12 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 5. Divisible groups Z-modules are the same as additive groups (abelian groups with group operation +). And we will see that injective Z-modules are the same as divisible groups. Definition 5.1. An additive group D is called divisible if for any x D and any positive integer n there exists y D so that ny = x.(Wesaythatx is divisible2 by n.) 2 Equivalently, multiplication by any n>0givesasurjectivemapD ⇣ D. Example 5.2. Q is divisible. Another interesting example is: p Zp Qp Zp the product of all cyclic groups of primeL order Zp modulo their direct sum. This is divisible since, for any n, we have p Zp p>n Zp ⇠= Qp Zp Qp>n Zp and multiplication by n is an isomorphismL ZLp ⇠= Zp for all primes p>n. Proposition 5.3. Any quotient of a divisible group is divisible. Proof. Suppose D is divisible and K is a subgroup. Then any element of the quotient D/K has the form x + K where x D.Thisisdivisiblebyanypositiven since, if ny = x then 2 n(y + K)=ny + K = x + K Therefore D/K is divisible. ⇤ Example 5.4. Q/Z is divisible. Theorem 5.5. Tfae (the following are equivalent) for any additive group D: (1) D is divisible. (2) If A is a subgroup of a cyclic group B then any homomorphism A D extends to B. ! (3) D is an injective Z-module. Proof. (2) (1): Take B = Z, A = nZ.Foranyx D,letf : A D be given by ) 2 ! f(an)=ax.So,f(n)=x.Extendthistof : Z D be an extension of f to Z.Then ! ny = f(n)=f(n)=x.So,D is divisible. (1) (2): We may assume A =0.IfA is a nontrivial subgroup of a cyclic group ) 6 B then the generator a0 of A is a multiple of the generator b0 of B: a0 = nb0.For any f : A D,letx = f(a ). Since D is divisible, there is y D so that ny = x. ! 0 2 Then there is a unique homomorphism f : B D sending b to y and f A = f since ! 0 | f(a0)=f(nb0)=ny = x. (3) (2) follows from the definition of injective module. (2) ) (3). Given that D satisfies (2), we will use Zorn’s Lemma to prove that D is injective.) Suppose that X is a submodule of Y and f : X D is a homomorphism. Then we want to extend f to all of Y .TouseZorn’sLemmawetakethesetofallpairs! MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 13 (C, g)whereX C Y and g is an extension of f (i.e., f = g X). This set is partially ✓ ✓ | ordered in an obvious way: (C, g) < (C0,g0)ifC C0 and g = g0 C.Italsosatisfiesthe ✓ | hypothesis of Zorn’s Lemma. Namely, any totally ordered subset (C↵,g↵)hasanupper bound: ( C↵, g↵). Zorn’s Lemma tells us that this set has a maximal element, say, (M,g). We[ just[ need to show that M = Y .Weshowthisbycontradiction. If M = Y then there is at least one element y Y which is not in M.LetB be the cyclic subgroup6 of Y generated by y.LetA = B2 M.Then,by(2),therestriction \ h = g A : A D extends to a homomorphism h : B D.Then(byLemmabelow) | ! ! g h : M B D induces a homomorphism ⊕ ⊕ ! g + h : M + B D ! which is equal to g on M and h on B.Then(M + B,g + h) > (M,g)contradictingthe maximality of (M,g). Therefore, M = Y and g : Y D is an extension of f : X D. ! ! So, D is an injective Z-module. ⇤ Lemma 5.6. Suppose that A, B are submodules of an R-module C and f : A X, g : B X are homomorphisms of R-modules which agree on A B. Then we! get a well-defined! homomorphism f + g : A + B X by the formula \ ! (f + g)(a + b)=f(a)+g(b). Proof. Well-defined means that, if the input is written in two di↵erent ways, the output is still the same. So suppose that a + b = a0 + b0.Thena a0 = b0 b A B.So, − − 2 \ f(a a0)=f(a) f(a0)=g(b0 b)=g(b0) g(b) − − − − by assumption. Rearranging the terms, we get f(a)+g(b)=f(a0)+g(b0)asdesired. ⇤ This lemma also follows from left-exactness of Hom: We have a short exact sequence of R-modules: jA ( j ) p +p 0 A B − B A B A B A + B 0. ! \ −−−! ⊕ −− − ! ! Since Hom (A B,X) = Hom (A, X) Hom (B,X)thisgivesaleftexactsequence: R ⊕ ⇠ R ⊕ R 0 Hom (A + B,X) Hom (A, X) Hom (B,X) Hom (A B,X). ! R ! R ⊕ R ! R \ Exactness of this sequence is equivalent to the Lemma. 14 MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 6. Injective envelope There is one very important fact about injective modules which is not covered in Lang’s book. This is the fact that every R-module M embeds in a minimal injective module which is called the injective envelope of M. This is from Jacobson’s Basic Algebra II. Definition 6.1. An embedding A, B is called essential if every nonzero submodule of B meets A.I.e.,C B,C =0 !A C =0. ✓ 6 ) \ 6 For example, Z , Q is essential because, if a subgroup of Q contains a/b,thenit ! contains a Z. Also, every isomorphism is essential. 2 Exercise 6.2. Show that the composition of essential maps is essential. Lemma 6.3. Suppose A B. Then ✓ (1) X B s.t. A X =0and A, B/X is essential. (2) 9C ✓ B maximal\ so that A C!is essential. 9 ✓ ✓ Proof. For (1) the set of all X B s.t. A X =0hasamaximalelementbyZorn’s Lemma. Then A, B/X must✓ be essential,\ otherwise there would be a disjoint sub- module of the form!Y/X and X Y,A Y =0contradictingthemaximalityofY .For ⇢ \ (2), C exists by Zorn’s Lemma. ⇤ Lemma 6.4. Q is injective i↵every short exact sequence 0 Q M N 0 ! ! ! ! splits. Proof. If Q is injective then the identity map Q Q extends to a retraction r : M Q giving a splitting of the sequence. Conversely,! suppose that every sequence as above! splits. Then for any monomorphism i : A, B and any morphism f : A Q we can form the pushout M in the following diagram! ! i A / B f f 0 ✏ j ✏ Q / M These morphisms form an exact sequence: f ( i ) (j, f ) A Q B − 0 M 0 −−! ⊕ −− − ! ! This implies that j is a monomorphism. [If j(x)=0then(j, f 0)(x, 0) = 0. So, there is an a A so that (x, 0) = (f(a),i(a)). But i(a)=0impliesa−=0.So,x = f(a)=0.] Since2j is a monomorphism there is a short exact sequence j 0 Q M coker j 0 ! ! ! ! We are assuming that all such sequences split. So, there is a retraction r : M Q ! (r j = id ). It follows that r f 0 : B Q is the desired extension of f : A Q: ◦ Q ◦ ! ! r f 0 i = r j f = id f = f ◦ ◦ ◦ ◦ Q ◦ MATH 131B: ALGEBRA II PART A: HOMOLOGICAL ALGEBRA 15 So, Q is injective. ⇤ Lemma 6.5. Q is injective if and only if every essential embedding Q, C is an isomorphism. ! Proof. ( )SupposeQ is injective and Q, C is essential. Then the identity map Q Q )extends to a retraction r : C Q whose! kernel is disjoint from Q and therefore ! ! must be zero making C ⇠= Q. ( ) Now suppose that every essential embedding of Q is an isomorphism. We want to show( that Q is injective. By the previous lemma it suffices to show that every short exact sequence j 0 Q M N 0 ! ! ! ! splits. By Lemma 6.3 there is a submodule X M so that jQ X =0andQ, M/X is essential. Then, by assumption, this map must✓ be an isomorphism.\ So, M =!Q X ⇠ ⊕ and the sequence splits proving that Q is injective. ⇤ Theorem 6.6. For any R-module M there exists an essential embedding M, Q with Q injective. Furthermore, Q is unique up to isomorphism under M. ! Proof. We know that there is an embedding M, Q where Q is injective. By Lemma ! 0 0 6.3 we can find Q maximal with M, Q, Q0 so that M, Q is essential. Claim: Q is injective. ! ! ! If not, there exists an essential Q, N.SinceQ0 is injective, there exists f : N Q0 extending the embedding Q, Q .! Since f is an embedding on Q,kerf Q =0.This! ! 0 \ forces ker f =0sinceQ, N is essential. So, f : N Q0 is a monomorphism. This contradicts the maximality! of Q since the image of N !is an essential extension of M in Q0 which is larger than Q. It remains to show that Q is unique up to isomorphism. So, suppose M, Q0 is ! another essential embedding of M into an injective Q0.ThentheinclusionM, Q0 ! extends to a map g : Q Q0 which must be a monomorphism since its kernel is disjoint ! from M. Also, g must be onto since g(Q)isinjectivemakingtheinclusiong(Q) , Q0 ! split which contradicting the assumption that M, Q0 is essential unless g(Q)=Q0. ! ⇤.