comparison of de Rham cohomology and etale cohomology

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.)?

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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.

http://www.numdam.org/item?id=PMIHES_1966_29_95_0

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I don't know how my link became mangled, but ignore the junk string before http. –  Hunter Brooks Dec 1 '10 at 23:44
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/numdam-bin/item?id=PMIHES_1966__29__95_0 –  B R Dec 2 '10 at 0:47
The most classical type examples are non-ordinary Enriques surfaces in characteristic 2. –  Torsten Ekedahl Dec 2 '10 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 $W/p$.)