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Nov 9, 2015 at 20:32 comment added Joe Silverman @DanielLoughran Right, I always forget those darn real places! But in the interest of consistency, I'd write them as $\frac12\mathbb Z/\mathbb Z$, so they look like subgroups of $\mathbb Q/\mathbb Z$.
Nov 9, 2015 at 15:42 comment added Daniel Loughran Does one not actually have $H^2(\text{Gal}(\bar K/K),\mathsf{A}_{\bar K}^*)=\prod_{v \text{ real}} \mathbb Z/2\mathbb Z \prod_{ v\nmid \infty} \mathbb Q/\mathbb Z$, where products are over places $v$ of $K$?
Nov 9, 2015 at 15:24 history answered Joe Silverman CC BY-SA 3.0