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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

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Duplicate:… –  Chandan Singh Dalawat May 8 '12 at 12:08
Here is a modified question which comes up from time to time, and which is open to the best of my knowledge. Trivially a smooth projective variety lifts over the Witt vectors if you allow yourself to remove a sufficiently ample Cartier divisor, or equivalently, if you allow yourself to modify your original projective scheme to a new projective scheme (which agrees with the original scheme on a nonempty open). Is the same true for modifications in codimension 2? –  Jason Starr May 8 '12 at 13:01
Dear Jakob, There is a celebrated paper of Deligne and Illusie (easily found on MathSciNet) which discusses obstructions to lifting from $\mathbb F_p$ to $\mathbb Z_p/p^2$. Regards, –  Emerton May 11 '12 at 3:55
Dear Matt, you have become so much of a $p$-adician that you think of $\mathbf{Z}/p^2$ as $\mathbf{Z}_p/p^2$. –  Chandan Singh Dalawat May 11 '12 at 4:05

3 Answers 3

up vote 9 down vote accepted

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.

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Maybe I'm being dense here, but why is $\mathbb{P}^1$ the only smooth projective curve over $\mathbb{Z}$? –  Akhil Mathew May 11 '12 at 0:03
(Ah, someone just pointed me to a result of Fontane discussed at…) –  Akhil Mathew May 11 '12 at 0:47
The theorem of Fontaine (and Abrashkin) says that for $n>0$, there is no abelian scheme of dimension $n$ over $\mathrm{Spec}(\mathbf{Z})$. The case $n=1$ (elliptic curves) was known earlier and is due to Tate, I believe. –  Chandan Singh Dalawat May 11 '12 at 2:59
I fail to see how this answers the title question. the ring of integers is not the ring of $p$-adic integers. –  Joël May 17 '12 at 13:51

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 $\mathbb{Z}_{p}$.

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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 .

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See also Ravi's earlier answer… –  Chandan Singh Dalawat May 11 '12 at 3:58

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