I have a basic question concerning comparison of different cohomology theories. Let $X$ be a projective smooth (or just proper smooth) variety over a separably closed field $k$ of characteristic $p,$ which is not liftable to characteristic zero. Is there any relation between the de Rham cohomology $H^n(X,\Omega^{\bullet}_X)$ and the $\ell$adic cohomology? For example, do they have the same dimension (over $k$ and $Q_l$ resp.)?
2 Answers
I believe the answer is no, that these two spaces need not have the same vector space dimension. Grothendieck here cites an example of Serre in a footnote on the last page; unfortunately, I don't have access to Serre's original paper at the moment.

$\begingroup$ I don't know how my link became mangled, but ignore the junk string before http. $\endgroup$ Commented Dec 1, 2010 at 23:44

1$\begingroup$ I believe there's supposed to be a double underscore surrounding 29 in the link (which I guess is why it appears italicized). (The paper is "On the de Rham cohomology of algebraic varieties"). Here's a fixed version: numdam.org/numdambin/item?id=PMIHES_1966__29__95_0 $\endgroup$– B RCommented Dec 2, 2010 at 0:47

2$\begingroup$ The most classical type examples are nonordinary Enriques surfaces in characteristic 2. $\endgroup$ Commented Dec 2, 2010 at 15:10
Often, yes. What always has the same dimension as $H^n_{et}(X,Q_l)$ is the rational crystalline cohomology $H^n_{cr}(X)\otimes K$ with coefficients in the fraction field $K$ of the Witt vectors $W$ of $k$. $H^n_{cr}(X)$ itself will have coefficients in $W$, and of course, have rank equal to the dimension of $H^n_{cr}(X)\otimes K$. But it might have torsion in general. On the other hand, there is an exact sequence $$0\rightarrow H^n_{cr}(X)\otimes_W k\rightarrow H^n(X,\Omega_X^{\cdot})\rightarrow H^{n+1}_{cr}(X)[p]\rightarrow 0$$ as in the universal coefficient theorem. This is because crystalline cohomology can be taken with any of the torsion coefficients $W/p^n$, and when you take it with coefficients in $W/p=k$, you get exactly De Rham cohomology. (One of the most important things to learn at the beginning about crystalline cohomology with $W/p^n$ coefficients is that it can be computed using the divided power De Rham complex associated to a smooth embedding over $W/p^n$, which reduces to the De Rham complex of $X$ itself when the coefficients are $W/p$.)
So you will get the same dimensions you want if enough of crystalline cohomology is torsionfree. All this is explained in introductory books, such as the one by Berthelot and Ogus, except the comparison with \'etale cohomology. That is perhaps explained in a paper by Katz and Messing from the 70's.

4$\begingroup$ If one also wants the proper case one needs to use de Jong's alterations on top of KatzMessing. $\endgroup$ Commented Dec 2, 2010 at 15:08