Let $X$ be an algebraic variety over a field $k$ and we consider the cohomological Brauer group $H^2_{et}(X,\mathcal{O}_X^*)$.

For any element $\alpha \in H^2_{et}(X,\mathcal{O}_X^*)$ and any closed point $x\in X$, my question is:

Could we always find a Zariski open neighborhood $U$ of $x$ (which depends on $\alpha$), such that $\alpha|_U$ is trivial in $H^2_{et}(U,\mathcal{O}_U^*)$?

If not,

  1. Is there any counter-example?
  2. Is the same statement true if we restrict ourselves to the Brauer group rather than the cohomological Brauer group, i.e. elements in $H^2_{et}(X,\mathcal{O}_X^*)$ that comes from an Azumaya algebra?

Edit: As pointed out by @Donu Arapura in the comments, the statement is trivially wrong if we do not assume that $k$ is algebraically closed. So we should assume $k$ is algebraically closed.

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    $\begingroup$ Take $X=Spec\, k$... $\endgroup$ Oct 30, 2017 at 2:12
  • $\begingroup$ @DonuArapura That makes sense. But what if we assume $k$ is algebraically closed? $\endgroup$ Oct 30, 2017 at 2:17
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    $\begingroup$ It's still the same answer. For example, choose a (cohomological) Brauer class that maps to a nonzero element of $H^2_{\operatorname{\acute et}}(\operatorname{Spec} k(X), \mathbb G_m)$; in particular it cannot be zero on any (nonempty) Zariski open. $\endgroup$ Oct 30, 2017 at 2:40
  • $\begingroup$ @R.vanDobbendeBruyn I roughly got your point but could you provide some more details? For example is it obvious that there always exists an element in $H^2_{et}(X,\mathcal{O}_X^*)$ which maps to a non-zero element in $H^2_{et}(\text{spec}K(X),\mathbb{G}_m)$? I'm sorry I'm not very familiar with this area. $\endgroup$ Oct 30, 2017 at 3:03
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    $\begingroup$ For regular varieties $X$, $Br(X) \rightarrow Br(K(X))$ is injective, see www-math.mit.edu/~poonen/papers/Qpoints.pdf 6.6.7 and references therein. So, you only need a regular surface with $H^2(X, G_m) \neq 0$. For example, take an elliptic fibration $E \rightarrow P^1$ whose quotient by the hyperelliptic involution is a hirzebruch surface $F$, branched over a curve $C \subset F$. Every nontrivial 2-torsion element of the Jacobian of $C$ yields a nontrivial element in the Brauer group of E, see arxiv.org/pdf/math/0408006.pdf Section 7 (using Theorem 6.2, see also 8.7). $\endgroup$ Oct 30, 2017 at 3:52

1 Answer 1


An explicit simple counter-example is the following: just take the quaternion algebra $(x,y)$ over $k(x,y)$, where $k$ is a field with $\mathrm{char}(k) \neq 2$. This is non-zero on any open subset of the affine plane (it is well defined on the complement of the coordinate axes).

One can see that it is non-zero as its residue along the coordinate axis $x=0$ is given by $y \in k(y)^{*}/k(y)^{*2}$, which is non-zero.


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