Lemma 10.96.10. Let $R$ be a ring. Let $I$ be an ideal of $R$. Let $M$ be an $I$-adically complete $R$-module, and let $K \subset M$ be an $R$-submodule. The following are equivalent

$K = \bigcap (K + I^ nM)$ and

$M/K$ is $I$-adically complete.

Lemma 10.96.10. Let $R$ be a ring. Let $I$ be an ideal of $R$. Let $M$ be an $I$-adically complete $R$-module, and let $K \subset M$ be an $R$-submodule. The following are equivalent

$K = \bigcap (K + I^ nM)$ and

$M/K$ is $I$-adically complete.

**Proof.**
Set $N = M/K$. By Lemma 10.96.1 the map $M = M^\wedge \to N^\wedge $ is surjective. Hence $N \to N^\wedge $ is surjective. It is easy to see that the kernel of $N \to N^\wedge $ is the module $\bigcap (K + I^ nM) / K$.
$\square$

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