When $X$ is a smooth proper variety over $\mathbb F_q$, we know by Deligne's theory of weights that the dimension of $H^i_{\operatorname{\acute et}}(\bar X, \mathbb Q_\ell)$ does not depend on $\ell$. In fact, we even get that Frobenius acts with the same characteristic polynomial.

Do we have a similar theorem when $X$ is no longer smooth and proper (e.g. $X$ smooth affine, or proper but singular)? How about if we assume resolution of singularities? (We could also assume other things like the Tate conjecture or a suitable subset of the standard conjectures, but that feels a bit like cheating.)

A related question: in the smooth proper case, do we know that there is some sort of canonical isomorphism $$H^i(\bar X,\mathbb Q_\ell) \otimes_{\mathbb Q_\ell} A \stackrel \sim \to H^i(\bar X,\mathbb Q_{\ell'}) \otimes_{\mathbb Q_{\ell'}} A$$ for some big 'period' ring $A$ containing $\mathbb Q_\ell$ and $\mathbb Q_{\ell'}$?

Remark. (Also remarked by Ben Webster) In characteristic $0$, the result is known by comparison to the singular cohomology of the associated analytic space. However, in positive characteristic there cannot exist a $\mathbb Q$-valued Weil cohomology theory to compare with. In fact, there does not even exist an $\mathbb R$-valued or a $\mathbb Q_p$-valued Weil cohomology theory. The argument is outlined in 2.2 of these notes, and is attributed to Serre.

An obvious workaround would be to lift everything to characteristic $0$ and use comparison there, but again no luck: Serre gave an example of a smooth projective variety that does not lift to characteristic $0$.

  • $\begingroup$ There's a pretty easy way to make a universal period ring, which is to take the ring generated by, for each variety, the $n \times n$ matrices over $\mathbb Q_{\ell} \otimes \mathbb Q_{\ell'}$ and add whatever relations you want. This construction is not completely crazy - a variant of it is supposed to get you the complex period ring, for instance. So the question becomes whether your relations force the ring to be $0$. I think depending on what relations you demand, this will be equivalent to some set of standard conjectures. $\endgroup$
    – Will Sawin
    Mar 18, 2016 at 5:07
  • $\begingroup$ For instance, if you demand cycle classes are preserved, you require that every cycle that is $\ell$-adically nomologically trivial is $\ell'$-adically homologically trivial, which as far as I know is not known. $\endgroup$
    – Will Sawin
    Mar 18, 2016 at 5:08
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    $\begingroup$ In "Miscellany on traces in l-adic cohomology: a survey" Illusie writes that for arbitrary varieties the $l$-independence of Betti numbers is unknown. Though, he mentions that over finite field the Euler characterstic defined via cohomology with compact support is $l$-independent because it is equal to the degree of the zeta-function. And he refers to G. Laumon, "Comparaison de caracteristiques d’Euler-Poincare en cohomologie l-adique" for the equality between compact support Euler characteristic and usual Euler characteristic. $\endgroup$
    – SashaP
    Mar 18, 2016 at 6:53
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    $\begingroup$ I am not certain how lifting to characteristic $0$ is supposed to help. Certainly the affine scheme $\mathbb{A}^1_{\mathbb{F}_p} = \text{Spec}\ \mathbb{F}_p[t]$ lifts to characteristic $0$. The issue is that it lifts both as $\mathbb{A}^1_{\mathbb{Z}_p}= \text{Spec}\ \mathbb{Z}_p[t]$ and as, say, $\text{Spec}\ \mathbb{Z}_p[t,u]/\langle u(pt-1)+1 \rangle$. The generic fibers of these two lifts have different cohomology. $\endgroup$ Mar 18, 2016 at 13:19
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    $\begingroup$ @JasonStarr: Ah, of course. I forgot that for the smooth and proper base change theorems to hold, the morphism needs to be, you know, smooth and proper. $\endgroup$ Mar 18, 2016 at 17:37

1 Answer 1


If you looks at Milne's notes, Chapter 21, you'll see that the comparison theorem with the singular cohomology of the $\mathbb{C}$-points has no dependence on properness; there is a smoothness hypothesis there, but Milne says it can be removed. Since independence of $\ell$ is true for singular cohomology, it must hold for etale cohomology as well.

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    $\begingroup$ So this is how you prove it in characteristic $0$; I should probably have added this to my post. However, in characteristic $p$ there is no equivalent of singular cohomology. In fact, it is known that there cannot exist an $\mathbb Q$-valued Weil cohomology theory in characteristic $p$. Also, not every variety can be lifted to characteristic $0$, so you cannot use the complex comparison theorem to deduce the characteristic $p$ one. $\endgroup$ Mar 18, 2016 at 4:52
  • $\begingroup$ in the singular case, I am not sure it is necessarily true that singular cohomology determines the etale cohomology. Could you clarify? $\endgroup$
    – user141498
    Jun 10, 2019 at 14:16

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