14
$\begingroup$

I have the following question.

Let $R$ be a commutative ring with unit, and let $\hat{R}$ denote its completion (w.r.t. any ideal $I$). Let $M$ be an $\hat{R}$-module. Is $M= N\otimes_R \hat{R}$ for some $R$-module $N$?

$\endgroup$
5
  • $\begingroup$ At first I wondered if $M=N$ might work, but I see now that being a module over $\hat{R}$ and being complete is probably different. I guess all we have to prove is that the image of the completion functor isn't onto the objects of $\hat{R}$-Mod $\endgroup$ – rschwieb Jun 8 '13 at 15:24
  • 1
    $\begingroup$ Dear @rschwieb, The functor $-\otimes_R\hat{R}$ is not the same as the completion functor, i.e., if $N$ i s not a finite $R$-module or $R$ is not Noetherian, the natural map $N\otimes_R\hat{R}\rightarrow\hat{N}$ need not be an isomorphism. Moreover the completion functor is not essentially surjective. For example, $\mathbf{Q}_p$ is not a $p$-adically complete $\mathbf{Z}_p$-module, so it is not the $p$-adic completion of any $\mathbf{Z}$-module. $\endgroup$ – Keenan Kidwell Jun 8 '13 at 17:21
  • 1
    $\begingroup$ I meant to add to my comment: But note that $\mathbf{Q}\otimes_\mathbf{Z}\mathbf{Z}_p=\mathbf{Q}_p$. $\endgroup$ – Keenan Kidwell Jun 8 '13 at 17:27
  • $\begingroup$ @KeenanKidwell Ah! Great, thank you for helping develop my intuition about it. I have not even studied all the rudiments. I will keep your examples in mind. $\endgroup$ – rschwieb Jun 8 '13 at 22:32
  • $\begingroup$ Intuitively, if $N$ is a finite dimensional vector space on $\mathbb K$, then $N\otimes_{\mathbb K}\mathbb K[[h]]$ is isomorphic to $N[[h]]$ and as counterexample one should take any $M$ which is not topologically free (ex.: M=hN[[h]]?) $\endgroup$ – Avitus Jun 9 '13 at 8:47
1
$\begingroup$

Let $A \rightarrow B$ be a faithfully flat morphism of rings, then the category of $A$-modules is equivalent to the category of $B$-modules with descent data (see here). In less precise words, a $B$-module is of the form $M \otimes_A B$ exactly when one can provide this descent data.

When $\hat{A}$ is the completion of a noetherian ring $A$ at an ideal $I$ contained in the Jacobson radical, then the map $A \rightarrow \hat{A}$ is faithfully flat. For example, this is satisfied for any noetherian local ring.

I'm not sure if it is reasonable to expect a more complete answer (i.e. not Noetherian or $I \not\subset \text{Jac}(A)$).

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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