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Let $X$ be a smooth proper geometrically integral scheme over $\overline{\mathbb F_p}$. Assume $X$ is the specialization of a smooth proper scheme over $\mathbb Z_p^{nr}$. Let $L$ be an ample line bundle on $X$.

I would like to show that $L^{\otimes p}$ lifts to characteristic zero. However, the obstruction to lifting $L$ lives in $\mathrm{H}^2(X,\mathcal O_X)$ which might be non-zero.

On the other hand, the obstruction space is an $\overline{\mathbb{F}_p}$-vector space. So the obstruction vanishes after multiplying with $p$.

Does this imply that $L^{\otimes p}$ lifts?

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    $\begingroup$ There are separate obstructions for lifting mod $p^2$, $p^3$ etc. So even if you can do the first lift, you could get stuck at the $p^2$ step, where the obstruction space is only a $\mathbb{Z}/p^2$ module. $\endgroup$
    – Donu Arapura
    Commented Feb 19, 2017 at 19:58
  • $\begingroup$ @DonuArapura Right! I forgot that. I'm still a bit confused though. Why is it then that $L$ will lift if $\mathrm{H}^2(X,\mathcal{O}_X)=0$? Your comment seems to suggest that the vanishing of this $H^2$ a priori only implies the liftability mod $p$. Am I missing something? $\endgroup$
    – George
    Commented Feb 19, 2017 at 20:06
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    $\begingroup$ This is described very clearly in any reference on infinitesimal deformation theory, e.g., the beginning of Koll'ar's "Rational Curves on Algebraic Varieties", Mike Artin's TIFR book on "Deformations of Singularities", etc. Deeper are Illusie's books on the cotangent complex. The short answer is that the functor is pro-represented by a complete local ring $W[[t_1,\dots,t_n]]/\langle f_1,\dots,f_r \rangle$ where $n$ equals $h^1(X,\mathcal{O}_X)$ and $r$ equals $h^2(X,\mathcal{O}_X)$. If $r$ equals $1$, for instance, $f_1$ might be $p^d$. $\endgroup$
    – Jason Starr
    Commented Feb 19, 2017 at 20:38
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    $\begingroup$ Another way to say this: if you managed to lift $L$ modulo $p^n$, then the obstruction to lifting modulo $p^{n+1}$ lie in $H^2(X,O_X)$. Just because you know that the obstruction is zero on a particular step, you don't know whether it will vanish on the next. On the other hand, if $H^2(X,O_X)=0$, all of the obstructions vanish. (Sorry Jason, did not mean to interrupt --- I did not realize you were about to add more.) $\endgroup$
    – t3suji
    Commented Feb 19, 2017 at 20:43
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    $\begingroup$ . . . So if $h^2(X,\mathcal{O}_X)$ is positive, then the obstruction to lifting modulo $p$, $p^2$, etc. might all vanish until we reach $p^d$. On the other hand, if $h^2(X,\mathcal{O}_X)$ equals $0$, then for every $d\geq 0$, for every ring homomorphism $W[[t_1,\dots,t_n]]\to W/p^dW$, this homomorphism lifts to $W[[t_1,\dots,t_n]]\to W$. $\endgroup$
    – Jason Starr
    Commented Feb 19, 2017 at 20:44

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This cannot always be done. If $X$ is a supersingular K3 surface then there are $22$ ample line bundles $L$ that give independent classes; if every $L^{\otimes p}$ lifted to char. zero, then you'd have a K3 surface in char. zero with $22$ independent line bundles, which is impossible. (Or $X$ could be any liftable unirational surface with $p_g(X)>0$.)

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