Proposition 16

Let $A$ be a filtered ring, ($A,$) its filtration, $E$ an $A$-module and ($E_n$) the filtration on $E$ derived from that on $A$ consisting of the $E_n = A_n E$. Suppose that the filtration ($A,$) is exhaustive and the module $E$ is finitely generated. If $i : E \to \hat{E}$ is the canonical mapping, then, for all $n \in \mathbf{Z}$,

In particular $\hat{E}$ is a finitely generated $\hat{A}$-module.

The equation $A_n E = E_n$ implies, by virtue of the continuity of the external law on the $A$-module $8, \hat{A}_n \hat{E} \subset 8$, and obviously $\hat{A}_n \hat{E} \supset \hat{A}_n i(E)$. By hypothesis there exists a surjective homomorphism $u : L \to E$, where $L = A_s^I$, $I$ being a finite set; let $L$ be given the product filtration, consisting of the $L_n = A_n^I$, which define on $L$ the product topology; then $\hat{L} = \hat{A}_s^I$ and $\hat{L}_n = \hat{A}_n^I$ (General Topology, Chapter II, § 3, no. 9, Corollary 2 to Proposition 18). Let $j : L \to \hat{L}$ be the canonical mapping and $(e_i)_{i \in I}$ the canonical basis of $L$; for an element $\sum_{i \in I} a_{ij}(e_i)$ (where $a_i \in \hat{A}$ for all $i \in I$) to belong to $\hat{L}_n$, it is necessary and sufficient that $a_i \in \hat{A}_n$ for all $i$; then $\hat{L}_n = \hat{A}_n . j(L)$. This being so, by definition $u(\hat{L}_n) = A_n E = E_n$ and hence $u$ is a strict morphism of $L$ onto $E$ (General Topology, Chapter III, § 2, no. 8, Proposition 24). Lemma 2 then shows that $\hat{u} : \hat{L} \to \hat{E}$ is a surjective strict morphism. As $\hat{L}_n$ is an open subgroup off, $\hat{u}(\hat{L}_n)$ is an open (and therefore closed) subgroup of $\hat{E}$; but $\hat{u}(\hat{L}_n) = \hat{A}_n \hat{u}(j(L)) = \hat{A}_n i(E)$ and, as $i(E_n) \subset A_n i(E) \subset \hat{A}_n i(E)$, finally $\hat{E}_n \subset \hat{A}_n i(E) \subset \hat{A}_n \hat{E} \subset \hat{E}_n$ and therefore $\hat{E}_n = \hat{A}_n \hat{E} = \hat{A}_n i(E)$; setting $n = 0$, we obtain the second formula of (25).