Bourbaki

Exercise 11, § 2

Algebra, II, 389

Bourbaki marks this one as harder than the rest.

Let A be a ring and I an A-module. I is called injective if, for every exact sequence E′→E→E′′E' \to E \to E'' of A-linear mappings, the sequence

Hom⁡(E′′,I)→Hom⁡(E,I)→Hom⁡(E′,I) \operatorname{Hom}(E'', I) \to \operatorname{Hom}(E, I) \to \operatorname{Hom}(E', I)
is exact.

(a) Show that the following properties are equivalent: (α) I is injective. (β) For every exact sequence 0→E′→E→E′′→00 \to E' \to E \to E'' \to 0 of A-linear mappings, the sequence

0→Hom⁡(E′′,I)→Hom⁡(E,I)→Hom⁡(E′,I)→0 0 \to \operatorname{Hom}(E'', I) \to \operatorname{Hom}(E, I) \to \operatorname{Hom}(E', I) \to 0
is exact.

(γ) For every A-module M and every submodule N of M, every A-linear mapping of N into I can be extended to an A-linear mapping of M into I. (δ) For every left ideal aa of A and every A-linear mapping f:a→If : a \to I, there exists an element b∈Ib \in I such that f(a)=abf(a) = ab for all a∈aa \in a. (ζ) I is a direct factor of every A-module containing it. (θ) For every A-module E which is the sum of I and a monogenous submodule, I is a direct factor of E.

(To prove that (δ) implies (γ), show that property (δ) implies that if E is an A-module and F a submodule of E such that E/FE/F is monogenous, then every linear mapping of F into I can be extended to a linear mapping of E into I. Then use Zorn’s Lemma. To prove that (θ) implies (δ), consider the A-module As×I=MA_s \times I = M, the submodule N of M consisting of the elements (a,−f(a))(a, -f(a)) for a∈aa \in a, and apply (θ) to the quotient module M/NM/N.)

(b) For a Z\mathbf{Z}-module E to be injective, it is necessary and sufficient that for all x∈Ex \in E and every integer n≠0n \neq 0, there exists y∈Ey \in E such that ny=xny = x. In particular, the Z\mathbf{Z}-modules Q\mathbf{Q} and Q/Z\mathbf{Q}/\mathbf{Z} are injective.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.