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Let $k$ be a finite field, $X_0$ be a smooth affine variety over $k$ and $C\rightarrow X$ a smooth projective family of curves of genus $\geq 2$.

By a result of Elkik we can always lift $X_0$ to a smooth affine scheme over $W(k)$.

It seems that, up to replace $X_0$ with an open subset, one can always lift $C_0\rightarrow X_0$ to a smooth projective family of curves $C\rightarrow X$.

Does anyone have a reference for this fact?

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    $\begingroup$ I'm pretty sure this is not always possible. Moret-Bailly constructed [MB,Exp. No. 8] a non-isotrivial family of semistable genus $2$ curves of compact type over $\mathbb P^1$, which is smooth over an affine open $U \subseteq \mathbb P^1$. But such a thing should not be liftable, because there are no non-isotrivial families of compact type curves over $\mathbb P^1$ in characteristic $0$. $\endgroup$ Nov 28, 2018 at 20:35
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    $\begingroup$ @R.vanDobbendeBruyn. It is always possible. It follows quite easily from smoothness of the stack $\mathcal{M}_g$ and the construction of "Chow forms", cf. Corollary 3.15 of the following (it takes a while to unpack the definitions in that section): arxiv.org/pdf/1703.08334v1.pdf $\endgroup$ Nov 28, 2018 at 20:43
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    $\begingroup$ For a locally closed immersion of $X_0$ in $\mathbb{A}^n_k$, consider the diagonal morphism from $X_0$ to $\mathbb{A}^n\times \mathcal{M}_g$. By Chow's Lemma for stacks (roughly), there exists a smooth morphism $M\to \mathcal{M}_g$ such that the generic point of $X_0$ lifts to $\mathbb{A}^n\times M$. Up to shrinking $X_0$, assume $X_0$ is a locally closed subscheme of $\mathbb{A}^n\times M$. Now realize $X_0$ as an open in a complete intersection, and deform the coefficients of the defining equations. Lifting affine varieties is much easier than lifting projective varieties. $\endgroup$ Nov 28, 2018 at 20:49
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    $\begingroup$ @Jason Starr Is the following correct then: Let $A\rightarrow \mathbb{P}^1$ be the MB family over $\mathbb{F}$, i.e., the Jacobian of the family Remy mentioned. Then there exists an open $U\subset \mathbb{P}^1$ such that the pair $A_U\rightarrow U$ lifts to $\mathcal{A_U}\rightarrow \mathcal{U}$ over $W(\mathbb{F})$? The reason I find this surprising is that the lifted pair will have to have bad reduction at some points at $\infty$, but when reduced mod $p$ the reduction is everywhere good. $\endgroup$
    – Raju
    Nov 28, 2018 at 21:33
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    $\begingroup$ @Raju. If you form a projective model of the lift of the family over $U$, then the closed fiber is reducible, and the Moret-Bailly pencil is one irreducible component. The closed fiber does intersect the boundary of the moduli space, but that intersection is disjoint from the irreducible component of the Moret-Bailly pencil. $\endgroup$ Nov 29, 2018 at 8:09

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I am just posting my comment as an answer. This type of lifting is "classical". In the article linked above with Chenyang Xu, we needed bounds on the lift, to deduce height bounds for some rational points. So we worked out in detail the classical result.

The result that I called "Chow's Lemma for Stacks" is not actually Chow's Lemma for Stacks (sorry). Rather, it is Théorème 6.3, p. 49 of the following.

MR1771927
Laumon, Gérard; Moret-Bailly, Laurent
Champs algébriques.
Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge, Volume 39.
Springer-Verlag, Berlin, 2000. xii+208 pp.
ISBN: 3-540-65761-4

For each field $k$ of finite characteristic $p$, denote by $W(k)$ a complete, local, Noetherian DVR with uniformizing element $p$ such that $W(k)/pW(k)$ equals $k$. Let $\mathcal{M}$ be a smooth stack over $\text{Spec}\ W(k)$. Let $X_0$ be a connected, smooth, quasi-projective $k$-scheme. By hypothesis, there exists a locally closed immersion of $k$-schemes, $$i:X_0 \to \mathbb{P}^n_k.$$ Let $\zeta:X_0\to \mathcal{M}$ be a $1$-morphism over $\text{Spec}\ W(k)$. By Théorème 6.3, up to replacing $X_0$ by a dense open subscheme, there exists a factorization of $\zeta$, $$X_0\xrightarrow{z} M \xrightarrow{\phi} \mathcal{M},$$ where $M$ is a smooth, quasi-projective $W(k)$-scheme, where $\phi$ is a smooth $1$-morphism over $W(k)$, and where $z$ is a $W(k)$-morphism.

So now we are reduced to lifting over $W(k)$ the locally closed subscheme $X_0$ in the closed fiber of the smooth, quasi-projective $W(k)$-scheme $$\mathbb{P}^n_{W(k)}\times_{\text{Spec}\ W(k)} M.$$ This is "classical". Since $X_0$ is smooth, the locally closed immersion is a regular embedding. Thus, up to shrinking $X_0$ further, this closed subscheme is an open subscheme of a complete intersection of degree $d$ ample hypersurfaces in $\mathbb{P}^n_{W(k)}\times_{\text{Spec}\ W(k)} M$ for all $d\geq d_0$. (You can spend a lot of time trying to optimize degrees in this argument, but the OP did not ask for an "optimal lifting", just some lifting.) The defining equations of those hypersurfaces have coefficients that are elements of $k$. By lifting those coefficients to $W(k)$, we can lift the hypersurfaces over $W(k)$. This lifts the complete intersection over $W(k)$.

As the commenters note, this clashes with intuition that lifts should preserve intersection numbers and other notions that definitely are preserved for lifts of projective schemes. However, this is a well-known "pathology" of lifts of affine schemes. We can find projective models of these lifts, but then the closed fiber will be reducible. Those properties suggested by our intuition typically do occur for the closed fiber of a projective model. However, the property may hold for some irreducible component of the closed fiber different from the component containing $X_0$.

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  • $\begingroup$ Thank you very much for the answer. Just one (stupid) question: Are you using somehow that $k$ is finite? $\endgroup$ Nov 29, 2018 at 14:29
  • $\begingroup$ No, I am not asking that $k$ is finite. I wrote the definition of $W(k)$ in that way to avoid discussion of a construction of $W(k)$ (that discussion is useful in some contexts, but not terribly relevant here). For this lifting result, we could replace $W(k)$ by any DVR, it need not even be complete (since we are just lifting coefficients of defining equations). $\endgroup$ Nov 29, 2018 at 14:31
  • $\begingroup$ Perfect, I will take a couple of days to familiarize with the answer and then I will accept it. Thank you $\endgroup$ Nov 29, 2018 at 14:56

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