Bourbaki

Exercise 13, § 3

Algebra, X, p. 182

Let M be an A-module. A minimal injective resolution of M is an injective resolution (e, l) of M such that l' is an injective envelope of Zn(l)Z^n(l) for all n⩾0n \geqslant 0. (a) Every A-module admits minimal injective resolutions; two such resolutions are isomorphic. (b) Let l, l' be two injective resolutions of M, l being minimal; let f:l→l′f : l \to l' and g:l′→lg : l' \to l be two morphisms of resolutions. Then f is injective, g is surjective and l' is the direct sum of subcomplexes lm (f) (isomorphic to l) and Ker (g) (with zero homology).

Solution

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Machine translation of the English, by gpt-5-6-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.