Bourbaki

Lemma 4

Tag 0DQF, permanent. Cite it as [Bourbaki, Tag 0DQF].

Algebra, chapter X, § 1. Compléments d’algèbre linéaire

Let M be a nonzero noetherian A-module, I an injective envelope of M. Then I possesses an indecomposable injective submodule.

One may obviously suppose that M is a submodule of I. Let N be a submodule of M, such that I is not an injective envelope of N and maximal for this property. According to remark 2 (X, p. 21), there exists a submodule I₁ of I which is an injective envelope of N; then I₁ is a direct factor in I, let J be a direct factor complement. One has J ≠ 0, let us prove that J is indecomposable. If J′ is a nonzero direct factor submodule of J, one has J′ ∩ M ≠ 0 and

(J′∩M)∩N⊂J′∩I1=0.(J' \cap M) \cap N \subset J' \cap I_1 = 0.

The submodule N′ = (J′ ∩ M) + N of M is direct sum of J′ ∩ M and N, hence strictly contains N. Moreover N′ is contained in the submodule J′ + I₁ of J, which is direct sum of J′ and I₁, hence injective. According to the maximal character of N, this implies J′ + I₁ = I, hence J′ = J, and J is indecomposable.

Let I\mathscr{I} denote the set (X, p. 21, cor. 1) of classes of indecomposable injective A-modules.

Recall (X, p. 17, cor. 1) that, if A is left noetherian, every A-module direct sum of injective A-modules is injective.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Nothing.

Translated with machine help, not yet reviewed by a person.