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Let $R$ be a profinite commutative ring, and let us take two ordinary $R$-modules, $M$ and $N.$Let $M^\wedge, N^\wedge$ be the profinite completion as $R$-modules, i.e we complete with respect to the submodules such that the quotient is finite. Consider the completed tensor product $M^\wedge \hat{\otimes}_R N^\wedge.$ Is it true that $M^\wedge \hat{\otimes}_R N^\wedge \cong (M \otimes_R N)^\wedge,$ i.e that the completed tensor product commutes with profinite completion?

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  • $\begingroup$ How are you defining the completed tensor product? $\endgroup$ – user113529 Sep 3 '17 at 15:53
  • $\begingroup$ @user43687 As in : stacks.math.columbia.edu/tag/0AMU $\endgroup$ – Twistediso Sep 3 '17 at 15:53
  • $\begingroup$ But then, by definition the completed tensor product is precisely the completion of the tensor product. $\endgroup$ – user113529 Sep 3 '17 at 16:04
  • $\begingroup$ @user43687: Well, there is something more to show, right? Namely, one must show that the submodules of the tensor product of the form $M' \otimes_R N + M \otimes_R N'$ where $M' \subset M, N' \subset N$ has the property that $M/M',N/N'$ has finite length is cofinal in the set of submodules of the tensor product so that the quotient has finite length. But this should be true, right? $\endgroup$ – Twistediso Sep 3 '17 at 16:10
  • $\begingroup$ No, there they are defining a fundamental system of open submodules for the topology on the tensor product, so there is no property to verify $\endgroup$ – user113529 Sep 3 '17 at 16:16

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