Let $X$ be a smooth projective variety over a finite field $\mathbb{F}_p$. What are the conditions for the existence of a projective variety $X'$ over $\mathbb{Q}_p$ such that $X$ is a special fiber of $X'$?
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$\begingroup$ Why the downvote? $\endgroup$– Yemon ChoiCommented Mar 15, 2013 at 7:10
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$\begingroup$ I suspect it is related to this: tea.mathoverflow.net/discussion/1552/string-of-down-votes $\endgroup$– JRNCommented Mar 15, 2013 at 7:40
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$\begingroup$ This might be standard, but a variety over $\mathbb{Q}_p$ with "special fiber ..." just means a scheme $\mathfrak{X}/\mathbb{Z}_p$ whose generic fiber is $X'$ and special fiber is $X$ right? There are lots of examples of when this can't happen. Sometimes they formally lift, but aren't projective. Sometimes the deformation theory is highly obstructed. I'm not sure there can ever be some general conditions on $X$ that would guarantee a lift. Are you interested in a more specific type of variety or at least of a particular dimension. $\endgroup$– MattCommented Mar 15, 2013 at 15:52
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$\begingroup$ @Matt: I am interested in curves and surfaces. We can assume that $\dim X'= \dim X+1$. $\endgroup$– Naga VenkataCommented Mar 15, 2013 at 19:40
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$\begingroup$ Alright, then I think you mean that $X'$ should be over $\mathbb{Z}_p$ otherwise the dimensions won't make any sense. Every smooth projective curve always lifts to characteristic $0$ because the obstructions to deforming both the curve and an ample line bundle lie in $H^2$ which vanishes since it is a curve. So by Grothendieck's Existence Theorem the formal lift algebraizes. Surfaces are more delicate. There are known results like every K3 surface lifts to characteristic $0$, but there are also known surfaces that do not lift. $\endgroup$– MattCommented Mar 15, 2013 at 23:11
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