Lemma 10.98.1. Let $I \subset A$ be a finitely generated ideal of a ring. Let $(M_ n)$ be an inverse system of $A$-modules with $I^ n M_ n = 0$. Then $M = \mathop{\mathrm{lim}}\nolimits M_ n$ is $I$-adically complete.

**Proof.**
We have $M \to M/I^ nM \to M_ n$. Taking the limit we get $M \to M^\wedge \to M$. Hence $M$ is a direct summand of $M^\wedge $. Since $M^\wedge $ is $I$-adically complete by Lemma 10.96.3, so is $M$.
$\square$

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