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Let $A$ be a commutative noetherian ring, $I$-adically complete (and separated) with respect to an ideal $I \subseteq A$.

Let $M$ be a finite $A$-module, and let $N$ be an $I$-adically complete $A$-module.

Is it true that $M\otimes_A N$ is also $I$-adically complete?

If $M$ is free this is clear. Is it true in general?

Thank you!

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Consider the natural transformation $(-) \otimes N \to \lim_n ((-) \otimes_A N \otimes_A A/I^n)$ of endofunctors of $A$-modules. Both functors preserve finite colimits, i.e. are linear and right exact. For the left one this is clear, and for the right one this follows from the general fact that inverse limits are right exact for surjective systems. Thus the locus where the natural transformation is an isomorphism is closed under finite colimits (you can also write this down using a five lemma argument, but it is more tedious). Since $N$ is assumed to be complete, $A$ lies in the locus, hence every finitely presented $A$-module. We don't need that $A$ is complete or that $A$ is noetherian.

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