# Modules over Completion

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 $N$?

• 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 – rschwieb Jun 8 '13 at 15:24
• 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. – Keenan Kidwell Jun 8 '13 at 17:21
• I meant to add to my comment: But note that $\mathbf{Q}\otimes_\mathbf{Z}\mathbf{Z}_p=\mathbf{Q}_p$. – Keenan Kidwell Jun 8 '13 at 17:27
• @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. – rschwieb Jun 8 '13 at 22:32
• 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]]?) – Avitus Jun 9 '13 at 8:47

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 $I \not\subset \text{Jac}(A)$).