46 Injective modules

Recall. If R is a ring with identity then an R-module P is projective iff one of the following equivalent conditions holds:

1) For any homomorphism f : P → N and an epimorphism g : M → N there is a homomorphism h: P → M such that the following diagram commutes:

P } } h } } f } } } ~} g  M / N

f g 2) Every short exact sequence 0 → N −→ M −→ P → 0 splits.

46.1 Proposition. Let R be a ring and let J be an R-module. The following conditions are equivalent.

1) For any homomorphism f : N → J and an monomorphism g : M → N there is a homomorphism h: M → J such that the following diagram commutes: J O `A A A A h f A A A A M g / N

f g 2) Every short exact sequence 0 → J −→ M −→ N → 0 splits.

Proof. Exercise.

185 46.2 Definition. An R-module J is an injective module if J satisfies one of the equivalent conditions of Proposition 46.1.

46.3 Theorem (Baer’s Criterion). Let R be a ring with identity and let J be an R-module. The following conditions are equivalent.

1) J is an injective module.

2) For every left ideal I C R and for every homomorphisms of R-modules ¯ ¯ f : I → J there is a homomorphism f : R → J such f|I = f.

Proof. 1) ⇒ 2) Given a homomorphism f : I → J we have a diagram

J O

f

 i I / R where i: I,→ R is the inclusion homomorphism. By the definition of an injective ¯ ¯ ¯ module there is a homomorphism f : R → J such that f = fi = f|I

2) ⇐ 1) Assume that J is an R-module satisfying 2). It is enough to show that if M is an R-module, N is a submodule of M, and f : N → J is an R- module homomorphism then there exists a homomorphism f¯: M → J such that ¯ f|N = f

Let S be a set of all pairs (K, fK ) such that (i) K is a submodule of M such that N ⊆ K ⊆ M

(ii) fK : K → J is a homomorphism such that fK |N = f

186 Define partial ordering on S as follows:

0 0 (K, fK ) ≤ (K , fK0 ) if K ⊆ K and fK0 |K = fK Check: assumptions of Zorn’s Lemma 29.10 are satisfied in S, and so S contains a maximal element (K0, fK0 ).

It will suffice to show that K0 = M. Assume, by contradiction, that K0 6= M, and let m0 ∈ M − K0. Define

I := {r ∈ R | rm0 ∈ K0} Check: I is an ideal of R and the map

g : I → J, g(r) = fK0 (rm0) is a homomorphism of R-modules. By the assumptions on J we have a homo- morphism g¯: R → J such that g¯|I = g. Define

K0 + Rm0 := {k + rm0 | k ∈ K, r ∈ R}

Check: K0 + Rm0 is a submodule of M and the map

0 0 f : K0 + Rm0 → J, f (k + rm0) = fK0 (k) +g ¯(r)

0 is a well defined homomorphism of R-modules such that f |N = f. This shows 0 that (K0 + Rm0, f ) ∈ S. We also have

0 (K0, fK0 ) < (K0 + Rm0, f )

This is impossible since by assumption (K0, fK0 ) is a maximal element in S.

46.4 Corollary. Let R be an integral domain and let K the field of fractions of R. Then K is an injective R-module.

Proof. Let I be an ideal of R and let f : I → K be a homomorphism of R- modules. For 0 6= r, s ∈ I we have

rf(s) = f(rs) = sf(r)

187 As consequence in K we have f(r)/r = f(s)/s for any 0 6= r, s ∈ I. Denote this element by a. Define

f¯: R → K, f¯(r) := ra ¯ ¯ Check: f is a homomorphism of R-modules and f|I = f.

By Baer’s Criterion (46.3) it follows that K is an injective R-module.

46.5 Example. Q is an injective Z-module.

46.6 Definition. Let R be an integral domain. An R-module M is divisible if for every r ∈ R − {0} and for every m ∈ M there is n ∈ M such that rn = m.

46.7 Theorem. If R is a PID then an R-module J is injective iff J is divisible.

Proof. Exercise.

46.8 Example.

Since Z is a PID injective Z-modules are divisible Z-modules (i.e. divisible abelian groups).

Exercise: an abelian group G is divisible iff G is isomorphic to a direct sum of copies of Q and Z(p∞) for various primes p.

46.9 Corollary. If R is a PID, J is an injective R-module and K is a submodule of J then J/K is injective.

188 Proof. Since J is divisible thus so is J/K (check!).

46.10 Note. If R is not a PID then a quotient of an injective R-module need not be injective.

Note. If R is a ring with identity then for any R-module M there exists an epimorphism of R-modules: f : P −→ M L where P is a projective module (take e.g. P = m∈M R).

46.11 Theorem. If R is a ring with identity then for any R-module M there exist a monomorphism j : M −→ J where J is an injective R-module.

46.12 Lemma. For any abelian group G there exists a monomorphism i: G −→ D where H is a divisible abelian group.

L Proof. We have an epimorphism f : g∈G Z → G which gives an isomorphism

=∼ M ϕ: G −→ Z/ Ker(f) g∈G L L Moreover, the monomorphism g∈G Z → g∈G Q induces a monomorphism M M ψ : Z/ Ker(f) −→ Q/ Ker(f) g∈G g∈G L We can take D := g∈G Q/ Ker(f) and i := ψϕ.

189 46.13 Note. Let G is an abelian group, let R let be a ring, and let HomZ(R,G) be the set of all homomorphisms of abelian groups ϕ: R → G.

Check: HomZ(R,G) is an R-module with pointwise addition and with multipli- cation by elements of R given by

(r · ϕ)(s) := ϕ(sr) for r, s ∈ R.

46.14 Lemma. If D is a divisible abelian group and R is a ring with identity then HomZ(R,D) in an injective R-module.

Proof. Exercise.

Proof of Theorem 46.11. Let M be an R-module. Consider M as an abelian group. By Lemma 46.12 we have a monomorphism of abelian groups

i: M −→ D where D is a divisible abelian group. Consider the induced map

i∗ : HomZ(R,M) → HomZ(R,D), i∗(ϕ) = i ◦ ϕ Check:

1) i∗ is a monomorphism.

2) i∗ is a homomorphism of R-modules.

Since M is an R-module we also have a map

f : M → HomZ(R,M), f(m)(r) = rm Check:

1) f is a monomorphism.

190 2) f is a homomorphism of R-modules.

As a consequence we obtain a monomorphism of R-modules

i∗f : M −→ HomZ(R,D)

Moreover, by Lemma 46.12 HomZ(R,D) is an injective R-module.

191 47 Exact functors

47.1 Definition. A chain complex of R-modules is a sequence of R-modules and R-homomorphisms

di+1 di di−1 ... −→ Mi+1 −→ Mi −→ Mi−1 −→ Mi−2 −→ ... such that didi+1 = 0 for all i.

47.2 Note. If M∗ = (Mi, di) is a chain complex then Im(di+1) ⊆ Ker(di).

47.3 Definition. If M∗ = (Mi, di) is a chain complex of R-modules then the i-th homology module of M∗ is the module

Hi(M∗) = Ker(di)/ Im(di+1)

Recall. A sequence

di+1 di di−1 M∗ = (... −→ Mi+1 −→ Mi −→ Mi−1 −→ Mi−2 −→ ...) is exact if Ker(di) = Im(di+1) for all i. Therefore M∗ is exact iff Hi(M∗) = 0 for all i.

47.4 Definition. Let R, S be rings. A functor F : R-Mod → S-Mod is exact if for every short exact sequence of R-modules

f g 0 −→ N −→ M −→ K −→ 0 the sequence F (f) F (g) 0 −→ F (N) −→ F (M) −→ F (K) −→ 0 is short exact.

192 47.5 Note. If F : R-Mod → S-Mod is a functor such that F (0) = 0 and

di+1 di M∗ = (... −→ Mi+1 −→ Mi −→ Mi−1 −→ ...) is a chain complex of R-modules then

F (di+1) F (di) F (M∗) = (... −→ F (Mi+1) −→ F (Mi) −→ F (Mi−1) −→ ...) is a chain complex of S-modules.

Moreover, the functor F is exact iff for every chain complex M∗ we have isomor- phisms ∼ F (Hi(M∗)) = Hi(F (M∗)) for all i.

47.6 Note. For a ring R and R-modules L, M let HomR(L, M) be the set of all R-module homomorphisms ϕ: L → M. Notice that HomR(L, M) is an abelian group (with respect to the pointwise addition of homomorphisms). Moreover, for any homomorphism of R-modules f : M → N the map

f∗ : HomR(L, M) → HomR(L, N), f∗(ϕ) = f ◦ ϕ is a homomorphism of abelian groups. This defines a functor

HomR(L, −): R-Mod −→ Ab

This functor is in general not exact. Take e.g. R = Z, L = Z/2Z. We have a short exact sequence of abelian groups:

·2 0 → Z −→ Z −→ Z/2Z → 0 On the other hand the sequence

0 → HomZ(Z/2Z, Z) −→ HomZ(Z/2Z, Z) −→ HomZ(Z/2Z, Z/2Z) → 0 ∼ ∼ is not exact since HomZ(Z/2Z, Z) = 0 and HomZ(Z/2Z, Z/2Z) = Z/2Z.

193 47.7 Proposition. Let R be a ring and let L be an R-module. If

f g 0 −→ N −→ M −→ K −→ 0 is a short exact sequence of R-modules then

f∗ g∗ 0 −→ HomR(L, N) −→ HomR(L, M) −→ HomR(L, K) is an exact sequence of abelian groups.

Proof. Exercise.

47.8 Theorem. Let R be a ring with identity and let P be an R-module. The functor HomR(P, −) is exact iff P is a projective module.

Proof. By Proposition 47.7 is suffices to show that P is a projective module iff for every epimorphism of R-modules g : M → K the map

g∗ : HomR(P,M) −→ HomR(P,K) is an epimorphism. This follows directly from Theorem 43.9.

47.9 Definition. Let C, D be categories. A contravariant functor F : C → D consists of

1) an assignment Ob(C) → Ob(D), c 7→ F (c)

2) for every c, c0 ∈ C a function

0 0 HomC(c, c ) → HomD(F (c ),F (c)), f 7→ F (f)

such that F (idc) = idF (c) and F (gf) = F (f)F (g).

194 47.10 Example. Let R be a ring and let L be an R-module. For any homo- morphism of R-modules f : M → N we have a map

∗ ∗ f : HomR(N,L) −→ HomR(M,L), f (ϕ) = ϕ ◦ f

Moreover, f ∗ is a homomorphism of abelian groups.

This defines a contravariant functor

HomR(−,L): R-Mod −→ Ab

47.11 Note. The functor HomR(−,L) is in general not exact. Take e.g. R = Z, L = Z. We have a short exact sequence of abelian groups:

·2 0 → Z −→ Z −→ Z/2Z → 0 On the other hand the sequence

0 → HomZ(Z/2Z, Z) −→ HomZ(Z, Z) −→ HomZ(Z, Z) → 0 is not exact.

47.12 Proposition. Let R be a ring and let L be an R-module. If

f g 0 −→ N −→ M −→ K −→ 0 is a short exact sequence of R-modules then

g∗ f ∗ 0 −→ HomR(K,L) −→ Hom(M,L) −→ Hom(M,L) is an exact sequence of abelian groups.

Proof. Exercise.

195 47.13 Theorem. Let R be a ring with identity and let J be an R-module. The functor HomR(−,J) is exact iff J is an injective module.

Proof. By Proposition 47.12 is suffices to show that J is an injective module iff for every monomorphism of R-modules f : N → M the map

∗ f : HomR(M,J) −→ HomR(N,J) is an epimorphism. This follows directly from Proposition 46.1.

47.14 Note. Let

di M∗ = (... −→ Mi −→ Mi−1 −→ ...) be a chain complex of R-modules. For any R-module L we have the induced chain complex of abelian groups

∗ di HomR(M∗,L) = (... −→ HomR(Mi−1,L) −→ HomR(Mi,L) −→ ...)

Homology groups of the complex HomR(M∗,L) are called cohomology groups of M∗ with coefficients in L. We denote:

i H (M∗,L) := Hi(HomR(M∗,L))

If L is an injective module then by Theorem 47.13 the functor HomR(−,L) is exact. By (47.5) in such case we have

i ∼ H (M∗,L) = HomR(Hi(M∗),L)

196