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3 the Hom Functors — Projectivity and Injectivity

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3 the Hom Functors — Projectivity and Injectivity 3 The Hom Functors — Projectivity and Injectivity. Our immediate goal is to study the phenomenon of category 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)-bimodule 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 exact sequence 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 ring. A left module RP is projective in case the covariant evaluation functor HomR(P, ):RMod → Ab is exact. Dually, a left module RQ is injective in case the contravariant duality 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 idempotent, 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-monomorphism 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 monomorphisms : 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 free module 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 projective module 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 ideal 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 cyclic module 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 .
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