Proposition 11

Let $N$ be an $A$-module. The following conditions are equivalent:

(i) $N$ is injective. (ii) $\operatorname{Ext}_A^i(M, N) = 0$ for every $A$-module $M$ and every integer $i > 0$.

(iii) $\mathrm{Ext}_A^1(M, N) = 0$ for every $A$-module $M$.

(iv) There exists an exact sequence

where $I$ is injective and where $\mathrm{Ext}_A^1(C, N) = 0$;

(v) $\mathrm{Ext}_A^1(M, N) = 0$ for every cyclic $A$-module $M$.

(i) $\Rightarrow$ (ii) : this is the corollary of Prop. 5 of X, p. 88.

(ii) $\Rightarrow$ (iii) $\Rightarrow$ (v) : this is trivial.

(iii) $\Rightarrow$ (iv) : this is clear since $N$ is a submodule of an injective module (X, p. 19, Cor. 3).

(iv) $\Rightarrow$ (i) : since $\mathrm{Ext}_A^1(C, N) = 0$, the canonical homomorphism

is surjective (X, p. 93, Corollary); there therefore exists an $A$-linear retraction of $u$ and $N$ is isomorphic to a direct factor of $I$, hence is injective (X, p. 16, Prop. 9).

(v) $\Rightarrow$ (i) : if $\alpha$ is an ideal of $A$, we have $\mathrm{Ext}_A^1(A/\alpha, N) = 0$; the canonical mapping $\mathrm{Hom}_A(A, N) \to \mathrm{Hom}_A(\alpha, N)$ is therefore surjective and $N$ is injective (X, p. 16, Prop. 10).