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May 12, 2022 at 12:25 comment added Nobody Now I get the point, thanks!
May 12, 2022 at 12:01 comment added Will Sawin This follows from the claim that $H^i (X_{\overline{\mathbb F_q}}, \mathbb F_\ell[t]/(t^r) ) = H^i (X_{\overline{\mathbb F_q}}, \mathbb F_\ell) \otimes_{\mathbb F_\ell} \mathbb F_\ell [t]/(t^r) $ which unraveled is just the statement that cohomology commutes with finite direct sums of coefficients, or, maybe transparently, tensor products with flat modules over the coefficient ring.
May 12, 2022 at 11:58 comment added Nobody @PiotrAchinger By definition, we have $$H^{i}(X\times_{\mathbb{F}_{q}}\overline{\mathbb{F}_{q}},\mathbb{F}_{\ell}((t))):=\left( \varprojlim_{r}H^{i}_{\text{\'et}}(X\times_{\mathbb{F}_{q}}\overline{\mathbb{F}_{q}},\mathbb{F}_{\ell}[t]/(t^{r}))\right) \otimes_{\mathbb{F}_{\ell}[[t]]}\mathbb{F}_{\ell}((t)) $$, why are they the same?
May 12, 2022 at 9:53 comment added Piotr Achinger Isn’t the cohomology group in question the same as $H^i(X\otimes_{\mathbb{F}_q}\overline{\mathbb{F}}_q, \mathbb{F}_\ell)\otimes_{\mathbb{F}_\ell}\mathbb{F}_\ell((t))$?
May 12, 2022 at 9:35 history asked Nobody CC BY-SA 4.0