Elementary Comutative 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 332.7. Change of rings 36

3. Exact sequences of modules 393.1. Exact sequences 393.2. The snake lemma 433.3. Exactness of Hom 483.4. Exactness of Tensor 493.5. Projective modules 503.6. Injective modules 523.7. Flat modules 54

4. Fraction constructions 574.1. Rings of fractions 574.2. Modules of fractions 584.3. Exactness of fractions 604.4. Tensor modules of fractions 624.5. Homomorphism modules of fractions 634.6. The polynomial ring is factorial 64

5. Localization 655.1. Prime ideals 655.2. Localization of rings 67

5

6 CONTENTS

5.3. Localization of modules 685.4. Exactness and localization 705.5. Flat ring homomorphisms 71

6. Finite modules 736.1. Finite Modules 736.2. Free Modules 756.3. Cayley-Hamiltons theorem 776.4. Nakayamas Lemma 786.5. Finite Presented Modules 806.6. Finite ring homomorphisms 83

7. Modules of finite length 857.1. Simple Modules 857.2. The Length 857.3. Artinian Rings 877.4. Localization 907.5. Local artinian ring 91

8. Noetherian modules 938.1. Modules and submodules 938.2. Noetherian rings 948.3. Finite type rings 958.4. Power series rings 978.5. Fractions and localization 988.6. Prime filtrations of modules 98

9. Primary decomposition 1019.1. Support of modules 1019.2. Ass of modules 1039.3. Primary modules 1069.4. Decomposition of modules 1069.5. Decomposition of ideals 108

10. Dedekind rings 11110.1. Principal ideal domains 11110.2. Discrete valuation rings 11210.3. Dedekind domains 113

Bibliography 117

Index 119

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.

Nielsen, University of Aarhus, Winter 2004

7

1A dictionary on rings and ideals

1.1. Rings

1.1.1. Definition. An abelian group is a setAwith an additionAA A, (a, b) 7a+ 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+ afor 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 : A C respects addition (a+ b) = (a) + (b).The projection pi : A A/B, a 7 a+B is a homomorphism. If (b) = 0 for allb B, then there is a unique homomorphism : A/B C such that = pi.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 = bafor 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 rings : R S is an additive group homomorphism respecting multiplication andidentity

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

An isomorphism is a homomorphism : R S having an inverse map 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 : R S is a ring homomorphism, then (0) = 0 and (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 groupsR1R2, ((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 R1R2 R1, (a1, a2) 7 a1 and R1R2 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

)ankbk

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 = a1.

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 integersZ is a domain. The units inZ are {1}. The rationalnumbersQ, the real numbersR and the complex numbersC are fields. The naturalnumbersN 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 : 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 {I} be a family of ideals in R. Then the additive subgroups

I and I are ideals.

Proof. The claim for the intersection is clear. Use the formulasb +

c =

(b + c) and ab =

ab 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) LetK be a field. InK 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 =

bB Rb

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

Rb = {ab|a R}A finite ideal is

(b1, . . . , bn) = Rb1 + +RbnProof. The righthand side is contained in the ideal RB. Moreover the righthandside is an ideal containing B, so equality.

1.2.6. Definition. Let : R S be a ring homomorphism. For an ideal J Sthe contracted ideal is 1(J) R and denoted J R. The kernel is the idealKer = 1(0). For an ideal I R the extended ideal is the ideal (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+ Iis 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 ab = a(b b) + b(a a) I . Therefore ab + I = ab + 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 pi : R R/I, a 7 a+ I is a ringhomomorphism.


1.2.9. Proposition. Let : R S be a ring homomorphism.(1) Let I Ker be an ideal. Then there is a unique ring homomorphism

: R/I S such that = pi.

R

pi

// S

R/I

==

(2) The homomorphism : R/Ker S is a ring isomorphism onto thesubring (R) of S.

R

pi

// (R)

R/Ker

::

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

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

1.2.10. Corollary. Let pi : R R/I be the projection. The map I 7 J =pi1(I ) gives a bijective correspondence between ideals I in R/I and ideals J inR containing I . Also I = pi(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 wherem 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

)apkbk = 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 productIJ = {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 ringK is a field if and only if any nonzero ring homomorphism

: K R is injective.(6) Letm,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 : R S is a ring homomorphism and Q S a primeideal then 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/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 isI = {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+nk=0

(m+ nk

)am+nkbk 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) Letm be a natural number. Show that Z/(m) is reduced ifm 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 thatI P .

1.4. Chinese remainders

1.4.1. Definition. Ideals I, J R are comaximal ideals if I + J = R.1.4.2. Proposition (Chinese remainder theorem). Let I1, . . . , Ik be pairwise co-maximal ideals in a ring R. Then(1) For a1, . . . , ak R there is a a R, such that aam 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 eachm

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 aam = +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) Subjectivity 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 Pnkk = Pn11 Pnkkand

R/Pn11 Pnkk R/Pn11 R/Pnkkis 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 . . . p

nkk be a factorization into different primes. Show that

Z/(pn11 pnkk ) Z/(pn11 ) Z/(pnkk )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 thatI,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 on P . An element in P maximal for inclusion is anirreducible principal ideal. A generator of an irreducible element is an irreducibleelement in R.

1.5.3. Definition. A domain R is a unique factorization domain if(1) Every irreducible element in P is a prime ideal.(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 irreducibles 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 cancellationm 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. Proposition. 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 of a set of elements in P is the greatest commondivisor and an infimum is the least common multiple.

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) = (pm11 ) . . . (p

mkk ) and (b) = (p

n11 ) . . . (p

nkk ) with mi, ni 0 then greatest

common divisor is (pmin(m1,n1)1 ) . . . (pmin(mk,nk)k ) and least common multiple is

(pmax(m1,n1)1 ) . . . (pmax(mk,nk)k ).

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 aprincipal ideal domain.

(4) Let Z[5] be the smallest subring ofC containing5. 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 + . . . amXm, a polynomial with an R being the nthcoefficient. Multiplication is given by XiXj = Xi+j extended by linearity. Ifg = b0 + b1X + . . . bnXn is an other polynomial, then

f + g = (a0 + b0) + (a1 + b1)X + + (ak + bk)Xk + . . .fg = a0b0 + (a0b1 + a1b0)X + + (a0bk + a1bk1 + + 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.(3) The constant polynomial 1 is the unique monic polynomial of degree 0 and

the identity in the polynomial ring.

1.6. POLYNOMIALS 17

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. Proposition. Let R be a domain, 0 6= f, d R[X] polynomials with dmonic. 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 : R S be a ring homomorphism. For any elementb S there is a unique ring homomorphism R[X] S extending and mappingX 7 b.Proof.

a0 + a1X + . . . amXm 7 (a0) + (a1)b+ . . . (am)bmis 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 : R S be a ring homomorphism and B S a subset.The ring generated over R by B is

R[B] = (R)[B] Sthe smallest subring of S containing (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 : R S be a ring homomorphism.(1) If b S and X is a family of variables, then there is a surjective ring

homomorphismR[X] R[b], X 7 b

making R[b] a factor ring of the polynomial ring R[X].


(2) If S is a finite type ring over R, then S is a factor ring of a polynomial ringin finitely many variables over R.

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

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

1.7. Roots

1.7.1. Definition. Let : 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(Xa)+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 highestm such that

f = q(X a)m

A root of multiplicitym = 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], thenm1 + +mk deg(f).1.7.6. Definition. The derivative of a polynomial f =

anX

n R[X] is

f =

nanXn1

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(Xa)2+ cX+d and by 1.7.7 f = q(Xa)2+2q(Xa) + c. I follows that a is a root of multiplicitym > 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 thatm1 + +mk deg(f).

(2) LetK be a field and let a1, . . . , an K. Show that the ideal (X1a1, . . . , Xnan)is maximal inK[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 withQ they constitute the prime fields.

1.8.2. Proposition. LetK be a field then the polynomial ringK[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 overK.

1.8.5. Example. (1) Let K be a field and f an irreducible polynomial in K[X].ThenK 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 dimensionalK-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 extensionK 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 fieldextensionK 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) LetK 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 seriesanX

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

bnX

n the sum and product areanX

n +

bnXn =

(an + bn)Xn

anXn

bnXn =

(k

ankbk)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. Proposition. 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. IfK is a field, thenK[[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 ringQ[[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 ringK[[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.

2Modules

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 multiplicationRM 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 = xfor all a, b R, x, y M . A submoduleM 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 f1 : N M which is also a homomorphism. Theidentity isomorphism is denoted 1M :M M .2.1.2. Lemma. Let R be a ring andM 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 andM be a module. The mapM 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 andM 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 onM 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 : R S be a ring homomorphism and N an S-module.The map

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

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

2.1.11. Definition. Let : 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 . The scalar multiplication is 2.1.10

RN N, (a, x) 7 ax = (a)xAn 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 for

the unique ring homomorphismZ 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 andM a module. The abelian group RMwith multiplication

(a+ x)(b+ y) = ab+ (ay + bx)is a ring. R is a subring andM 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 : R S be a ring homomorphism. Show that 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 andM a module. Let N be a family of submod-ules. Then the additive subgroups

N and

N are submodules.

Proof. Use the formulasx +

y =

(x + y) and a

x =

ax to

conclude thatN is a submodule. If x, y N for all , then x + y, ax N

for all , soN 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, ofM containing Y . The moduleM 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 =

yY 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 andM 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

RM/N M/N, (a, x+N) 7 ax+Nis well defined and a scalar multiplication, 2.1.1.

Proof. If x+N = y+N then xy N and so a(xy) = axay 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 groupM/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 = R1R2 be the product ring 1.1.4. There is a bijective(up to natural isomorphism) correspondence.(1) IfM1 is an R1-module andM2 is an R2-module, thenM = M1 M2 is an

R-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 submodulesM M,N N , then f1(N ) M and f(M ) N aresubmodules.

Proof. If x, y f1(N ) then f(x + y) = f(x) + f(y) N and for a Rf(ax) = af(x) N so x+y, ax f1(N ) proving f1(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 = f1(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. Proposition. 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

26 2. MODULES

2.3.6. Corollary. Let p :M M/N be the projection onto a factor module. Themap L 7 L = p1(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(p1(L)) = L. If N Lis a submodule of M then clearly L p1(p(L)). Moreover if x p1(p(L))then p(x) = p(y) for some y L and therefore y x N . It follows thatL = p1(p(L)) and the correspondence is bijective. Inclusions are easily seen tobe preserved. Also easily p(L1 + L2) = p(L1) + p(L2) and p1(L1 L2) =p1(L1) p1(L2) 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/Lgiven 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

""FFF

FFFF

Fg // L

Cok f

gOO

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 2.3.101x : RM,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 ) = yY Ann(y) is the ideal ofelements

Ann(Y ) = {a R|ay = 0, for all y Y }2.3.12. Proposition. Let R be a ring and Y a subset of a moduleM .(1) Ann(Y ) = Ann(RY ).(2) a Ann(M) if and only if aM = 0.(3) If modulesM 'M then Ann(M) = Ann(M ).(4) The induced homomorphism

1x : R/Ann(x) Rx, a+Ann(x) 7 axis an isomorphism.

Proof. (1) If a Ann(Y ) then a biyi = biayi = 0 giving the not so obviousAnn(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)thenM is an R/I-module with the scalar multiplication

R/I M M, (a+ I, x) 7 axThat 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)

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.

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 f1(N1 +N2) 6= f1(M1) + f1(M2).

(3) Let R be a ring andM a module. Show thatM 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 (M) a family of modules. The productM is the abelian group of all families (x), x M with term wise ad-

dition. The settingr(x) = (rx)

is a scalar multiplication on

M. The direct sum

M is the subgroupof

M consisting of families with only finitely many nonzero terms. This is asubmodule.

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 andM a family of modules. By 2.4.1 there aremodules and homomorphisms(1) The direct product is

M.

(2) The projections p :

M M are the homomorphisms p((x)) =x .

(3) The direct sum is

M. Elements in

M are written as finite sumsx.

(4) The injections i :M

M are the homomorphisms given by i(x) =(x), where x = x and x = 0, 6= .

2.4.3. Proposition. Let R be a ring andM a family of modules.(1) Given a family of homomorphisms f : L M, then there exists a unique

homomorphism f : LM such that f = p f .L

f @@@

@@@@

@f //

M

p{{vvvvvvvvv

M

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

M L such that g = g i.M

g // L

M

i

ddHHHHHHHHH g

>>~~~~~~~~

Proof. (1) f(y) = (f(y)) is the unique homomorphism. (2) g(x) =

g(x)

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

2.4. SUM AND PRODUCT 29

2.4.4. Definition. A family of submodules M M constitutes a direct sum ifany element x M has a unique finite representation

x =

x, x M

2.4.5. Proposition. The following conditions are equivalent(1) The familyM M constitutes a direct sum.(2) The natural homomorphism

M

M

is an isomorphism.(3) For all

M 6=

M = 0

Proof. (1) and (2) are equivalent. If (1) is true and x =

6= x M 6=M, then x

6= x = 0. Therefore by uniqueness x = 0 and (3)

is true. Conversely if (3) is true and

x = 0, then x =

6= x M

6=M = 0. for any . This shows uniqueness and therefore (1) is

true.

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

R 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

ay, where a R, y Y .

The standard basis of

R consists of the elements e, where each is a familywith 0 for 6= and exactly 1 at index .2.4.7. Proposition. A module is free if and only if it admits a basis. If y is a basisof F then there is an isomorphism

f :

R F

where f(a) =

ay.

Proof. Given a free module f :

R F then y = f(e) is a basis. Con-versely given a basis y F then 1y : R F 2.3.11 is a family of homomor-phisms giving a homomorphism f :

R F by 2.4.3. As f(

a) =

ay

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 basis

consisting 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 Zorns 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 y. For a moduleM anda family of elements x M there is a unique homomorphism g : F M suchthat g(y) = x given by g(

ay) =

ax.

Proof. The basis y F gives the isomorphism 2.4.7 f :

R F . Thefamily 1x : R M gives a homomorphism g :

R M by 2.4.3. Then

g = g f1.2.4.12. Corollary. LetM be an R-module and

M R the free module with basis

ex indexed by x M . The homomorphismM

RM,

axex 7

ax x

is surjective identifyingM 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 ,m2n2

arenonzero numbers in two submodules, then n1m2m1n1 = n2m1

m2n2

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) LetM be a finite family of modules. Show thatM =

M,

(3) Let N M be a family of submodules modules. Show thatM/

N '

M/N

and that M/

N '

M/N

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

RHomR(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 a rings. A functor is a construction T , which toR-modulesM,N associates S-modules T (M), T (N) and a 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 goes

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 functors transform isomorphisms into iso-morphism.Given functors T, T a natural homomorphism is a family M : T (M) T (M)of homomorphisms, such that for each f : M N the following diagram com-mutes

T (M)

T (f)

M // T (M)

T (f)

T (N)N // T (N)

In the contravariant case the diagram is

T (M)M // T (M)

T (N)

T (f)

OO

N // T (N)

T (f)

OO

A natural isomorphism is a natural homomorphism such that each M 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)is a functor.

(2) The construction

M 7 HomR(M,N), f 7 Hom(f, 1N )is a contravariant functor

32 2. MODULES

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)and

Hom(f f, 1N ) = Hom(f, 1N ) Hom(f , 1N )showing the conditions on compositions.

2.5.6. Corollary. Let a R give scalar multiplications aM , aN .(1)

Hom(aM , 1N ) = Hom(1M , aN ) : HomR(M,N) HomR(M,N) f 7 afis scalar multiplication aHomR(M,N).

(2) The map R HomR(M,M), a 7 aM is a homomorphism.2.5.7. Example. Let R = R1R2 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 andM a family of modules. For any moduleN there are natural isomorphisms(1) HomR(

M, N) '

HomR(M, N)

(2) HomR(N,

M) '

HomR(N,M)

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

M N isuniquely determined by the family g = g i :M N . (2) A homomorphismf : N M is uniquely determined by the family f = p f : N M.2.5.9. Lemma. Let R be a ring andM,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).2.5.10. 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 evxProof. 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 andM 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.

2.6. TENSOR PRODUCT MODULES 33

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

M = HomR(M,R)

If f :M N is a homomorphism, then the dual homomorphism isf = 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 evxwhere evx(f) = f(x) 2.5.10.

Proof. This is a special case of 2.5.11

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

M Mis 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 integersm,n.(3) Let I R be an ideal andM a module. Show thatHomR(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.

2.6. Tensor product modules

2.6.1. Definition. Let R be a ring andM,N modules. The tensor productmodule

M R N = F/F is the factor module F/F , where F = MNR 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) ontoM R N is x y = (x, y) + F .2.6.2. Remark. The relations are interpreted.(1) There are identities inM R N

(x1 + x2) y = x1 y + x2 y, x (y1 + y2) = x y1 + x y2ax y = a(x y) = x ay

(2) The map :M N M R N, (x, y) 7 x y

has partial maps x 7 xy :M MRN and y 7 xy : N MRNthat are all homomorphisms.

34 2. MODULES

(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. Proposition. Given a map :M N L such that the partial maps x 7(x, y) : M L and y 7 (x, y) : N L are homomorphisms. Then thereexists a unique homomorphism u :M R N L such that u(x y) = (x, y).

M N

&&MMMMM

MMMMMM

M // M R N

u

L

Proof. By 2.6.1M RN = F/F . The homomorphism 2.4.11 F K, (x, y) 7(x, y) has F in the kernel. 2.3.5 gives the homomorphism u.

2.6.4. Remark. Two homomorphisms u, v :MRN L are equal if u(xy) =v(x y) for all x M,y N .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)Proof. The map south-east

M Nfg

// M R N

M N // M R N

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

2.6.6. Definition. f g :M R N M R N is the induced homomorphism2.6.5.

2.6.7. Proposition. Let R be a ring. The constructions(1)

M 7M R N, f 7 f 1N(2)

N 7M R N, g 7 1M gare functors.

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.

2.6. TENSOR PRODUCT MODULES 35

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

aM 1N = 1M aN :M R N M R N, x y 7 a(x y)is scalar multiplication aMRN .

2.6.9. Example. Let R be a ring.(1) Then there is an isomorphism

RR R ' R, a b 7 ab(2) LetM,N,L be modules. Composition of maps gives a homomorphism

HomR(N,L)R HomR(M,N) HomR(M,L), g f 7 g fby 2.5.3. This is a natural homomorphism in each variable.

(3) For a moduleM composition

HomR(M,M)R HomR(M,M) HomR(M,M), g f 7 g fgives 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) M R R 'M, x a 7 ax(2) (M R N) ' N RM, x y 7 y x(3) (M R N)R L 'M R (N R L), (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 mapM RRM, 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 mapM N M R (N R L), (x, y) 7 x (y z) induces thehomomorphism z :M RN M R (N RL), xy 7 x (y z) by 2.6.3.Finally the map (MRN)LMR(NRL), (xy, z) 7 z(xy). inducesthe homomorphism (MRN)RL 'MR(NRL), (xy)z 7 x(yz)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 andM a family of modules. For any moduleN there is a natural isomorphism

(

M)R N '

(M R N)

giving the identification (x) y =

(x y).

Proof. 2.4.3 and 2.6.3 give a pair of inverse homomorphisms. Fix y N . Thefamily g : M

(M R N), x 7 x y induces by 2.4.3 a homomor-

phism gy :M

(M R N). The map (

M) N

(M R

N), (x, y) 7 gy(

x) induces by 2.6.3 the homomorphism (

M) R

N (MRN), (x)y 7xy. The family i1N :MRN (M)R N induces by 2.4.3 the inverse.

36 2. MODULES

2.6.12. Example. Let R be a ring, F a free module with basis y and G a freemodule with basis z . Then F R G is a free module with basis y z .2.6.13. Proposition. LetR be a ring andM,N,Lmodules. 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))[ gProof. 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 homomorphismM RN L, x y 7 g(x)(y)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 thatQZ 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 andM a module. Show thatM/PM R

M/QM = 0.

2.7. Change of rings

2.7.1. Proposition. (1) Let : R S be a ring homomorphism and N an Smodule. The restriction scalars 2.1.11 viewing N as an R-module through, R N N, (a, x) 7 ax = (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-modulesM 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 byis an S-scalar multiplication.

Proof. For fixed b S the map M N M R N, (x, y) 7 x by inducesthe homomorphism b : M R N M R N, x y 7 x by by 2.6.3. Thisgives a well defined scalar multiplication SM RN M RN, (b, xy) 7b(x y).2.7.3. Definition. Let R S be a ring homomorphism andM an R-module. Thechange of ring S-module isM R S with S-scalar multiplication 2.7.2

S M R S M R S, (b, x c) 7 x bc

2.7. CHANGE OF RINGS 37

2.7.4. Proposition. The constructionM 7M R S

andf :M M 7 f 1S :M R S M R S

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

Proof. This is clear from 2.7.2 and 2.6.7.

2.7.5. Proposition. Let R S be a ring homomorphism,M an R-module andNan S-modules. Then there is a natural isomorphism of S-modules.

M R S S N 'M R N, x b y 7 x byProof. The homomorphism v : S S N N, b y 7 by is an isomorphism,2.6.10. The homomorphism 1M v :M R SS N 'M RN is an R-moduleisomorphism, 2.6.7. The identity x bc y = x b cy proves this to be anS-module homomorphism.

2.7.6. Proposition. Let R S be a ring homomorphism,M an R-module andNan S-modules. Then there is a natural isomorphism

HomR(M,N) ' HomS(M R S,N), f 7 (x b 7 bf(x))Proof. A given f is mapped to the compositeMR N R S N which is anS-homomorphism. Given a homomorphism g :M R S N then the compositeM M R S N is an R-homomorphism and an inverse to the first givenmap.

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 2.1.1.2.7.8. Definition. Let R S be a ring homomorphism andM 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.

Proof. This is clear from 2.7.7 using 2.5.4.

2.7.10. Proposition. Let R S be a ring homomorphism and M an R-moduleand N 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.

38 2. MODULES

(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-moduleM 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 bcextended by linearity. R S R S, r 7 r 1 = 1 r is the natural ringhomomorphism.

2.7.13. Proposition. Let , : R S, S and , : S, S T give a commu-tative diagram of ring homomorphisms, = . Then b b 7 (b)(b)is the unique homomorphism making the following diagram commutative.

R

// S

555

5555

5555

5555

5

S //

))SSSSSSS

SSSSSSSS

SSSSS S R S

##T

Proof. This is clear by 2.6.3.

2.7.14. Example. Let R S be a ring homomorphism. ThenR[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 : 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 (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 ]

3Exact 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 gFor 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 onM the sequence

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

is exact.

39

40 3. EXACT SEQUENCES OF MODULES

3.1.4. Proposition. The 0-sequence

0 // Mf // N

g // L

is exact if and only if the following equivalent statements are satisfied.(1) f is an isomorphism onto Ker g.(2) 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

hOO

h>>||||||||

Proof. (1) This is clearly equivalent with exactness. (2) Assume the sequence ex-act. Imh Ker g = Im f, so by (1) put h = f1 h. Assume (2) satisfied andapply it to Ker g M to see that (1) is satisfied.3.1.5. Proposition. The 0-sequence

Mf // N

g // L // 0

is exact if and only if the following equivalent statements are satisfied.(1) The factor homomorphism 2.3.9 g : Cok f L induced by g is an isomor-

phism.(2) 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) The equivalence follows from 2.3.9. (2) Assume the sequence exact. By2.3.5 there is k : Cok f K such that k p = k. By (1) put k = k g1.Assume (2) satisfied and apply it to N Cok f to see that (1) is satisfied.3.1.6. Proposition. Let

M // N // L

be a family of exact sequences. Then there are exact sequences:(1) The sum

M //N //

L

(2) The product M //

N //

L

Proof. Clear since kernel and image is calculated componentwise.

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. EXACT SEQUENCES 41

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) LetM N be a submodule, then there is a short exact sequence0 // M // N // N/M // 0

(3) For scalar multiplication with nonzero divisor a R onM 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

and0 // Im f // N // Cok f // 0

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

0 // Ker aM // MaM // aM // 0

and0 // 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 vProof. (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 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 gProof. 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. Proposition. 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

g f = 0, u f = 1M , g v = 1L, f u+ v g = 1NIf the sequence is split exact then

0 // Lv // N

u // N // 0

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

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 // L // 0

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

3.2. THE SNAKE LEMMA 43

3.1.16. Example. Let Zi is the family of modules each a copy of Z indexed by thenatural 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 2

k for all k. Iffk = (0, . . . , 0, 2nk, . . . ) +

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 72 // Z/(4) // Z/(2) // 0

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

0 // Z/(2)1 73 // 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 : 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 (z) =x + Imu Coku.

Proof. Assume g(y) = z and f (x) = v(y). There is x M with f(x) = yy.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 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 is schematically

Kerw

N

v

g // L

M

f // N

Coku

z_

y_

// z

x_

// v(y)

(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*+/.

Cokuf // Cok v

g // Cokw

Proof. By construction of it is clear that the sequence is a 0-sequence: If y Ker v then to calculate (g(y)) the choice v(y) = 0 gives g = 0. Alsof ((z)) = v(y) + Im v shows that f = 0. Exactness at Ker v and Cok vare clear. Given z Kerw such that (z) = 0. By 3.2.2 choose y, g(y) = zand x, f (x) = v(y). Then (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 (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*+/.

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*+/.

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 L x 7(x,x)// M/N M/L(x,y) 7xy// 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) 7xy // 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 5where 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 asfollows

M1

u1

// M2

u2

// Cok f1

u3

// 0

0 // Ker f 2 // M 2 // Im f 2 // 0

0 // Cok f2

u3

// M4

u4

// Im f4

u5

// 0

0 // Ker f 4 // M 4 // M 5

0 // Im f2

u3

// M3

u3

// Cok f2

u3

// 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 thatKeru3 = 0 and Coku3 = 0 and u3 is therefore an isomor-phism.

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

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

0 // Ker f // Ker g f f // Ker g*+/.

Cok fg // Cok g f // Cok g // 0


Proof. Look at the two diagrams

0

// M

f

1 // M

gf

// 0

0 // Ker g // Ng // L

M

gf

f // N

g

// Cok f

// 0

0 // L1 // L // 0

By the snake lemma the sequences

0 // Ker f // Ker g f*+/. =f

Ker g // Cok f // Cok g f

Ker g f // Ker g // Cok f*+/. =g

Cok g f // Cok g // 0are exact and overlap to give the windmill sequence.

3.2.10. Remark. The windmill is

Ker g f //

$$III

IIII

IIKer g

%%LLLLL

LLLLL

zzvvvvvvvvvv

Ker f //

99ssssssssssM

f //

gf @@@@

@@@@

N //

g

Cok f

0

OO

L

))RRRRR

RRRRRR

RRRR

uullllll

llllll

llll

Cok g

eeLLLLLLLLLLLCok g foo

3.2.11. 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. Proposition. Given an exact sequence

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.3.3.2. Proposition. Given a sequence

0 // Mf // N

g // L

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

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

Then the original sequence is exact.

Proof. For K = M the identity 1M is mapped to g f 1M = 0 so it is a 0-sequence. Assume f(x) = 0. TakeK = R then 1x is mapped to f 1x = 1f(x) =0. Therefore 1x = 0 and so x = 0. That insures that f is injective. Assumeg(y) = 0. Take K = R then 1y is mapped to g 1y = 1g(y) = 0. There exists ahomomorphism h : RM such that f h = 1y. By 2.5.12 h = 1x and thereforef(x) = y. The original sequence is now proven exact.


Mf // N

g // L // 0

and 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.3.3.4. 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.

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.

3.4. EXACTNESS OF TENSOR 49

3.3.5. Proposition. Given a sequence

0 // Mf // N

g // L // 0

The following are equivalent.(1) The sequence is split exact.(2) For anyK the following sequence is exact

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

(3) For anyK 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 3.3.2 and3.1.11 the original sequence is split exact. (3) (1): Let K = M , then there is aretraction to f . By 3.3.4 and 3.1.11 the original sequence is split exact.

3.3.6. 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


Mf // N

g // L // 0

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

M R K // N R K // LR K // 0Proof. LetK be any module. By 3.3.4 it is enough to see that the sequence

0 // HomR(LR 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 3.3.3.


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 RM // K R N // K R L // 0Proof. This follows from the functor properties 3.1.14.

3.4.3. Proposition. Let I R be an ideal. For any moduleM , the homomorphismM R R/I M/IM, x a+ I 7 ax+ IM

is an isomorphism.

Proof. x+ IM 7 x 1 is an inverse.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 + Jis 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/IkMis an isomorphism.

Proof. By Chinese remainders 1.4.2

R/I1 Ik R/I1 R/Ikis an isomorphism. Tensor withM and use 3.4.3 to get the isomorphism.

3.4.6. Exercise. (1) Calculate Z/(m)Z Z/(n) for all integersm,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 2Z1Z/(2) : ZZZ/(2) ZZZ/(2)

is not injective.

3.5. Projective modules

3.5.1. Definition. AnR-module F is a projective module if for any exact sequenceN L 0 the sequence

HomR(F,N) HomR(F,L) 0is exact.

3.5.2. Proposition. A module F is projective if and only if any surjective homo-morphismM 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 = KerNF L, (y, z) 7 g(y)h(z) and pN :M N, pF :M F the

3.5. PROJECTIVE MODULES 51

projections, then g pN = hpF . 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

3.5.3. Corollary. A short exact sequence

0 // Mf // N

g // F // 0

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

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

3.5.5. Example. Let I R be an ideal. If R/I is projective, then there is a ringdecomposition R/I R ' R.3.5.6. Proposition. A direct summand in a projective module is projective.

Proof. Let F F be projective and g : M F surjective. By 3.5.2 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.

3.5.7. 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 3.5.2 aprojective factor module has a section, and is therefore by 3.1.10 a direct summand.

3.5.8. Proposition. Let F be a family of projective modules, then the direct sum F is a projective module.

Proof. Let N L be surjective. Then by 2.5.8HomR(

F, N) HomR(

F, L)

is the product HomR(F, N)

HomR(F, L)

which is surjective by 3.1.6. SoF is projective.

3.5.9. 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.3.5.10. 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.


3.5.11. Proposition. Any moduleM admits an exact sequence

F M 0where 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.

3.5.12. 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. AnR-moduleE is an injective module if for any exact sequence0M N the sequence

HomR(N,E) HomR(M,E) 0is exact.

3.6.2. 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

3.6.3. Corollary. A short exact sequence

0 // Ef // N

g // L // 0

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

3.6.4. Example. Let I R be an ideal. If I is injective, then there is a ringdecomposition R/I R ' R.3.6.5. 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 3.6.2 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.3.6.6. Proposition. LetE be a family of injective modules, then the direct product

E is an injective module

3.6. INJECTIVE MODULES 53

Proof. LetM N be injective. Then by 2.5.8HomR(N,

E) HomR(M,

E)

is the product HomR(N,E)

HomR(M,E)

which is surjective by 3.1.6. SoE is injective.

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

is exact.

Elementary Comutative Algebra

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