Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $X'$?

share|improve this question
Why the downvote? –  Yemon Choi Mar 15 '13 at 7:10
I suspect it is related to this: tea.mathoverflow.net/discussion/1552/string-of-down-votes –  Joel Reyes Noche Mar 15 '13 at 7:40
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. –  Matt Mar 15 '13 at 15:52
@Matt: I am interested in curves and surfaces. We can assume that $\dim X'= \dim X+1$. –  Naga Venkata Mar 15 '13 at 19:40
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. –  Matt Mar 15 '13 at 23:11
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.