Lemma 10.96.12. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

$R$ is $I$-adically complete,

$\bigcap _{n \geq 1} I^ nM = (0)$, and

$M/IM$ is a finite $R/I$-module.

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

Lemma 10.96.12. Let $R$ be a ring. Let $I \subset R$ be an ideal. Let $M$ be an $R$-module. Assume

$R$ is $I$-adically complete,

$\bigcap _{n \geq 1} I^ nM = (0)$, and

$M/IM$ 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 10.96.1 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 10.96.11 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.
$\square$

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