Bourbaki

Theorem 3

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

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

Let I be an injective A-module.

a) If I is the injective envelope of a noetherian A-module M, I is direct sum of a finite family of indecomposable (injective) submodules.

b) If A is left noetherian, I is direct sum of a family of indecomposable (injective) submodules.

c) If I is direct sum of indecomposable (injective) submodules, there exists one and only one family of cardinals (aE)E∈I(a_E)_{E \in \mathscr{I}}, such that I is isomorphic to

⨁E∈IE(aE).\bigoplus_{E \in \mathscr{I}} E^{(a_E)}.

Let us first note that c) follows from prop. 14 (X, p. 21) and from VIII, § 1, No. 7, theorem 2. Let us prove a).

Let N be a submodule of M whose injective envelopes are direct sum of a finite family of indecomposable submodules, and maximal for this property (there exists one since M is noetherian). According to remark 2 (X, p. 21), there exists a submodule I₁ of I which is an injective envelope of N. If I₁ = I, the proof is complete; otherwise, let JJ be a direct factor complement of I1I_1 in II. Then JJ is the injective envelope of the noetherian module J∩MJ\cap M (loc. cit.), hence possesses a nonzero indecomposable injective submodule J′J' (lemma 4). Then I1+J′I_1+J' is injective, direct sum of a finite family of indecomposable submodules, and the injective envelope of the submodule (I1+J′)∩M(I_1+J')\cap M of MM which strictly majorizes NN, whence a contradiction.

Suppose AA is left noetherian and let us prove b). Let XX be the set sum of the sets Hom⁡A(E,I)\operatorname{Hom}_A(E,I) for E∈JE\in\mathscr{J}. To each subset YY of XX associate an AA-module EYE_Y and an AA-homomorphism fY:EY→If_Y:E_Y\to I in the following way: YY is the sum of a family (Y(E))E∈J(Y(E))_{E\in\mathscr{J}} where Y(E)⊂Hom⁡A(E,I)Y(E)\subset\operatorname{Hom}_A(E,I), one sets

EY=⨁E∈JE(Y(E)) E_Y=\bigoplus_{E\in\mathscr{J}}E^{(Y(E))}

and the component of ff on the direct factor of EYE_Y corresponding to the element yy of Y(E)⊂Hom⁡A(E,I)Y(E)\subset\operatorname{Hom}_A(E,I) is y:E→Iy:E\to I. Let YY be a subset of XX, such that fYf_Y is injective and that YY is maximal for this property (such a subset exists according to E, III, p. 20); it suffices to prove that fYf_Y is bijective. Otherwise, let JJ be a direct factor complement of the injective submodule Im⁡(fY)\operatorname{Im}(f_Y) of II; since JJ is nonzero, it possesses a nonzero noetherian submodule (because A is supposed noetherian), hence also a nonzero injective submodule J′J' which is the injective envelope of a noetherian module. According to a), J′J' is direct sum of a finite nonempty family of indecomposable submodules. There therefore exists a finite nonempty subset Y′Y' of XX such that fY′f_{Y'} maps bijectively EY′E_{Y'} onto J′J'. Since Im⁡(fY)∩J′=0\operatorname{Im}(f_Y)\cap J'=0, one has Y∩Y′=∅Y\cap Y'=\varnothing and fY∪Y′f_{Y\cup Y'} is injective; this contradicts the maximal character of YY and completes the proof.

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

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