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Let $X$ be a smooth projective variety over a finite field, with a closed immersion to some other smooth projective variety $S$, with $S$ liftable.

Suppose there exists an fpqc cover $S'\to S$, such that $X\times_SS'$ can be lifted to mixed characteristic.

Can $X$ be lifted to mixed characteristic?

For example, suppose $S = \mathbb{P}^n$, and $X\to S$ is a projective embedding. Take $S^{\rm perf}$, the perfection of $S$ (inverse limit along the absolute Frobenius map). Then $S' := S^{\rm perf}\to S$ is an fpqc cover.

Say $X_{S'}$ is liftable to mixed characteristic. Is $X$? Equivalently, do obstructions to liftability of $X$ give obstructions to liftability of $X_{S'}$?

My guess is that the answer is no. If anyone sees an obstruction to liftability of $X_{S'}$ coming from non-liftability of $X$, that is what I am looking for.

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    $\begingroup$ No; take $X=S$ to be any non-liftable smooth variety and $S'$ any affine cover of $X$. Smooth affines always lift, so this gives a counterexample. $\endgroup$
    – Daniel Litt
    Commented May 15, 2018 at 2:29
  • $\begingroup$ @DanielLitt $S$ is assumed to be liftable. I forgot this in the general question, but this is clear in the example, since $\mathbb{P}^n$ is liftable $\endgroup$
    – user124171
    Commented May 15, 2018 at 2:34
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    $\begingroup$ Ok, then take X any smooth non-liftable projective variety, S to be projective space, and S’ any affine cover of projective space... $\endgroup$
    – Daniel Litt
    Commented May 15, 2018 at 2:36
  • $\begingroup$ @DanielLitt How does this give a counterexample to the OP's question about the choice $S' = (\mathbf{P}^n)^{perf}\to S$? I agree yours gives a counterexample to the more general question. $\endgroup$
    – user95222
    Commented May 15, 2018 at 2:46
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    $\begingroup$ The map $X^{\rm{pf}}\to X_{S^{\rm{pf}}}$ over $S^{\rm{pf}}$ is not an isomorphism when $X\ne S$ (assume $X$ and $S$ are connected, so irreducible): $X^{\rm{pf}}$ is reduced but the closed subscheme $X\times_S S^{\rm{pf}}$ of $S^{\rm{pf}}$ is non-reduced when $X\ne S$. Indeed, since $F_S$ is faithfully flat it suffices to show $X\times_{S,F_S} S$ is non-reduced if $X\ne S$. This is finite flat radiciel of degree $p^{\dim S}$ over $X$, and if reduced it would (by irreducibility) be integral with function field inside $k(X)^{1/p}$, so $p^{\dim X}= [k(X)^{1/p}:k(X)]\ge p^{\dim S}$, contradiction. $\endgroup$
    – nfdc23
    Commented May 15, 2018 at 5:14

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