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_ nin 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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