Rank of $\mathbb{Z}_{p}$-module $H_{et}^{i}(X,\mathbb{Z}_{p}(r))$

Question

I want to ask the following question.

Let $X$ be a smooth projective variety of dimension $d$ over $p$-adic field $k$ ( i.e. finite extension of $\mathbb{Q}_{p}$). Is it true that etale cohomology $H_{et}^{i}(X,\mathbb{Z}_{p}(r))$ is finite generated $\mathbb{Z}_{p}$-module for any integer $i$ and $r$, where the group $H_{et}^{i}(X,\mathbb{Z}_{p}(r))$ is by definition equal to $ \varprojlim_n H_{et}^{i}(X,\mathbb{Z}/p^{n}(r))$ and $ \mathbb{Z}/p^{n}(r)$ is the Tate-twist.

In case of finite generation, how the rank (number of copies of $\mathbb{Z}_{p}$) of the group $H_{et}^{i}(X,\mathbb{Z}_{p}(r))$ depends upon $X$. Is this depends upon the dimension $d$. Can someone tell me reference where I can see such statements. Thanks in advance for any help.

Answer 1

The first question has a positive answer. This is essentially a combination of two very different finiteness results and quite specific to local fields. In what follows, I will explain the argument for $r=0$; the general case is absolutely similar.

Let me redefine etale cohomology as $\mathrm{R}\Gamma(X, \mathbf{Z}_\ell):=\mathrm{R}\lim_n \mathrm{R}\Gamma(X, \mathbf{Z}/\ell^n\mathbf{Z})$, and $\mathrm{H}^i(X,\mathbf{Z}_\ell):=\mathrm{H}^i(\mathrm{R}\Gamma(X, \mathbf{Z}_\ell))$. We will show later that this recovers the formulas you suggested.

To get the desired finiteness and comparison results, we will need two preliminary results.

Lemma 1: Let $k$ be a separably closed field, and $X$ a finite type $k$-scheme. Then $\mathrm{R}\Gamma(X, \mathbf{Z}/\ell^n\mathbf{Z})\in D_{coh}(\mathbf{Z}/\ell^n\mathbf{Z})$.

This is a consequence of a more general result that $\mathrm{R}f_*$ preserves constructible sheaves for a morphism $f\colon X \to Y$ between finite type $k$-schemes. We apply this result to the structure morphism $f\colon X \to \operatorname{Spec} k$ to get the desired finiteness.

Lemma 2: Let $K$ be a finite extension of $\mathbf{Q}_p$, and $M$ is a finite $\mathbf{Z}/\ell^n\mathbf{Z}$-module with a continuous action of $G_K$. Then $\mathrm{R}\Gamma_{\mathrm{cont}}(G_K, M)\in D_{coh}(\mathbf{Z}/\ell^n\mathbf{Z})$.

This is Proposition 5.2/14 in Serre's book ``Galois Cohomology'' (essentially boils down to the finiteness of $K^\times/K^{\times, \ell}$).

Corollary 3: Let $K$ be a finite extension of $\mathbf{Q}_p$ and $X$ a finite type $K$-scheme. Then $\mathrm{R}\Gamma(X, \mathbf{Z}/\ell^n\mathbf{Z})\in D_{coh}(\mathbf{Z}/\ell^n\mathbf{Z})$.

Proof: First of all, we use the composition of derived functors to see that
$$ \mathrm{R}\Gamma(X, \mathbf{Z}/\ell^n\mathbf{Z}) \simeq \mathrm{R}\Gamma(\operatorname{Spec} K, \mathrm{R}f_*\mathbf{Z}/\ell^n\mathbf{Z}), $$ where $f\colon X\to \operatorname{Spec} K$ is the structure morphism. Now we can use the identification of 'etale cohomology of $\operatorname{Spec} K$ with continuous group cohomology of $G_K$ to rewrite the latter expression as follows: $$ \mathrm{R}\Gamma(\operatorname{Spec} K, \mathrm{R}f_*\mathbf{Z}/\ell^n\mathbf{Z}) \simeq \mathrm{R}\Gamma_{\mathrm{cont}}(G_K, \mathrm{R}\Gamma(X_{\overline{K}}, \mathbf{Z}/\ell^n\mathbf{Z})). $$

The complex $\mathrm{R}\Gamma(X_{\overline{K}}, \mathbf{Z}/\ell^n\mathbf{Z})$ has finite cohomology by Lemma 1. So Lemma 2 (and the Grothendieck spectral sequence) implies that $$ \mathrm{R}\Gamma(X, \mathbf{Z}/\ell^n\mathbf{Z}) \simeq \mathrm{R}\Gamma_{\mathrm{cont}}(G_K, \mathrm{R}\Gamma(X_{\overline{K}}, \mathbf{Z}/\ell^n\mathbf{Z})) \in D_{coh}(\mathbf{Z}/\ell^n\mathbf{Z}). $$

Theorem 4: Let $K$ be a finite extension of $\mathbf{Q}_p$ and $X$ a finite type $K$-scheme. Then $\mathrm{H}^i(X, \mathbf{Z}_\ell)$ is a finite $\mathbf{Z}_\ell$-module and the natural morphism $\mathrm{H}^i(X, \mathbf{Z}_\ell) \to \lim_n \mathrm{H}^i(X, \mathbf{Z}/\ell^n\mathbf{Z})$ is an isomorphism.

Proof: We start by proving the latter claim. Since $\mathrm{R}^i\lim=0$ for $i>1$, the Grothendieck's spectral sequence boils down to short exact sequences: $$ 0\to \mathrm{R}^1\lim_n \mathrm{H}^{i-1}(X, \mathbf{Z}/\ell^n\mathbf{Z}) \to \mathrm{H}^i(X, \mathbf{Z}_\ell)\to \lim_n \mathrm{H}^i(X, \mathbf{Z}/\ell^n\mathbf{Z})\to 0. $$

Corollary 3 implies that all groups $\mathrm{H}^{i-1}(X, \mathbf{Z}/\ell^n\mathbf{Z})$ are finite. So the $\mathrm{R}^1\lim$ term vanish by the Mittag-Leffler criterion. This already implies that $\mathrm{H}^i(X, \mathbf{Z}_\ell)\simeq \lim_n \mathrm{H}^i(X, \mathbf{Z}/\ell^n\mathbf{Z})$.

Now there are different ways to finish the proof. Here is the one I like.

Define $K_n$ to be the kernel of $\mathrm{H}^i(X, \mathbf{Z}_\ell)\to\mathrm{H}^i(X, \mathbf{Z}/\ell^n\mathbf{Z})$. An isomorphism $\mathrm{H}^i(X, \mathbf{Z}_\ell)\simeq \lim_n \mathrm{H}^i(X, \mathbf{Z}/\ell^n\mathbf{Z})$ implies that $\cap_n K_n=\{0\}$. On the other hand, a long exact sequence associated to a distinguished triangle (it requires little work to construct it) $$ \mathrm{R}\Gamma(X, \mathbf{Z}_\ell)\xrightarrow{\ell^n} \mathrm{R}\Gamma(X, \mathbf{Z}_\ell) \to \mathrm{R}\Gamma(X, \mathbf{Z}/\ell^n\mathbf{Z}) $$ implies that $K_n=\ell^n\mathrm{H}^i(X, \mathbf{Z}_\ell)$ and $\mathrm{H}^i(X, \mathbf{Z}_\ell)/\ell \subset \mathrm{H}^i(X, \mathbf{F}_\ell)$ is finite.

In particular, we conclude that $\mathrm{H}^i(X, \mathbf{Z}_\ell)$ is $\ell$-adically separated. But on the other hand, $\mathrm{R}\Gamma(X, \mathbf{Z}_\ell)$ is derived $\ell$-adically complete (as a derived limit of $\ell^n$-torsion complexes), and so its cohomology modules are derived complete as well. Therefore, Tag091T implies that $\mathrm{H}^i(X, \mathbf{Z}_\ell)$ is (classically) $\ell$-adically complete.

Finally, Nakayama's lemma and finiteness of $\mathrm{H}^i(X, \mathbf{Z}_\ell)/\ell$ imply finiteness of $\mathrm{H}^i(X, \mathbf{Z}_\ell)$.