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 ) of exerc. 11 a) of II, p. 184, by reasoning as for proving that implies 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, an essential homomorphism. Now prove that there exists an injective homomorphism , such that 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 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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