Timeline for Is any element in $H^2_{et}(X,\mathcal{O}_X^*)$ locally trivial in the Zariski topology?
Current License: CC BY-SA 3.0
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| Oct 30, 2017 at 9:58 | answer | added | Daniel Loughran | timeline score: 6 | |
| Oct 30, 2017 at 3:52 | comment | added | Aaron Landesman | 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). | |
| Oct 30, 2017 at 3:03 | comment | added | Zhaoting Wei | @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. | |
| Oct 30, 2017 at 2:40 | comment | added | R. van Dobben de Bruyn | 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. | |
| Oct 30, 2017 at 2:19 | history | edited | Zhaoting Wei | CC BY-SA 3.0 |
add the condition of algebraically closed on the base field $k$.
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| Oct 30, 2017 at 2:17 | comment | added | Zhaoting Wei | @DonuArapura That makes sense. But what if we assume $k$ is algebraically closed? | |
| Oct 30, 2017 at 2:12 | comment | added | Donu Arapura | Take $X=Spec\, k$... | |
| Oct 30, 2017 at 2:08 | history | asked | Zhaoting Wei | CC BY-SA 3.0 |