Lemma 64.14.6. Consider the situation of Theorem 64.14.4 and let $\ell $ be a prime number invertible in $k$. Then

**Sketch of proof.**
Observe first that the assumption makes sense because $H^ i(C, \underline{\mathbf{Z}/\ell ^ n \mathbf{Z}})$ is a free $\mathbf{Z}/\ell ^ n \mathbf{Z}$-module for all $i$. The trace of $\varphi ^*$ on the 0th degree cohomology is 1. The choice of a primitive $\ell ^ n$th root of unity in $k$ gives an isomorphism

compatibly with the action of the geometric Frobenius. On the other hand, $H^1(C, \mu _{\ell ^ n}) = J[\ell ^ n]$. Therefore,

Moreover, $H^2(C, \mu _{\ell ^ n}) = \mathop{\mathrm{Pic}}\nolimits (C)/\ell ^ n\mathop{\mathrm{Pic}}\nolimits (C) \cong \mathbf{Z}/\ell ^ n \mathbf{Z}$ where $\varphi ^*$ is multiplication by $\deg \varphi $. Hence

Thus we have

and the corollary follows from Theorem 64.14.4. $\square$

## 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Your email address will not be published.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.

## Comments (0)