Theorem 63.3.9. 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 $F_{X_{\bar k}}^ f = \pi _ X \circ \mathop{\mathrm{Spec}}(\text{frob}_ k) = \mathop{\mathrm{Spec}}(\text{frob}_ k) \circ \pi _ X$. $\square$

There are also:

• 12 comment(s) on Section 63.3: Frobenii

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).