Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
add comment

2 Answers 2

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.

share|improve this answer
add comment

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

share|improve this answer
2  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.