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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