Theorem 2

Let M be an A-module. a) M possesses injective envelopes. b) If (I, i) and (J, j) are two injective envelopes of M, there exists an isomorphism f : I → J such that f ∘ i = j.

It should be noted that the homomorphism f whose existence is asserted in b) is not uniquely determined in general.

a) We may suppose that M is a submodule of an injective A-module E (X, p. 19, cor. 3). Consider the set F, ordered by inclusion, of the submodules I of E, containing M, and such that the canonical injection i : M → I satisfies property (E). Since F is inductive, it possesses a maximal element (E, III, p. 20); let I be a maximal element of F. It suffices to prove that I is a direct factor submodule of E. Let N be a submodule of E such that N ∩ I = 0, and maximal for this property (such an N exists by loc. cit.). The composite homomorphism

where u and v are the canonical homomorphisms, is injective; since E is injective, there therefore exists a homomorphism w : E/N → E such that w ∘ v ∘ u = u, that is to say w ∘ v(x) = x for x ∈ I. Put I′ = Im(w) = Im(w ∘ v) and let i′ : M → I′ be the canonical injection. Then I ⊂ I′; to complete the proof, it suffices to prove that i′ satisfies condition (E): this implies that I = I′ (maximal character of I) and hence that w ∘ v is a projector of E onto I.

Let therefore P be a submodule of I′ such that P ∩ M = 0. We have P = w ∘ v(Q), where Q is a submodule of E containing N; moreover

hence Q ∩ I = 0 since i : M → I possesses property (E). By the maximal character of N, this implies Q = N, that is to say v(Q) = 0, hence P = 0, which was to be proved.

b) Let (I, i) and (J, j) be two injective envelopes of M. Since J is injective, there exists a homomorphism f : I → J such that f ∘ i = j; we have

hence Ker f = 0 and f is injective. Then f(I) is an injective submodule of J, hence a direct factor; since j satisfies (E), this implies f(I) = J and f is bijective.