The Stacks project

Theorem 64.3.9. Let $\mathcal{F}$ be an abelian sheaf on $X_{\acute{e}tale}$. Then for all $j\geq 0$, $\text{frob}_ k$ acts on the cohomology group $H^ j(X_{\bar k}, \mathcal{F}|_{X_{\bar k}})$ as the inverse of the map $\pi _ X^*$.

Proof. The composition $X_{\bar k} \xrightarrow {\mathop{\mathrm{Spec}}(\text{frob}_ k)} X_{\bar k} \xrightarrow {\pi _ X} X_{\bar k}$ is equal to $F_{X_{\bar k}}^ f$, hence the result follows from the baffling theorem suitably generalized to nontrivial coefficients. Note that the previous composition commutes in the sense that $F_{X_{\bar k}}^ f = \pi _ X \circ \mathop{\mathrm{Spec}}(\text{frob}_ k) = \mathop{\mathrm{Spec}}(\text{frob}_ k) \circ \pi _ X$. $\square$


Comments (0)

There are also:

  • 12 comment(s) on Section 64.3: Frobenii

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 03SV. Beware of the difference between the letter 'O' and the digit '0'.