Lemma 15.94.5. Let $I$ be an ideal of a Noetherian ring $A$. Let $M$ be a derived complete $A$-module. If $M/IM$ is a finite $A/I$-module, then $M = \mathop{\mathrm{lim}}\nolimits M/I^ nM$ and $M$ is a finite $A^\wedge$-module.

Proof. Assume $M/IM$ is finite. Pick $x_1, \ldots , x_ t \in M$ which map to generators of $M/IM$. We obtain a map $A^{\oplus t} \to M$ mapping the $i$th basis vector to $x_ i$. By Proposition 15.94.2 the derived completion of $A$ is $A^\wedge = \mathop{\mathrm{lim}}\nolimits A/I^ n$. As $M$ is derived complete, we see that our map factors through a map $q : (A^\wedge )^{\oplus t} \to M$. The module $\mathop{\mathrm{Coker}}(q)$ is zero by Lemma 15.91.7. Thus $M$ is a finite $A^\wedge$-module. Since $A^\wedge$ is Noetherian and complete with respect to $IA^\wedge$, it follows that $M$ is $I$-adically complete (use Algebra, Lemmas 10.97.5, 10.96.11, and 10.51.2). $\square$

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