The Stacks project

Proof. Omitted. $\square$

Proof. This is sketched below. $\square$

Proof. Note that we have already proved this (REFERENCE) when $\dim X \leq 1$. The general case follows easily from that case together with the proper base change theorem. $\square$

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,

1. 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 $D(\mathbf{Z}/\ell ^ n\mathbf{Z})$,

2. 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]),

3. 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

4. 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. $\square$

Proof. Since $\mathbf{Z}/\ell ^ n\mathbf{Z}$ is a local ring and $K_ n$ is perfect, each $K_ n$ can be represented by a finite complex $K_ n^\bullet$ of finite free $\mathbf{Z}/\ell ^ n \mathbf{Z}$-modules such that the map $K_ n^ p \to K_ n^{p+1}$ has image contained in $\ell K_ n^{p+1}$. It is a fact that such a complex is unique up to isomorphism. Moreover $\pi _ n$ can be represented by a morphism of complexes $\pi _ n^\bullet : K_ n^\bullet \to K_ n^\bullet$ (which is unique up to homotopy). By the same token the isomorphism $\varphi _ n : K_{n+1} \otimes _{\mathbf{Z}/\ell ^{n+1}\mathbf{Z}}^{\mathbf{L}} \mathbf{Z}/\ell ^ n\mathbf{Z}\to K_ n$ is represented by a map of complexes

In fact, $\varphi _ n^\bullet$ is an isomorphism of complexes, thus we see that

• there exist $a, b\in \mathbf{Z}$ independent of $n$ such that $K_ n^ i = 0$ for all $i\notin [a, b]$, and

• the rank of $K_ n^ i$ is independent of $n$.

Therefore, the module $K_\infty ^ i = \mathop{\mathrm{lim}}\nolimits _ n \{ K_ n^ i, \varphi _ n^ i\}$ is a finite free $\mathbf{Z}_\ell$-module and $K_\infty ^\bullet$ is a finite complex of finite free $\mathbf{Z}_\ell$-modules. By induction on the number of nonzero terms, one can prove that $H^ i\left(K_\infty ^\bullet \right) = \mathop{\mathrm{lim}}\nolimits _ n H^ i\left(K_ n^\bullet \right)$ (this is not true for unbounded complexes). We conclude that $H_\infty ^ i = H^ i\left(K_\infty ^\bullet \right)$ is a finite $\mathbf{Z}_\ell$-module. This proves ii. To prove the remainder of the lemma, we need to overcome the possible noncommutativity of the diagrams

However, this diagram does commute in the derived category, hence it commutes up to homotopy. We inductively replace $\pi _ n^\bullet$ for $n\geq 2$ by homotopic maps of complexes making these diagrams commute. Namely, if $h^ i : K_{n+1}^ i \to K_ n^{i-1}$ is a homotopy, i.e.,

then we choose $\tilde h^ i : K_{n+1}^ i\to K_{n+1}^{i-1}$ lifting $h^ i$. This is possible because $K_{n+1}^ i$ free and $K_{n+1}^{i-1}\to K_ n^{i-1}$ is surjective. Then replace $\pi _ n^\bullet$ by $\tilde\pi _ n^\bullet$ defined by

With this choice of $\{ \pi _ n^\bullet \}$, the above diagrams commute, and the maps fit together to define an endomorphism $\pi _\infty ^\bullet = \mathop{\mathrm{lim}}\nolimits _ n\pi _ n^\bullet$ of $K_\infty ^\bullet$. Then part i is clear: the elements $t_ n = \sum (-1)^ i \text{Tr}\left(\pi _ n^ i |_{K_ n^ i}\right)$ fit into an element $t_\infty$ of $\mathbf{Z}_\ell$. Moreover

where the last equality follows from the fact that $\mathbf{Q}_\ell$ is a field, so the complex $K_\infty ^\bullet \otimes \mathbf{Q}_\ell$ is quasi-isomorphic to its cohomology $H^ i(K_\infty ^\bullet \otimes \mathbf{Q}_\ell )$. The latter is also equal to $H^ i(K_\infty ^\bullet )\otimes _{\mathbf{Z}}\mathbf{Q}_\ell = H_\infty ^ i \otimes \mathbf{Q}_\ell$, which finishes the proof of the lemma, and also that of Theorem 64.20.5. $\square$