Given a smooth projective variety $X / \mathbb F_p$, can one find a smooth projective $\mathcal X / \mathbb Z_p$ such that $\mathcal X \times_{\mathbb Z_p} \mathbb F_p = X$? (or similarly with $\mathbb Z$ instead of $\mathbb Z_p$).
Thank you!or
[Math] Lift smooth projective varieties over $\mathbb F_p$ to $\mathbb Z_p$
ag.algebraic-geometry
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EDIT 7/15/14 I was just looking back at this old answer, and I don't think I ever answered the stated question. I can't delete an accepted answer, but I'll point at that, as far as I can tell, the Vakil reference I give also only address the question of deforming $X$ over $\mathbb{Z}_p$, not of embedding it in some larger flat family over $\mathbb{Z}_p$.
EDIT Oops! David Brown points out below that I misread the question. I was answering the question of finding a smooth scheme which does not deform in a smooth family over Z_p.
Well, to make up for that, I'll point to some references which definitely contain answers. Look at section 2.3 of Ravi Vakil's paper Murphy's Law in algebraic geometry: Badly-behaved deformation spaces for some history, and several good references. Moreover, Ravi describes how to build an explicit cover of P^2 in characteristic p which does not deform to characteristic 0. Basically, the idea is to take a collection of lines in P^2 which doesn't deform to characteristic 0 and take a branched cover over those lines. For example, you could take that p^2+p+1 lines that have coefficients in F_p.
Picard number $1$ is the most frequent case among all varieties, so you cannot expect a classification. It's quite the opposite, you might stand a chance to classify those (within some class) that have Picard number larger than $1$. For instance a general $K3$ surface has Picard number $1$ and the locus of those with a given Picard number becomes smaller as the Picard number increases.
If the Picard number is larger than $1$, that usually means that the variety admits some non-trivial maps which gives you a handle on them or a starting point if you will. If the Picard number is $1$, it is hard to get any traction to get some way to study the object.
On the other hand that means that you can get lots of examples with Picard number $1$. Chances are, if you choose a variety at random it will have Picard number $1$. You can get lots of examples that are not complete intersections by the simple observation that for a complete intersection of dimension $d$, the middle cohomology groups of the structure sheaf vanish, that is, $H^i(X,\mathscr O_X)=0$ for $0<i<d$. This is actually another way to see that an abelian variety of dimension at least $2$ cannot be a complete intersection.
In particular, you can find lots of examples among surfaces. Surfaces with $H^1(X,\mathscr O_X)\neq 0$ are known as irregular, so any irregular surface with Picard number $1$ gives and example that you want. One way to ensure that the Picard number is $1$ is to make sure that $\mathrm{rk}\, H^2(X,\mathbb{Z})=1$. In other words, any surface with $q \ne 0$ and $b_2=1$ gives you an example.
Of course, you would want an explicit example. Unfortunately, I can't think of one at the moment, but I am fairly certain, that a general surface with $H^1(X,\mathscr O_X)\neq 0$ has Picard number one, so that should give you plenty of examples. In fact, one could argue that that's why I can't give an explicit one, because they are the general ones (and anything explicit is not general).
Best Answer
Of course not. The only smooth projective curve over $\mathbf{Z}$ is $\mathbf{P}_1$, and for every prime $p$ there is a smooth projective curve over $\mathbf{F}_p$ other than $\mathbf{P}_1$.
Addendum A bit of googling brings up the overview of Serre's work written on the occasion of his getting the Abel prize. It is mentioned there that recently (2005) Serre has improved his 1961 result mentioned in the comment below
Reference I've come across these notes by Yi Ouyang of a course by Luc Illusie on Topics in Algebraic Geometry ; the final section deals with Serre's example.