1
$\begingroup$

Let $(R, \mathfrak{m}) $ be a local $k$- algebra and $\Omega^{1}_{R}$ denote the module of Kahler differential. Does the canonical map $ \Omega^{1}_{R} \otimes_{R} \hat{R} \rightarrow \Omega^{1}_{\hat{R}}$ is an isomorphism?, where $\hat{R}$ is the completion of $R$ with respect to $\mathfrak{m}$ and $k$ is any field of characteristics $0$.

If not, under what condition on $R$ turn this map to be isomorphism?. Is smoothness of $R$ will suffices to conclude this?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ No, this is already false for $R = k[t]_{(t)}$, $\widehat{R} = k[[t]]$, because the equation $d(\sum a_n t^n) = \sum n a_n t^{n-1}$ holds in the target but not in general in the source (N.B. your map goes in the wrong way). That's why for complete rings one usually works with continuous differentials rather than Kahler differentials. $\endgroup$
    – Piotr Achinger
    Commented May 11, 2020 at 7:21
  • $\begingroup$ Ok got it. Can you tell me what is $\Omega^{1}_{k[[x]]}$? $\endgroup$
    – Sunny
    Commented May 11, 2020 at 12:40
  • $\begingroup$ nothing particularly useful. We know that $k[[x]]$ is the increasing union of smooth $k[x]$-algebras (by Neron desingularization) which seems to show that $\Omega^1_{k[[x]]}$ is free of infinite rank. I don't know what you want to do with it but it's probably the wrong object to look at. $\endgroup$
    – Piotr Achinger
    Commented May 11, 2020 at 13:02

0

Reset to default

You must log in to answer this question.