Proposition 14

Let I be a nonzero injective A-module. The following conditions are equivalent: (i) I is indecomposable (VII, § 4, no 8, def. 3); (ii) 0 is not the intersection of two nonzero submodules of I; (iii) I is the injective envelope of all its nonzero submodules; (iv) the ring $\mathrm{End}_A(I)$ is local (VIII, § 1, no 4, def. 4).

(i) $\Rightarrow$ (iii) : let M be a nonzero submodule of I. By Remark 2 above, there exists a submodule I’ of I, which is an injective envelope of M. Since I’ is nonzero and a direct factor in I, we have $I = I'$ if I is indecomposable.

(iii) $\Rightarrow$ (ii) : let E and F be two submodules of I, such that $E \cap F = 0$. If $E \neq 0$, then I is an injective envelope of E by (iii), hence « $E \cap F = 0$ » implies « $F = 0$ ».

(ii) $\Rightarrow$ (i) : this is trivial.

(iv) $\Rightarrow$ (i) : this results from VIII, § 1, no 6, prop. 13.

(i) $\Rightarrow$ (iv) : suppose I is indecomposable. First note that every injective endomorphism $f$ of I is bijective (since $f(I)$ is then a nonzero direct factor submodule of I). Moreover, every endomorphism $f$ of I, whose restriction to a nonzero submodule E of I is injective, is injective (indeed, since (i) $\Rightarrow$ (iii), I is an injective envelope of E, hence « $E \cap \mathrm{Ker}\ f = 0$ » implies « $\mathrm{Ker}\ f = 0$ »). This being so, let $f$ be a noninvertible element of $\mathrm{End}_A(M)$; by VIII, § 1, no 4, prop. 9, it is a matter of proving that $1 - f$ is invertible. Since $f$ is not injective, we have $\mathrm{Ker}\ f \neq 0$; since the restriction of $1 - f$ to $\mathrm{Ker}\ f$ is injective, $1 - f$ is injective, hence bijective.