Bourbaki

Exercise 18, § 2

Algebra, II, p. 186

Bourbaki marks this one as harder than the rest.

(a) For an A-module I to be injective, now prove that it is necessary and sufficient that I admit no essential extension distinct from itself (now prove that this condition implies condition δ\delta) of exerc. 11 a) of II, p. 184, by reasoning as for proving that θ\theta implies δ\delta in this exercise). (b) Let I be an injective A-module. For a submodule E of I to be injective, it is necessary and sufficient that E admit no essential extension distinct from itself and contained in I (using exerc. 15 e), now prove that this condition implies that E is a direct factor of I). (c) Let I be an injective A-module, E a submodule of I, h:E→Fh : E \to F an essential homomorphism. Now prove that there exists an injective homomorphism j:F→Ij : F \to I, such that j∘hj \circ h is the canonical injection of E into I.

(d) Let I be an A-module, E a submodule of I. Now prove that the following properties are equivalent: α) I is injective and is essential extension of E. β) I is injective and for every injective homomorphism j of E into an injective A-module J, there exists an injective homomorphism of I into J extending j. γ) I is injective, and is the smallest injective submodule of I containing E. δ) I is essential extension of E and for every essential homomorphism h : E → F, there exists an injective homomorphism j : F → I such that j∘hj \circ h is the canonical injection of E into I. (Use c) and a)). When these equivalent conditions are satisfied, I is called an injective envelope of E; I is then also an injective envelope of every submodule of I containing E. (e) Let I be an injective A-module, E a submodule of I. Now prove that every maximal element of the set of essential extensions of E contained in I is an injective envelope of E (use b)). In particular, every A-module admits an injective envelope (cf. II, p. 185, exerc. 14). (f) If I, I' are two injective envelopes of the same A-module E, now prove that there exists an isomorphism of I onto I' leaving invariant the elements of E.

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.