Lift smooth projective varieties over $\mathbb F_p$ to $\mathbb Z_p$?

Question

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

Answer 1

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

by showing that if the variety can be lifted (as a flat scheme) to a local ring $A$, then $p.A=0$. The basic idea consists in transposing the problem to the context of finite groups.

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.

Answer 2

Follow-up to Chandan Singh Dalawat's excellent answer: smooth curves over $\mathbb{F}_p$ can always be lifted to over $\mathbb{Z}_p$. But there are smooth projective surfaces over $\mathbb{Z}/p^n$ that cannot be lifted to $\mathbb{Z}_p$, see Link .

Answer 3

It is not an easy task to give necessary and sufficient conditions on $X$ to lift it to $\mathbb{Z}_{p}$. As additional references you can check "Finite Group Schemes, local moduli for abelian varieties, and lifting problems" by Oort, F. and "Deformations and liftings of finite commutative group schemes" by Oort F. and Mumford, D.

In the case where $X$ is an ordinary abelian variety the answer is affirmative, and indeed all such possible lifts are classified by Serre-Tate theory on deformations. You can find a very clear proof and interpretation of this theory in "Serre-Tate local moduli" by Katz, M.

Let me try to summarize it:

Take an ordinary abelian variety $X$ over $k=\mathbb{F}_{p}$. Then for each $n$ the subgroup scheme $X[p^{n}]$ has a unique (up to a unique isomorphism) connected-etale seuqence

$0 \rightarrow X[p^{n}]^{0} \rightarrow X[p^{n}] \rightarrow X[p^{n}]^{et} \rightarrow 0$ (1)

where both $X[p^{n}]^{0}$ and $X[p^{n}]^{et}$ have order $p^{nd}$ ($d$ is the dimension of $X$). Indeed since $k$ is perfect this sequence splits. Now take an Artin local ring $A$ with residue field $k$ (in your case take the ring $W_{m}(k)$=$\mathbb{Z}/p^{m}\mathbb{Z}$). Since the kernel of the map $A \rightarrow k$ consists of nilpotent elements, there is a unique (up to an isomorphism) etale group scheme $Y_{n}$ over $A$ such that $Y_{n} \otimes_{A} k = X[p^{n}]^{et}$ (This is one of Grothendieck's theorems). Now consdier the dual abelian variety $\hat{X}$ of $X$ and do the same thing for it. The theory of Cartier dulaity implies that $\hat{X}[p^{n}]^{et}$ and $X[p^{n}]^{0}$ are duals of each other. So this means that there is a unique connected subgroup scheme $Z_{n}$ over $A$ such that $Z_{n} \otimes_{A} k=X[p^{n}]^{0}$. Now take any extension of $Y_{n}$ by $Z_{n}$, i.e. a subgroup scheme $G_{n}$ over $A$ given with an exact sequence

$0 \rightarrow Z_{n} \rightarrow G_{n} \rightarrow Z_{n} \rightarrow 0$ (2)

This sequence lifts (1). But now we can find such extension for all $n$, i.e. we have lifted the connected-etale sequence of $p$-divisible groups

$0 \rightarrow X[p^{\infty}]^{0} \rightarrow X[p^{\infty}] \rightarrow X[p^{n}]^{\infty} \rightarrow 0$

over $k$ to a similar sequence over $A$. Now the "general" Serre-Tate theorem implies that there is a unique abelian variety $\mathbb{X}$ over $A$ which lifts $X/k$ and has the chosen $p$-divisible group $(G_{n})^{n}$. So we identified the liftings of $X/k$ to $A$ with the group of the extensions of $Y_{\infty} = (Y_{n})^{n}$ by $Z_{\infty} = (Z_{n})^{n}$.

Now let $A=A_{m}$$=W_{m}(k)$ and $G=G_{m}$. We can repeat this process for all $m$ in a compatible way, i.e. if $r > m$ we can choose $G_{r}$ in a such a way that $G_{r} \otimes_{A_{r}}$ $A_{m} =G_{m}$ because Grothendieck's theorem, Cartier duality and Serre-Tate theorem does not require a field or an Artin ring (or even a Noetherian ring) as a base. Now take the limit (!) of these abelian varieties to get a an abelian variety over $\mathbb{Z}_{p}$.