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Mar 14, 2021 at 21:26 vote accept curious math guy
Mar 14, 2021 at 21:22 answer added user175951 timeline score: 7
Mar 14, 2021 at 21:13 comment added Peter Scholze Even if $\ell\neq \mathrm{char} k$, there is no reason that $H^i(X_{k^{\mathrm{sep}}},\mathbb F_\ell)$ is semisimple, because of the reduction modulo $\ell$ issue. In fact, I think this representation can be pretty arbitrary. The same ought to happen mod $p$.
Mar 14, 2021 at 20:36 history asked curious math guy CC BY-SA 4.0