3 The Hom Functors Projectivity and Injectivity.

Transcription

1 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 R U S. This bimodule determines two Hom functors defined by Hom R ( R U S, ):RMod SMod and Hom R (, R U S ):RMod ModS Hom R ( R U S, ): R M Hom R ( R U S, R M) and Hom R (, R U S ): R M Hom R ( R M, R U S ) for each R M in RMod, and Hom R ( R U S,f):ϕ f ϕ and Hom R (f, R U S ):ψ ψ f for all f : M N in RMod and all ϕ Hom R (U, M) and all ψ Hom R (N,U). One readily checks that both of these are additive, that Hom R ( R U S, ) is covariant and Hom R (, R U S ) is contravariant. In general, these functors are not exact, but each is left exact. That is, for each short exact sequence in RMod the sequence is exact in SMod, and 0 Hom(U, K) Hom(U,f) 0 Hom(N,U) Hom(g,U) 0 K f M g N 0 Hom(U, M) Hom(U,g) Hom(U, N) Hom(M,U) Hom(f,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 Theorem. If R U S is a bimodule, then the functors Hom R ( R U S, ):RMod SMod and Hom R (, R U S ):RMod ModS are both additive and left exact with Hom R ( R U S, ) covariant and Hom R (, R U S ) contravariant.

2 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 fixed, 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 R P is projective in case the covariant evaluation functor Hom R (P, ) :RMod Ab is exact. Dually, a left module R Q is injective in case the contravariant duality functor Hom R (,Q):RMod Ab is exact. We defined projective and injective left modules the definition for right modules should be clear. Also, note that although we defined these in terms of functors into Ab, exactness holds more generally. For one example, if S is another ring, then R P S is projective as a left R-module iff Hom R (P, ) :RMod SMod is exact. You can easily check out this and the other cases. The first 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 find 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, first, let 0 e R be an idempotent, so ere is a ring with identity e. Define a function Λ e : RMod eremod by Λ e : M em for each R M and Λ e (f) :ex ef(x) for each f : M N in RMod. Then one easily checks that Λ e is an additive, exact, covariant functor. The following Lemma then describes a natural isomorphism of the functors Λ e and Hom R (Re, ) 3.2. Lemma. Let e R be a non-zero idempotent. Then for each R M there is a natural ereisomorphism ρ M : em Hom R (Re, M), defined by ρ M (ex) :ae aex for all ex em and ae Re. Proof. Clearly, each ρ M (ex) isanr-linear map from Re to M and the map ρ M : ex ρ M (ex) is additive. So let ere ere, ex em, and ae Re. Then ρ M (erex)(ae) = aerex =(aere)(ex) = ρ M (ex)(aere) = [ereρ M (ex)](ae), (where the last equality comes from the way ere acts on Hom R (Re, M)), so that ρ M is R-linear. For each ϕ Hom R (Re, M) we see that ϕ = ρ M (eϕ(e)), so ρ M is epic, and if ex Ker ρ M, then

3 24 Section 3 ex = ρ M (ex)(e) =0,soρ M is monic. Finally, let f : M N in RMod and consider the following diagram: em ef en ρ M ρ N f Hom(Re, M) Hom(Re, N) Then for each ex em and ae Re f (ρ M (ex))(ae) = f(aex) =aef(ex) = ρ N (ef(ex))(ae), so the diagram commutes and ρ is natural. Since the functor Λ e is exact, this means that Hom R (Re, ) is exact and Re is projective. In particular, the regular module R R is projective. Now a module R P is projective precisely when the functor Hom R (P, ) is exact. Equivalently, R P is projective iff for every R-epimorphism M g 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 h f g M N 0 Dually, a module R Q is injective provided the functor Hom R (,Q) is exact. So, equivalently, R Q is injective iff for every R-monomorphism 0 N g M, every R-homomorphism f : N Q factors through g; that is, there exists a homomorphism h : M Q with f = h g. Q f h 0 g N M 3.3. Proposition. In RMod, (1) A coproduct (P, (i α ) Ω ) of left R modules (P α ) Ω is projective iff each P α is projective; (2) A product (Q, (p α ) Ω ) of left R modules (Q α ) Ω is injective iff each Q α is injective.

4 Non-Commutative Rings Section 3 25 Proof. We ll prove (1) and leave the dual proof to you. Moreover, it will suffice to assume that P is the coproduct P = Ω P α with coordinate monomorphisms ι α : P α P. Then also let π α be the coordinate projection π α : P P α. (= ) Consider the diagram P π α P α h h f g M N 0 with g an epimorphism and f a homomorphism. Since P is projective and fπ α : P N, there is a homomorphism h : P M for which fπ α = gh. Define h by h = h ι α : P α M. Then gh = gh ι α = fπ α ι α = f and so P α is projective. ( =) Consider the diagram ι α P α P h h α f g M N 0 where g is an epimorphism and f : P N is a homomorphism. Since P α is projective and fι α : P α N, there exists an h α : P α M with gh α = fι α. Then h = Ω h α : P M and (gh f)ι α = ghι α fι α = gh α fι α = 0 for each α Ωsogh = f and P is projective. Now by Lemma 3.2, the regular module R R is projective, so it follows from Proposition Corollary. Every free module is projective. Proof. By Lemma 3.2, the regular modules R R and R R are projective Proposition. For a module R P the following are equivalent: (a) R P is projective; (b) Every epimorphism M g P 0 splits; (c) R P is isomorphic to a direct summand of a free module. Proof. (a) = (b). Consider the diagram P h 1 P g M P 0 with exact row. If it commutes, then gh =1 P and the epimorphism g is split.

5 26 Section 3 (b) = (c). There is some free module R F 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 R M there is a projective module R P 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 find a dual to Corollary 3.6. We begin with a really nice test for injectivity Lemma. [The Injective Test Lemma.] A module R Q is injective iff 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 satisfies the stated condition. As a first 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, first note that right multiplication by x, ρ = ρ x : R Rx is an epimorphism and ρ (N) =I is a left ideal of R. Now consider the following diagram: 0 inc I R ρ I ρ 0 N inc Rx f Q By hypothesis there is a homomorphism h : R Q such that h I = fρ I. Since Ker ρ I, we have Ker ρ Ker h, so that h lifts to some g : Rx Q with h = gρ. But then for all a I, we have gρ(a) =h(a) =fρ(a),sog N = f, as claimed. Now consider the diagram Q f 0 N inc M

6 Non-Commutative Rings Section 3 27 Let F be the set of all R homomorphisms g such that R N D g = Dom g R M and g : D g Q with g N = f. Then F is a poset w.r.t. the relation g g iff both D g D g and g D g = g. By Zorn there exists a maximal g F. Let D = D g. 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 defined 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 finding some injective abelian groups. More generally, let R be a PID. A module R Q is divisible in case for each 0 d R, dq = Q. For example, the Z modules (= abelian groups) Q and Z p are divisible Lemma. A module R Q over a PID R is injective iff 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 iff q = f(d) =g(d) =dg(1) iff q dq. So by the Injective Test Lemma 3.7 R Q is injective iff 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 abelian group lies in at least one injecive Corollary. If M is an abelian group then there is some divisible group D and a monomorphism M D. Proof. There is an epimorphism from the free abelian group 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 Lemma. Let R be a ring. Then for every divisible abelian group D, the R-module Hom Z (R R,D) is injective. Proof. The proof uses one version of Adjoint Associativity ; namely, if R and S are rings and if there are modules R M, S N R, S K, then there is an isomorphism natural in each variable Hom R (M,Hom S (N,K)) = Hom S (N R M,K). Now R R is projective, so the functor R R ( ) is exact. Also Z D is injective, so the functor Hom Z (,D)

7 28 Section 3 is exact. Thus, the functor Hom Z (R R ( ),D) is exact. Then by Adjoint Associativity, the functor Hom R (, Hom Z (R, D)) is exact. Therefore, Hom Z (R, D) is injective. Now, at last. we have the important dual to Corollary Corollary. For every module R M there exists some injective R Q 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(R R,D) is injective, and clearly, RM = Hom R (R, M) Hom Z (R, M) Hom Z (R, D). And with that we can obtain a (partial) dual to the important Proposition Proposition. A module R Q is injective iff every monomorphism 0 Q M splits. Proof. ( ) Consider the diagram Q 1 h Q 0 f Q M If it commutes, then hf =1 Q, and the monomorphism f is split. ( ) By Corollary 3.11, there is an injective module E with Q E. So if the inclusion map Q E splits, then E = Q Q 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 characteristic property of finitely generated projectives. We begin with an arbitrary left R-module R M. Then its R-dual M = Hom R (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 (x 1,...,x n )inm and (f 1,...,f n )inm is called a dual basis for M in case for each x M, n x = (xf i )x i. i=1

8 Non-Commutative Rings Section 3 29 As an important example of such a dual basis, let R F be a free module with free basis (x 1,...,x n ). (See Exercise 2.4.) Then this free basis determines a unique dual basis for F. Indeed, for each x F there exists a unique element (a 1,...,a n ) R (n) with n x = a i x i. i=1 Then for each i =1,...,n the map p i : F R defined by p i : x a i is a linear functional and the pair (x 1,...,x n ) and (p 1,...,p n ) forms a dual basis for F. And now the important Lemma. [Dual Basis Lemma.] An R-module R P is finitely generated projective iff it has a finite dual basis. Proof. (= ) Since P is finitely generated there is a free module F with free basis (x 1,...,x n ) and an epimorphism ϕ : F P. Since P is also projective, this map splits and there is a monomorphism ψ : P F with (we ll write these functions on the right) ψϕ =1 P. By the above example, there is a dual basis (x 1,...,x n ), (p 1,...,p n ) for F. But then clearly, (x 1 ϕ,...,x n ϕ), (ψp 1,...,ψp n ) is a dual basis for P. ( =) Let (y 1,...,y n ), (f 1,...,f n ) be a dual basis for P. Then y 1,...,y n spans P, and there is a free module F with free basis (x 1,...,x n ) and an epimorpihsm ϕ : F P such that ϕ : x i y i for i =1,...,n. But there is a homomorphosm ψ : P F via n n ψ : x = (xf i )x i (xf i )y i. Then ψϕ =1 P and ϕ splits; thus, P is projective. i=1 i=1 Exercises Let e R be a non-zero idempotent and consider the functor Λ e : RMod eremod. (a) Show that Λ e is exact. (b) Show that there is a natural isomorphism η : er R ( ) Λ e. (c) Deduce that every projective module is flat. [A module M R is flat in case the functor M R ( ):RMod Ab is exact.] 3.2. Prove that if e, f R are non-zero idempotents, then Hom R (Re, Rf) = ere erf frf = HomR (fr,er).

9 30 Section Let R Q 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 Products of projective modules need not be projective. Indeed, prove that the abelian group Z N is not free. [Hint: Consider the subgroup M of Z N consisting of all the sequences (x 1,x 2,...) Z N such that each power of 2 divides all but finitely many of the terms. If Z N is free, then so is M. But then consider 2M.] 3.5. Prove that every coproduct of injective left R-modules is injective iff R is left noetherian. [Hint: Use the Injective Test Lemma. ( ) Every left ideal I of R is finitely generated, so if f : I Qα, then the image is in a finite sum of the Q α s. ( ) Suppose there is an infinite ascending chain I 1 <I 2 < of left ideals. For each n, let Q n be an injective module containing R/I n. LetI = n=1i n and let f : I Q = n=1 Q n be defined by π n f(a) =a + I n. Show that this cannot be extended to R.] 3.6. A ring R is (left) self-injective if the regular module R R 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 commutative ring whose lattice of ideals is a finite chain, say 0 <I 1 < <I n = R. Prove that R is self-injective There is a version of the Dual Basis Lemma more general than that given in Lemma Indeed, let R M be an R-module. A pair of indexed sets (x α ) α ΩinM and (f α ) α ΩinM is a dual basis for M in case for each x M, xf α = 0 for almost all α Ω and x = α Ω(xf α )f α. Prove that a module R P is projective iff it has a dual basis.