ Ass(M).

9.3.7. Theorem. Let R be a noetherian ring and M a module. The followingconditions are equivalent:(1) M = 0.(2) Supp(M) = ∅.(3) Ass(M) = ∅.(4) MP = 0 for all prime P ∈ Ass(M).

Proof. (4) ⇒ (1): If M 6= 0, then by 8.6.2 and 9.3.2 there is P ∈ Ass(M) ⊂Supp(M).


9.3.8. Corollary. Let R be a noetherian ring and f : M → N a homomorphism.The following conditions are equivalent:(1) f is injective.(2) fP is injective for all prime ideals P ∈ Ass(M).

Proof. (2)⇒ (1): Ker fP = 0 for all P ∈ Ass(Ker f) ⊂ Ass(M). Conclusion by9.3.7.

9.3.9. Theorem. Let R be a noetherian ring and M a module.(1) a ∈ R is a nonzero divisor on M if and only if a /∈ ∪P∈Ass(M)P .(2) The set of zero divisors on M is

∪P∈Ass(M)P

Proof. aM is injective if and only if aMPis injective for all P ∈ Ass(M), 9.3.8.

This happens when a /∈ ∪P∈Ass(M)P .

9.3.10. Proposition. Let U ⊂ R be a multiplicative subset and M a module. De-note ι : R→ U−1R and aι : Spec(U−1R)→ Spec(R).

(1) The map aι−1 defines an inclusion

Ass(M) ∩ {P ∈ Spec(R)|P ∩ U = ∅} → Ass(U−1M)P 7→ PRP

(2) If R is noetherian, then the map (1) is a bijective correspondence.

Proof. Let P be a prime ideal. For any homomorphism R/P → M there is acommutative diagram

R/P

��

// M

��U−1R/P // U−1M

(1) If P ∈ Ass(M) and P ∩ U = ∅ then U−1P ∈ Ass(U−1M). (2) If U−1P ∈Ass(U−1M), thenP∩U = ∅. 8.2.9 gives an isomorphismU−1 HomR(R/P,M) 'HomU−1R(U−1R/P,U−1M), so there is a diagram as above and R/P → M isinjective. It follows that P ∈ Ass(M).

9.3.11. Proposition. Let R be a noetherian ring and M a module. Then any min-imal prime P ∈ Supp(M) is contained in Ass(M).

Proof. Assume P ∈ Supp(M) is minimal. Then the RP -module MP has supportexactly in the maximal ideal, so {PRP } = Ass(MP ). Conclusion by 9.3.10.

9.3.12. Definition. A non minimal prime ideal in Ass(M) is an embedded primeof M .

9.3.13. Theorem. Let R be a noetherian ring and M a finite module. Then theassociated primes Ass(M) is a finite set.

Proof. Follows immediately from 9.3.4 and 8.1.6. Let 0 = M0 ⊂ · · · ⊂Mn = Mbe a filtration with factors Mi/Mi−1 ' R/Pi. Then Ass(M) ⊂ {Pi}.

9.3.14. Corollary. Let R be a noetherian ring and M a finite module. The follow-ing conditions are equivalent:(1) M has finite length.

9.3. ASS OF MODULES 131

(2) Supp(M) consists of maximal ideals.(3) Ass(M) consists of maximal ideals.(4) Ass(M) = Supp(M).

Proof. Most is just a restatement. (4) ⇒ (2): Use that the minimal ideals in thesupport are maximal.

9.3.15. Lemma. Let (R,P ) be a local ring M a module. Then P ∈ Ass(M) ifand only if HomR(k(P ),M) 6= 0.

Proof. k(P ) is simple, so a homomorphism k(P )→M is either 0 or injective.

9.3.16. Proposition. Let R be a noetherian ring and M a module. For a prime Pthe following conditions are equivalent:

(1) P ∈ Ass(M).(2) PRP ∈ Ass(MP ).(3) HomRP

(k(P ),MP ) 6= 0.

Proof. (1)⇔ (2): 9.3.10. (2)⇔ (3): 9.3.15.

9.3.17. Theorem. Let R be a noetherian ring and M a finite module. For anymodule N

Ass(HomR(M,N)) = Supp(M) ∩Ass(N)

Proof. By 8.2.9 HomR(M,N)P ' HomRP(MP , NP ). So reduce to the case

where (R,P) is local. Now

HomR(k(P ),HomR(M,N)) = HomR(M,HomR(k(P ), N))

= Homk(P )(M ⊗R k(P ),HomR(k(P ), N))

Conclusion by Nakayama’s lemma 6.4.1 and 9.3.15.

9.3.18. Corollary. Let R be a noetherian ring and M a finite module. For a primeP the following conditions are equivalent:(1) P ∈ Supp(M).(2) P ∈ Ass(HomR(M,R/P )).

9.3.19. Proposition. Let R be a noetherian ring and F a finite module. AssumerankF ⊗R k(P ) = n for all primes P . Then F is locally free (projective) if andonly if FP is free for all P ∈ Ass(R).

Proof. LetQ be a maximal ideal and 0→ K → RnQ → FQ → 0 exact. Ass(K) ⊂

Ass(R), so KP = 0 for all P ∈ Ass(K). By 9.3.7 K = 0.

9.3.20. Theorem. Let (R,P ) be a noetherian local ring. If there is a nonzero finiteinjective module E, then R is artinian.

Proof. Let Q be a prime and f : R/Q → E. If a ∈ P\Q then aR/Q is injective,so there is f ′ : R/Q→ E such that f = f ′ ◦ aR/Q. That is P HomR(R/Q,E) =HomR(R/Q,E), so by Nakayama’s lemma 6.4.1 HomR(R/Q,E) = 0 if Q 6= P .By 9.3.7 0 6= HomR(R/P,E) ⊂ HomR(R/Q,E), so Q = P . R is artinian by8.6.6.

9.3.21. Exercise. (1) Show that

Ass(Z/(n)) = {(p)|p prime dividing n}


(2) Show that

Ass(K[X,Y ]/(X) ∩ (X2, Y 2)) = {(X), (X,Y )}and point out an embedded prime.

(3) Let I ⊂ R be an ideal such that√I = I . Show that R/I has no embedded prime

ideals.(4) Let I, J ⊂ R be a ideals such that JRP ⊂ IRP for all P ∈ Ass(R/I). Show that

J ⊂ I .

9.4. Primary modules

9.4.1. Definition. A submodule N ⊂ M is a primary submodule or more pre-cisely P -primary if Ass(M/N) = {P}. An ideal is a primary ideal if it is aprimary submodule of the ring.

9.4.2. Proposition. A prime ideal P ⊂ R is a P -primary ideal.

Proof. By 9.3.2 Ass(R/P ) = {P}.

9.4.3. Proposition. Let R be a noetherian ring and M a finite module. For aproper submodule N ⊂M the following conditions are equivalent:

(1) N ⊂M is primary for some prime(2) The set of zero divisors on M/N is contained in the radical

√Ann(M/N).

(3) For any zero divisor a on M/N there is a power an ∈ Ann(M/N).

Proof. (1)⇒ (2),(3): Supp(M/N) = V (P ) by 9.3.11. So by 9.1.6√

Ann(M/N) =P is the set of zero divisors by 9.3.9. (2),(3)⇒ (1): By 9.3.7 Ass(M/N) 6= ∅. IfP1, P2 ∈ Ass(M/N) then by 9.3.9 the zero divisors inP1∪P2 ⊂

√Ann(M/N) ⊂

P1 ∩ P2, so P1 = P2.

9.4.4. Corollary. Let R be a noetherian ring and I ⊂ R a proper ideal. Thefollowing conditions are equivalent:

(1) I ⊂ R is primary for some prime.(2) Any zero divisor in the ring R/I is nilpotent.(3) If ab ∈ I for some b /∈ I then some power an ∈ I .

9.4.5. Corollary. Let R be a noetherian ring.(1) If an ideal I ⊂ R is P -primary then

√I = P .

(2) If the radical√I = P is a maximal ideal, then I ⊂ R is P -primary.

(3) A finite power Pn ⊂ R of a maximal ideal is P -primary.

Proof. (1) By 9.3.11 Supp(R/I) = V (P ), so√I = P . (2) SuppR/I = V (I) =

{P}. By 9.3.2 and 9.3.7 Ass(R/I) = {P}. (3)√Pn = P .

9.4.6. Proposition. Let R be a noetherian ring and M a module.(1) If N,N ′ ⊂M are P -primary, then N ∩N ′ is P -primary.(2) If N ⊂ M is P -primary and M/N is finite, then Ann(M/N) ⊂ R is P -

primary and√

Ann(M/N) = P .

Proof. (1) From 9.3.5 Ass(M/N ∩ N ′) ⊂ Ass(M/N) ∪ Ass(M/N ′). (2) Letx1, . . . , xn be nonzero generators of M/N , then R/Ann(xi) ' Rxi ⊂ M/Nshows that Ann(xi) isP -primary. By (1) Ann(M/N) = ∩i Ann(xi) isP -primary.√

Ann(M/N) = P by 9.4.5.

9.4. PRIMARY MODULES 133

9.4.7. Definition. If i : M → U−1M is the canonical homomorphism to themodule of fractions and N ′ ⊂ U−1M is a submodule, then abuse the notationN ′ ∩M = i−1(N ′).

9.4.8. Theorem. Let R be a noetherian ring and M a module. Suppose U ⊂ R isa multiplicative subset and P is a prime ideal.

(1) If U ∩ P 6= ∅ and N ⊂M is P -primary, then U−1N = U−1M .(2) If U ∩ P = ∅ and N ⊂ M is P -primary, then U−1N ⊂ U−1M is PU−1R-

primary and N = U−1N ∩M .(3) If N ′ ⊂ U−1M is PU−1R-primary, then N ′ ∩M ⊂ M is P -primary and

N ′ = U−1(N ′ ∩M).

Proof. From 9.3.10: (1) Ass(U−1M/N) = ∅. (2) Ass(U−1M/N) = {PRP }.Now Ass(U−1N ∩ M/N) ⊂ Ass(M/N) = {P}. But (U−1N ∩ M/N)P =NP ∩MP /NP = 0, so N = U−1N ∩M by 9.3.7. (3) PU−1R ∩ R = P ∈Ass(M/N ′ ∩M) is the only associated prime disjoint from U . Let N ′ ∩M ⊂N ⊂ M be such that N/N ′ ∩M ' R/Q. If Q ∩ U 6= ∅, then U−1N = N ′ andtherefore N = N ′ ∩M is a contradiction, so N ′ ∩M ⊂M is P -primary.

9.4.9. Corollary. Let R be a noetherian ring. Suppose U ⊂ R is a multiplicativesubset and P is a prime ideal.

(1) If U ∩ P 6= ∅ and I ⊂ R is P -primary, then IU−1R = U−1R.(2) If U ∩ P = ∅ and I ⊂ R is P -primary, then IU−1R ⊂ U−1R is PU−1R-

primary and I = IU−1R ∩R.(3) If I ′ ⊂ U−1R is PU−1R-primary, then I ′ ∩ R ⊂ R is P -primary and I ′ =

(I ′ ∩R)U−1R.

9.4.10. Corollary. Let R be a noetherian ring and P a prime ideal.(1) If P ∈ Supp(M) of a finite module, then PnMP ∩M ⊂ M is a P -primary

submodule.(2) PnRP ∩R ⊂ R is a P -primary ideal.

Proof. (1) By Nakayama’s lemma 6.4.1 Supp(MP /PnMP ) = {PRP }. Then

PnMP ⊂MP is PRP -primary. Conclusion by 9.4.8.

9.4.11. Example. Let R = Z[X] and P = (p,X) for some prime number p. ThenR/P ' Z/(p), so P is a maximal ideal. The ideal Q = (p2, X) satisfies strictinclusions P 2 ⊂ Q ⊂ P . So

√Q = P gives that Q is P -primary, but Q is not a

power of P .

9.4.12. Example. Define the subring R = Z[pX,X2] ⊂ Z[X] for some primenumber p.

(1) R = {∑aiX|p|a2i+1}.

(2) The ideal P = (pX,X2) = {∑aiX|p|a2i+1, a0 = 0} has R/P ' Z, so P

is a non maximal prime ideal.(3) The ideal Q = (q, pX,X2) = {

∑aiX|p|a2i+1, q|a0} has R/Q ' Z/(q), so

for q a prime number P ⊂ Q is a maximal ideal.(4) The ideal P 2 = (p2X2, pX3, X4) = {

∑aiX|p|a2i+1, a0 = a1 = 0, p2|a2}

has p · pX2 = p2X2 ∈ P 2, pX2 /∈ P 2 and pn /∈ P 2, so by 9.4.4 P 2 is notprimary.

9.4.13. Exercise. (1) Let K be a field. Show that (X2, Y ) ⊂ K[X,Y ] is (X,Y )-primary.


(2) Let p be a prime number. Show that (pk) ⊂ Z is a primary ideal.

9.5. Decomposition of modules

9.5.1. Definition. A submodule L ⊂ M has a primary decomposition if thereexist a family Ni ⊂M of Pi-primary submodules, such that

L = N1 ∩ · · · ∩Nk

A primary decomposition is a reduced primary decomposition if Pi 6= Pj for i 6= jand no Ni can be excluded.

9.5.2. Lemma. Let R be a noetherian ring and L ⊂M submodule with a primarydecomposition. Then by intersection there is a reduced primary decomposition.

Proof. By 9.4.6 intersection of P -primary submodules is P -primary, so replacemore P -primary factors by their intersection.

9.5.3. Lemma. Let R be a noetherian ring and M a finite module.(1) For each Pi ∈ Ass(M) there is a submodule Ni ⊆ M such that Ass(Ni) =

Ass(M)− {Pi} and Ass(M/Ni)) = {Pi}.(2) The intersection ⋂

i

Ni = 0

(3) The module M injects

0→M →⊕

i

M/Ni

Proof. The submodule Ni is given by 9.3.6. Ass(∩Ni) = ∅, so conclusion by9.3.7.

9.5.4. Theorem. Let R be a noetherian ring and M a finite module. A propersubmodule L ⊂M has a reduced primary decomposition

L = N1 ∩ · · · ∩Nk

with Ni Pi-primary. For any such(1) The primes are determined

Ass(M/L) = {P1, . . . , Pk}(2) There is an inclusion

0→M/L→⊕

i

M/Ni

Proof. Apply 9.5.3 to M/L to get a reduced primary decomposition. By (2)Ass(M/L) ⊂ {P1, . . . , Pk}. For i there is an inclusion ∩j 6=iNj/L ' ∩j 6=iNj +Ni/Ni ⊂ M/Ni. It follows since the decomposition is reduced that {Pi} ∈Ass(∩j 6=iNj/L) ⊂ Ass(M/L). So {P1, . . . , Pk} ⊂ Ass(M/L).

9.5.5. Proposition. Let R be a noetherian ring and M a finite module. If

L = N1 ∩ · · · ∩Nk

is a reduced primary decomposition of L ⊂M and Pi is one of the minimal primesin Ass(M/L), then

Ni = LPi ∩M

9.5. DECOMPOSITION OF MODULES 135

and therefore uniquely determined.

Proof. Since Pi is minimal, NjPi= MPi for j 6= i. By 9.4.8 follows that Ni =

NPi ∩M = LPi ∩M .

9.5.6. Proposition. LetR be a noetherian ring andM a finite module. Let L ⊂Msuch that M/L 6= 0 has finite length. If Ass(M/L) = {P1, . . . , Pk}, then there isa reduced primary decomposition

L = N1 ∩ · · · ∩Nk

where

Ni = LPi ∩M

and an isomorphism

M/L '⊕

i

M/Ni

Proof. This follows from 9.5.4, 9.5.5 and 9.2.6. Since M/Ni have finite lengthSupp(M/Ni) = {Pi} and the conditions in 9.2.6 are satisfied.

9.5.7. Proposition. Let R be a noetherian ring and M a finite length module. IfAss(M) = {P1, . . . , Pk}, then there is a reduced primary decomposition

0 = P1nM ∩ · · · ∩ Pk

nM

and exact sequences


There are isomorphisms

M '⊕

i

MPi '⊕

i

M/PinM


9.5.8. Proposition. LetR be a noetherian ring andM a finite module. Let L ⊂Mhave a reduced primary decomposition

L = N1 ∩ · · · ∩Nk

where Ni is Pi-primary. Assume U to be a multiplicative subset disjoint fromexactly P1, . . . , Pm. Then

U−1L = U−1N1 ∩ · · · ∩ U−1Nm

is a reduced primary decomposition.


9.5.9. Exercise. (1) Describe the primary decomposition over a field.


9.6. Decomposition of ideals

9.6.1. Theorem. Let R be a noetherian ring. A proper ideal I has a reducedprimary decomposition

I = Q1 ∩ · · · ∩Qk

with Qi Pi-primary. For any such(1) The primes are determined

Ass(R/I) = {P1, . . . , Pk}(2) There is an inclusion

0→ R/I →⊕

i

R/Qi

Proof. This is a case of 9.5.4.

9.6.2. Proposition. Let R be a noetherian ring. If

I = Q1 ∩ · · · ∩Qk

is a reduced primary decomposition andPi one of the minimal primes in Ass(R/I),then

Qi = IRPi ∩Rand therefore uniquely determined.


9.6.3. Definition. Let P be a prime ideal. The symbolic power of P is

P (n) = R ∩ PnRP

9.6.4. Lemma. Let P ⊂ R be a prime ideal in a noetherian ring.(1) If P is a maximal ideal then P (n) = Pn is P -primary.(2) PnRP ⊂ RP is PRP -primary.(3) P (n) ⊂ R is P -primary.

Proof. (1) By 9.4.5 Pn is primary and by 9.4.9 Pn = P (n). (2) This follows from(1). (3) This follws from 9.4.10.

9.6.5. Proposition. Let R be a noetherian ring and let

Pn = Q1 ∩ · · · ∩Qk

be a reduced primary decomposition of a power of a prime ideal P . If Q1 is P -primary, then

Q1 = P (n)


9.6.6. Proposition. Let I ⊂ R be a proper ideal in a noetherian ring such thatR/I is artinian. If Ass(R/I) = {P1, . . . , Pk}, then there is a reduced primarydecomposition

I = P(n)1 ∩ · · · ∩ P (n)

k

and an isomorphismR/I '

⊕i

R/P(n)i


9.6. DECOMPOSITION OF IDEALS 137

9.6.7. Proposition. Let R be an artinian ring. If Ass(R) = {P1, . . . , Pk}, thenthere is a reduced primary decomposition

0 = Pn1 ∩ · · · ∩ Pn

k

and exact sequences0→ Pn

i → R→ RPi → 0There are isomorphisms

R '⊕

i

RPi '⊕

i

R/Pni


9.6.8. Proposition. Let a proper ideal I ⊂ R in a noetherian ring have a reducedprimary decomposition

I = Q1 ∩ · · · ∩Qk

where Qi is Pi-primary. Assume U to be a multiplicative subset disjoint fromexactly P1, . . . , Pm. Then

IU−1R = Q1U−1R ∩ · · · ∩QmU

−1R

is a reduced primary decomposition.


9.6.9. Proposition. Let R be a noetherian ring and I an ideal. Then√I = P1 ∩ · · · ∩ Pk

is the primary decomposition and the elements in Ass(R/√I) = {P1, . . . , Pk} are

minimal primes over I .

Proof. By 9.3.13 there are only finitely many minimal primes over I . By 5.1.8√I

is the intersection of these and by 9.6.1 this is a reduced primary decompositiondetermining the set Ass(R/

√I).

9.6.10. Corollary. Let R be a reduced noetherian ring. Then all elements inAss(R) are minimal primes. That is, there are no embedded primes.

9.6.11. Corollary. LetR be a noetherian ring. The following statements are equiv-alent:

(1) R is reduced.(2) RP is a field for all P ∈ Ass(R).(3) RP is a domain for all P ∈ Ass(R).

Proof. (1)⇒ (2): 0 = P1 ∩ · · · ∩ Pk where Pi are minimal. Then 0 = PiRPi isthe decomposition by 9.6.8, so the maximal ideal is zero. (3)⇒ (1): Ass(

√0) ⊂

Ass(R) and√

0P =√

0 ⊂ RP for all P ∈ Ass(R). So√

0 = 0 by 9.3.7.

9.6.12. Example. Let R be a unique factorization domain. A factorization intopowers of different irreducible primes is a reduced primary decomposition of aprincipal ideal.

(1) Let (p) be a prime divisor. The homomorphism 1pn−1 : R → R/(pn) fits toan exact sequence 0 → R/(p) → R/(pn) → R/(pn−1) → 0. By 9.3.3 andinduction Ass(R/(pn)) ⊂ Ass(R/(p)) ∪Ass(R/(pn−1)) = {(p)} gives thatany power (pn) is a (p)-primary ideal.


(2) If (a) = (pn11 ) . . . (pnk

k ) is a prime factorization then

(a) = (pn11 ) ∩ · · · ∩ (pnk

k )

0→ R/(a)→ R/(pn11 )× · · · ×R/(pnk

k )

is a primary decomposition and Ass(R/(a)) = {(p1), . . . , (pk)}.9.6.13. Exercise. (1) Let I ⊂ R be an ideal. Show that if P =

√I is a maximal ideal,

then I is a P -primary ideal.(2) Let I ⊂ R be an ideal. Show that if I contains a power of a maximal ideal P , then I

is a P -primary ideal.(3) Let K be a field and I = (X2, XY ) ⊂ K[X,Y ]. Show that

√I = (X), but I is not

(X)-primary.

10

Dedekind rings

10.1. Principal ideal domains

10.1.1. Lemma. Let R be a domain. The set of elements x ∈M in a module withAnn(x) 6= 0 is a submodule.

Proof. If ax = by = 0 then ab(x+ y) = 0. Now use that R is a domain.

10.1.2. Definition. LetR be a domain. An element x ∈M in a module is a torsionelement if Ann(x) 6= 0. By 10.1.1 the set of torsion elements is a submodule

T (M) = {x ∈M |Ann(x) 6= 0}of M , the torsion submodule. If T (M) = 0 then M is a torsion free module. IfT (M) = M then M is a torsion module.

10.1.3. Lemma. Let R be a domain and M a module.(1) M is torsion free if and only if any nonzero a ∈ R is a nonzero divisor on M .(2) The factor M/T (M) is torsion free.(3) Let K be the fraction field of R.

T (M) = Ker(M →M(0)) = Ker(M →M ⊗R K)

(4) If U ⊂ R is multiplicative, then U−1T (M) = T (U−1M).(5) T is a left exact functor: if 0 → M → N → L is exact, then 0 → T (M) →

T (N)→ T (L) is exact.

Proof. (1) x ∈ T (M) ⇔ Ann(x) 6= 0. (2) Let x + T (M) ∈ M/T (M) and0 6= a ∈ R such that ax ∈ T (M). Then bax = 0 for some b 6= 0. It follows thatx + T (M) = 0. (3) From 4.4.1 M → M ⊗R K is the canonical homomorphismM →M(0) to the fractions by all nonzero denominators, so the kernel is T (M) by4.2.7. (4) By (3) U−1T (M) = Ker(U−1M → U−1M ⊗K) = T (U−1M). (5)If f : M → N is injective and f(x) ∈ T (N), then x ∈ T (M), so f(T (M)) =Ker(T (N)→ T (L)).

10.1.4. Corollary. Let R be a domain and F a flat module. Then F is torsion free.

Proof. If 0 6= a ∈ R and F flat, then aF = aR ⊗ 1F is injective.

10.1.5. Corollary. Let R be a domain and M a module. The following conditionsare equivalent:

(1) M is torsion free.(2) MP is torsion free for all prime ideals P .(3) MP is torsion free for all maximal ideals P .

Proof. By 10.1.3 T (M)P = T (MP ). Conclusion by the local-global principle5.4.1.

139

140 10. DEDEKIND RINGS

10.1.6. Corollary. Let R be a noetherian domain and M a module. The followingconditions are equivalent:

(1) M is a torsion module.(2) (0) /∈ Supp(M).(3) (0) /∈ Ass(M).

Proof. (1)⇔ (2): By 10.1.3 T (M) = Ker(M →M(0)). (2)⇔ (3): By 9.3.11 theset of primes Ass(M) and Supp(M) have the same minimal elements.

10.1.7. Proposition. Let R be a principal ideal domain. A submodule of a finitefree module is free.

Proof. Let F ⊂ Rn be a submodule. If n = 1 F is a principal ideal and free.Let n > 1 and p : Rn → R be the last projection. Then p(F ) is a principal idealand free. By induction F ∩ Ker p ⊂ Rn−1 is free, so the split exact sequence0→ F ∩Ker p→ F → p(F )→ 0 gives F ' F ∩Ker p⊕ p(F ) is free.

10.1.8. Proposition. Let R be a principal ideal domain.(1) A torsion free module is flat.(2) A finite torsion free module is free.(3) A finite torsion module has finite length.

Proof. (1) For a nonzero ideal (a) ⊂ R the composite R ' (a) → R is aM .For a torsion free F the homomorphism aF : F ' (a) ⊗R F → F is injective.So F is flat by 3.7.12. (2) By (1) a finite torsion free module F is flat. By 8.5.7F is projective and therefore isomorphic to a submodule of a finite free module.Conclusion by 10.1.7. (3) By 10.1.6 there are only maximal ideals in the support.Conclusion by 9.2.13.

10.1.9. Proposition. Let R be a principal ideal domain. A finite module M de-composes

M = T (M)⊕ Fas a direct sum of the torsion submodule and a finite free submodule F .

Proof. By 10.1.8 M/T (M) is free, so 0 → T (M) → M → M/T (M) → 0 issplit exact. This gives M/T (M) ' F ⊂M .

10.1.10. Proposition. Let R be a principal ideal domain. A finite torsion moduleM has a decomposition

M =⊕

(p)∈Ass(M)

Mp

where (p) is a prime divisor and

Mp = {x ∈M |pnx = 0, for some n}

Proof. This comes from the primary decomposition 9.5.7. M ' ⊕pM/(pn)M .If x ∈ Mp and q 6= p then R = (p, qn) ⊂ Ann(x + (qn)M). Therefore Mp =Ker(M → ⊕q 6=pM/(qn)M) 'M/(pn)M .

10.1.11. Proposition. Let R be a principal ideal domain and (p) a prime divisor.A finite torsion module M such that M = Mp has decomposition

M = R/(pn1)⊕ · · · ⊕R/(pnk)

where n1 ≥ · · · ≥ nk.

10.2. DISCRETE VALUATION RINGS 141

Proof. Let n1 = n be such that pn ∈ Ann(M), but pn−1x 6= 0 for some x ∈ M .The short exact sequence 0 → Rx → M → M/Rx → 0 of R/(pn)-modulesis split exact, since R/(pn) ' Rx is an injective module 7.6.11. Conclusion byinduction on `R(M).

10.1.12. Theorem. Let R be a principal ideal domain and M a finite module.Then M has decomposition

M = R/(pn11 )⊕ · · · ⊕R/(pnk

k )⊕Rn

where (p1), . . . , (pk) are not necessarily distinct prime divisors.

10.1.13. Example. (1) A finite abelian group is isomorphic to

Z/(pn11 )⊕ · · · ⊕ Z/(pnk

k )

where p1, . . . , pk are not necessarily distinct prime numbers.(2) A finite dimensional C vector space with a linear map to itself is by 2.1.13 a

C[X] module

C[X]/(X − λ1)n1 ⊕ · · · ⊕C[X]/(X − λk)nk

where λi are the not necessarily distinct eigenvalues. Choosing C-basis 1, (X−λ), . . . , (X − λ)n−1 in C[X]/(X − λ)n, then the matrix of multiplication byX is a lower Jordan block 0BBB@

λ1 λ

. . .1 λ

1CCCAJust use the formulas X(X − λ)i = (X − λ)i+1 + λ(X − λ)i. All together

this gives the Jordan block decomposition.

10.1.14. Exercise. (1) Show that a nonzero finite Z-submodule of Q is a free moduleof rank 1.

(2) Show that a finite torsion Z-module is a finite group.(3) Let K be a field. Show that a finite torsion K[X]-module is a finite K vector space.(4) Let R be a noetherian domain such that every nonzero prime ideal is maximal. Let

M be a finite module. Show that the torsion submodule T (M) has finite length.

10.2. Discrete valuation rings

10.2.1. Definition. A local principal ideal domain, which is not a field, is a dis-crete valuation ring. A generator of the maximal ideal is a local parameter or auniformizing parameter.

10.2.2. Proposition. Let (R, (p)) be a discrete valuation ring. Then⋂n

(pn) = 0

Proof. See Krull’s intersection theorem 8.5.5.

10.2.3. Proposition. Let (R,P ) be a noetherian local domain. The followingstatements are equivalent:

(1) R is a discrete valuation ring.(2) P is a principal ideal.(3) `R(P/P 2) = 1.


Proof. (1)⇒ (2)⇔ (3): Nakayama’s lemma 6.4.1. (2)⇒ (1): Let P = (p) and0 6= a R, by 10.2.2 there is n such that a ∈ (pn) − (pn+1). Since any ideal isfinitely generated it follows that a nonzero ideal is of the form (pn).

10.2.4. Proposition. Let (R, (p)) be a discrete valuation ring. Any nonzero idealis of the form (pn) for a unique n = 0, 1, 2, . . . .

Proof. (p) is the only prime divisor. Conclusion by unique factorization 1.5.6.

10.2.5. Corollary. Let (R, (p)) be a discrete valuation ring. Any nonzero elementin the fraction field K of R has a unique representation upn where u is a unit andn ∈ Z.

10.2.6. Definition. Let K be a field. A surjective map v : K\{0} → Z satisfying(1) v(xy) = v(x) + v(y)).(2) If x+ y 6= 0, then v(x+ y) ≥ min(v(x), v(y)).

is a valuation on K.

10.2.7. Proposition. Let v be a valuation on a field K.(1) The subset R = {x|v(x) ≥ 0} ∪ {0} is subring a of K.(2) The units in R are {x|v(x) = 0}.(3) Any element p, v(p) = 1 generates the same prime divisor (p).(4) (R, (p)) is a discrete valuation ring.

Proof. (1) Calculate v(±1) = 0, so ±1 ∈ R. If v(x), v(y) ≥ 0 then v(xy) ≥ 0and v(x + y) ≥ 0, so x, y ∈ R ⇒ xy, x − y ∈ R. Therefore R is a subring. (2)v(x) = 0 ⇒ v(x−1) = 0. (3), (4) If v(p) = v(q) = 1 then pq−1 is a unit, so(p) = (q). If 0 6= x ∈ R with v(x) = n > 0 then x = (xp−1)p ∈ (p), so (p) is theunique maximal ideal.

10.2.8. Definition. The ring in 10.2.7 is the discrete valuation ring of v.

10.2.9. Proposition. Let (R, (p)) be a discrete valuation ring with fraction fieldK. The map

v : K\{0} → Z, upn 7→ n

is a valuation and R is the discrete valuation ring of v.

Proof. By 10.2.4 v is well defined. Calculate upmvpn = uvpm+n and if m ≤ n,upm+vpn = (u+vpn−m)pm to see that v is a valuation. ClearlyR is the valuationring of v.

10.2.10. Proposition. Let (R, (p)) be a discrete valuation ring. A finite moduleMhas decomposition

M = R/(pn1)⊕ · · · ⊕R/(pnk)⊕Rn

where n1 ≥ · · · ≥ nk > 0.


10.2.11. Example. Let (k[[X]], (X)) be the power series ring over a field. A finitemodule is isomorphic to

k[[X]]/(Xn1)⊕ · · · ⊕ k[[X]]/(Xnk)⊕ k[[X]]n

where n1 ≥ · · · ≥ nk > 0.

10.3. DEDEKIND DOMAINS 143

10.2.12. Exercise. (1) LetK be a field. Show that the subringK[[X2, X3]] ⊂ K[[X]]is not a discrete valuation ring.

10.3. Dedekind domains

10.3.1. Definition. A noetherian domain R, which is not a field, is a Dedekinddomain if all local rings RP at nonzero prime ideals are discrete valuation rings.

10.3.2. Proposition. Let R be a Dedekind domain.(1) Any nonzero prime ideal is maximal.(2) If U ⊂ R is multiplicative, then U−1R is a field or a Dedekind domain.

Proof. (1) Prime ideals in a nonzero prime ideal P correspond to prime ideals inthe discrete valuation ring RP . So (0), P are the only primes and P is maximal.(2) This is clear from 5.2.13.

10.3.3. Proposition. Let R be a noetherian domain which is not a field. The fol-lowing conditions are equivalent:

(1) R is a Dedekind domain.(2) Every nonzero proper ideal in R is a product of finitely many maximal ideals.(3) `R(P/P 2) = 1 for all maximal ideals.

Proof. (1) ⇒ (2): By 9.6.4 the primary ideals P (n) = Pn. Conclusion by 9.6.5and Chinese remainders 1.4.2. (2)⇒ (1): Assume (R,P ) is local. By Nakayama’slemma 6.4.1 there is p ∈ P\P 2. Since (p) = Pn it follows that (p) = P andR is a discrete valuation ring. (1) ⇔ (3): 10.2.3 and the isomorphism P/P 2 'PRP /P

2RP .

10.3.4. Proposition. Let R be Dedekind domain.(1) If R is a unique factorization domain then it is a principal ideal domain.(2) If R has only finitely many maximal ideals then it is a principal ideal domain.

Proof. (1) Any nonzero prime ideal is principal. Conclusion by 10.3.3. (2) LetP, P2 . . . , Pk be the finitely many maximal ideals. Choose a ∈ P\P 2∪P2 · · ·∪Pk,5.1.3. Then (a) is P primary. By 10.3.3 (a) = Pn, so (a) = P . As all maximalideals are principal, conclusion by 10.3.3.

10.3.5. Proposition. Let R be Dedekind domain. An ideal I is generated by atmost two elements.

Proof. Let Ass(R/I) = {P1, . . . , Pk} and U = R\P1 ∪ · · · ∪ Pk, then by 10.3.4U−1R is a principal ideal domain. By 10.1.6 U−1R ' U−1I , so choose by 8.2.9a homomorphism f : R → I such that U−1f is an isomorphism. Then f isinjective and the ideal f(R) = (a) ⊂ I satisfies: Pi /∈ Supp(I/(a)) for any i.Therefore Ass(I/(a)) ∩ V (I) = ∅. Let Q1, . . . , Qm ∈ Ass(I/(a)) and chooseb ∈ I\Q1∪· · ·∪Qm. By 9.3.9 b is a nonzero divisor on I/(a) and therefore bI/(a)

is an isomorphism as I/(a) has finite length, 7.2.6. It follows that I = (a, b).

10.3.6. Theorem. Let R be Dedekind domain.(1) A torsion free module is flat.(2) A finite torsion free module is projective.(3) Any ideal is projective.


(4) Let F be a finite torsion free module. Then there is a number n and an idealI such that

F ' Rn ⊕ I(5) A finite torsion module has finite length.

Proof. (1), (2), (3), (5) These follow from 10.1.6, 6.5.16. (4) Let R have fractionfield K and assume rankK F ⊗R K = n + 1. Choose a nonzero homomorphismF → R and get by induction on n, F ' I1 ⊕ · · · ⊕ In+1 for nonzero ideals Ijin R. It suffices to treat the case n = 1. Let Ass(R/I1) = {P1, . . . , Pk} andU = R\P1∪ · · ·∪Pk, then by 10.3.4 U−1R is a principal ideal domain. By 10.1.8U−1I2 ' U−1R, so choose by 8.2.9 a homomorphism f : I2 → R such thatU−1f is an isomorphism. Then f is injective and the ideal f(I2) ⊂ R satisfies:Pi /∈ Ass(R/f(I2)) for any i. It follows that I1 + f(I2) = R. Conclusion by asplit exact sequence

0→ I → I1 ⊕ I2 → R→ 0

10.3.7. Proposition. LetR be a Dedekind domain. A finite moduleM decomposes

M = T (M)⊕ Fas a direct sum of the torsion submodule and a finite torsion free submodule F .

Proof. By 10.3.6 the projection M →M/T (M) splits.

10.3.8. Proposition. Let R be a Dedekind domain. A finite torsion module M hasa decomposition

M ' R/Pn11 ⊕ · · · ⊕R/P

nkk

where P1, . . . , Pk are not necessarily distinct maximal ideals.

Proof. From 9.5.7 M ' ⊕PMP and by 10.2.10 the RP torsion module MP '⊕iRP /P

niRP ' ⊕iR/Pni .

10.3.9. Theorem. Let R be a Dedekind domain and M a finite module. Then Mhas a decomposition

M = R/Pn11 ⊕ · · · ⊕R/P

nkk ⊕R

n ⊕Q1 · · ·Ql

where P1, . . . , Pk, Q1, . . . , Ql are not necessarily distinct maximal ideals.

10.3.10. Example. R = Z[X]/(X2−5) is a noetherian domain. The ideal (2, X+1) is a prime ideal in Z[X] andX2−5 = (X+1)2−2(X+1)−4 ∈ (2, X+1), sothis corresponds to a prime ideal P ⊂ R. Calculate P/P 2 = (2, X+1)/(4, 2(X+1), (X+1)2, X2−5) = (2, X+1)/(4, 2(X+1), (X+1)2) which has a nontrivialsubmodule (2, (X+1)2)/(4, 2(X+1), (X+1)2). P/P 2 is not simple, so by 10.3.3R is not a Dedekind domain.

10.3.11. Exercise. (1) Show that the ring Z[√−5] is a Dedekind domain.

(2) Show that the ring Z[√

5] is not a Dedekind domain.

Bibliography

A. Altman and S. Kleiman, Introduction to Grothendieck duality theory, Springer-Verlag 1970.M. Atiyah and I. Macdonald, An introduction to commutative algebra, Addison-Wesley 1969.N. Bourbaki, Algébre, Hermann 1942-.N. Bourbaki, Algébre commutative, Hermann-Masson 1961-.M. Brodmann and R. Sharp, Local cohomology: an algebraic introduction with geometric applica-tions, Cambridge University Press 1997.W. Bruns and J. Herzog, Cohen-Macaulay rings, Cambridge University Press 1993.H. Cartan and S. Eilenberg, Homological algebra, Princeton 1956.J. Dieudonné, Cours de géométrie algébrique, Presses Universitaires de France 1974.D. Eisenbud, Commutative algebra with a view toward algebraic geometry, Springer-Verlag 1996.E.G. Evans and P. Griffith, Syzygies, Cambridge University Press 1985.R. Fossum, The divisor class group of a Krull domain, Springer-Verlag 1973.W. Fulton, Algebraic curves, Benjamin inc. 1969.W. Fulton, Intercection theory, Springer-Verlag 1984.R. Gilmer, Multiplicative ideal theory, Marcel Dekker 1972.A. Grothendieck, Sur quelques points d’algégre homologique, Tôhoku Math, Journ., 9, 1957.A. Grothendieck, Eléments de géométrie algébrique, IHES 1960-67.A. Grothendieck, Séminaire de géométrie algébrique, IHES 1960-67.A. Grothendieck, Local cohomology, Springer-Verlag 1967.R. Hartshorne, Residues and duality, Springer-Verlag 1966.R. Hartshorne, Algebraic geometry, Springer-Verlag 1977.M. Hochster, Topics in homological theory of modules over commutative rings, American Mathe-matical Society 1974.H.C. Hutchins, Examples of commutative rings, Polygonal Publishing House 1981.B. Iversen, Generic local structure in commutative algebra, Springer-Verlag 1974.B. Iversen, Cohomology of sheaves, Springer-Verlag 1986.B. Iversen, Local rings, Aarhus 1974-80.I. Kaplansky, Commutative rings, Allyn and Bacon 1970E. Kunz, Introduction to commutative algebra, Birkhäuser 1980.J.P. Lafon, Algébre commutative, Hermann 1977.S. Lang, Algebra, Addison-Wesley 1965.H. Matsumura, Commutative algebra, Benjamin inc. 1970.J. Milne, http://www.jmilne.org.D. Mumford, Introduction to algebraic geometry, Red and 1 inch thick.D. Mumford, Algebraic geometry I, complex projective varieties, Springer-Verlag 1976.M. Nagata, Local rings, Interscience Publ. 1962.D.G. Northcott, An introduction to homological algebra, Cambridge University Press 1960.D.G. Northcott, Lessons on rings, modules and multiplicities, Cambridge University Press 1968.C. Peskine, An algebraic introduction to complex projective geometry, Cambridge University Press1997.C. Peskine and L. Szpiro, Dimension projective finie et cohomologie locale, IHES 1973.M. Reid, Undergraduate commutative algebra, Cambridge University Press 1995.P.C. Roberts, Homological invariants of modules over commutative rings, Les Presses del’Université de Montréal 1980.P.C. Roberts, Multiplicities and Chern classes in local algebra, Cambridge University Press 1998.J-P. Serre, Algébre locale - multiplicités, Springer-Verlag 1965.J-P. Serre, Faisceaux algébrique cohérents, Ann. of Math., 61, 1955.

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146 BIBLIOGRAPHY

I. Shafarevich, Basic algebraic geometry, Springer-Verlag 1974.R. Sharp, Steps in commutative algebra, Cambridge University Press 1990.W.V. Vasconcelos, Computational methods in commutative algebra and algebraic geometry,Springer-Verlag 1998.B.L. van der Waerden, Algebra, Springer-Verlag 1967.O. Zariski and P. Samuel, Commutative algebra, van Nostran 1958-60.

Index

(B), 11(R, P ), 75(aij), 89(xi), 891M , 211x, 27A/B, 9Aij , 89I : J , 11IJ , 11IM , 23IS, 11J ∩R, 11M(P ), 77M/N , 24M [ 1

u], 67

M∨, 33MP , 77N ′ ∩M , 133N : L, 27P -primary, 132P (n), 136R/I , 11RY , 23R[B], 17R[X], 16R[[X]], 20R[ 1

u], 64

RP , 75T (M), 139U−1M , 65U−1R, 63U−1f , 66V (B), 123XB , 123Ann, 27Ass, 128Cok, 25HomR, 30Im, 25Ker, 11, 25Spec, 123Supp, 126det, 89

C, 10Fp, 19N, 10Q, 10R, 10Z, 10ι, 63`R, 102⊕, 28⊗, 35⊗R, 34Q

, 28rankR, 91√

I , 14aM , 22b|a, 15c(f), 71char, 12eα, 29ei, 87evx, 33f ′, 18f(P ), 77f∨, 33fP , 77k(P ), 75o(f), 20v :, 142aφ, 1230-sequence, 41

abelian group, 9addition, 9alternating:, 90annihilator, 27artinian module, 105artinian modules, 105artinian ring, 107artinian rings, 107ass of modules, 128associated map, 123associated prime ideal, 128associative, 9, 21

basis, 29

147

148 INDEX

bilinear, 21binomial formula, 10

canonical homomorphism, 65canonical ring homomorphism, 63Cayley-Hamilton’s theorem, 91change of ring, 37change of rings, 37characteristic, 12Chinese remainder theorem, 14Chinese remainders, 14coefficient, 16, 20cofactor matrix, 89cokernel, 25colon ideal, 11, 27comaximal ideals, 14commutative, 9composition series, 102constants, 16content of polynomial, 71contracted ideal, 11contravariant, 31cosets, 9Cramer’s rule, 90

decomposable, 30decomposition of ideals, 136decomposition of modules, 134Dedekind domain, 143Dedekind domains, 143degree, 16derivative, 18determinant, 89direct product, 28direct sum, 28, 29discrete valuation ring, 141discrete valuation ring of v, 142discrete valuation rings, 141distributive, 9divisible module, 58division, 15domain, 10dual homomorphism, 33dual module, 33

embedded prime, 130essential extension, 59evaluation, 33evaluation map, 17exact functor, 50, 51exact sequence, 41exact sequences, 41exactness of fractions, 67exactness of hom, 50exactness of tensor, 53extended ideal, 11

factor group, 9

factor module, 24factor ring, 11faithfully flat, 83faithfully flat ring homomorphism, 84faithfully flat ring homomorphisms, 83field, 10field extension, 19fields, 19finite field extension, 19finite ideal, 11finite length, 102finite module, 87finite modules, 87finite presented module, 94finite presented modules, 94finite ring extension, 99finite ring homomorphism, 99finite ring homomorphisms, 99finite type ring, 17finite type rings, 117finitely generated ring, 17five lemma, 48flat module, 60flat modules, 60flat ring homomorphism, 81flat ring homomorphisms, 81fraction field, 64free module, 29free modules, 89Frobenius homomorphism, 12functor, 31

Gauss’ lemma, 71generated, 23going-down, 85going-up, 100Gorenstein ring, 112greatest common divisor, 16

Hilbert’s basis theorem, 117homomorphism, 9, 21homomorphism module, 30homomorphism modules, 30homomorphism modules of fractions, 70

ideal, 11ideal generated by, 11ideal product, 11ideals, 11idempotent, 15identity, 9, 21identity isomorphism, 9, 21image, 25indecomposable, 30induced module, 39injections, 28injective envelope, 59injective module, 56

INDEX 149

injective modules, 56irreducible component, 125irreducible element, 15irreducible principal ideal, 15irreducible space, 125irreducible subset, 125isomorphism, 9, 21

Jacobson radical, 77

kernel, 11, 25kernel and cokernel, 25Krull’s intersection theorem, 117, 120Krull’s theorem, 73

leading coefficient, 16least common multiple, 16left exact contravariant functor, 50left exact functor, 50length, 102linear map, 22local artinian ring, 110local parameter, 141local ring, 75local ring homomorphism, 75localization, 109localization of modules, 77localization of noetherian rings, 120localization of rings, 75localized homomorphism, 77localized module, 77localized ring, 75locally free module, 79

maximal ideal, 13minimal prime, 126minimal prime ideal, 74minor, 89module, 21module of fractions, 65modules and homomorphisms, 21modules of fractions, 65monic polynomial, 16monomial, 16multilinear:, 89multiplication, 9multiplication of principal ideals, 15multiplicative subset, 63multiplicity, 18

Nakayama’s lemma, 93natural homomorphism, 31natural isomorphism, 31negative, 9nilpotent, 14nilradical, 14noetherian module, 113noetherian ring, 115noetherian rings, 115

noncommutative ring, 9nontrivial idempotent, 15nonzero divisor, 10, 22normed:, 90notherian modules, 113

order, 20

polynomial, 16polynomial ring, 16polynomials, 16power series, 20power series ring, 20power series rings, 118primary decomposition, 134primary ideal, 132primary modules, 132primary submodule, 132prime divisor, 15prime element, 15prime fields, 19prime filtrations of modules, 121prime ideal, 13prime ideals, 13, 73principal ideal, 11principal ideal domain, 15principal ideal domains, 139principal open subsets, 123product ring, 10projection, 9, 24projections, 28projective module, 55projective modules, 55proper ideal, 11

radical, 14rank, 91reduced, 14reduced primary decomposition, 134reflexive module, 33residue field, 75residue homomorphism, 77restriction of scalars, 22retraction, 43right exact contravariant functor , 50right exact functor, 50ring, 9ring extension, 9ring generated, 17ring of fractions, 63rings, 9rings of fractions, 63root, 18roots, 18

scalar multiplication, 21, 22section, 43semi-local ring, 128

150 INDEX

short exact sequence, 43simple module, 101simple modules, 101simple root, 18snake homomorphism, 46snake lemma, 47spectrum, 123split exact sequence, 44standard basis, 29subfield, 19subgroup, 9submodule, 21submodule generated, 23submodules and factor modules, 23subring, 9sum and product, 28support, 126support of modules, 126symbolic power, 136

tensor modules of fractions, 69tensor product, 34tensor product modules, 34tensor product ring, 39the length, 102The local-global principle, 79the polynomial ring is factorial, 71the snake lemma, 45torsion element, 139torsion free module, 139torsion module, 139torsion submodule, 139total ring of fractions, 64

uniformizing parameter, 141unique factorization, 15unique factorization domain, 15unit, 10

valuation, 142vector space, 22

windmill lemma, 49

Zariski topology, 123zero, 9zero divisor, 10, 22zero ideal, 11zero module, 21zero submodule, 21

elementary commutative algebra

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