Lemma 63.14.6. Consider the situation of Theorem 63.14.4 and let $\ell$ be a prime number invertible in $k$. Then

$\sum \nolimits _{i = 0}^2 (-1)^ i \text{Tr}(\varphi ^* |_{H^ i (C, \underline{\mathbf{Z}/\ell ^ n \mathbf{Z}})}) = V(\varphi ) \mod \ell ^ n.$

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

$H^ i(C, \underline{\mathbf{Z}/\ell ^ n \mathbf{Z}}) \cong H^ i(C, \mu _{\ell ^ n})$

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

\begin{eqnarray*} \text{Tr}(\varphi ^* |_{H^1 (C, \underline{\mathbf{Z}/\ell ^ n \mathbf{Z}})})) & = & \text{Tr}_ J (\varphi ^*) \mod \ell ^ n \\ & = & \text{Tr}_{\mathbf{Z}/\ell ^ n \mathbf{Z}} (\varphi ^* : J[\ell ^ n] \to J[\ell ^ n]). \end{eqnarray*}

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

$\text{Tr} (\varphi ^*|_{H^2 (C, \underline{\mathbf{Z}/\ell ^ n \mathbf{Z}})}) = \deg \varphi .$

Thus we have

$\sum _{i = 0}^2 (-1)^ i \text{Tr}(\varphi ^* |_{H^ i (C, \underline{\mathbf{Z}/\ell ^ n \mathbf{Z}})}) = 1 - \text{Tr}_ J(\varphi ^*) + \deg \varphi \mod \ell ^ n$

and the corollary follows from Theorem 63.14.4. $\square$

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