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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).