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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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let m be a fitered module wit filteration`{m(n)} such that m is complete with respect to the that every submodule of m is complete with respect to the induced filtration

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Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good. You may find that answers which are easier to read are also easier to understand. Thanks! – Tom Oldfield Dec 17 '12 at 5:41
This is not an answer. If you have a question, then ask a question. – Did Dec 17 '12 at 7:11

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