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Lemma about completion

Thanks to Kestutis Cesnavicius
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aisejohan committed Dec 18, 2018
1 parent 401dc38 commit e98dcec9337f75a3fa607c2f8d018603ceb2801e
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  1. +31 −0 formal-spaces.tex
@@ -801,6 +801,37 @@ \section{Topological rings and modules}
Then $b$ and $c$ are mutually inverse as they are on a dense subset.
\end{proof}

\begin{lemma}
\label{lemma-completion-adic-star}
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 $n \geq 1$.
\end{lemma}

\begin{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 $M^\wedge = \lim M/I^nM'$ contains
$(M')^\wedge = \lim M'/I^nM'$
as an open submodule and the topology on $(M')^\wedge$ is
$I$-adic by Algebra, Lemma \ref{algebra-lemma-hathat-finitely-generated}.
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 \ref{lemma-ses} 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.
\end{proof}

\begin{definition}
\label{definition-toplogy-tensor-product}
Let $R$ be a topological ring. Let $M$ and $N$ be linearly

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