Is simplicity preserved under completion of the base ring?

Let $(A,\mathfrak{m})$ be a noetherian local ring and $R$ be an $A$-algebra, which is finitely generated generated as an $A$-module (module finite $A$-algebra). Let $\widehat{A}$ be the $\mathfrak{m}$-adic completion and define $\widehat{R}=R\otimes_A \widehat{A}$.

Do we have a 1:1-correspondence between the isomorphism classes of simple left $R$-modules and the isomorphism classes of simple left $\widehat{R}$-modules, induced by the map $S \rightarrow S\otimes_A \widehat{A}$?

Since $R$ is module-finite over $A$, we have that $\widehat{\mathfrak{m}}\widehat{R}\\subset rad(\widehat{R})$. So given a simple $\widehat{R}$-module $S$, then $\widehat{\mathfrak{m}}$ annihilates $S$, that is $S$ is in fact a simple $R\otimes_A\widehat{A}/\widehat{\mathfrak{m}}$-module. But $\widehat{A}/\widehat{\mathfrak{m}}=A/\mathfrak{m}$. So $S$ is a simple $R/\mathfrak{m}R$-module. But since $\mathfrak{m}\subset rad(R)$, this implies that $S$ is a simple $R$-module.

Now given a simple $R$-module $M$, then this is a finite length $A$-module, so the map $M\rightarrow M\otimes_A\widehat{A}$ is an isomorphism. But is this also an isomorphism of $\widehat{R}$-modules?

Or is this eventually not true? Is there literature about the behaviour of $R$-modules under completion of the base ring?

I ask this, because it is often easier to find simple modules if the base ring $A$ is complete, e.g. if $A/\mathfrak{m}$ is algebraically closed and $R$ is an Azumaya algebra, then $\widehat{R}\cong M_s(\widehat{A})$, and the simple modules for $\widehat{R}$ are easy to see, but for $R$?

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Are you asking if the cardinalities of the iso classes of simple left $R$-modules and simple left $\hat R$-modules are the same? Or are you asking if the map $M \mapsto \hat R \otimes_R M$ induces such a bijection? – Tilman May 9 '11 at 21:23
@Tilman: the latter, I should think. – Pete L. Clark May 9 '11 at 21:31
Yes, thanks. That is what i meant, i modified the question and added some of my thoughts on the problem. – TonyS May 10 '11 at 14:44
Since $R$ is module finite over $A$, $\widehat{R} \cong \widehat{A} \otimes_A R$. Hence $\widehat{A} \otimes_A M \cong \widehat{A} \otimes_A (R \otimes_R M) \cong (\widehat{A} \otimes_A R) \otimes_R M \cong \widehat{R} \otimes_R M$. So the natural map $M \to \widehat{R} \otimes_R M$ that sends $m \to 1 \otimes m$ is also an isomorphism, and it's $R$-linear because $r.(1 \otimes m) = r\otimes m = 1 \otimes rm$. – user91132 May 10 '11 at 14:54
Ah, okay. So in general $\widehat{R}$ is the completion of $R$ with respect to $rad(R)$? And in our our case this is isomorphic to $\widehat{A}\otimes_A R$. I didn't know that. You can post this as an answer, then i can click this to be the accepted answer :-). – TonyS May 10 '11 at 15:50