Exercise 18, § 2
Algebra, II, 391
Bourbaki marks this one as harder than the rest.
(a) For an A-module I to be injective, show that it is necessary and sufficient that I admit no essential extension distinct from itself (show that this condition implies condition (δ) of Exercise 11(a), arguing as for the proof that (θ) implies (δ) in that 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 Exercise 15(e)), show 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 and an essential homomorphism. Show that there exists an injective homomorphism such that is the canonical injection of E into I.
(d) Let I be an A-module and E a submodule of I. Show that the following properties are equivalent: (α) I is injective and is an essential extension of E. (β) I is injective and for every injective homomorphism of E into an injective A-module J, there exists an injective homomorphism of I into J extending . (γ) I is injective and is the smallest injective submodule of I containing E. (δ) I is an essential extension of E and for every essential homomorphism , there exists an injective homomorphism such that is the canonical injection of E into I. (Use (c) and (a).) When these equivalent conditions hold, 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 and E a submodule of I. Show 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. Exercise 14).
(f) If I, I' are two injective envelopes of the same A-module E, show that there exists an isomorphism of I onto I' leaving invariant the elements of E.
Solution
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