Proposition 2

Let $A$ be a noetherian ring, $\alpha$ an ideal of $A$, $E$ an $A$-module and $M$ a sub-$A$-module of $E$ annihilated by $\alpha$. The following conditions are equivalent: (i) $E$ is an injective envelope of $M$ ; (ii) the $A/\alpha$-module $E_1$ is an injective envelope of the $A/\alpha$-module $M$, the module $E$ is the union of the $E_n$ and the canonical mapping $\beta_E$ is bijective.

Suppose condition (i) is satisfied. The $A/\alpha$-module $E_1$ is injective (Lemma 1, a)), and contains $M$; since every sub-$A/\alpha$-module of $E_1$ is a sub-$A$-module of $E$, $E_1$ is an injective envelope of the $A/\alpha$-module $M$. By Lemma 1, the union of the $E_n$ is an injective sub-$A$-module of $E$ containing $M$, hence equal to $E$. Since $E$ is injective, for every $n \geq 0$ one has an exact sequence

since $\mathrm{Hom}_A(A/\alpha^m, E)$ is identified with $E_m$ for every $m$ and the canonical injection of $\mathrm{Hom}_A(\alpha^n/\alpha^{n+1}, E_1)$ into $\mathrm{Hom}_A(\alpha^n/\alpha^{n+1}, E)$ is bijective, one deduces that the canonical homomorphism $\beta_E$ is bijective, whence (ii).

Suppose (ii) is satisfied. Let $e : M \to I$ be an injective envelope of $M$. Since $I$ is injective, there exists an $A$-linear mapping $\varphi : E \to I$ extending $e$. But $\varphi$ maps $E_n$ into $I_n$ for every $n$, hence induces homomorphisms gr^{a}(\varphi) : \operatorname{gr}^{a}(E) \to \operatorname{gr}^{a}(I) \text{ and } \varphi_1 : E_1 \to I_1 \text{ rendering the diagram commutative}

Since $E_1$ and $I_1$ are injective envelopes of the $A/\alpha$-module $M$, the homomorphism $\varphi_1$ is bijective; since $\beta_E$ and $\beta_I$ are bijective, it follows that $\operatorname{gr}^{a}(\varphi)$ is bijective. This implies, by induction on $n$, that $\varphi$ induces a bijection of $E_n$ onto $I_n$ for every $n \geq 1$; hence $\varphi$ is bijective, which entails (i).