# Crystalline cohomology over arbitrary perfect ground fields

In preparing a set of notes, I realize there is a simple question for which I cannot find an easy answer; likely due to simple ignorance. Any introductory text on etale cohomology notes that supersingular elliptic curves with endomorphism giving a quaternion algebra provided a representation theoretic obstruction for a Weil cohomology theory valued in $\mathbf{Q}_p$ (or $\mathbf{Z}_p$). Although such curves can be defined over $\mathbf{F}_p$, their endomorphism algebra is defined over $\mathbf{F}_{p^2}$.

Crystalline cohomology is a Weil cohomology theory for smooth projective varieties over perfect fields, which a priori includes $\mathbf{F}_p$. This seems to suggest that when the ground field is $\mathbf{F}_p$, this is Weil cohomology theory taking values in $W(\mathbf{F}_p) \cong \mathbf{Z}_p$, appearing to violate the obstruction above. However, in those examples values should probably be in $W(\mathbf{F}_{p^2})$ instead. Grothendieck's Dix Exposes article "Crystals and the de Rham Cohomology of Schemes" states that coefficient ring of a $p$-adic cohomology should be in some extension of $\mathbf{Q}_p$ but is unclear what extensions are forced. The Wikipedia article also suggests to work with varieties over algebraically closed fields which again resolves the issue, but in all the surveys and notes I've seen, it seems to be vague on any limitations of the choice perfect ground field to guarantee a Weil cohomology theory.

Can someone point out what if any limits should be imposed on the ground fields or why someone super cleaver can't come up with other examples, like the stated elliptic curves, which limit the ground field?

• I don't think it's right to say that crystalline cohomology "violates the obstruction above". The obstruction is to a Weil cohomology theory with $\mathbb Z_p$-coefficients for varieties an algebraically closed field of characteristic $p$, or over any field containing $\mathbb F_{p^2}$. – Will Sawin Aug 24 '17 at 18:35
• The reason someone super clever can't come up with other examples that limit the ground field is precisely that crystalline cohomology exists and gives a Weil cohomology theory with $W(\mathbb F_q)$-coefficients for smooth projective varieties over $\mathbb F_q$. – Will Sawin Aug 24 '17 at 18:39
• Thanks @WillSawin, I modified the language. I understand that there is no obstruction. The question is how to explain the appearance of one as $\mathbf{F}_p$ is not prohibited. This at least raises the question of what ground fields are allowed for crystalline cohomology to give a Weil cohomology theory, which my cursory search has shown none, but this seems to suggest that it is a $\mathbf{Z}_p$-valued Weil cohomology theory for at least some varieties. – lemiller Aug 24 '17 at 18:52
• I meant above, of course, that the search showed no limitation on the ground field. – lemiller Aug 24 '17 at 19:07
• I think the literature shows, not just that there is no limitation on the ground field, but that crystalline cohomology is a Weil cohomology theory over every perfect ground field (valued in the ring of Witt vectors for that field). On the other hand, crystalline cohomology is not a $\mathbb Z_p$-valued cohomology theory for any varieties except those over $\mathbb F_p$. You're right that it's possible that there exists a $\mathbb Z_p$-valued cohomology theory for some class of varieties over some field $\mathbb F_{p^n}, n>1$, but it would be different from crystalline cohomology. – Will Sawin Aug 24 '17 at 19:16