Theorem 64.20.5. Let $X$ be a separated scheme of finite type over a finite field $k$ and $\mathcal{F}$ be a $\mathbf{Q}_\ell $-sheaf on $X$. Then $\dim _{\mathbf{Q}_\ell }H_ c^ i(X_{\bar k}, \mathcal{F})$ is finite for all $i$, and is nonzero for $0\leq i \leq 2 \dim X$ only. Furthermore, we have

**Proof.**
We explain how to deduce this from Theorem 64.20.4. We first use some étale cohomology arguments to reduce the proof to an algebraic statement which we subsequently prove.

Let $\mathcal{F}$ be as in the theorem. We can write $\mathcal{F}$ as $\mathcal{F}'\otimes \mathbf{Q}_\ell $ where $\mathcal{F}' = \left\{ \mathcal{F}'_ n\right\} $ is a $\mathbf{Z}_\ell $-sheaf without torsion, i.e., $\ell : \mathcal{F}'\to \mathcal{F}'$ has trivial kernel in the category of $\mathbf{Z}_\ell $-sheaves. Then each $\mathcal{F}_ n'$ is a flat constructible $\mathbf{Z}/\ell ^ n\mathbf{Z}$-module on $X_{\acute{e}tale}$, so $\mathcal{F}'_ n \in D_{ctf}(X, \mathbf{Z}/\ell ^ n\mathbf{Z})$ and $\mathcal{F}_{n+1}' \otimes ^{\mathbf{L}}_{\mathbf{Z}/\ell ^{n+1}\mathbf{Z}} \mathbf{Z}/\ell ^ n\mathbf{Z} = \mathcal{F}_ n'$. Note that the last equality holds also for standard (non-derived) tensor product, since $\mathcal{F}'_ n$ is flat (it is the same equality). Therefore,

the complex $K_ n = R\Gamma _ c\left(X_{\bar k}, \mathcal{F}_ n'\right)$ is perfect, and it is endowed with an endomorphism $\pi _ n : K_ n\to K_ n$ in $D(\mathbf{Z}/\ell ^ n\mathbf{Z})$,

there are identifications

\[ K_{n+1} \otimes ^{\mathbf{L}}_{\mathbf{Z}/\ell ^{n+1}\mathbf{Z}} \mathbf{Z}/\ell ^ n\mathbf{Z} = K_ n \]in $D_{perf}(\mathbf{Z}/\ell ^ n\mathbf{Z})$, compatible with the endomorphisms $\pi _{n+1}$ and $\pi _ n$ (see [Rapport 4.12, SGA4.5]),

the equality $\text{Tr}\left(\pi _ X^* |_{K_ n}\right) = \sum _{x\in X(k)} \text{Tr}\left(\pi _ x |_{(\mathcal{F}'_ n)_{\bar x}}\right)$ holds, and

for each $x\in X(k)$, the elements $\text{Tr}(\pi _ x |_{\mathcal{F}'_{n, \bar x}}) \in \mathbf{Z}/\ell ^ n\mathbf{Z}$ form an element of $\mathbf{Z}_\ell $ which is equal to $\text{Tr}(\pi _ x |_{\mathcal{F}_{\bar x}}) \in \mathbf{Q}_\ell $.

It thus suffices to prove the following algebra lemma. $\square$

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