
Lemma 79.4.6. Let $R$ be a topological ring. Let $M$ be a topological $R$-module. Let $I \subset R$ be a finitely generated ideal. Assume $M$ has an open submodule whose topology is $I$-adic. Then $M^\wedge$ has an open submodule whose topology is $I$-adic and we have $M^\wedge /I^ n M^\wedge = M/I^ nM$ for all $n \geq 1$.

Proof. Let $M' \subset M$ be an open submodule whose topology is $I$-adic. Then $\{ I^ nM'\} _{n \geq 1}$ is a fundamental system of open submodules of $M$. Thus $M^\wedge = \mathop{\mathrm{lim}}\nolimits M/I^ nM'$ contains $(M')^\wedge = \mathop{\mathrm{lim}}\nolimits M'/I^ nM'$ as an open submodule and the topology on $(M')^\wedge$ is $I$-adic by Algebra, Lemma 10.95.3. Since $I$ is finitely generated, $I^ n$ is finitely generated, say by $f_1, \ldots , f_ r$. Observe that the surjection $(f_1, \ldots , f_ r) : M^{\oplus r} \to I^ n M$ is continuous and open by our description of the topology on $M$ above. By Lemma 79.4.5 applied to this surjection and to the short exact sequence $0 \to I^ nM \to M \to M/I^ nM \to 0$ we conclude that

$(f_1, \ldots , f_ r) : (M^\wedge )^{\oplus r} \longrightarrow M^\wedge$

surjects onto the kernel of the surjection $M^\wedge \to M/I^ nM$. Since $f_1, \ldots , f_ r$ generate $I^ n$ we conclude. $\square$

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