# Which interesting characterestic zero field $E$ (e.g a pseudofinite field) can support a Weil cohomology?

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 $$C$$ ?

• 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). – schematic_boi May 11 at 8:11
• @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 $\mathbf{F}_p$ – John P. May 11 at 15:45
• 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. – John P. May 11 at 15:45
• @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). – zzy May 11 at 15:48