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Nov 25, 2017 at 6:29 comment added user19475 Yes, I forgot the infinite places.
Nov 25, 2017 at 5:04 comment added user87684 This is true when $X$ is a smooth projective curve over a finite field by a Thm of Grothendieck, and if $X$ is smooth projective geometrically irreducible over a finite field by fairly elementary arguments, but it's not quite true if $X$ is the spectrum of the ring of integers of an algebraic number field $K$. $\text{Br}(\mathscr{C})$ should surject onto $H^1(X_{\rm\acute{e}t}, \text{Pic}^0_{\mathscr{C}/X})$ (actually onto its image into $H^1(X_{\rm\acute{e}t}, \text{Pic}_{\mathscr{C}/X})$, which is off from the Tate-Shafarevich group by a finite group of exponent $2$ if $K$ has a real place.)
Nov 7, 2017 at 15:21 history answered user19475 CC BY-SA 3.0