MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 9 4.3. proof of lemmas. There are four lemmas to prove: (1) Aproductofinjectivemodulesisinjective. (2) HomR(M, RR^) ⇠= HomZ(M,Q/Z)(isomorphismofleftR-modules) (3) Q/Z is an injective Z-module. (4) M M ^^. ✓ Assuming (3) T = Q/Z is injective, the other three lemmas are easy: Proof of (4) Lemma 4.5. AnaturalembeddingM M ^^ is given by the evaluation ! map ev which sends x M to ev : M ^ T which is evaluation at x: 2 x ! evx(')='(x). This is easily seen to be additive: evx+y(')='(x + y)='(x)+'(y)=evx(')+evy(')=(evx + evy)('). The right action of R on M gives a left action of R on M ^ which gives a right action of R on HomZ(M ^,T)sothat(evxr)(')=evx(r'). This implies that evaluation x evx is a right R-module homomorphism: 7! evxr(')='(xr)=(r')(x)=evx(r')=(evxr)('). Finally, we need to show that ev is a monomorphism. In other words, for every nonzero element x M we need to find some additive map ' : M T so that evx(')='(x) =0. To do this2 take the cyclic group C generated by x ! 6 C = kx : k Z { 2 } This is either Z or Z/n.Inthesecondcaseletf : C T be given by ! k f(kx)= + Z Q/Z n 2 This is nonzero on x since 1/n is not an integer: f(x)= 1 + =0+ .IfC then n Z Z ⇠= Z let f : C T be given by 6 ! k f(kx)= + Z 2 1 Then, f(x)= 2 + Z is nonzero in Q/Z.SinceT is Z-injective, f extends to an additive map f : M T .So,ev is nonzero since ev (f)=f(x)=f(x) =0.So,ev : M M ^^ ! x x 6 ! is a monomorphism. ⇤ Proof of (1): products of injectives are injective. Suppose that J↵ are injective. Then we want to show that Q = J↵ is injective. Let p↵ : Q J↵ be the projection map. Suppose that A B and f : A Q is any morphism. Then! we want to extend f to B. ⇢ Q ! Since each J↵ is injective each composition f↵ = p↵ f : A J↵ extends to a morphism f : B J .I.e.,f A = f for all ↵.Bydefinitionoftheproductthere◦ ! ↵ ! ↵ ↵| ↵ exists a unique morphism f : B Q = J so that p f = f for each ↵.So, ! ↵ ↵ ◦ ↵ p f A = f A = f : A J ↵ ◦ | Q↵| ↵ ! ↵ The uniquely induced map A J is f.Therefore,f is an extension of f to B as ! ↵ required. ⇤ Q 10 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA Proof of (2): HomR(M, RR^) ⇠= HomZ(M,Q/Z). To prove this we will give a 1-1 correspondence and show that it (the correspondence) is additive. We also show that this map is an isomorphism of left R-modules. If f Hom (M, R^) then f : M R^,anR-linear map, which means that for 2 R R ! R each x M we get f(x) RR^ = HomZ(R, Q/Z), i.e., f(x):R Q/Z.Furthermore, for r 2R, x M,wehave2 f(xr)=f(x)r or f(xr)(s)=f(x)(rs).! In particular we can 2 2 evaluate this at 1 R.Thisgives'(f):M Q/Z by the formula 2 ! '(f)(x):=f(x)(1) This defines a mapping ' : HomR(M, RR^) Hom (M,Q/Z) ! Z We need to show that this is additive, i.e., '(f + g)='(f)+'(g) This is an easy calculation which follows from the way that f + g is defined, namely, addition of function is defined “pointwise” which means that (f + g)(x)=f(x)+g(x) by definition. So, x M, 8 2 '(f + g)(x)=(f + g)(x)(1) = [f(x)+g(x)](1) = f(x)(1) + g(x)(1) = '(f)(x)+'(g)(x)=['(f)+'(g)](x). We would also like to know that ' is left R-linear, i.e., we want to show '(rf)=r'(f). This follows from the fact that 1r = r = r1! '(rf)(x)=[rf(x)](1) = f(x)(1r) [r'(f)](x)='(f)(xr)=f(xr)(1) = f(x)(r1) 1 Finally we need to show that ' is a bijection. To do this we find the inverse '− = . For any additive g : M Q/Z let (g):M RR^ be given by (g)(x):R Q/Z: ! ! ! [ (g)(x)](r):=g(xr) Since this is additive in all three variables, is additive and (g)isadditive.Wealso need to check that (g)isahomomorphismsofrightR-modules, i.e., that (g)(xr)= (g)(x)r. This is an easy calculation: [ (g)(xr)](s)=g((xr)s)=g(xrs) [ (g)(x)r](s)=[ (g)(x)](rs)=g(x(rs)) The verification that is the inverse of ' is also straightforward: For all f 2 HomR(M, RR^)wehave ('(f))(x)(r)='(f)(rx)=f(rx)(1) = [rf(x)](1) = f(x)(1r)=f(x)(r) So, ('(f)) = f.Similarly,forallg Hom (M,Q/Z)wehave: 2 Z '( (g))(x)=[ (g)(x)](1) = g(x1) = g(x) So, '( (g)) = g. ⇤ I will do the remaining Lemma (3) (injectivity of Q/Z)nexttime. MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 11 4.4. Examples. First, a trivial example: 4.4.1. fields. Suppose that R = k is a field. Since all short exact sequences of vector spaces split, we have: Theorem 4.7. In both categories mod-k and Mod-k all objects are both projective and injective. 4.4.2. polynomial ring. Let R = Z[t], the integer polynomial ring in one generator. This is a commutative Noetherian ring. It has dimension 2 since a maximal tower of prime ideal is given by 0 (t) (t, 2) ⇢ ⇢ These ideals are prime since the quotient of R by these ideals are domains (i.e., have no zero divisors): R/0=Z[t],R/(t)=Z are domains and R/(t, 2) = Z/(2) = Z/2Z is a field, making (2,t)intoamaximalideal. Proposition 4.8. A Z[t]-module M is the same as an abelian group together with an endomorphism M M given by the action of t. A homomorphism of Z[t]-modules f : M N is an additive! homomorphism which commutes with the action of t. ! Proof. IwillusethefactthatthestructureofanR-module on an additive group M is the same as a homomorphism of rings ' : R End(M). When R = Z[t], this homomorphism is given by its value on t since '!(f(t)) = f('(t)). For example, if f(t)=2t2 + 3 then '(f(t)) = '(2t2 +3)=2'(t) '(t)+3id = f('(t)) ◦ M Therefore, ' is determined by '(t) End (M)whichisarbitrary. 2 Z ⇤ For R = Z[t], what do the injective R-modules look like? We know that Q = RR^ is injective. What does that look like? Q = HomZ(Z[t], Q/Z) But Z[t]isafreeabeliangrouponthegenerators1,t,t2,t3, .Therefore,anelement ··· f Q, f : Z[t] Q/Z is given uniquely by the sequence 2 ! f(1),f(t),f(t2),f(t3), Q/Z ···2 Multiplication by t shifts this sequence to the left since (ft)(ti) f(tti)=f(ti+1). This proves the following. − Theorem 4.9. The injective Z[t]-module Q = Z[t]^ is isomorphic to the additive group of all sequences (a0,a1,a2, ) of elements ai Q/Z with the action of t given by shifting to the left and dropping the··· first coordinate. I.e.,2 t(a ,a ,a , )=(a ,a , ). 0 1 2 ··· 1 2 ··· The word “isomorphism” is correct here because these are not the same set..