Lemma 10.96.10 (031A)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.96: Completion
Lemma 10.96.10 (cite)

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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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