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Let $R$ be a Noetherian regular $k$-algebra (where $k$ any field of char = 0) of dimension greater than 0. Is it true that $H^{0}_{dR}(R \lvert k) = k$?.

More generally we could ask, Is it true that $H^{0}_{dR}(R \lvert k)$ is finite $k$-algebra?

In other words I am asking suppose a regular function $P \in R$ such that its derivative is zero i.e. $d(P) = 0$ in $\Omega^{1}_{R}$, then does $P$ necessarily belongs to $k$? where, $d : R \rightarrow \Omega^{1}_{R}$ is the Kahler derivation.

I know this holds for polynomial algebra and I thought it should be consistent with the fact in manifold that a smooth function which derivative is zero must be constant.

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  • $\begingroup$ Probably you want to assume $Spec(R)$ is connected? $\endgroup$
    – Sam Gunningham
    Commented Sep 21, 2020 at 12:34
  • $\begingroup$ Sure we can assume that as this is also needed in manifold case as well. $\endgroup$
    – Sunny
    Commented Sep 21, 2020 at 12:36
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    $\begingroup$ Take $k=\mathbb{Q}$ and $R$ be any finite extension . $\endgroup$
    – Mohan
    Commented Sep 21, 2020 at 15:19
  • $\begingroup$ Ok well you are right. But in some sense these are kind of trivial counter example which I don't want. May be I should be precise about my question and should say dimension of R is greater than 0. $\endgroup$
    – Sunny
    Commented Sep 21, 2020 at 15:30
  • $\begingroup$ OK, building on Mohan's comment, take $R=\mathbb{Q}(i)[x]$. A bit more precision perhaps? $\endgroup$
    – Donu Arapura
    Commented Sep 21, 2020 at 15:44

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