1

0. MODULES OVER COMMUTATIVE RINGS

E. L. Lady

August 5, 1998

The assumption in this book is that the reader is either a student of abelian grouptheory who knows the bare minimum of commutative ring theory or a commutativering theorist whose knowledge of abelian group theory is essentially nil. In this chapterwe valiantly attempt to present a quick and dirty summary of the commutative ringtheory which will be relevant for what follows and which might not be included in thenormal basic graduate algebra course, as well as all the abelian group theory (in disguisedform) which is needed to study finite rank torsion free groups and finite rank torsion freemodules over dedekind domains. Although most of the results in this chapter are wellknown, your author has found it hard to track some of them down in standard references,so it seems worthwhile to gather them all together here.

By inference, then, it might seem plausible that this book is accessible to a reader whoknows no abelian group theory and almost no commutative ring theory, provided that heknows at least a little non-commutative ring theory, at least enough to be comfortablewhen encountering the terms “left” and “right.” However it is doubtful that this bookwould be a good choice of reading matter for such a reader.

proposition 0.1. (1) A vector space over an infinite field cannot be the union of a finitenumber of proper subspaces.

(2) If p1, . . . , pn are prime ideals in a commutative ring and a is an ideal such thata ⊆ ⋃n1 pi , then a ⊆ pi for some i .

proof: (1) Suppose that V = V1 ∪ · · · ∪ Vn with Vi V . We may assume wlog that Vis not a union of any proper subset of {V1, . . . , Vn} . Thus we may choose v1 ∈ V1 withv1 /∈ V2 ∪ · · · ∪ Vn and v2 ∈ V2 with v2 /∈ V1 . Then for any k 6= 0 ∈ K , kv1 + v2 /∈ V1 ∪ V2

so kx + y ∈ Vi for i ≥ 3. Since K is infinite and there are only finitely many Vi , itfollows that there exist distinct k, k′ ∈ K such that kv1 + v2, k

′v1 + v2 ∈ Vi for somei ≥ 3. But it then follows that v1 ∈ Vi , a contradiction.

(2) We may assume without loss of generality that a is not contained in the union ofany proper subset of {p1, . . . , pn} . Thus there exist ai ∈ a with ai ∈ pi and ai /∈ pj forj 6= i . Then a2 · · ·an ∈ pi for i > 1 and a2 · · ·an /∈ p1 since p1 is prime. If n > 1,let x = a1 + a2 · · ·an . Then x ∈ a but x /∈ ⋃n1 pi , a contradiction. Thus n = 1 anda ⊆ p1 . X

2

LOCALIZATION AND GLOBALIZATION. If S is a multiplicative set in acommutative ring R and M is an R-module then S−1M is defined in a well knownfashion. We will call S−1M the localization of M with respect to S . If S is thecomplement of a prime ideal p , we write Mp = S−1M . (Some other authors prefer toreserve the term “localization” for the case where S is the complement of a prime ideal.)It is well known that Rp is a local ring (or “quasi-local,” some would say) whose uniquemaximal ideal is pRp . It is also well known that if M is noetherian then S−1M isnoetherian. There is a canonical map θ : M → S−1M with θ(m) = m/1 and θ(m) = 0 ifand only if sm = 0 for some s ∈ S .

proposition 0.2. For any R-module M and multiplicative set S , S−1M ≈ S−1R⊗RM .

proof: The maps ϕ : S−1R ⊗R M → S−1M and ψ : S−1M → S−1R ⊗R M given by

ϕ(rs⊗R m

)=rm

sand ψ

(ms

)=

1s⊗m are inverse to each other. X

proposition 0.3. If M and N are R-modules and S is a multiplicative set then

(1) HomS−1R(S−1M, S−1N) = HomR(S−1M, S−1N) = HomR(M,S−1N) .(2) S−1M ⊗S−1R S

−1N = S−1M ⊗R S−1N .

Furthermore, HomR(M,S−1N) and S−1M ⊗R S−1N are S−1R-modules.

proof: (1) For ϕ ∈ HomR(S−1M, S−1N) , m ∈M , r ∈ R , s, s′ ∈ S ,

ϕ(rs

m

s′)

=s

sϕ(rmss′)

=r

sϕ(smss′)

=r

sϕ(ms′).

Thus every R-linear map from S−1M to S−1N is in fact S−1R-linear. Furthermore,every R-linear map from M to S−1N extends uniquely to a map from S−1M to S−1N .

(2) This follows from the fact that for m ∈ S−1M , n ∈ S−1N , r ∈ R and s ∈ S , thefollowing holds in S−1M ⊗R S−1N :

m⊗ rn

s=sm

s⊗ rn

s=rm

s⊗ sn

s=rm

s⊗ n. X

proposition 0.4. If M is a finitely generated module over a noetherian ringR then for every R-module N and multiplicative set S , the obvious mapS−1 HomR(M,N) → HomR(S−1M,S−1N) is an isomorphism.

proof: The “obvious map” is the one that takes ϕ/s ∈ S−1 HomR(M,N) to the

homomorphism given bym

s′7→ ϕ(m)

ss′. This is clearly an isomorphism when M = R . It

then follows that it is an isomorphism when M = Rt for t < ∞ . Now if M is finitelygenerated then there is a surjection ε : Rt → M for some t < ∞ . Since R is noetherian,Ker ε is also finitely generated and we thus get an exact sequence Rs → Rt →M → 0.

3

Since localization preserves exactness, applying Hom( , N) and localizing with respect toS then yields a commutative diagram

0 → HomR(M,N) −−−−→ S−1 HomR(Rt, N) −−−−→ S−1 HomR(Rs, N)y y≈y≈

0 → HomR(S−1M,S−1N) −−−−→ HomR(S−1Rt, S−1N) −−−−→ HomR(S−1Rs, S−1N)where the two right-hand vertical maps are isomorphisms. It follows that the left-handvertical map is also an isomorphism, proving the result. X

Note: It is essential here that M be finitely generated and R noetherian (or, moregenerally, that M be finitely presented). A lot of things in the chapters that follow wouldbecome trivialized if Proposition 0.4 were true when M is not finitely generated. (See, inparticular, Chapter 9.)

proposition 0.5. Let M, N, P be modules over a commutative ring R .

(1) If m1, m2 ∈M , then m1 = m2 if and only if m1/1 = m2/1 ∈Mm for all maximalideals m .

(2) M = 0 if and only if Mm = 0 for all maximal ideals m .(3) Suppose that N, P ⊆ M . Then N = P if and only if Nm = Pm for all maximal

ideals m .(4) If ϕ ∈ HomR(M,N) then ϕ is monic [epic] if and only if ϕm : Mm → Nm is monic

[epic] for all maximal ideals m .(5) A sequence M → N → P is exact if and only if the induced sequence

Mm → Nm → Pm is exact for all maximal ideals m .(6) If M is a submodule of a vector space over the quotient field F of R , then

M =⋂pMp .

warning. Mm ≈ Nm for all maximal ideals m does not, in general, imply that M ≈ N .

corollary 0.6. Chinese Remainder Theorem. Let a1, . . . , an be ideals in a commutativering R such that ai + aj = R for all i 6= j . Let M be an R-module. Then

M⋂aiM

≈n⊕1

M

aiM.

proof: For each k there is an obvious map M/⋂

aiM → M/akM and therefore there isa map M/

⋂aiM →⊕

M/aiM . Since ai+aj = R for k 6= j , a prime ideal p can containat most one ai , and (aiM)p = Mp if ai 6⊆ p . Thus if ak ⊆ p then (

⋂n1 aiM)

p= akMp

and Mp/aiMp = 0 for i 6= k . Therefore for all p ,

M

akMp=(

M⋂n1 aiM

)p

→(

n⊕1

M

aiM

)p

=(

M

akM

)p

is an isomorphism, so the result follows from Proposition 0.5. X

4

proposition 0.7. A finitely generated projective module M over a local ring R is free.In fact, if m is the maximal ideal in R and m1, . . . , mt ∈ M are such that the cosetsm1, . . . , mt are a basis for M/mM as a vector space over R/m , then m1, . . . , mt are abasis for M .

proof: Let m be the unique maximal ideal in R . Choose m1, . . . , mt ∈ M so thatthe cosets m1, . . . , mt are a basis for the vector space M/mM over the field R/m . Letϕ : Rt → M be defined by ϕ(r1, . . . , rt) =

∑rimi . It follows easily from Nakayama’s

Lemma that ϕ is surjective. Since M is a projective module, ϕ splits, so Rt = K ⊕ Lwith K = Kerϕ and L ≈ M . Then K is finitely generated. Since ϕ induces anisomorphism from Rt/mRt to M/mM , it follows that K/mK ⊕ L/mL ≈ M/mM . Theseare finite dimensional vector spaces over the field R/m and comparing dimensions yieldsK/mK = 0. Thus K = 0 by Nakayama’s Lemma. Thus ϕ is monic and hence anisomorphism. X

Projective modules over a local ring are free even if not finitely generated, but theproof is much more difficult.

proposition 0.8. A finitely generated module M over a noetherian ring is projective ifand only if Mp is a free Rp-module for all prime ideals p .

proof: ( ⇒ ): Using the criterion that projective modules are just the direct summandsof free modules, it is easy to see that the localization of a projective R-module at p is aprojective module over Rp . It then follows from Proposition 0.7 that this localization is afree Rp-module.

( ⇐ ): Suppose now that M is finitely generated and for all p , Mp is a freeRp-module. To show that M is projective one must show that for every surjectionϕ : X → Y , the induced map ϕ∗ : HomR(M,X) → HomR(M,Y ) is surjective. ByProposition 0.5, it suffices to prove that for all maximal ideals p , the localized map(HomR(M,X))p → (HomR(M,Y ))p is surjective. But since M is finitely generated, byProposition 0.4 and Proposition 0.3 there are natural isomorphisms yielding the followingcommutative diagram:

(HomR(M,X))p −−−−→ (HomR(M,Y ))py≈y≈

HomRp(Mp, Xp) −−−−→ HomRp

(Mp, Yp) −−−−→ 0

where the bottom map is surjective since Mp is a projective Rp-module . Thus(HomR(M,X))p −→ (HomR(M,Y ))p is a surjection, proving the result. X

remark. The hypothesis that M be finitely generated is essential here. In laterchapters we will see many examples of non-finitely generated non-projective modules Msuch that Mp is a free Rp-module for all prime ideals p . Such modules will be calledlocally free. (See Chapter 9 in particular.)

5

COMAXIMAL IDEALS.

definition 0.9. Local Property.

proposition 0.10. Let a, b, c ⊆ R .(1) If a + b = R then a ∩ b = ab .(2) If a + b = R and a + c = R , then a + (b ∩ c) = R .(3) If a + b = R then for any n ≥ 1 , an + bn = R .

proof: In each case the hypothesis and conclusion are “local properties,” i. e. it sufficesto prove in (1) that ap ∩ bp = apbp for all prime ideals p . Thus wlog we may assumethat R is a local ring.

(1) & (3) Since a + b = R , if R is a local ring it follows that a = R or b = R , If, say,b = R then a ∩ b = a ∩R = a = aR = ab . Furthermore an + bn = an +R = R .

(2) Assuming again that R is a local ring, either a = R or b = c = R . Both cases aretrivial. X

proposition 0.11. If a + b = R then a ⊕ b ≈ R ⊕ ab .

proof: By Lemma 0.10 the usual short exact sequence

0 → a ∩ b → a ⊕ b → a + b → 0

becomes0 → ab → a ⊕ b → R→ 0,

which splits because R is free. Thus a ⊕ b ≈ R ⊕ ab . X

ASSOCIATED PRIMES. One of the most basic concepts in abelian group theoryis that of the order of an element. And for any abelian group G , one is particularlyinterested in the set of prime numbers p such that G contains an element of order p .For modules over a commutative ring, the analogous concepts are the annihilator of anelement and the set of associated primes of a module.

If M is an R-module and m ∈ M we will write annM for the annihilator of M andannm for the annihilator of m . I. e.

annm = {r ∈ R | rm = 0}annM = {r ∈ R | (∀m ∈M) rm = 0} =

⋂m∈M

annm.

.

lemma 0.12. Let m ∈ M and let p be a prime ideal. Then m/1 6= 0 ∈Mp if and only ifannm ⊆ p .

proof: In fact, m/1 = 0 if and only if sm = 0 for some s /∈ p , i. e. if and only ifannm 6⊆ p . X

6

definition 0.13. We say that a prime ideal p is an associated prime for M if thereexists m ∈M such that p = annm .

We write AssM (sometimes called the assassinator of M ) for the set of associatedprimes for M .

The French call AssM “l’assassin de M .” A literal translation of this would be“the murderer of M ,” or “the killer of M .” Since this sounds much more violent inEnglish than it does in French, English speaking mathematicians have invented the word“assassinator” to translate the French.

lemma 0.14. A prime ideal p is an associated prime for M if and only if M contains asubmodule isomorphic to R/p .

proof: For m ∈ M , if p = ann(m) then R/p is isomorphic to the cyclicsubmodule Rm . X

proposition 0.15. If p is a prime ideal then AssR/p = {p} .

proof: Clearly p ∈ AssR/p . If q ∈ AssR/p then R/p contains a submodule Xisomorphic to R/q . But then X is cyclic, i. e. is a principal ideal in the integral domainR/p . Hence X ≈ R/p and q = annX = annR/p = p . X

For p to be an associated prime of M it is important that we have p = annm forsome m , not merely p ⊆ annm . It is also important that p be a prime ideal. It is notobvious, however, that there always exist ideals satisfying both of these requirements. Inother words, it is not a priori obvious that if M 6= 0 then AssM will be non-empty. Fornoetherian rings, however, this does turn out to be the case. (See Proposition ** below.)

example 0.16. Let R = Z .(1) If p is a non-zero prime number, then (p) ∈ AssM if and only if M contains an

element of order p .(2) 0 ∈ AssM if and only if M contains a torsion free element, i. e. an element m

with rm 6= 0 for all r 6= 0 ∈ Z .

We say that a module M over an integral domain R is torsion if for every m ∈ Mthere exists r 6= 0 ∈ R such that rm = 0. We say that M is torsion free if nonon-trivial submodule of M is torsion, or, equivalently,

(∀m ∈M) (∀r ∈ R) m 6= 0 & r 6= 0 ⇒ rm 6= 0.

proposition 0.17. Let M be a non-trivial module over a noetherian integral domain R .Then

(1) M is torsion if and only if 0 /∈ AssM .(2) M is torsion free if and only if AssM = {0} .

7

proof: (1) M is torsion if and only if there does not exist an element m ∈ M whoseannihilator is the zero ideal.

(2) M is torsion free if and only if the annihilator of every non-trivial element is thezero ideal. X

If annM = 0 then we say that M is faithful. In abelian group theory, a group Gwhich is not faithful as a Z-module is called bounded.

Clearly a module which is not faithful is torsion, however the converse is far from true.For instance, with R = Z , the module Q/Z is torsion but also faithful.

A classical theorem in abelian group theory implies that every module over a dedekinddomain which is not faithful is a direct sum of cyclic modules. Therefore it is only thefaithful torsion modules which have any real interest in abelian group theory.

The following theorem is known as the Krull Intersection Theorem. In abelian grouptheory terms, it says that a finitely generated torsion free module has no (non-trivial)elements of infinite p-height , for any prime ideal p .

proposition 0.18. If M is a finitely generated torsion free module over a noetherianintegral domain and p is any prime ideal then

⋂∞1 pkM = 0 .

note: Counter-examples abound when M is not finitely generated. For instance, ifR = Z , M = Q , and p is any non-trivial prime ideal then pM = M so

⋂pkM = M 6= 0.

lemma 0.19. Suppose that M is a module over a commutative noetherian ring R andlet m 6= 0 ∈M . Let S be a multiplicative set in R such that S ∩ annm = ∅ . Then thereexists p ∈ AssM such that annm ⊆ p and S ∩ p = ∅ . In fact, if q is any prime idealsuch that annm ⊆ q then there exists p ∈ AssM with annm ⊆ p ⊆ q .

proof: Since annm 6= R then by Zorn’s Lemma there exist ideals q maximal withrespect to the properties annm ⊆ q and q ∩ S = ∅ . We claim that any such ideal mustbe prime. In fact, if r, s /∈ q then S ∩ (r) + q 6= ∅ and S ∩ (s) + q 6= ∅ . Thus there existrx+ q1, sy + q2 ∈ S , with x, y ∈ R , q1, q2 ∈ q . Then rsxy + rxq2 + syq1 + q1q2 ∈ S andsince S ∩ q = ∅ it follows that rs /∈ q .

Let q be such a prime ideal. Since R is noetherian, among the ideals p such thatp ⊆ q and p = ann rm for r ∈ R and rm 6= 0, there exist ones maximal withthis property. Let p = ann rm be such an ideal. Clearly annm ⊆ ann rm = p .We claim that p is prime. In fact, if r1r2 ∈ p and r2 /∈ p then r2rm 6= 0 andp ⊆ p + (r1) ⊆ ann(r2rm) ⊆ q (why?). By the maximality of p we conclude thatp + (r1) = p , i. e. r1 ∈ p . This shows that p is prime and since p = ann rm , thusp ∈ AssM . X

8

corollary 0.20. If R is noetherian then

AssM = ∅ ⇐⇒ M = 0.

proof: If M = 0 then clearly AssM = ∅ . Conversely, if M 6= 0 then there existsm 6= 0 ∈M . Thus by Lemma 0.19 there exists p ∈ AssM . X

There are actually occasions when the most economical way of proving that a moduleM over a noetherian ring is trivial is by proving that AssM = ∅ .

proposition 0.21. If m ∈M then m = 0 if and only if m/1 = 0 ∈Mp for all p ∈ AssM .

proof: If m 6= 0 then by Lemma 0.19 there exists p ∈ AssM with annm ⊆ p . ByLemma 0.12, m/1 6= 0 ∈Mp . X

definition 0.22. The support of an R-module M is defined to beSuppM = {p |Mp 6= 0} .

proposition 0.23. If M is finitely generated then

SuppM = {p | p is prime and p ⊇ annM}.

proof: If annM 6⊆ p then sM = 0 for some s /∈ p so Mp = 0, i. e. p /∈ SuppM .Now suppose that m1, . . . , mt is a finite set of generators for M . Let p be a prime

ideal with annM =⋂t

1 annmi ⊆ p . Since p is prime, annmi ⊆ p for some i . Thusfor all s /∈ p , s /∈ annmi so smi 6= 0. Hence mi/1 6= 0 ∈ Mp so that Mp 6= 0 andp ∈ SuppM . X

example 0.24. Let R = Z and M = Q/Z . M is not a finitely generated module. Theideal 0 is prime and annM = 0 but 0 /∈ SuppM . In fact, for every m ∈ M there existss /∈ (0) such that sm = 0. Thus the localization of M at the zero ideal is trivial.

note. By the Localization-Globalization Theorem (Theorem 0.5),M = 0 ⇐⇒ SuppM = ∅ .

We have the following variation on Nakayama’s Lemma.

proposition 0.25. If M is a finitely generated R-module and p ∈ SuppM thenpM 6= M .

proof: If pM = M then pMp = Mp . Considering M as an Rp-module (necessarilyfinitely generated), Nakayama’s Lemma then implies Mp = 0, i. e. p /∈ SuppM . X

9

corollary 0.26. AssM ⊆ SuppM . Conversely, if q ∈ SuppM then there existsp ∈ AssM with p ⊆ q .

proof: If p ∈ AssM then M contains a submodule M1 isomorphic to R/p and so0 6= (M1)p ⊆Mp , so p ∈ SuppM .

Now suppose that q ∈ SuppM . Then there exists m ∈ M such that m/1 6= 0 ∈ Mq .Thus by Lemma 0.12, annm ⊆ q . Therefore by Lemma 0.19, there exists p ∈ AssM withp ⊆ q . X

corollary 0.27. If AssM consists of maximal ideals then SuppM = AssM .

In particular, if G is a bounded (i. e. not faithful) abelian group, then Corollary 0.27combined with Proposition FinGenSupp says that AssG consists of the prime divisorsof the smallest positive integer n such that nG = 0. From this we easily derive the factthat if G is finite, then AssG consists of those primes dividing the order of G . (This isessentially just Cauchy’s Theorem.)

proposition 0.28. (1) SuppS−1R S−1M = { pS−1R | p ∈ SuppM & p ∩ S = ∅ }

(2) AssS−1R S−1M = { pS−1R | p ∈ AssM & p ∩ S = ∅ }

(3) AssR S−1M = { p | p ∈ AssM & p ∩ S = ∅ } .

proposition 0.29. If N ⊆M then AssN ⊆ AssM ⊆ AssN ∪ AssM/N .

proof: If p ∈ AssN then p = annn for n ∈ N and clearly p ∈ AssM . Now supposethat p ∈ AssM . Then M contains a submodule M1 isomorphic to R/p . If M1 ∩N 6= 0then p ∈ AssM1 ∩ N ⊆ AssN since AssM1 = {p} by Proposition 0.15. On the otherhand, if M1 ∩N = 0 then M/N contains the submodule (M1 +N)/N ≈ M1 ≈ R/p , sop ∈ AssM/N . X

proposition 0.30. If M is a finitely generated module over a commutative noetherianring then AssM is a finite set. (note: In general, SuppM will not be finite.)

proof: M is noetherian, thus there exists a submodule N of M maximal with respectto the property that AssN is finite. Furthermore N 6= 0 since if p ∈ AssM then Mcontains a submodule N ′ isomorphic to R/p and by Proposition 0.15, AssR/p = {p} .If N M applying this reasoning to M/N shows that there is a submodule N ′′ of Mwith N N ′′ such that AssN ′′/N is finite and non-empty. But by Proposition 0.29,AssN ′′ ⊆ AssN ∪ AssN ′′/N , contradicting the maximality of N . Therefore N = M andAssM is finite. X

A module M is called p-primary if AssM = {p} . (Somewhat inconsistently, asubmodule N of M is sometimes called a p-primary submodule if M/N is p-primary.This is a historical accident due to the fact that the term p-primary comes both fromabelian group theory and ideal theory. The second usage will not occur in this book.)

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Proposition0.15 shows that for any prime ideal p , R/p is p-primary. The followingproposition shows how to generalize in case p is maximal.

proposition 0.31. If p is a maximal ideal then for all n > 0 the cyclic module R/pn isp-primary.

proof: If p′ is any prime then by Proposition 0.23, p′ ∈ SuppR/pn if and only if pn ⊆ p′

i. e. if and only if p ⊆ p′ . Since p is maximal it then follows that SuppR/pn = {p} . Since∅ 6= AssR/pn ⊆ SuppR/pn it follows that AssR/pn = {p} . X

If p is not maximal then R/pn need not be p-primary. For dedekind domains,however, which are the rings of interest in this book, all non-trivial prime ideals aremaximal.

proposition 0.32. If p is minimal in AssM then Mp is p-primary.

proof: By Proposition 0.28, AssMp = {p′ ∈ AssM | p′ ⊆ p} = {p} . X

In abelian group theory, we know that an abelian group G is p-primary for someprime number p if and only if the order of every element of G is a power of p . Theanalogous characterization for modules over a commutative ring is the following:

proposition 0.33. If R is noetherian and M is an R-module and p is a prime ideal,the following conditions are equivalent:

(1) M is p-primary.(2) The natural map θ : M →Mp is monic and

(∀m ∈M) (∃k ≥ 1) pkm = 0.

proof: (1) ⇒ (2): Suppose that M is p-primary, i. e. AssM = {p} . Then byLemma 0.12, if m 6= 0 ∈ M , then m/1 6= 0 ∈ Mp . Therefore θ : M → Mp is monic. Nowlet r 6= 0 ∈ p and let S = {rk | k ≥ 1} . By Proposition0.28, p /∈ AssS−1M ⊆ {p} .Therefore AssS−1M = ∅ and so S−1M = 0. This means that for each m ∈ M thereexists rk ∈ S such that rkm = 0. Since this is true for all r ∈ p and p is finitelygenerated (because R is noetherian) it then follows easily that pk

′m = 0 for some k′ .

(2) ⇒ (1): Now suppose the stated conditions hold. Since M → Mp is monic, byProposition 0.29, AssM ⊆ AssMp , so it suffices to prove that AssMp = {p} . Thereforewe may assume wlog that M = Mp . Now let q ∈ AssM . Then q ⊆ p and q = annmfor some m . By assumption, for some k , pk ⊆ annm = q Then p ⊆ q since q is prime.Therefore q = p , showing that M is p-primary. X

definition 0.34. We say that r ∈ R is a zero divisor on M if rm = 0 for somem 6= 0 ∈M .

11

proposition 0.35. If R is noetherian then⋃{p | p ∈ AssM} is the set of elements in R

which are zero divisors on M .

proof: By definition, r ∈ R is a zero divisor on M if and only if r ∈ annm for somem 6= 0 ∈ M . If this is the case then by Lemma 0.19, r ∈ p for some p ∈ AssM .Conversely, if p ∈ AssM then pm = 0 for some m 6= 0 ∈M , so p consists of zero divisorson M . X

corollary 0.36. If R is noetherian then⋃{p | p ∈ AssR} equals the set of zero divisors

in R .

MODULES WITH FINITE LENGTH. Henceforth we will assume that the ring Ris noetherian.

In abelian group theory, an important role is played by the finite groups. A number ofimportant definitions and theorems in the theory of finite rank torsion free groups have todo with the condition that a certain quotient G/H is finite. (For instance, if G is a finiterank torsion free group and H ⊆ G , then H is quasi-equal to G if and only if G/H isfinite.)

For many commutative rings R , there do not exist non-trivial R-modules which arefinite as sets. Instead, the appropriate condition is that a module have finite length.

definition 0.37. A module M over a ring R is said to have finite length if and only ifit has a composition series

0 = M0 M1 . . . M` = M

where each quotient Mi/Mi−1 is a simple module, and in this case we define lengthM tobe the length ` of this composition series. The Jordan-Holder Theorem asserts thatlengthM is independent of the particular composition series.

A module has finite length if and only if it is both noetherian and artinian.

theorem0.38. If R is a noetherian ring and M is an R-module such that all theassociated primes of M are maximal, then AssM = SuppM and for each p ∈ AssM , thecanonical map M → Mp is surjective and Mp is isomorphic to the p-primary componentof M , i. e. to {m ∈ M | (∃n) pnm = 0} . Furthermore M is the direct sum of itsp-primary components.

proof: By Theorem 0.5, to show that M → Mp is surjective, it suffices to provethat for every maximal ideal m , the induced map Mm → (Mp)m is surjective. Butsince p is maximal, if m 6= p then m /∈ AssMp = SuppMp so that (Mp)m = 0 soMm → (Mp)m is surely surjective. And if m = p then Mm → (Mp)m is essentiallythe identity map. Likewise this shows that for each maximal ideal m , the obvious mapMm →

(⊕p∈AssM Mp

)m

is an isomorphism, showing that M → ⊕AssM Mp is an

isomorphism.

12

Now for given p ∈ AssM , define a functor F by F(X) = {x ∈ X | (∃n) pnx = 0} .(Later, we will frequently write X [p∞] = F(X) .) Since p is maximal, by Proposition 0.28,Mp is p-primary, so by Proposition 0.33, F(Mp) = Mp . On the other hand,for p 6= q ∈ AssM then F(Mq) = 0 since if x ∈ F(Mq) then (∃n) pnx = 0and by Proposition 0.33, (∃n) qnx = 0. Taking n as the larger of these two, byProposition 0.10 pn + qn = R and Rx = (pn + qn)x = 0 so x = 0. It follows thatF(M) = F (

⊕AssM Mp) = Mp . X

note: When R = Z the preceding theorem simply gives us the familiar result thatevery torsion abelian group (finitely generated or not) is a direct sum of its p-primarycomponents.

Recall that in general, one cannot conclude that two modules M and N areisomorphic from the fact that Mp ≈ Np for every prime ideal p . However it followsfrom Theorem 0.38 that this is the case if AssM (and consequently AssN ) consists ofmaximal ideals.

corollary 0.39. If AssM consists of maximal ideals and Mp ≈ Np for every maximalideal p , then M ≈ N .

proof: If Mp ≈ Np for every maximal ideal p then this will in fact be true forevery prime ideal p . It then follows that AssN = AssM . But by Theorem 0.38,M ≈⊕AssM Mp and N ≈⊕AssN Np . It follows that M ≈ N . X

proposition 0.40. If M is an artinian module then AssM is finite and consists ofmaximal ideals.

proof: If p ∈ AssM then M contains a submodule isomorphic to R/p . Thus if M isartinian then R/p is artinian. But an artinian integral domain is a field. Thus p is amaximal ideal.

It now follows from Theorem 0.38 that M ≈ ⊕AssM Mp , and that Mp 6= 0 for all

p ∈ AssM . Since M is artinian, this direct sum can have only finitely many components.Thus AssM is finite. X

theorem0.41. A module M over a noetherian ring R has finite length if and only if itis finitely generated and AssM consists of only maximal ideals.

proof: ( ⇒ ): If M has finite length then it is certainly noetherian, hence finitelygenerated. It is also artinian so by Proposition0.40, all its associated primes are maximal.

( ⇐ ): If M is finitely generated then it is noetherian, so there exists a submoduleN of M which is maximal with respect to the property of having finite length. Nowif AssM consists of maximal ideals then by Proposition0.27, AssM = SuppM .Since SuppM/N ⊆ SuppM , then SuppM/N consists of maximal ideals and soSuppM/N = AssM/N . If p ∈ AssM/N then M/N contains a submodule N1/N

13

isomorphic to R/p (where N ⊆ N1 ). Since p is maximal, N1/N is simple and since Nhas finite length it follows that N1 has finite length, contradicting the maximality of N .ThusAssM/N = ∅ so M/N = 0 and M has finite length. X

proposition 0.42. Let M have finite length and I = annM . Then AssM consists ofthe prime ideals containing I . Furthermore lengthM ≥ lengthR/I .

proof: By Corollary 0.27 and Proposition 0.23, AssM = SuppM and SuppM consistsof those prime ideals containing annR/I = I .

Thus AssM = AssR/I and so by Theorem 0.41, R/I has finite length. ByProposition 0.38, M ≈⊕Mp and R/I ≈⊕(R/I)p . It thus suffices to prove that for eachp ∈ AssM , lengthMp ≥ length(R/I)p . In other words, we may suppose that R is a localring and I = pk , where pk−1M 6= 0. Then for i < k , applying Nakayama’s Lemma topiM shows that pi+1M piM . Then the series

0 = pkM pk−1M . . . pM M

with length k can be refined to a composition series, so thatk = lengthR/I ≤ lengthM . X

lemma 0.43. Let R be a noetherian local ring with maximal ideal m . Let E be the

injective envelope of R/m . Then the functor HomR( , E) restricted to the category offinite-length R-modules preserves length. Conversely, if M is an R-module such thatHomR(M,E) has finite length, then M has finite length.

proof: Consider first a module S with length 1 (i. e. a simple module). Then S ≈ R/m(so E is isomorphic to the injective envelope of S ). Now if ϕ ∈ HomR(S,E) , thenϕ(S) is simple (unless trivial) and ϕ(S) ∩ (R/m) 6= 0 since E is an essential extensionof R/m . Thus ϕ(S) = R/m , since both these modules are simple. From this we seethat HomR(S,E) ≈ HomR(R/m, R/m) ≈ R/m , so that length HomR(S,E) = 1.Since the functor HomR( , E) preserves (or, more precisely, reverses) short exactsequences, it now follows by induction that for all R-modules M with finite length,length HomR(M,E) = lengthM .

Now suppose that M is an R-module such that HomR(M,E) has finite length.Consider first the case where M is finitely generated. M contains submodules N suchthat M/N has finite length (for instance, N = M ). By the preceding, for such an N ,HomR(M/N,E) has the same length as M/N . But the surjection M → M/N induces amonomorphism HomR(M/N,E) → HomR(M,E) , so lengthM/N ≤ lengthHomR(M,E) .Thus there is a bound on the length of M/N for those N such that M/N has finitelength. But if M/N has finite length and N 6= 0, then since N is noetherian, it containsa maximal proper submodule N ′ and lengthM/N ′ = 1 + lengthM/N . Thereforeeventually we must have N = 0, so that M has finite length.

If M is not finitely generated and HomR(M,E) has finite length, consider afinitely generated submodule M ′ of M . The inclusion map from M ′ to M induces

14

a surjection HomR(M,E) → HomR(M ′, E) (because E is injective). ThereforeHomR(M ′, E) has finite length and so M ′ has finite length by the preceding paragraph,and in fact the lengths of the finitely generated submodules of M are all bounded bylength HomR(M,E) . Furthermore, if M ′′ is also finitely generated and M ′ M ′′ thenlengthM ′′ > lengthM ′ . It follows that there exists a maximally finitely generatedsubmodule of M . But clearly this is only possible if this submodule is M itself.Therefore M is finitely generated and hence, by the preceding paragraph, has finitelength. X

theorem0.44. Let R be a commutative noetherian local ring and let m be its uniquemaximal prime ideal. The following conditions are equivalent:

(1) R is artinian.(2) R has finite length (as an R-module).(3) m is the only prime ideal in R .(4) AssR = {m} .(5) mk = 0 for some positive integer k .(6) The injective envelope E of R/m is finitely generated.(7) There exists a finitely generated injective R-module.

proof: (2) ⇒ (1): Clear.(1) ⇒ (4): By Proposition 0.40.(4) ⇒ (5): If AssM = {m} then for some k , mk1 = 0 ∈ R so mk = 0.(5) ⇒ (2): For each i from 0 to k , mi/mi+1 is finitely generated and a vector space

over the field R/m , hence is a finite-length R-module. Thus by induction, R/mi hasfinite length. If mk = 0, this shows that R has finite length.

(5) ⇒ (3): If mk = 0 and p is a prime ideal, then mk ⊆ p so m ⊆ p . Since m ismaximal, p = m .

(3) ⇒ (4): Clear since AssR 6= ∅ .(2) ⇔ (6): If E = E(R/m) is finitely generated, then it has finite length, since all its

associated primes are maximal. And conversely, if it has finite length then it is certainlyfinitely generated. But E ≈ HomR(R,E) . Therefore the preceding lemma implies that Ehas finite length if and only if R has finite length.

(6) ⇒ (7): Clear.(7) ⇒ (6): Let M be a non-trivial finitely generated injective module and let

p ∈ AssM . Then M contains a copy of R/p . Since R/p is an essential submodule ofRp/pRp , M therefore contains a submodule isomorphic to Rp/pRp . Therefore Rp/pRpis finitely generated. Now let x /∈ p . Then multiplication by x is an isomorphism onRp/pRp . If x ∈ m , this contradicts Nakayama’s Lemma. Therefore x /∈ m and weconclude that p = m . This shows that m ∈ AssM , and so M contains a submoduleisomorphic to R/m . Since M is injective, it thus contains a copy of E = E(R/m) .Therefore E is finitely generated.

(6) ⇒ (4) [Alternate proof]: Since AssE = {m} , it follows that for each e ∈ E ,mke = 0 for some k . Thus if E is finitely generated, then mkE = 0 for some k . It

15

now suffices to show that E is faithful in order to conclude that mk = 0. Supposein fact, by way of contradiction, that rE = 0 for some r 6= 0 ∈ R . The principalideal (r) is isomorphic to R/a , where a = ann r . Since a ⊆ m , there is a surjectionϕ : (r) → R/m (so ϕ(r) 6= 0). Since E = E(R/m) , this extends to a map ϕ : R → E .Then ϕ(r) = rϕ(1) = 0 since rE = 0. This is a contradiction, showing that no such rexists and therefore E is faithful. Since mkE = 0, it follows that mk = 0. X

corollary 0.45. A commutative noetherian ring is artinian if and only if every primeideal is maximal (including the zero ideal, if applicable). If this is the case, then R is afinite product of local rings each of which has a unique prime ideal.

RANK-ONE PROJECTIVE MODULES.

proposition 0.46. Let R be an integral domain with quotient field F and let P be asR-submodule of F . Then the following conditions are equivalent:

(1) P is projective.(2) There exist elements p1, . . . , pn ∈ P and f1, . . . , fn ∈ F such that fiP ⊆ R for all

i and∑fipi = 1 .

(3) There exists a submodule M of F such that MP = R .

Furthermore in this case P is generated by p1, . . . , pn .

proof: (1) ⇒ (2): If P is projective then it is a summand of a free module R(I) . Thenthere exist maps σ : P → R(I) and π : R(I) → P such that πσ = 1P . Localizing at thezero ideal, σ extends to a map σ0 : F → F (I) and π to a map π0 : F (I) → F . For eachi ∈ I let fi be the ith coordinate of σ0(1) and let pi = π0(ei) , where ei is the canonicalith basis vector of F (I) . Then the composition of σ0 with the projection of F (I) ontothe ith coordinate is given by x 7→ fix . Since this composition maps P into R it followsthat fiP ⊆ R . Furthermore since π is given by

∑yiei 7→

∑yipi . Then the equation

π0σ0(1) = 1 translates to∑fipi = 1. This sum can have only finitely many non-trivial

terms, and at this point we can replace I by the finite set of i ∈ I such that fipi 6= 0.(2) ⇒ (3): Given fi and pi as indicated, let M be the submodule of F generated by

f1, . . . , fn . Then clearly MP ⊆ R . But 1 =∑fipi ∈ MP so MP = R . Note also that

p1, . . . , pn generate P since for p ∈ P , we have p = p∑fipi =

∑(fip)pi and all fip ∈ R .

(3) ⇒ (2): If MP = R then 1 ∈MP so there exist fi ∈M , pi ∈ P with∑fipi = 1.

Furthermore for all i , fiP ⊆MP = R .(2) ⇒ (1): Given fi ∈ F and pi ∈ P with the indicated property, map P into

Rn by σ : p 7→ (f1p, . . . , fnp) and map Rn to P by π : (r1, . . . , rn) 7→ ∑ripi . Then

πσ(p) =∑pfipi = p1 = p . Thus σ is a split monomorphism and P is a summand of a

free module, hence is projective. X

Note that the module M above is uniquely determined by P . In fact, ifMP = M ′P = R then M = MR = MPM ′ = RM ′ = M ′ . We write M = P−1 and saythat the rank-one module P is invertible.

16

SUBMODULES OF DIRECT SUMS. If W is a dedekind domain (see below) withquotient field Q , then every torsion free W -module with rank at most r is isomorphicto a submodule of Qr . Thus if we had a good theory for submodules of direct sumsK1 ⊕ · · · ⊕Kr , we would completely understand finite torsion torsion free W -modules .

When the direct sum has only two factors, we have the following theorem:

theorem0.47. Let K and N be R-modules , let κ and ν be the projections of K ⊕Nonto K and N , and let K1 ⊆ K2 ⊆ K and N1 ⊆ N2 ⊆ N . Then the submodules M ofK ⊕N such that M ∩K = K1 and M ∩N = N1 and κ(M) = K2 and ν(M) = N2 are inone-to-one correspondance with the isomorphisms ϕ from K2/K1 to N2/N1 . Specifically,such an isomorphism ϕ corresponds to the inverse image in K ⊕N of the submodule

{(x, ϕ(x) | x ∈ K/K1}

of K/K1 ⊕N/N1 .

proof: ****Unfortunately, the value of this theorem is drastically limited because it does not

provide us a good way of determining whether or not two submodules of K ⊕ N areisomorphic.

DEDEKIND DOMAINS.For my own perverse reasons, I will generally use the symbol W rather than R when

referring to a dedekind domain. As I now start to develop the concepts needed to definea dedekind domain, I will now start calling my commutative ring W . The quotient fieldof W will be denoted by Q .

proposition 0.48. Let W be an integral domain with quotient field Q . The followingconditions are equivalent:

(1) If q ∈ Q and q satisfies a monic polynomial f(X) ∈W [X ] , then q ∈W .(2) If W ′ is a subring of Q with W ⊆ W ′ and W ′ is a finitely generated W -module,

then W ′ = W .(3) if W ′ is a subring of Q with W ⊆ W ′ and wW ′ ⊆ W for some w 6= 0 ∈ W , then

W ′ = W .

proof: **** (This is a very standard result.)

definition 0.49. An integral domain satisfying the conditions of Proposition 0.49 is calledintegrally closed (in its quotient field).

Using the terminology we will introduce in Chapter 3, condition (3) in Proposition 0.48says that W is not quasi-equal to any subring of Q properly containing it.

17

lemma 0.50. If W is integrally closed then so is S−1W for every multiplicative set S .

proof: Let Q denote the quotient field of W and let q ∈ Q be integral over S−1W andlet f ∈ S−1W [X ] be a monic polynomial satisfied by q . Let d be the degree of f and lets ∈ S be a common denominator for the coefficients of f . Then sdf(q) = 0 and one seesthat sq satisfies a monic polynomial in W [X ] . Thus if W is integrally closed, sq ∈W . Itfollows that q ∈ S−1W . Thus S−1W is integrally closed. X

lemma 0.51. A commutative ring W is integrally closed if and only if Wp is integrallyclosed for all prime ideals p .

proof: ( ⇒ ): By Lemma 0.50.( ⇐ ): Let q ∈ Q be integral over W . Then a fortiori q is integral over each Wp . If all

Wp are integrally closed, then q ∈ ⋂Wp = W (c. f. Proposition 0.*). X

proposition 0.52. Unique factorization domains are integrally closed. In particular,principal ideal domains are integrally closed.

proof: If W is a UFD with quotient field Q and q ∈ Q and f(q) = 0 where f ismonic in W [X ] , then X − q divides f in Q[X ] . By Gauss’s Lemma, X − q ∈ W [X ] , soq ∈W . X

definition 0.53. A local principal ideal domain is called a discrete valuation ring.

proposition 0.54. A discrete valuation ring is a maximal proper subring of its quotientfield.

proof: Let W be a discrete valuation ring and let (p) be the unique maximal ideal inW . By Proposition0.18,

⋂∞1 (pk) = 0 so for each r 6= 0 ∈ W there exists a largest k with

r ∈ (pk) . Thus r = upk with k ≥ 0 and u /∈ (p) , so u is invertible since (p) is the uniquemaximal ideal in W . From this we see easily that each non-trivial element of the quotientfield Q has the form x = upk where u is invertible in W and k is an integer. If x /∈ Wthen k < 0.

It now follows that any subring of Q strictly containing W must contain pk for somek < 0, and hence contains pnk for all n ≥ 1. We then see that it contains all elementsupm for any integer m and invertible element u of W . In other words, the subringcontains all of Q . X

proposition 0.55. A local integral domain with only one non-trivial prime ideal isintegrally closed if and only if it is a discrete valuation ring.

proof: ( ⇐ ): A discrete valuation ring is a principal ideal domain hence a uniquefactorization domain, hence by Proposition0.52 is integrally closed.

( ⇒ ): **** X

18

proposition 0.56. An integral domain W is integrally closed if and only if it is theintersection of all the valuation rings on its quotient field which contain it.

proof: ( ⇒ ):( ⇐ ): This follows from a more general result that the intersection of any family

of integrally closed subrings of a field Q is integrally closed. In fact, suppose thatW =

⋂Wi and each Wi is integrally closed. Let q ∈ Q be integral over W . Then it

is integral over each Wi . Hence q ∈ Wi for all i since each Wi is integrally closed.Therefore q ∈W , showing that W is integrally closed. X

proposition 0.57. If Q′ is a separable extension of Q and W is integrally closed, thenthe integral closure W ′ of W in Q′ is a finitely generated W -module.

proof: ****

proposition 0.58. Let W be an integral domain. The following conditions areequivalent:

(1) W is a principal ideal domain.(2) All ideals in W are free W -modules .(3) All finitely generated torsion free W -modules are free.

proof: (3) ⇒ (2): Clear.(2) ⇒ (1): If r 6= s ∈ W then r and s are linearly dependent since sr − rs = 0. Thus

an ideal in W which is free must have a basis consisting of at most one element, hence beprincipal.

(1) ⇒ (3): **** X

theorem0.59. Let W be an integral domain with quotient field Q . The followingconditions are equivalent:

(1) Every ideal in W is projective.(2) Every non-trivial ideal of W is invertible.(3) W is noetherian, integrally closed, and all non-trivial prime ideals are maximal.(4) W is noetherian and Wp is a discrete valuation ring for every non-trivial prime

ideal p .

proof: (1) ⇔ (2): By Proposition 0.46 ideals are projective if and only if they areinvertible.

Note now that by Proposition 0.46, projective ideals are finitely generated. Hence (1)implies that W is noetherian.

(1) ⇒ (4): The ideals in Wp have the form ap , where a is an ideal in W . Byhypothesis a is projective, so by Proposition 0.8 ap is a free Wp-module. Thus all idealsof Wp are free, so that Wp is a local principal ideal domain, i. e. a discrete valuation ring.

19

(4) ⇒ (3): To see that W is integrally closed, it suffices by Lemma 0.51 to see thatWp is integrally closed for all primes p . But this is true if Wp is a discrete valuation ring,since principal ideal domains are integrally closed. Now let p be a prime ideal in W . Theprime ideals contained in p correspond to the prime ideals of Wp . Since Wp is a discretevaluation ring, its only prime ideals are pWp and 0. Thus there are no non-trivial primeideals strictly contained in p , so p has height one. It follows that all prime ideals of Ware maximal.

(3) ⇒ (1): Let a be an ideal in W . Since W is noetherian, a is finitely generated.Hence by Proposition 0.8, to prove a projective it suffices to prove that ap is a freeWp-module for all primes p . But since Wp is a principal ideal domain, ap is in factfree. X

definition 0.60. An integral domain W satisfying the equivalent properties ofTheorem 0.59 is called a dedekind domain.

proposition 0.61. If W is a dedekind domain and S a multiplicative set in W , thenS−1W is a dedekind domain.

proof: ****

proposition 0.62. If a is a non-trivial ideal in a dedekind domain W then there areonly finitely many prime ideals containing a and W/a has finite length.

proof: Since a = annW/a 6= 0, 0 /∈ AssW/a . Thus by Theorem 0.59, all the associatedprimes of W/a are maximal. Thus by Proposition 0.23 the prime ideals containing a areprecisely those in SuppW/a = AssW/a . Hence there are only finitely many of them.Since W/a is finitely generated (in fact cyclic), it thus follows from Theorem 0.41 thatW/a has finite length. X

For the remainder of this book, W will denote a dedekind domain and Qwill denote its quotient field. Furthermore the term “prime ideal” will be usedhenceforth to mean a non-trivial prime.

The following proposition is crucial to the study of finite rank torsion free modules overdedekind domains. It is curiously hard to track down in standard references.

proposition 0.63. If W ′ is an integral domain and W ⊆ W ′ ⊆ Q , then W ′ is adedekind domain and is the localization of W with respect to some multiplicative set S .

proof: Let X be the set of prime ideals p of W which survive in W ′ , i. e. such thatpW ′ 6= W ′ or, equivalently, W ′

p 6= Q . Let S be the set-theoretic complement in W of⋃X p . Now if s ∈ S and p′ is a prime ideal in W ′ then W ∩ p′ ∈ X so s /∈ W ∩ p′

so s /∈ p′ since s ∈ W . Thus s does not belong to any prime ideal of W ′ and hence isinvertible in W ′ . Therefore S−1W ⊆W ′ .

20

To prove that S−1W = W ′ it suffices to prove that S−1Wp = W ′p for all prime ideals p

of W . If p /∈ X then S−1Wp = Q = W ′p . But if p ∈ X then Wp ⊆ S−1Wp ⊆ W ′

p Q .Since Wp is a discrete valuation ring, by Proposition 0.54 it is a maximal subring of Q , soS−1Wp = W ′

p . It follows that S−1W = W ′ . X

corollary 0.64. If W ′ is a subring of Q containing W and M, N are W ′-modules ,then HomW ′(M,N) = HomW (M,N) and M ⊗W ′ N = M ⊗W N .

proof: By Proposition 0.63, W ′ = S−1W for some multiplicative set S . Thus the resultreduces to Proposition 0.3. X

proposition 0.65. If G is a torsion module over a dedekind domain, then for every primep the p-primary component of G can be canonically identified with the localization Gp .Furthermore G =

⊕Gp .

proof: Immediate from 0.38 since all primes in AssG are necessarily maximal. X

proposition 0.66. If every p-primary component of a finitely generated torsionW -module G is cyclic, then G is cyclic.

proof: By Proposition 0.65, if G is torsion then G ≈ ⊕Gp and if G is finitely

generated then there can be only finitely many non-trivial summands in this expression.Now if each Gp is cyclic, choose a generator gp for each Gp . Then gp/1 = 0 ∈ Gq forq 6= p . It follows that if g =

∑gp , then the image of g in Gp is a generator for all p . It

then follows from Proposition0.5 that g generates G , so G is cyclic. X

proposition 0.67. A finitely generated torsion module over a dedekind domain W hasfinite length and is a finite direct sum of cyclic modules of the form W/pk , for variousprime ideals p .

proof: If G is torsion then 0 /∈ AssG . Since all non-trivial primes in a dedekinddomain are maximal, it follows from Theorem 0.41 that G has finite length. Furthermore,since all the associated primes of G are necessarily maximal, by Proposition 0.38,G ≈⊕AssGGp . It thus suffices to prove that each Gp is a direct sum of cyclic modulesW/pk . But we may think of Gp as a Wp-module, and by Proposition 0.59 Wp is aprincipal ideal domain, so this is a standard result. X

We will prove in Chapter 1 that every finitely generated torsion free module overa dedekind domain is a direct sum of ideals, hence in particular is projective. It thenfollows immediately that the torsion submodule of any W -module is a direct summand ofthat module.

21

proposition 0.68. Every non-trivial ideal in a dedekind domain W can be uniquelywritten as a finite product of prime ideals.

proof: Let a be a non-trivial ideal in W . Then by Proposition 0.62, W/a has finitelength, so SuppW/a is finite, which means simply that there are only finite many primeideals containing a . Furthermore if p is one of these primes then by Proposition 0.*aWp = pepWp for some positive integer ep . If we now write a′ = pe11 · · · pen

n , wherep1, . . . , pn are the primes containing a then for every prime p , aWp = a′Wp sincepiWpj

= Wpjfor j 6= i . Thus by Proposition0.5, a = a′ . X

proposition 0.69. If 0 6= b ⊆ a ⊆W then a/b ≈ W/a−1b ≈ b−1a/W . In particular, a/bis a cyclic W -module.

proof: At first sight, this result seems self-evident. For instance, one simply multipliesboth the numerator and denominator of a/b by a−1 to get W/a−1b . This is not quitevalid, however, mostly because a−1 is a set and not an element. This reasoning doeswork, however, if a is a principal ideal.

By Theorem 0.38, a/b ≈ ⊕(a/b)p . Thus to see that a/b ≈ W/a−1b it suffices to

see that (a/b)p ≈ (W/a−1b)p for all prime ideals p . Therefore we may localize at p .Hence we may suppose that W is a local dedekind domain, and therefore a principal idealdomain. But then the result is easy, as indicated above. X

proposition 0.70. A semi-local dedekind domain W is a principal ideal domain.

proof: It suffices to see that every prime ideal p is principal. Let p1, . . . , pn bethe finite set of non-trivial prime ideals in W . Since these are maximal ideals,piWpj

= Wpjand also pi + pj = W for i 6= j . It follows that the Jacobson radical

of W is J =⋂pi = p1 · · · pn (c. f. Proposition0.10). Let x ∈ p1 with x /∈ p2

1 . Bythe Chinese Remainder Theorem, the canonical surjection W

≈−→ ⊕W/p2

i mappingan element w onto the sum of its respective residue classes induces an isomorphismζ : W/J2 ≈−→ ⊕

W/p2i . Thus there exists w ∈ W such that ζ(w) = (x, 1, . . . , 1). Then

wWpi= Wi = p1Wpi

for i 6= 1 and since Wp1 is a principal ideal domain one sees thatwWp1 = p1Wp1 . Thus by Proposition 0.5, p1 = (w) . X

proposition 0.71. If a is a non-trivial ideal in W and w1 6= 0 ∈ a then there existsw2 ∈ a such that a is generated by w1 and w2 . Furthermore there exists q ∈ Q suchthat a−1 is generated by 1 and q .

proof: Let b = (w1) ⊆ a . Then by Proposition 0.69, a/b is cyclic. If w1 ∈ a is arepresentative for a generator of a/b , then a is generated by w1 and w2 .

Likewise, a−1/W ≈ W/a is cyclic, and if q ∈ a−1 is a representative for a generator ofa−1/W , then a−1 is generated by 1 and q . X

22

proposition 0.72. If a and b are non-trivial ideals in a dedekind domain W then thereexists an ideal a′ with a′ ≈ a such that a′ + b = W .

proof: Choose a 6= 0 ∈ a . Let c = aa−1 ⊆ W . Now by Proposition 0.71, c is generatedby bc together with an additional element x . Now multiplying the equation

c = xW + bc

by c−1 = a/a yieldsW = c−1c = xc−1 + b =

xa

a+ b.

But clearly ax/a ≈ a . X

proposition 0.73. If a and b are ideals in a dedekind domain W then

(1) a ⊕ b ≈W ⊕ ab .(2) a ⊕ a−1 ≈W ⊕W .

proof: (1) Let c = a + b . Then c−1a, c−1b ⊆ c−1(a + b) = W and c−1a + c−1b = W .Thus by Proposition0.11, c−1a ⊕ c−1b ≈ W ⊕ c−2ab . Thus a ⊕ b ≈ c ⊕ c−1ab . ****

proposition 0.74. If a is an ideal in a dedekind domain W , then there exists an ideal bsuch that a ⊕ b = W ⊕W .

proof: By Proposition 0.71, a is generated by two elements. Hence there exists asurjection W ⊕W → a . Since a is projective this must split, so that W ⊕W ≈ a ⊕ bfor some W -module b , which is necessarily projective. Localizing at the zero ideal yieldsQ ⊕ Q = a0 ⊕ b0 , so that b is isomorphic to a non-trivial projective submodule of Q .Then if d is a common denominator for a set of generators for b , then b ≈ db ⊆ R , so bis isomorphic to an ideal. X

We say that G is divisible if wG = G for every w 6= 0 ∈ W . (In other words, onecan divide an element in G by any non-zero element of W : If g ∈ G and w 6= 0 ∈ W ,then there exists x ∈ G such that wx = g . Note however that x is not necessarily uniqueunless G is torsion free.)

proposition 0.75. A W -module G is flat if and only if it is torsion free and injective ifand only if it is divisible.

proof: Over any integral domain, a flat module is torsion free since ****. On the otherhand, any module G is the direct union of its finitely generated submodules and if G istorsion free then by Proposition 0.** these finitely generated submodules are projective.Hence a torsion free module G is a direct union of flat submodules, hence is flat.

Now over any integral domain, an injective module is divisible since ****. On the otherhand, to say that G is divisible is to say that ****. X

If a is an ideal of W we will denote the module W/a by W (a) . For any W -module Gwe will write G[a] to denote {g ∈ G | a g = 0} . For a prime ideal p we will writeG[p∞] =

⋃∞1 G[pn] . This is simply the p-primary component of G as defined in

Theorem 0.38.

23

proposition 0.76. If p is a prime ideal, then there is an embeddingγk : W (pk) →W (pk+1) such that the image of γk is then W (pk+1)[pk] = pW (pk+1) .proof: By Proposition 0.**, W (pk) ≈ p−k/W and W (pl+1) ≈ p−k−1/W . Nowp−k ⊆ p−k−1 , and this inclusion then induces an embedding p−k/W → p−k−1/W .The image of this embedding is p(p−k−1/W ) , and this can also be seen to be(p−k−1/W )[pk] . X

For us, the significance of Proposition 0.76 is that it provides a chain ofmonomorphisms

W (p) � W (p2) � W (p3) � · · · .We can then take the direct limit, which we denote by W (p∞) :

W (p∞) = lim→ W (pk).

proposition 0.77. The following three modules are isomorphic:

(1) W (p∞)(2) The p-primary component of Q/W .(3) Q/Wp .

proof: The p-primary component of Q/W is equal to⋃∞

1 p−k/W and byProposition 0.69, p−k/W ≈ W (pk) . Furthermore, the inclusion p−k/W → p−k−1/Wcorresponds precisely to the embedding W (pk) → W (pk+1) indicated above. Thus thep-primary component of Q/W is isomorphic to W (p∞) .

On the other hand, by Proposition 0.**, the p-primary component of Q/W isisomorphic to (Q/W )p = Qp/Wp = Q/Wp . X

proposition 0.78. For each prime p , W (p∞) is isomorphic to the injective envelopeof W (p) . Every injective W -module is a direct sum of modules isomorphic to W (p∞)(for various p) and to Q .

proof: ****

proposition 0.79. If G is a p-primary module, where p is a non-zero prime ideal, and iflengthG[p] = 1 , then either G ≈W (pk) for some k <∞ or G ≈W (p∞) .proof: We claim first that G is an essential extension of G[p] . In fact, if N is anon-zero submodule of G then ∅ 6= AssN ⊆ AssG = {p} . Thus there exists n ∈ N withannn = p . Then 0 6= n ∈ N ∩G[p] .

It follows that the injective envelope of G is the same as the injective envelope of G[p] .But since lengthG[p] = 1, G[p] ≈ W (p) . Thus the injective envelope of G is isomorphicto W (p∞) .

To finish, it suffices to prove that every non-trivial proper submodule of W (p∞) isisomorphic to W (pk) for some k . In fact, **** X

24

proposition 0.80. If p is a non-trivial prime ideal and G is a reduced p-primary modulesuch that G/pG has finite length, then G has finite length.

proof: ****

THE p-ADIC COMPLETION. One of the most useful tools for understanding thedeep structure of finite rank torsion free modules is the concept of completion. If M isa finite rank torsion free module and p is a prime ideal, then we can define the p-adiccompletion either as the inverse limit of the quotient modules M/pnM or as the setof Cauchy sequences from M with respect to the p-adic topology. Really there is nodifference between the two constructions. Since most readers will be familiar with theconcept in any case, we give a quick and dirty construction:

For any prime ideal (or even any non-trivial ideal) p of R and any R-module Mthere exists the p-adic filtration on M : M ⊇ pM ⊇ p2M ⊇ . . . . By taking thisfamily of submodules as a neighborhood basis at 0, M becomes a topological module.(A neighborhood basis at a point m ∈ M consists of the family of cosets m + pnM ,n = 1, 2, . . . .) This topology is called the p-adic topology on M .

We write p∞M =⋂∞

1 pkM and note that p∞M is the closure of 0 in M with respectto the p-adic topology.

For convenience, I will choose definition of the completion which is quick and dirty,although it’s not the most useful way to think about it.

definition 0.81. Let M be a finite rank torsion free W -module. We define the p-adiccompletion of M to be the submodule M of

∏∞1 M/pkM consisting of those sequences

(mk)∞1 ∈∏∞1 M/pkM such that (∀k)mk+1 ≡ mk (mod pkM) .

The p-adic completion will be important in this book because of its algebraicproperties. The topology will not be of any real relevance. However, for the record, wewill give a few topological properties.

If we give each module M/pkM the discrete topology, then∏∞

1 M/pkM has aninduced product topology. The relative topology that M inherits is called the inverselimit topology. A sequence {xt} in M converges to an element x ∈ M with respectto the inverse limit topology if and only if the sequence converges in each separatecoordinate. Since the topology on the modules M/pkM is descrete, this means thatif where xt = {mtk}∞1 and x = {mk} , then limt→∞ xt = x if and only if for each k ,mtk = mk for sufficiently large t .

(This is related to the “finite topology” on EndRM as defined by Jacobson if we thinkof EndRM as a submodule of MM .)

proposition 0.82. The inverse limit topology on M is the same as the p-adic topology.M is a complete topological module in this topology. If θ : M → M is the canonical map,then θ(M) is topologically dense in M .

25

proof: The neighborhood system at 0 in the inverse limit topology has a basis consistingof those submodules Un consisting of elements whose first n coordinates are zero. Nowsince the first n coordinates live in M/pkM for k ≥ n , it follows that pnM ⊆ Un . Onthe other hand, since the sequences in M satisfy the condition mn+k ≡ mk (mod pn) ,it follows that if mr = 0 for r ≤ n then mr ∈ pnM for all r . Thus Un ⊆ pkM , so theinverse limit topology and the p-adic topology are the same.

It seems clear that the p-adic completion is a contrived device to simplify the proofsof certain theorems rather than a concept of intrinsic interest in its own right. There isno natural geometric way of visualizing the p-adic numbers intuitively, and there are nonatural real-world applications which can be naturally represented by p-adic numbers.

However in terms of ring theory, there is one way in which the p-adic completion arisesin a fairly intrinsic way.

proposition 0.83. If p is a (non-zero) prime in a dedekind domain, then

EndW W (p∞) ≈ Wp .

proof: *****

As an important application of this, we can construct the famous Pontrayginmodule.

For convenience, suppose that W is a discrete valuation ring and let p be it uniquenon-trivial prime. We will construct a submodule M of Q ⊕ Q containing W ⊕Wsuch that M ∩ (Q ⊕ 0) = W ⊕ 0 and M ∩ (0 ⊕ Q) = 0 ⊕W , and πi(M) = Q . ByTheorem 0.47, such a submodule will be given by an isomorphism ϕ from Q/W to itself.By Proposition 0.77, Q/W ≈ W (p∞) , and therefore by Proposition 0.83, ϕ correspondsto an invertible element α ∈ Wp .

More specifically, M will be the inverse image in Q⊕Q of the module

{(x, ϕ(x)) | x ∈ Q/W} = {(x, αx) | x ∈ Q/W} ⊆ Q/W ⊕Q/W .

Now x ∈ Q/W can be written as a cosetw

pk+W for some w ∈ W and k ≥ 0. Using

this, we will describe M in the following way:

M = { (w, αw)pk

| w ∈W, k ≥ 0 } .

We need to make sense of this, since α ∈ Wp and so αw /∈W .For given k , there exists ak ∈ W such that α ≡ ak (mod pkWp) . It follows that

α

pk

andakpk

have the same image in Q/W . Now if ϕ is the automorphism of Q/W given

by ϕ(x) = αx , then ϕ(w

pk+ W ) =

αw

pk+W =

akw

pk+ W . Thus we can describe the

Pontryagin module corresponding to α ∈ Wp as

{ (w, akw)pk

| w ∈W, k ≥ 0 } .

26

proposition 0.84. (1) M is a torsion free Rp-module.

(2) M/pM ≈M/pM .

(3) M is reduced but is q -divisible for every prime ideal q 6= p .

(4) The p-adic topology on M is the same as the topology it inherits as a subspaceof∏∞

1 M/pkM if the modules M/pnM are considered discrete spaces and∏∞1 M/pkM is given the product topology.

(5) M is a closed subspace of∏∞

1 M/pkM .

(6) The diagonal map from M to∏∞

1 M/pkM embeds M/p∞M as a pure dense

submodule of M .

proof: (1) The factors M/pkM are all Rp-modules , hence so is∏∞

1 M/pkM . And if(mk) ∈ ∏∞

1 M/pkM satisfies the compatibility conditions mn+1 ≡ mk (mod pkM) andr ∈ Rp , then (rmk) will also satisfy these conditions. Now suppose (mk) 6= 0 ∈ M . Thenfor some n , mk 6= 0 ∈M/pkM . ****

(2) For k ≥ 1 identify∏∞k M/pkM with the submodule of

∏∞1 M/pkM consisting

of sequences (mk) ∈ M such that mk = 0 for n ≤ k . Then the family of subsets∏∞k M/pkM forms a neighborhood basis at 0 for

∏∞1 M/pkM . Thus it will suffice to

show that pkM = M ∩∏∞k M/pkM . ****

(3) It suffices to see that for any fixed k , the set of (mk) ∈ M satisfying mk+1 ≡ mk

(mod pkM) is a closed subset of M . But this set is the kernel of a continuous map fromM into the discrete module M/pkM .

(4) The diagonal map sends m ∈ M to the “constant sequence” (mk) where for all n ,mk is the coset m +M/pkM ∈ M/pkM . Clearly this is a homomorphism and we seeimmediately that its kernel is p∞M . X

proposition 0.85. If 0 →M1 →M2 →M3 → 0 is exact then so is0 → M1 → M2 → M3 → 0 .

proof: ****

It follows from Proposition 0.85 that an element of M can be represented as an infiniteseries . . . .

In order to say wear and tear on your author’s brain, for the remainder of this sectionwe will restrict attention to modules over a dedekind domain W . As before, Q willdenote its quotient field.

proposition 0.86. Wp is a commutative ring and QWp is a field.

proposition 0.87. If M is a p-primary W -module , then M has a unique Wp-modulestructure compatible with its W -module structure.

27

proof: The crucial fact here is that if M is p-primary then for every m ∈ M thereexists n (dependent on m , of course) such that pnm = 0. Now consider a sequencer = {rk} ∈ R ⊆ ∏R/pkR .; We will define rm = rnm . This is actually independent of n ,provided that pnm = 0 because of the condition that rn+k ≡ rn (mod pn) . From this, itis easy to see that all the conditions are satisfied for M to be an R-module.

If W is the ring of integers and M has finite rank, then all M/pnM are finite and soM is compact by Tychonof’s Theorem. In general, if p is a maximal ideal we can onlysay that the M/pnM have finite length. But it still seems intuitively plausible that Mwould have all the algebraic properties that one would expect of a compact module. Infact, abelian group theorists will recognize that M belongs to the class of “algebraicallycompact” modules, which are characterized by the following Proposition 0.* below. Theproof is a simple consequence of the fact that Wp is the inverse limit of the family ofmodules W/pkW . What this means is given by the following proposition:

proposition 0.88. Let πn : Wp → W/pkW and πn : W/pn+1W → W/pkW be thecanonical maps. Let M be a W -module and for each n ≥ 1 let ϕn : M → R/pkW .

Suppose that for every n , πnϕn+1 = ϕn . Then there exists a unique ϕ : M → Wp suchthat ϕn = πnϕ for all n .

proof: ****

corollary 0.89. The map Hom(M, Wp) → Hom(M,W/pW ) given by ϕ 7→ π1ϕ issurjective.

proof: ****

proposition 0.90. If p is a maximal ideal and M is a pure submodule of any (not

necessarily torsion free) W -module X , then M is a summand of X .

proof: ****

One should not expect that M is a finite rank W -module, except in the trivial caseR = R . However we have the following:

proposition 0.91. If M is a finite rank torsion free W -module then M is a finite ranktorsion free Wp-module . Furthermore, if p is maximal then M is a free Wp-module .

proof: ****

ADDITIVE FUNCTORS. It is an absolute prerequisite for much of this book thatthe reader be comfortable with the ideas of categories and natural transformations. Allthat your author can do to help with this is to present here a few topics which may seem,from the prospective of a standard graduate algebra course, a bit specialized.

28

The categories of modules (whether right or left) over a commutative ring, alongwith almost all of the other categories used in this book, are what are called additivecategories. This means that if ϕ , ψ ∈ Hom(X, Y ) then ϕ + ψ and ϕ − ψ are defined,and these operators make Hom(X, Y ) into an abelian group. Furthermore, it is requiredthat composition of functors be bilinear, i. e. in the same notation if α ∈ Hom(W,X) andβ ∈ Hom(Y, Z) then (ϕ+ψ)α = ϕα+ψα and β(ϕ+ψ) = βϕ+βψ . Furthermore, any twoobjects X and Y in these categories have a direct sum X ⊕ Y , which is simultaneouslya product and a coproduct. Finally, an additive category always has a zero object 0,analogous to the trivial module in the category of modules over a ring and characterizedby the property that Hom(0, X) = Hom(X, 0) = 0 for all objects X .

Essentially, the point of an additive category is that one may use all of the instinctsand habits one has learned from working with modules over a ring without gettinginto trouble. The one exception is that for some of the categories used in this bookstatements that involve kernels or cokernels, or the word “epimorphism,” may not workexactly the way one thinks they should. This does not cause problems if one simply usescommon sense. However for technical reasons it has seemed better to avoid the term“epimorphism” in favor of the safer term “surjection.”

The following examples may be skipped for now by anyone who is confused rather thanenlightened by them.

examples 0.92.

(1) For an integral domain R , we can consider the category oftorsion free R-modules, i. e. those modules M such that(∀r ∈ R)(∀m ∈M) r 6= 0 &m 6= 0 ⇒ rm 6= 0. This is an additive categoryand it causes no special problems unless one has been trained so thoroughly incategory theory that one no longer has common sense. For instance if K is thequotient field of R and ϕ : R → K is the inclusion map, then within the categoryin question (assuming K 6= R) it would not make sense to talk about the cokernelof ϕ , since the cokernel in the usual sense, i. e. K/R , is not torsion free and hencedoes not exist within the category. Now it turns out that within the categoryof torsion free modules over R ϕ does indeed have a cokernel, namely the zeromodule, and that ϕ is an epimorphism. However only a bloody-minded foolwould think in those terms. Which is why in this book we will avoid the word“epimorphism.”

(2) In Chapter 3 we will introduce a category which is absolutely fundamental tothe study of torsion free modules. Namely the category of “torsion free modulesunder quasi-homomorphisms.” The objects in this category are simply thetorsion free modules, but instead of using ordinary homomorphisms we definethe morphisms from M to N to be the elements of KHom(M,N) , whereKHom(M,N) denotes the localization of Hom(M,N) at the zero ideal (or, if oneprefers, K ⊗R Hom(M,N)). Thus morphisms in this category are not (usually)homomorphisms from M to N in the customary sense. Nonetheless, the set ofmorphisms KHom(M,N) is an abelian group, and this is an additive category,

29

with the direct sum of two modules being simply their ordinary direct sum. Wewill see in Chapter 3 that this category presents no particular problems, and thatthe way to approach it is simply to forget that there’s anything strange about itand pretend that one is actually in the category from part (1) above.

Unlike “epimorphism,” the word “exact” is too important for us to be able to livewithout. Fortunately, exactness poses no real conflict here between category theory andcommon sense. Exactness in the category of torsion free R-modules (Example 0.92 (1))means exactly what is does for the category of all R-modules . (So that – for what it’sworth – although as noted above the inclusion map R → K is an epimorphism in thecategory of torsion free modules, the sequence R → K → 0 is not exact in the category.)And we can avoid confusion when working in the category of torsion free modules underquasi-homomorphisms (Example 0.92 (2)) by the simple expedient of using the term“quasi-exact” rather than “exact.”

As Peter Freyd has pointed out, the whole reason one defines categories in the firstplace is in order to define functors. And the reason that we have made a fuss about theconcept of an additive category is not because it itself has any crucial importance in whatfollows, but because we want to be able to define the concept of an additive functor.

definition 0.93. A functor F between additive categories C and D is called an additivefunctor if the function it induces from HomC(C1, C2) to HomD(F (C1), F (C2)) is ahomomorphism. In other words F (ϕ+ ψ) = F (ϕ) + F (ψ) .

Additive functors are to additive categories what continuous maps are to topologicalspaces, or what homomorphisms are to modules. In other words, just about any functorthat anyone ever uses between two additive categories and in particular virtually anyfunctor used in this book, is additive. The concept is worth noting, however, because of acouple of useful propositions.

proposition 0.94. An additive functor preserves direct sums.

proof: ****

Actually, the converse is also true: A functor between additive categories is additive ifand only if it preserves direct sums (see [Freyd], for instance). However this converse isnot especially useful, at least not for us.

proposition 0.95. If R is a commutative ring and F is an additive functor from thecategory of R-modules into the category of abelian groups, then for every R-module M ,the abelian group F (M) has a canonical R-module structure, so that from practicalpurposes one may as well think of F as a functor from the category of R-modules intoitself.

proof: ****

30

There is a wonderful bit of black magic which is probably the main reason fordiscussing all of this at all.

proposition 0.96. If S and T are additive functors from the category of R-modulesto some other additive category, and if η : S → T is a natural transformation, and ifηR : S(R) → T (R) is an isomorphism, then ηP : S(P ) → T (P ) is an isomorphism forevery finitely generated projective module P . Furthermore if S and T preserve arbitrary(i. e. infinite) coproducts, then ηP is an isomorphism for all projective modules P .

proof: ****

definition 0.97. If C and D are categories and if F : C → D and G : D → Care functors, then we say that F is a left adjoint to G and that G is a rightadjoint to F is HomD(F (C), D) is naturally isomorphic to HomC(C,G(D)) forpair of objects C ∈ C and D ∈ D . In other words, there exists an isomorphismη : HomD(F (C), D) ≈−→ HomC(C,G(D)) such that whenever γ : C → C ′ and δ : D → D′

are morphisms in C and D then the following diagrams commute:

HomD(F (C′), D)F (γ)∗−−−−→ HomD(F (C), D)

η

y η

yHomC(C′, G(D))

γ∗−−−−→ HomC(C,G(D))

HomD(F (C), D) δ∗−−−−→ HomD(F (C), D′)

η

y η

yHomC(C,G(D))

G(δ∗)−−−−→ HomC(C,G(D′)).

The concept of a pair of adjoint functors seems initially mysterious and technical.However adjoint pairs are actually quite pervasive in module theory. The best wayof understanding the concept is by thinking of the pair of functors F = ⊗RM andG = HomR(M, ) where M is a module over a commutative ring R .

proposition 0.98. Let R be a commutative ring and M an R-module. Let F and Gbe the functors from the category of R-modules into itself given by F (X) = X ⊗R Mand G(Y ) = HomR(M,Y ) . Then F is a left adjoint to G , i. e. there exist a naturalisomorphisms η : HomR(X ⊗RM, Y ) −→ HomR(X, HomR(M,Y )) .proof: For ϕ ∈ HomR(X ⊗R M, Y ) define η(ϕ) : X → HomR(M,Y ) byη(ϕ)(x) : m 7→ ϕ(x ⊗m) . It is left to the intrepid reader to check that (1) η(ϕ) thusdefined is an R-linear homomorphism; (2) η itself is an R-linear map, in particular thatη(rϕ) = rη(ϕ) ; (3) that η is an isomorphism, i. e. one-to-one and surjective; and (4) thatη is natural in the sense described above. X

31

corollary 0.99. If R′ is an overring of R and P is an R′-module thenHomR′(R′ ⊗RM,P ) ≈ HomR(M,P ) .

(This is used in Chapter 4.)

proof: ****

If F is a left adjoint to G , then in particular for any X in C and Y

in D we get isomorphisms η : Hom(F (X), F (X)) ≈−→ Hom(X,GF (X))and η−1 : Hom(G(X), G(X)) ≈−→ Hom(FG(Y ), Y ) . We thus get mapsδ = η(1F (X)) : X → GF (X) and σ = η−1(1G(X)) : FG(Y ) → Y . It is easyto check that δ and σ are in fact natural transformations. They are usually notisomorphisms.

Just about the only time we will use δ and σ is in the case of the tensor productand Hom functors. They are “natural” not only in the technical sense, but also in theordinary usage of the word. The map

δ : X → HomR(M, M ⊗R X)

is given by δ(x)(m) = m⊗ x and

σ : M ⊗R HomR(M,Y ) → Y

is given by σ(m⊗ ϕ) = ϕ(m) . We sometimes refer to σ here as the “evaluation map.”These two innocent little natural transformations will play a crucial role in Chapter 9,

after making a brief appearance in Chapter 4. One can easily check that the isomorphismη : HomR(X ⊗R M, Y ) → HomR(X, HomR(M,Y )) described in the proof ofProposition 0.99 sends ϕ ∈ HomR(X ⊗RM, Y ) to the map given by the composition

Xδ−→ HomR(M,M ⊗R X)

ϕ∗−→ HomR(M,Y )

and that η−1 sends ψ ∈ HomR(X, HomR(M,Y )) to the map given by the composition

X ⊗RM ψ⊗R1M−−−−−→ HomR(M,Y ) ⊗RM σ−→ Y.