Lemma 63.14.6. Consider the situation of Theorem 63.14.4 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 63.14.4. $\square$

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