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Let's consider the category of smooth projective varieties over a fixed characteristic $p>0$ algebraically closed field $k$. For a Weil cohomology theory with coefficient field $E$, by definition it shall satisfy finiteness property, Poincare duality, Kunneth formula, existence of cycle maps, weak and hard Lefschetz theorem.

Consider the class $\mathcal {C}$ of characterestic zero fields $E$ that support a Weil cohomology. Not all fields of zero characteristic can be a coeffiecient field, as is discussed in Serre’s example of a supersingular elliptic curve over $k$. We know

  • $\mathbb R, \mathbb Q_p \notin \mathcal{C}$.
  • For $E_1 \hookrightarrow E_2$, $E_1 \in \mathcal {C} \Rightarrow E_2 \in \mathcal {C}$ .

  • $\mathbb Q_l \in \mathcal {C}$ $(l \not = p)$ .

  • $W(k)[1/p] \in \mathcal {C}$.

More interestingly, we also know some pseudofinite field lie in $\mathcal C$, namely the ultraproduct of $\mathbb F_l$ $(l \not =p)$ using a non-principal ultrafilter, see "a new Weil cohomology theory" by Ivan Tomasic. And there is a dicussion of Weil II for such Weil cohomology theory, see https://webusers.imj-prg.fr/~anna.cadoret/Weil2Ultra_OberwolfachReports.pdf

So my question is, can we describe other interesting fields in $C$ ?

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  • $\begingroup$ a minor remark not directly relevant to the question: if you are over the prime field of characteristic $p>0$ (so not an algebraically closed field), then crystalline cohomology is a Weil cohomology with $\mathbb{Q}_p$-coefficients. There was an answer by SashaP on MO explaining why Serre's objection does not apply (and also a comment by Will Sawin, I think). $\endgroup$ – schematic_boi May 11 at 8:11
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    $\begingroup$ @zzy No. The comment says “if you are over the prime field of char $p>0$”. The prime field of char $p>0$ is $\mathbf{F}_p$, and $W(\mathbf{F}_p)=\mathbf{Z}_p$, so $W(\mathbf{F}_p)[1/p]=\mathbf{Q}_p$. The comment only says that if $X$ is defined over $\mathbf{F}_p$, then crystalline cohomology takes value in finite dimensional $\mathbf{Q}_p$-vector spaces as a Weil cohomology. There’s no inconsistency with Serre’s objection, since supersingular elliptic curves have lots of endomorphisms that are not defined over $\mathbf{F}_p$ and the ring of those that are defined over $\mathbf{F}_p$ $\endgroup$ – John P. May 11 at 15:45
  • $\begingroup$ is usually much smaller than an order in a quaternion algebra, and so can act on a $2$-dimensional $\mathbf{Q}_p$-vector space. Of course, if $E$ is such a curve, defined over $\mathbf{F}_p$, and you call $E’$ its base change to $\mathbf{F}_{p^2}$ and $K := W(\mathbf{F}_{p^2})[1/p]$, then $\text{End}_{\mathbf{F}_{p^2}}(E’)$ is indeed an order in a quaternion algebra (nonsplit at $p$ and $\infty$) and so it cannot act on $\mathbf{Q}_p^{\oplus 2}$. On the other hand, $H^1_{\rm crys}(E’) = K^{\oplus 2}$, so no problem. $\endgroup$ – John P. May 11 at 15:45
  • $\begingroup$ @JohnP OK, thank you for this remark, I am a little careless... Anyway, my interest lies in algebraically closed fields, because it's more geometric (we have all endomorphisms). $\endgroup$ – zzy May 11 at 15:48

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