Proposition 12

Let $A$ be a filtered ring, $E$ a filtered $A$-module and $(A_n)$ and $(E_n)$ the respective filtrations on $A$ and $E$. Suppose that $A$ is complete and the filtration $(E_n)$ is exhaustive and separated. Let $(x_i)_{i \in I}$ be a finite family of elements of $E$ and for $i \in I$ let $n(i)$ be an integer such that $x_i \in E_{n(i)}$; finally let $\xi_i$ be the class of $x_i$ in $\mathrm{gr}_{n(i)}(E)$. Then, if $(\xi_i)$ is a system of generators of the $\mathrm{gr}(A)$-module $\mathrm{gr}(E)$, $(x_i)$ is a system of generators of the $A$-module $E$.

In the $A$-module $L = A'$, let $L_0$ denote the set $(a_i)$ such that $a_i \in A_{n - n(i)}$ for all $i \in I$; if $p$ and $q$ are the least and greatest of the $n(i)$, then $A^I_{n-q} \supset L_0 \supset A^I_{n-p}$ and the topology defined on $L$ by the definition $(L_n)$ is the same as the product topology; hence $L$ is a complete filtered $A$-module. As $L$ is free, there exists an $A$-linear mapping $u : L \to E$ such that $u((a_i)) = \sum_{i \in I} a_i x_i$ and it is obviously compatible with the filtrations; we must prove that $u$ is surjective and for this it is sufficient, by virtue of Corollary 2 to Theorem 1, no. 8, to show that

is surjective or also that, for all $x \in E_n$, there exist a family $(a_i)$ such that $a_i \in A_{n-n(i)}$ for all $i \in I$ and $x \equiv \sum_{i \in I} a_i x_i \pmod{E_{n+1}}$. Let $\xi$ be the class of $x$ in $\mathrm{gr}_n(E)$; since the $\xi_i$ generate the $\mathrm{gr}(A)$-module $\mathrm{gr}(E)$, there exist $\alpha_i \in \mathrm{gr}(A)$ such that $\xi = \sum_{i \in I} \alpha_i \xi_i$ and we may assume that $\alpha_i \in \mathrm{gr}_{n-n(i)}(A)$ by replacing if need be $\alpha_i$ by its homogeneous component of degree $n - n(i)$. Then $\alpha_i$ is the image of an element $a_i \in A_{n-n(i)}$ and the family $(a_i)$ has the required property.