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Let $R=\mathbb{C}[[x,y]]$ be a power series ring in two variables (or maybe more generally a strictly Henselian local ring) with maximal ideal $\mathfrak{m}$.

What is $H^*_{et}(\operatorname{Spec}(R)\setminus\{\mathfrak{m}\}, \mathbb{G}_m)$?

My motivation is this: I'm trying to understand the extent to which etale cohomology of $\mathbb{G}_m$ resembles the cohomology of $\mathcal{O}_X^*$ in the complex-analytic setting. For example, one might expect that in the case above, $H^2$ ought to be $\mathbb{Z}$.

I'd be satisfied with an answer that computes $H^2$.

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    $\begingroup$ The algebraic cohomology of $\mathbb G_{\rm m}$ is very different from the analytic cohomology of $\mathcal O^*$. For a regular scheme the former is always torsion in degree at least 2 (this is a well-known result of Grothendieck), whereas the analytic cohomology of a complex manifold tends to contain positive-dimensional $\mathbb Q$-vector spaces. $\endgroup$
    – Angelo
    May 9, 2013 at 5:35

1 Answer 1

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With your particular choice of $R$, the $H^2$ is $0$. More generally, if $R$ is a strictly Henselian regular local ring of dimension $2$, then by the purity for the Brauer group (in this particular case it is known and due to, I believe, Grothendieck; for a proof see Grothendieck "Le groupe de Brauer II", Prop. 2.3) $H^2_{et}(R \setminus \{ \mathfrak{m} \}, \mathbf{G}_m) = H^2_{et}(R, \mathbf{G}_m) = 0$. The first equality is due to the purity because you're removing a closed subscheme of codimension $2$; the second equality is because $R$ is strictly Henselian and $\mathbf{G}_m$ is smooth: by Grothendieck "Le groupe de Brauer III" appendix, Thm. 11.7 2), the cohomology can thus be computed over the separably closed residue field, where it vanishes.

More generally assuming that $R$ is a strictly Henselian regular local ring of dimension $\ge 2$, the $H^2$ is always supposed to be $0$ by the aforementioned purity and the argument as above. This is open (as far as I know), although many cases are known, including $\dim R \le 3$. For a brief survey on precisely this question see Gabber "On purity for the Brauer group" in Oberwolfach report No. 37/2004.

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    $\begingroup$ The general case of a strictly Henselian regular local ring of dimension $\geq 2$ is now solved by the author of the answer above, see arxiv.org/abs/1711.06456, Theorem 5.3. $\endgroup$ Aug 18, 2022 at 12:32

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