# The Stacks Project

## Tag 031D

Lemma 10.95.12. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

1. $R$ is $I$-adically complete,
2. $\bigcap_{n \geq 1} I^nM = (0)$, and
3. $M/IM$ is a finite $R/I$-module.

Then $M$ is a finite $R$-module.

Proof. Let $x_1, \ldots, x_n \in M$ be elements whose images in $M/IM$ generate $M/IM$ as a $R/I$-module. Denote $M' \subset M$ the $R$-submodule generated by $x_1, \ldots, x_n$. By Lemma 10.95.1 the map $(M')^\wedge \to M^\wedge$ is surjective. Since $\bigcap I^nM = 0$ we see in particular that $\bigcap I^nM' = (0)$. Hence by Lemma 10.95.11 we see that $M'$ is complete, and we conclude that $M' \to M^\wedge$ is surjective. Finally, the kernel of $M \to M^\wedge$ is zero since it is equal to $\bigcap I^nM = (0)$. Hence we conclude that $M \cong M' \cong M^\wedge$ is finitely generated. $\square$

The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 21817–21827 (see updates for more information).

\begin{lemma}
\label{lemma-finite-over-complete-ring}
Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module.
Assume
\begin{enumerate}
\item $R$ is $I$-adically complete,
\item $\bigcap_{n \geq 1} I^nM = (0)$, and
\item $M/IM$ is a finite $R/I$-module.
\end{enumerate}
Then $M$ is a finite $R$-module.
\end{lemma}

\begin{proof}
Let $x_1, \ldots, x_n \in M$ be elements whose images in $M/IM$ generate
$M/IM$ as a $R/I$-module. Denote $M' \subset M$ the $R$-submodule
generated by $x_1, \ldots, x_n$. By Lemma \ref{lemma-completion-generalities}
the map $(M')^\wedge \to M^\wedge$ is surjective.
Since $\bigcap I^nM = 0$ we see in particular that $\bigcap I^nM' = (0)$.
Hence by Lemma \ref{lemma-when-finite-module-complete-over-complete-ring}
we see that $M'$ is complete, and we conclude that $M' \to M^\wedge$
is surjective. Finally, the kernel of $M \to M^\wedge$ is
zero since it is equal to $\bigcap I^nM = (0)$.
Hence we conclude that $M \cong M' \cong M^\wedge$
is finitely generated.
\end{proof}

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