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Lately I've looked at $f$-adic completions of commutative rings. I had posted two questions regarding the topic on math.SE which didn't receive any attention and I think they might be fit for mathoverflow. If not so, please feel free to tell me.

1.) Let $A$ a commutative ring, $f \in A$ a non zerodivisor. Let $\hat{A}$ the $f$-adic completion of $A$. Let $G$ be an $\hat{A}$-Module without $f$-torsion.

How can we show that $\operatorname{Tor}_1^A(\hat{A},G_f/G)=0$?

I have already shown $\hat{A} \otimes_A A_f/A \cong \hat{A}_f/ \hat{A}$

Of course (as in my linked question), it holds again that $\varinjlim G/f^n G = G_f/G$, therefore it would suffice to show $\operatorname{Tor}_1^A(\hat{A}, G/f^n G) =0$ for all $n \geq 0$.

2.) Let $A$ a commutative ring, $f \in A$ not a zerodivisor and $\hat{A}$ the $f$-adic completion of $A$ (i.e. the completion along the ideal $(f)$).

For $A$ noetherian it is well-known that $\hat{A}$ is flat as an $A$-algebra. If $A$ is not noetherian however, this is not true in general.

Are there (explicit or general) examples where $A$ is not noetherian but its $f$-adic completion $\hat{A}$ is a flat $A$-algebra?

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    $\begingroup$ You must be reading the Beauville--Laszlo paper "Un lemma de descente". Go beyond the "abridged English version" at the start of the paper and read the more detailed French version (= the actual paper). There you'll find Lemme 3(a) which answers your vanishing question. For your other question, a trivial example is when $A = \widehat{A}$ (but not noetherian), such as $A = R[\![z]\!]$ with $f = z$ and $R$ not noetherian. Slightly less trivial example: take $A$ to be the integral closure of a discrete valuation ring $A_0$ in an algebraic closure of its fraction field, $f$ a uniformizer of $A_0$. $\endgroup$
    – user76758
    Mar 25, 2014 at 9:11
  • $\begingroup$ Yes, I've been reading this paper. But why is it clear (in the proof of lemma 3(a)), that after tensoring the map $j$ stays injective? Am I missing something basic? Also thanks for your examples! I'll have a look at them. $\endgroup$
    – Louis
    Mar 25, 2014 at 10:58
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    $\begingroup$ The short exact sequence of interest to you consists entirely of $A/f^n A$-modules, so tensoring it against $\widehat{A}$ over $A$ is "the same" as tensoring it against $\widehat{A}/f^n \widehat{A}$ over $A/f^n A$. So you just need to check that the natural map $A/f^n A \rightarrow \widehat{A}/f^n \widehat{A}$ is an isomorphism, which follows from the fact that $f$ is not a zero-divisor in $A$. (You can just as well rename $f^n$ as $f$ for this purpose too.) $\endgroup$
    – user76758
    Mar 27, 2014 at 18:41

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