Question

If you want to compute crystalline cohomology of a smooth proper variety $X$ over a perfect field $k$ of characteristic $p$, the first thing you might want to try is to lift $X$ to the Witt ring $W_k$ of $k$. If that succeeds, compute de Rham cohomology of the lift over $W_k$ instead, which in general will be much easier to do. Neglecting torsion, this de Rham cohomology is the same as the crystalline cohomology of $X$.

I would like to have an example at hand where this approach fails: Can you give an example for

A smooth proper variety $X$ over the finite field with $p$ elements, such that there is no smooth proper scheme of finite type over $\mathbb Z_p$ whose special fibre is $X$.

The reason why such examples have to exist is metamathematical: If there werent any, the pain one undergoes constructing crystalline cohomology would be unnecessary.

Answer 1

This paper of Serre gives an example (I've justed pasted I. Barsotti's math-sci review). (The paper can be found in Serre's "Collected Works vol. II 1960-1971)

Serre, Jean-Pierre Exemples de variétés projectives en caractéristique $p$ non relevables en caractéristique zéro. (French) Proc. Nat. Acad. Sci. U.S.A. 47 1961 108--109.

An example of a non-singular projective variety $X_0$, over an algebraically closed field $k$ of characteristic $p$, which is not the image, $\text{mod}\,p$, of any variety $X$ over a complete local ring of characteristic 0 with $k$ as residue field. The variety $X_0$ is obtained by selecting, in a 5-dimensional projective space $S$, and for $p>5$, a non-singular variety $Y_0$ which has no fixed point for an abelian finite subgroup $G$ with at least 5 generators of period $p$, of the group $\Pi(k)$ of projective transformations of $S$, but which is transformed into itself by $G$; then $X_0=Y_0/G$. The reason for the impossibility is that $\Pi(K)$, for a $K$ of characteristic 0, does not contain a subgroup isomorphic to $G$. {Misprint: on the last line on p. 108 one should read $s(\sigma)=\exp(h(\sigma)N)$.}