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Let $k$ be a field of characteristic 0 and let $X$ be a $k$-variety.

For each prime $l$ invertible in $k$ we can associate to $X$ its $l$-adic étale cohomology $E_l(X)$ which is a graded-commutative $\mathbb{Q}_l$-algebra. It is known that some aspects of $E_l(X)$ do not depend on choice of $l$, e.g., the dimensions of the graded pieces.

Is it possible to view the collection $\{{E_l(X)}\}_l$ adelically somehow, say after completion, as completions of a $\mathbb{Q}$-algebra $E(X)$? Of the rational Chow ring of X perhaps?

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    $\begingroup$ Since you are assuming $char k=0$, there is no harm in assuming it embeds into $\mathbb{C}$ by shrinking $k$ if need be. Choose an embedding. Then $E(X)= H^*(X(\mathbb{ C}), \mathbb{Q})$ (singular cohomology) will do what you want by Artin's comparison theorem. $\endgroup$
    – Donu Arapura
    Commented Apr 14, 2019 at 17:36
  • $\begingroup$ @DonuArapura I think the funny part is when $\mathrm{char}\,k\neq0$. For good primes, is there still going to be a functorially constructed common vector space whose profinite completions give $l$-adic cohomology groups? $\endgroup$
    – user137767
    Commented Apr 14, 2019 at 17:51
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    $\begingroup$ @StepanBanach It's known than in positive char, there is no Weil cohomology with coefficients in $\mathbb{Q}$. I think that's what you're asking. $\endgroup$
    – Donu Arapura
    Commented Apr 14, 2019 at 20:44
  • $\begingroup$ @DonuArapura yes, you are right. I had certain reservations about the canonicity of a $\mathbb{Q}$-form of a $\mathbb{Q}_l$-vector space. I was ignorant of the fact that $\mathbb{Q}_l$ does not have non-trivial automorphisms Since this is true, functoriality for $l$-adic cohomology would imply functoriality for the putative $\mathbb{Q}$-vector space. The standard argument about supersingular elliptic curve then still applies. $\endgroup$
    – user137767
    Commented Apr 14, 2019 at 21:39

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It is not possible to define a $\mathbb{Q}$-cohomology underlying the $\ell$-adic etale cohomology over a finitely generated field $k$ of characteristic zero because the absolute Galois group of $k$ acts on the $\ell$-adic cohomology through an infinite, hence uncountable, quotient. On the other hand, it is possible to define an adelic cohomology: take etale cohomology with $\mathbb{Z}/N\mathbb{Z}$ coefficients, pass to the inverse limit over $N$, and tensor with $\mathbb{Q}$. This is done routinely.

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  • $\begingroup$ Do you know of an example in the literature which uses this construction? $\endgroup$
    – Vik78
    Commented Feb 26 at 7:41

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