MATH 131B: ALGEBRA II PART A: 5 Recall that in an additive finite are isomorphic to finite products. They are called finite sums. But the “sum” is defined to be the . We use matrix notation for between direct sums. For example: fg ' = : X Y A B hk !  is the unique 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. Suciently many projectives. Theorem 3.4. The categories Mod-R and mod-R have suciently 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 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 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 : 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 . C The dicult theorem we need to prove is the following: Theorem 3.12. The category Mod-R has suciently 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 : 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. 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 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 ✓ ✓ F ^. ⇤