Commutative algebra - Department of Mathematical Sciences - old ...

Proof. Let M → N be injective. Then by 2.5.8
is the product
3.6. INJECTIVE MODULES 53
HomR(N, Eα) → HomR(M, Eα)
HomR(N, Eα) → HomR(M, Eα)
which is surjective by 3.1.6. So Eα is injective.
3.6.7. Proposition. A module E is injective, if for any ideal I ⊂ R
is exact.
HomR(R, E) → HomR(I, E) → 0
Proof. Let f : E → L be an injective homomorphism. The set of submodules
f(E) ⊂ L ′ ⊂ L and retractions u ′ : L ′ → E is nonempty and inductively ordered.
By Zorn’s lemma a maximal L ′ , u ′ exists. If L/L ′ = 0 choose a y ∈ L\L ′ . The
homomorphism Ann(y + L ′ ) → E, a ↦→ u ′ (ay) extends to u ′′ : R → E by
hypothesis. The setting L ′ + Ry → E, x + ay ↦→ u ′ (x) + u ′′ (a) is a well defined
retraction. This contradicts maximality.
3.6.8. Definition. Let R be a domain. M is a divisible module if scalar multiplication
with a nonzero a ∈ R is surjective.
3.6.9. Proposition. (1) An injective module over a domain is divisible.
(2) A divisible module over a principal ideal domain is injective.
(3) Over a field any module is injective.
Proof. (1) Let E be an injective module and a = 0. Choose x ∈ E and look
at 1x : R → E. Scalar multiplication aR is injective, so there is an extension
1y : R → E such that 1x = 1y ◦ aR. Then ay = x. (2) Let E be divisible. To
extend f : (a) → E, choose x ∈ E such that ax = f(a), then 1x : R → E, 1 → x
extends f. (3) This is clear.
3.6.10. Proposition. Let R → S be a ring homomorphism and E an injective
R-module. The induced module HomR(S, E) is an injective S-module.
Proof. Let M → N be an injective homomorphism of S-modules. Then by
2.7.10 HomS(N, HomR(S, E)) → HomS(M, HomR(S, E)) is HomR(N, E) →
HomR(M, E) being surjective since E is an injective R-module.
3.6.11. Lemma. (1) Q and Q/Z are divisible and therefore injective Z modules.
(2) The homomorphism 2.5.11
is injective.
x ↦→ evx : M → HomZ(HomZ(M, Q/Z), Q/Z)
Proof. (1) Clear by 3.6.9. (2) Let x = 0 and choose h : Z/ Ann(x) → Q/Z
nonzero. Now Z/ Ann(x) Zx ⊂ M so extend h to h ′ : M → Q/Z. Then
evx(h ′ ) = h(1) = 0.
3.6.12. Proposition. Any module M admits an exact sequence
0 → M → E
where E is an injective module. That is, any module is a submodule of an injective
module.