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Jun 25, 2017 at 3:29 history edited user19475 CC BY-SA 3.0
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Jun 25, 2017 at 2:59 history edited user19475 CC BY-SA 3.0
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Jun 24, 2017 at 16:10 comment added user19475 And in other dimensions?
Jun 24, 2017 at 15:42 comment added Daniel Loughran mathoverflow.net/questions/185630/why-is-there-no-brauer-scheme
Jun 24, 2017 at 14:01 comment added abx If $X$ is a smooth complex projective variety, $H^2_{\rm ét}(X,\mathbb{G}_m)$, the Brauer group of $X$, is isomorphic to $(\mathbb{Q}/\mathbb{Z})^{n}$ for some $n$ ($=b_2-\rho $ for the experts). It is certainly not "representable by a scheme".
Jun 24, 2017 at 13:30 history asked user19475 CC BY-SA 3.0