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Let $ \mathbb{F}_{q} $ be a finite field of characteristic $ p $ and $ \overline{\mathbb{F}_{q}} $ be an algebraic closure of $ \mathbb{F}_{q} $. Let $ X $ be a smooth projective variety over $ \mathbb{F}_{q} $. Let $ \ell\neq p $ be a prime and let $ \mathbb{F}_{\ell} $ be a finite field of order $ \ell $. Deligne's integrality theorem states that the eigenvalues of the geometric Frobenius acting on $ \ell $-adic cohomology $ H^{i}(X\times_{\mathbb{F}_{q}}\overline{\mathbb{F}_{q}},\mathbb{Q}_{\ell}) $ are algebraic integers, cf. Corollarie 5.5.3 in

[Deligne, P.:Expos´e XXI in SGA 7, Lect. Notes Math. vol. 340, 363-400, Berlin Heidelberg New York Springer 1973].

We can also consider theory of etale sheaves with $ \mathbb{F}_{\ell}((t)) $-coefficients, which should be parallel to the theory of $ \ell $-adic sheaves. Namely, we first consider etale sheaves of $ \mathbb{F}_{\ell}[t]/(t^{r}) $-modules, and then define the category of $ \mathbb{F}_{\ell}[[t]] $-sheaves as the appropriate $ 2 $-limit. Finally, we define the category of $ \mathbb{F}_{\ell}((t)) $ by inverting $ t $. In this setting, we may ask the following question in the spirit of the above Deligne's integrality theorem:

Is it true that the eigenvalues of the geometric Frobenius actiong $ \mathbb{F}_{\ell}((t)) $-adic cohomology $H^{i}(X\times_{\mathbb{F}_{q}}\overline{\mathbb{F}_{q}},\mathbb{F}_{\ell}((t))) $ are algebraic over the prime field of $ \mathbb{F}_{\ell}((t)) $, i.e. they are in the finite subfield $ \mathbb{F}_{\ell}$ of $ \mathbb{F}_{\ell}((t)) $?

I'm not sure if the question I'm describing is in the correct formulation. Any comments and references would be highly appreciated.

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    $\begingroup$ Isn’t the cohomology group in question the same as $H^i(X\otimes_{\mathbb{F}_q}\overline{\mathbb{F}}_q, \mathbb{F}_\ell)\otimes_{\mathbb{F}_\ell}\mathbb{F}_\ell((t))$? $\endgroup$
    – Piotr Achinger
    Commented May 12, 2022 at 9:53
  • $\begingroup$ @PiotrAchinger By definition, we have $$H^{i}(X\times_{\mathbb{F}_{q}}\overline{\mathbb{F}_{q}},\mathbb{F}_{\ell}((t))):=\left( \varprojlim_{r}H^{i}_{\text{\'et}}(X\times_{\mathbb{F}_{q}}\overline{\mathbb{F}_{q}},\mathbb{F}_{\ell}[t]/(t^{r}))\right) \otimes_{\mathbb{F}_{\ell}[[t]]}\mathbb{F}_{\ell}((t)) $$, why are they the same? $\endgroup$
    – Nobody
    Commented May 12, 2022 at 11:58
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    $\begingroup$ This follows from the claim that $H^i (X_{\overline{\mathbb F_q}}, \mathbb F_\ell[t]/(t^r) ) = H^i (X_{\overline{\mathbb F_q}}, \mathbb F_\ell) \otimes_{\mathbb F_\ell} \mathbb F_\ell [t]/(t^r) $ which unraveled is just the statement that cohomology commutes with finite direct sums of coefficients, or, maybe transparently, tensor products with flat modules over the coefficient ring. $\endgroup$
    – Will Sawin
    Commented May 12, 2022 at 12:01
  • $\begingroup$ Now I get the point, thanks! $\endgroup$
    – Nobody
    Commented May 12, 2022 at 12:25

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