Lemma 110.10.1. Let $I \subset R$ be an ideal of a ring. Let $M$ be an $I$-adically complete $R$-module and let $N \subset M$ be a submodule. The following are equivalent
$N$ is closed in $M$,
$N = \bigcap _{n \geq 1} (N + I^ n M)$,
$M/N$ is $I$-adically complete
If $I$ is finitely generated, these conditions imply that $N$ is $I$-adically complete.
Proof. The equivalence of (1) and (2) follows from Formal Spaces, Lemma 87.4.2. The equivalence of (2) and (3) is Algebra, Lemma 10.96.10. Assume $I$ is finitely generated and (1) holds. Let $N^\wedge $ be the $I$-adic completion of $N$. Since $M$ is $I$-adically complete we have a factorization
of the inclusion map. Since $N^\wedge \to M$ is continuous and $N$ is dense in $N^\wedge $ and $N$ is closed in $M$, we see that $N^\wedge \to M$ factors through $N$. Thus $N^\wedge = N \oplus C$ for some $A$-module $C$. Thus $N$ is a direct summand of the $I$-adically complete module $N^\wedge $, see Algebra, Lemma 10.96.3 (this is where we use that $I$ is finitely generated), and hence $N$ is $I$-adically complete. $\square$
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