Lemma 1

Let $A$ be a ring, $a$ an ideal of $A$, and $I$ an $A$-module. For every integer $n \geqslant 0$, let $I_n$ denote the a-submodule of $I$ consisting of the elements annihilated by $a^n$.

a) Suppose that the $A$-module $I$ is injective. Then the $A/a^n$-module $I_n$ is injective for every $n \geqslant 0$.

b) Suppose that the ring $A$ is noetherian, and that for every $n \geqslant 0$ the $A/a^n$-module $I_n$ is injective. Then the union of the $I_n$ is an injective $A$-module.

a) The $A/a^n$-module $I_n$ is isomorphic to $\operatorname{Hom}_A(A/a^n, I)$, which is injective (A, X, p. 18, Prop. 11).

b) Let $J$ be the union of the $I_n$, $b$ an ideal of $A$ and $f : b \to J$ an $A$-homomorphism. It is a matter (A, X, p. 16, Prop. 10) of proving that there exists an element $x$ of $J$ such that one has $f(b) = bx$ for every $b \in b$. Since $b$ is of finite type, there exists an integer $n$ such that one has $f(b) \subset I_n$, that is, $f(a^n b) = 0$. By Cor. 2 of Prop. 1 of III, § 3, No. 1, there exists an integer $m \geqslant n$ such that $a^m \cap b \subset a^n b$, hence $f(a^m \cap b) = 0$. Then $f$ induces an $A/a^m$-linear mapping of $b/(a^m \cap b)$ into $I_m$; since the $A/a^m$-module $I_m$ is injective, there exists an element $x$ of $I_m$ such that one has $f(b) = bx$ for every $b \in b$, whence b).

Let $\alpha$ be an ideal of $A$; in what follows we agree to put $\alpha^n = A$ for every integer $n \leq 0$. Let $E$ be an $A$-module. For every $n \in \mathbf{Z}$, let $E_n$ denote the submodule of $E$ consisting of the elements annihilated by $\alpha^n$; let $\mathrm{gr}^\alpha(E)$ be the graded $A$-module of type $\mathbf{Z}$ such that $\mathrm{gr}^\alpha(E)_m = E_{-m+1}/E_{-m}$ for every integer $m$. The module $\mathrm{gr}^\alpha(E)_m$ is zero for $m \geq 1$, and $\mathrm{gr}^\alpha(E)_0$ is identified with $E_1$. Let $\mathrm{gr}(A)$ denote the graded ring associated with $A$ for the $\alpha$-adic filtration: one has $\mathrm{gr}(A)_n = \alpha^n/\alpha^{n+1}$ for every $n \in \mathbf{Z}$. Let $n$ and $m$ be integers. Passing to quotients from the bilinear mapping $(a, x) \mapsto ax$ of $\alpha^n \times E_{-m+1}$ into $E_{-m-n+1}$, one deduces an $A/\alpha$-bilinear mapping

which defines on $\mathrm{gr}^\alpha(E)$ a structure of graded $\mathrm{gr}(A)$-module. For every $n \in \mathbf{Z}$, from the $A/\alpha$-bilinear mapping $\alpha_{n,-n} : \mathrm{gr}(A)_n \times \mathrm{gr}^\alpha(E)_{-n} \longrightarrow E_1$ one deduces an $A/\alpha$-linear mapping $\beta_{E,n} : \mathrm{gr}^\alpha(E)_{-n} \longrightarrow \mathrm{Hom}_{A/\alpha}(\mathrm{gr}(A)_n, E_1)$; the mappings $\beta_{E,n}$ are the components of a homomorphism of graded $A/\alpha$-modules, called canonical,

For $a \in \mathrm{gr}(A)$, $x \in \mathrm{gr}^\alpha(E)$, $\beta_E(x)(a)$ is by definition the component in $\mathrm{gr}^\alpha(E)_0 = E_1$ of the element $ax$ of $\mathrm{gr}^\alpha(E)$. It follows that $\beta_E$ is $\mathrm{gr}(A)$-linear when one endows $\mathrm{Hom}_{\mathrm{gr}_{A/\alpha}}(\mathrm{gr}(A), E_1)$ with the structure of $\mathrm{gr}(A)$-module defined by the formula $(bf)(a) = f(ab)$ for $a, b$ in $\mathrm{gr}(A)$ and $f$ in $\mathrm{Hom}_{\mathrm{gr}_{A/\alpha}}(\mathrm{gr}(A), E_1)$.