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Home , Divisible group, Injective module

3 The Hom — Projectivity and Injectivity.

Our immediate goal is to study the phenomenon of equivalence, and that we shall do in the next Section. First, however, we have to be in control of the so-called Hom functors and projective modules. Later in the term, the duals of projective modules, the injective modules, will play a crucial role. So in this Section we will treat the basics of these two types of modules.

We begin with a pair of rings R and S aand an (R, S)- RUS. This bimodule determines two Hom functors

HomR(RUS, ):RMod → SMod and HomR( ,R US):RMod → ModS dened by

HomR(RUS, ):R M 7→ HomR(RUS,R M) and HomR( ,R US):R M 7→ HomR(RM,R US) for each RM in RMod, and

HomR(RUS,f):ϕ 7→ f ϕ and HomR(f,R US): 7→ f for all f : M → N in RMod and all ϕ ∈ HomR(U, M) and all ∈ HomR(N,U). One readily checks that both of these are additive, that HomR(RUS, ) is covariant and HomR( ,R US) is contravariant. In general, these functors are not exact, but each is left exact. That is, for each short

f g 0 → K → M → N → 0 in RMod the sequence

Hom(U,f) Hom(U,g) 0 → Hom(U, K) → Hom(U, M) → Hom(U, N) is exact in SMod, and

Hom(g,U) Hom(f,U) 0 → Hom(N,U) → Hom(M,U) → Hom(K, U) is exact in ModS. We do not want to take the time to prove these facts here. Although exactness at M is a bit of a challenge, the proof is pretty strightforward diagram chasing. Proofs are available in standard references; see, for example, [1] and [6]. In the interest of completeness, we’ll record all of this formally.

3.1. Theorem. If RUS is a bimodule, then the functors

HomR(RUS, ):RMod → SMod and HomR( ,R US):RMod → ModS

are both additive and left exact with HomR(RUS, ) covariant and HomR( ,R US) contravariant. Non-Commutative Rings Section 3 23

Since the notation for these functors is more than a little cumbersome, it is fairly common, when the bimodule U is xed, to abbreviate

M = Hom(U, M),f = Hom(U, f) and M = Hom(M,U),f = Hom(f,U).

These Hom functors need not be exact, but as we shall see the modules U for which they are exact play a very important role in our study. Let R be a . A left RP is projective in case the covariant evaluation HomR(P, ):RMod → Ab is exact. Dually, a left module RQ is injective in case the contravariant functor HomR(,Q):RMod → Ab is exact. We dened projective and injective left modules — the denition for right modules should be clear. Also, note that although we dened these in terms of functors into Ab, exactness holds more generally. For one example, if S is another ring, then RPS is projective as a left R-module i HomR(P, ):RMod → SMod is exact. You can easily check out this and the other cases.

The rst thing we want to determine is whether any of these modules exist. And, once we learn that there are plenty of them, we’d like to nd out just how nice, or not, they may be. Although these concepts are homologically dual, we cannot automatically take every true statement about projective modules, turn it around, and have a true statement about injective modules. So to some extent we have to treat them separately. We’ll begin with projectives.

But, rst, let 0 =6 e ∈ R be an , so eRe is a ring with identity e. Dene a function

e : RMod → eReMod by e : M 7→ eM for each RM and e(f):ex 7→ ef(x) for each f : M → N in RMod. Then one easily checks that e is an additive, exact, covariant functor. The following Lemma then describes a natural isomorphism of the functors e and HomR(Re, )

3.2. Lemma. Let e ∈ R be a non-zero idempotent. Then for each RM there is a natural eRe- isomorphism

M : eM → HomR(Re, M),

dened by M (ex):ae 7→ aex for all ex ∈ eM and ae ∈ Re.

Proof. Clearly, each M (ex)isanR-linear map from Re to M and the map M : ex 7→ M (ex) is additive. So let ere ∈ eRe, ex ∈ eM, and ae ∈ Re. Then

M (erex)(ae)=aerex =(aere)(ex)

= M (ex)(aere)

=[ereM (ex)](ae),

(where the last equality comes from the way eRe acts on HomR(Re, M)), so that M is R-linear.

For each ϕ ∈ HomR(Re, M) we see that ϕ = M (eϕ(e)), so M is epic, and if ex ∈ Ker M , then 24 Section 3

ex = M (ex)(e)=0,soM is monic. Finally, let f : M → N in RMod and consider the following diagram:

ef - eM eN

M N ? ? f Hom(Re, M) - Hom(Re, N)

Then for each ex ∈ eM and ae ∈ Re

f(M (ex))(ae)=f(aex)=aef(ex)

= N (ef(ex))(ae),

so the diagram commutes and is natural.

Since the functor e is exact, this means that HomR(Re, ) is exact and Re is projective. In particular, the regular module RR is projective.

Now a module RP is projective precisely when the functor HomR(P, ) is exact. Equivalently, RP g is projective i for every R-epimorphism M → N → 0, every R-homomorphism f : P → N factors through g; that is, there exists a homomorphism h : P → M with f = g h.

p P p p h p p p p f ©p ? g- - MN 0

Dually, a module RQ is injective provided the functor HomR(,Q) is exact. So, equivalently, RQ is g injective i for every R- 0 → N → M, every R-homomorphism f : N → Q factors through g; that is, there exists a homomorphism h : M → Q with f = h g.

Q p p 6Ip p h f p p p p - g- 0 NM

3.3. Proposition. In RMod,

(1) A coproduct (P, (i)) of left R–modules (P) is projective i each P is projective;

(2) A product (Q, (p)) of left R–modules (Q) is injective i each Q is injective. Non-Commutative Rings Section 3 25

Proof. We’ll prove (1) and leave the dual proof to you. Moreover, it will suce to assume that P ` → is the coproduct P = P with coordinate : P P . Then also let be the coordinate projection : P → P.

(=⇒) Consider the diagram - P P h h0 f ? © ? g- - MN 0 with g an epimorphism and f a homomorphism. Since P is projective and f : P → N, there is 0 0 0 a homomorphism h : P → M for which f = gh . Dene h by h = h : P → M. Then 0 gh = gh = f = f and so P is projective.

(⇐=) Consider the diagram - P P h h f ? © ? g- - MN 0 where g is an epimorphism and f : P → N is a homomorphism. Since P is projective and f : ` → → → P N, there exists an h : P M with gh = f. Then h = h : P M and

(gh f) = gh f = gh f = 0 for each ∈ sogh = f and P is projective.

Now by Lemma 3.2, the regular module RR is projective, so it follows from Proposition 3.3

3.4. Corollary. Every is projective.

Proof. By Lemma 3.2, the regular modules RR and RR are projective.

3.5. Proposition. For a module RP the following are equivalent:

(a) RP is projective; g (b) Every epimorphism M → P → 0 splits;

(c) RP is isomorphic to a direct summand of a free module.

Proof. (a) =⇒ (b). Consider the diagram

P h 1P © ? g- - MP 0 with exact row. If it commutes, then gh =1P and the epimorphism g is split. 26 Section 3

(b) =⇒ (c). There is some free module RF and some epimorphism F → P → 0. Apply (b).

(c) =⇒ (a). By Corollary 3.4 every free module is projective and by Proposition 3.3 every direct summand of a is projective.

Since every module is a factor of a free module, we have trivially the following important

3.6. Corollary. For every left R–module RM there is a projective module RP and an R- epimorphism P → M → 0.

The above results for projective modules all have duals for injective modules. There is one small hitch. The notion dual to that of a free module is not quite so obvious. So we will postpone discussion of these “injective cogenerators”. Fortunately, though, we can fairly easily nd a dual to Corollary 3.6. We begin with a really nice test for injectivity.

3.7. Lemma. [The Injective Test Lemma.] A module RQ is injective i for every left I

of R and every R–homomorphism f : I → Q there is a homomorphism g : R → Q with g|I = f.

Proof. The direction (=⇒) is trivial. For the other direction, (⇐=), asume that Q satises the stated condition. As a rst step we prove the claim

Claim. If N is a submodule of a Rx, and if f : N → Q is an R–homomorphism, then there is a homomorphism g : Rx → Q with g|N = f.

To prove the claim, rst note that right multiplication by x, = x : R → Rx is an epimorphism and (N)=I is a left ideal of R. Now consider the following diagram: - inc- 0 IR

|I ? ? - inc- 0 NRx

f ? Q

By hypothesis there is a homomorphism h : R → Q such that h|I = f|I . Since Ker I, we have Ker Ker h, so that h lifts to some g : Rx → Q with h = g. But then for all a ∈ I, we have g(a)=h(a)=f(a),sog|N = f, as claimed.

Now consider the diagram Q 6 f - inc- 0 NM Non-Commutative Rings Section 3 27

Let F be the set of all R–homomorphisms g such that RN Dg = Dom g R M and g : Dg → Q with 0 0 g|N = f. Then F is a poset w.r.t. the relation g g i both Dg Dg0 and g |Dg = g. By Zorn there exists a maximal g ∈F. Let D = Dg. We claim D = M. Indeed, suppose x ∈ M\D. Consisder the cyclic submodule Rx. Let K = Rx ∩ D Rx. By the Claim there is an h : Rx → Q with h|K = g|K . Let y ∈ D and a ∈ R with ax + y =0,soax = y ∈ K. Then h(ax)+g(y)=g(ax) g(y) = 0. That is, there is an R–homomorphism g : Rx + D → Q dened by g(ax + y)=h(ax)+g(y), contrary to the maximality of g.

Now we can begin to prove that injectives exist; we begin by nding some injective abelian groups.

More generally, let R be a PID. A module RQ is divisible in case for each 0 =6 d ∈ R, dQ = Q.For example, the Z–modules (= abelian groups) Q and Zp∞ are divisible.

3.8. Lemma. A module RQ over a PID R is injective i it is divisible.

Proof. Let d ∈ R be non-zero. For each q ∈ Q there is an R-homomorphism f : Rd → Q with f(d)=q. Then f has an extension g : R → Q i q = f(d)=g(d)=dg(1) i q ∈ dQ. So by the

Injective Test Lemma 3.7 RQ is injective i Q = dQ for each non-zero d ∈ R.

So there do exist some injective abelian groups. Now we prove that there exist a lot of them, enough so that every lies in at least one injecive.

3.9. Corollary. If M is an abelian group then there is some D and a monomorphism M → D.

Proof. There is an epimorphism from the Z(M) onto M, so that for some K Z(M), M = Z(M)/K Q(M)/K = D a divisible group.

But for an arbitrary ring R each of these divisible groups produces an injective R-module, so the injectives are starting to pile up.

3.10. Lemma. Let R be a ring. Then for every divisible abelian group D, the R-module

HomZ(RR,D) is injective.

Proof. The proof uses one version of “Adjoint Associativity”; namely, if R and S are rings and if there are modules RM, SNR, SK, then there is an isomorphism natural in each variable

HomR(M,HomS(N,K)) = HomS(N R M,K).

Now RR is projective, so the functor RR ( ) is exact. Also ZD is injective, so the functor HomZ( ,D) 28 Section 3

is exact. Thus, the functor HomZ(R R ( ),D) is exact. Then by Adjoint Associativity, the functor

HomR( , HomZ(R, D)) is exact. Therefore, HomZ(R, D) is injective.

Now, at last. we have the important dual to Corollary 3.6.

3.11. Corollary. For every module RM there exists some injective RQ and a monomorphism M → Q.

Proof. By Corollary 3.9 the abelian group M can be embedded in some divisible group D. But then by Lemma 3.10, Hom(RR,D) is injective, and clearly,

RM = HomR(R, M) HomZ(R, M) HomZ(R, D).

And with that we can obtain a (partial) dual to the important Proposition 3.5.

3.12. Proposition. A module RQ is injective i every monomorphism 0 → Q → M splits.

Proof. (⇒) Consider the diagram

Q 6@I h 1Q @ @ - f- 0 Q M

If it commutes, then hf =1Q, and the monomorphism f is split.

(⇐) By Corollary 3.11, there is an E with Q E. So if the inclusion map Q,→ E splits, then E = Q Q0 and by Proposition 3.3, Q is injective.

We conclude this Section with a brief return to projective modules. The point is to discuss an important property of nitely generated projectives. We begin with an arbitrary left

R-module RM. Then its R-dual M = HomR(M,R) is a right module. The elements of this R-dual M are often called linear functionals. And, here, we will write them as right operators. Thus, for each x ∈ M, r ∈ R, and f ∈ M we have

(rx)f = r(xf) ∈ R.

A pair of sequences (x1,...,xn)inM and (f1,...,fn)inM is called a dual for M in case for each x ∈ M, Xn x = (xfi)xi. i=1 Non-Commutative Rings Section 3 29

As an important example of such a dual basis, let RF be a free module with free basis (x1,...,xn). (See Exercise 2.4.) Then this free basis determines a unique dual basis for F . Indeed, for each x ∈ F (n) there exists a unique element (a1,...,an) ∈ R with

Xn x = aixi. i=1

Then for each i =1,...,n the map pi : F → R dened by pi : x 7→ ai is a linear functional and the pair (x1,...,xn) and (p1,...,pn) forms a dual basis for F . And now the important

3.13. Lemma. [Dual Basis Lemma.] An R-module RP is nitely generated projective i it has a nite dual basis.

Proof. (=⇒) Since P is nitely generated there is a free module F with free basis (x1,...,xn) and an epimorphism ϕ : F → P . Since P is also projective, this map splits and there is a monomorphism

: P → F with (we’ll write these functions on the right) ϕ =1P . By the above example, there is a dual basis (x1,...,xn), (p1,...,pn) for F . But then clearly, (x1ϕ,...,xnϕ), (p1,...,pn) is a dual basis for P .

(⇐=) Let (y1,...,yn), (f1,...,fn) be a dual basis for P . Then y1,...,yn spans P , and there is a free module F with free basis (x1,...,xn) and an epimorpihsm ϕ : F → P such that ϕ : xi 7→ yi for i =1,...,n. But there is a homomorphosm : P → F via

Xn Xn : x = (xfi)xi 7→ (xfi)yi. i=1 i=1

Then ϕ =1P and ϕ splits; thus, P is projective.

Exercises 3.

3.1. Let e ∈ R be a non-zero idempotent and consider the functor e : RMod → eReMod.

(a) Show that e is exact.

(b) Show that there is a natural isomorphism : eR R ( ) → e.

(c) Deduce that every projective module is at. [A module MR is at in case the functor

M R ( ):RMod → Ab is exact.]

3.2. Prove that if e, f ∈ R are non-zero idempotents, then

HomR(Re, Rf) =eRe eRffRf = HomR(fR,eR). 30 Section 3

3.3. Let RQ be a divisible module over a PID, R.

(a) Prove that every factor module Q/K of Q is divisible.

(b) Must every submodule of Q be divisible? Explain.

(c) Suppose that 0 → A → B → C → 0 is a short exact sequence over R. If both A and C are divisible, must B be divisible? Explain.

3.4. Products of projective modules need not be projective. Indeed, prove that the abelian group ZN is N N not free. [Hint: Consider the M of Z consisting of all the sequences (x1,x2,...) ∈ Z such that each power of 2 divides all but nitely many of the terms. If ZN is free, then so is M. But then consider 2M.]

3.5. Prove that every coproduct of injective left R-modules is injective i R is left noetherian. [Hint: Use the Injective Test Lemma. (⇐) Every left ideal I of R is nitely generated, so if f : I → ` Q, then the image is in a nite sum of the Q’s. (⇒) Suppose there is an innite ascending chain I

3.6. A ring R is (left) self-injective if the regular module RR is injective.

(a) Let R be a PID and let I be a non-trivial ideal of R. Prove that R is self-injective.

(b) More generally, let R be a whose lattice of ideals is a nite chain, say

0

3.7. There is a version of the Dual Basis Lemma more general than that given in Lemma 3.13. Indeed, let RM be an R-module. A pair of indexed sets (x) ∈ inM and (f) ∈ inM is a dual

basis for M in case for each x ∈ M, xf = 0 for almost all ∈ and X x = (xf)f. ∈

Prove that a module RP is projective i it has a dual basis.