MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 5 Recall that in an additive category finite coproducts are isomorphic to finite products. They are called finite sums. But the “sum” is defined to be the coproduct. We use matrix notation for morphisms between direct sums. For example: fg ' = : X Y A B hk ! is the unique morphism satisfying
(1) pA ' jX = f : X A (2) p ' j = g : Y !A A Y ! (3) pB ' jX = h : X B (4) p ' j = k : Y !B B Y ! 3.1. Su ciently many projectives. Theorem 3.4. The categories Mod-R and mod-R have su ciently many projec- tives, i.e., for every R-module M there is a projective R-module P which maps onto M and, if M is finitely generated, we can also choose P to be f.g.
Remark 3.5. The other two categories in our list: tor-Z and (Z)donothaveenough projectives. In fact they have no nonzero projective objects! IP added a proof below.
Proposition 3.6. The categories tor-Z and (Z) have no nonzero projective objects. P The proof uses two facts which are so important that I call them “Theorems” instead of “Lemmas”. Theorem 3.7. Any direct summand of a projective object is projective. Proof. Suppose that P = X Y is projective. Then we show that X is projective: Given any epimorphism f : A B and morphism g : X B,wegetthemorphism [g, 0] : X Y B.SinceX B!is projective, there is a lifting! h : X Y A of g 0 to A.Then h !j : X Y is a lifting of g to A since f h j =[g, 0] !j = g. X ! X X ⇤ Theorem 3.8. Any epimorphism p : X P where P is projective splits, i.e., there is a morphism s : P X, called a section,! so that p s = id . ! X Proof. Since P is projective and p : X P is an epimorphism, id : P P lifts to X. ! P ! X > s 9 p
idP ✏ P / P I.e., p s = id as claimed. P ⇤ Proof. In class we went through the proof that (1) The morphism f : Z Z given by multiplication by 2 is an epimorphism in the ! category (Z)off.g.freeabeliangroups. (2) This morphismP does not have as section. 6MATH131B:ALGEBRAIIPARTA:HOMOLOGICALALGEBRA
By Theorem 3.8, Z is not projective. The objects of (Z)are,bydefinition,isomorphic P to Zn for some n.Ifthiswereprojective,byTheorem3.7,Z would be projective, which is a contraction. Therefore, (Z) has no projective objects (except 0 which is always projective). P In the category of finite abelian groups, the object Zn is not projective since the epimorphism Zn2 Zn ! has no section (left as an exercise). But, every finite abelian group is a direct sum of cyclic groups. So, tor-Z has no nonzero projective objects. ⇤ Getting back to the good cases: Suppose that is an additive category with the following two properties: C (1) has enough projectives, i.e., for every object X,thereisanepimorphismP X whereC P is projective. ! (2) Every morphism in has a kernel (in ). C C Then each object X has a projective resolution: 0 X d0 P d1 P d2 P 0 1 2 ··· constructed as follows. P is a projective object with an epimorphism d : P X.By 0 0 0 ! assumption (2), this has a kernel K0.LetP1 be a projective object with epimorphism P K =kerd .(Inthediagramd : P P is the composition P K P .) 1 ! 0 0 1 1 ! 0 1 ⇣ 0 ⇢ 0 By induction on n we have an epimorphism Pn Kn 1.LetKn be the kernel of this ⇣ morphism and let Pn+1 be a projective object with an epimorphism Pn+1 ⇣ Kn+1.
Warning! In a general additive category, Kn 1 may not be the cokernel of Kn Pn. ⇢ Theorem 3.9. In the category Mod-R of all R-modules and, for R right Noetherian, the category mod-R of f.g. right R-modules, every object has a projective resolution. 3.2. Injective objects. Definition 3.10. An object Q of any additive category is injective if, for any monomorphism A B,anymorphismA Q extends to BC.I.e.,i↵ ! ! Hom (B,Q) Hom (A, Q) 0 C ! C ! is exact. As before this is equivalent to: Theorem 3.11. Q is injective if and only if Hom ( ,Q) is an exact functor. C The di cult theorem we need to prove is the following: Theorem 3.12. The category Mod-R has su ciently many injectives. I.e., every R- module embeds in an injective R-module (usually not finitely generated). As in the case of projective modules, since every morphism has a cokernel, this the- orem will tell us that every R-module M has an injective co-resolution which is an exact sequence: 0 M Q Q Q ! ! 0 ! 1 ! 2 !··· where each Qi is injective. MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 7 4. Injective modules I will go over Lang’s proof (XX 4) that every R-module M embeds in an injective § module Q.LangusesthedualM ^ of a module. (Another dual M _ is discussed in III 6.) § 4.1. dual module. Definition 4.1. The dual of a right R-module M is defined to be the left R-module
M ^ := HomZ(M,Q/Z) with left R-action given by (r')(x)='(xr) for all ' M ^,r R. 2 2 M ^ is, in general, not finitely generated. Proposition 4.2. Duality is a left exact functor
()^ : Mod-R R-Mod ! which is additive and takes sums to products:
^ M↵ ⇠= M↵^ ⇣M ⌘ Y Proof. We already saw that the hom functor HomZ( ,X)isleftexactforanyabelian group X.Itisalsoobviouslyadditive which means that (f + g)] = f ] + g] for all f,g : N M. I.e., the duality functor induces a homomorphism (of abelian groups): ! Hom (N,M) Hom (M ^,N^) R ! Z Duality also takes sums to products since a homomorphism f : M X ↵ ! is given uniquely by its restriction to eachM summand: f : M X and the f can all ↵ ↵ ! ↵ be nonzero. (So, (f↵)↵ is in the product not the sum.) ⇤ 4.2. Constructing injective modules. In order to get an injective right R-module we need to start with a left R-module. Theorem 4.3. Suppose F is a free left R-module. (I.e., F = R is a direct sum of R copies of R considered as a left R-module). Then F ^ is an injective right R-module. This theorem follows from the following lemma. Lemma 4.4. (1) A product of injective modules is injective. (2) HomR(M, RR^) ⇠= HomZ(M,Q/Z) (isomorphism of left R-modules) (3) Q/Z is an injective Z-module. Proof of the theorem. Lemma (3) implies that Hom ( , Q/Z)isanexactfunctor.(2) Z implies that HomR(M, RR^)isanexactfunctor.Therefore,RR^ is an injective R- module. Since duality takes sums to products,
F ^ =( RR)^ = RR^ is projective by (1) proving the Theorem.M Y ⇤ 8MATH131B:ALGEBRAIIPARTA:HOMOLOGICALALGEBRA We need one more lemma to prove the main theorem. Then we have to prove the lemmas. Lemma 4.5. Any right R-module is naturally embedded in its double dual:
M M ^^ ✓ Assume this 4th fact for a moment. Theorem 4.6. Every right R-module M can be embedded in an injective right R-module.
Proof. Let F be a free left R-module which maps onto M ^:
F M ^ 0 ! ! Since duality is left exact we get:
0 M ^^ F ^ ! ! By Lemma 4.5, we have M M ^^ F ^.So,M embeds in the injective module ✓ ✓ F ^. ⇤