Lemma 10.96.11 (031B)—The Stacks project

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Table of contents
Part 1: Preliminaries
Chapter 10: Commutative Algebra
Section 10.96: Completion
Lemma 10.96.11 (cite)

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Lemma 10.96.11. Let $R$ be a ring. Let $I$ be an ideal of $R$ . Let $M$ be an $R$ -module. If (a) $R$ is $I$ -adically complete, (b) $M$ is a finite $R$ -module, and (c) $\bigcap I^ nM = (0)$ , then $M$ is $I$ -adically complete.

Proof. By Lemma 10.96.1 the map $M = M \otimes _ R R = M \otimes _ R R^\wedge \to M^\wedge$ is surjective. The kernel of this map is $\bigcap I^ nM$ hence zero by assumption. Hence $M \cong M^\wedge$ and $M$ is complete. $\square$

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