Let $R$ be a discrete valuation ring with algebraically closed residue field $k$. Let $K:=\mathrm{Frac}(R)$ the fraction field of $R$. Suppose $K$ is of characteristic zero. Denote by $\overline{K}$ the algebraic closure of $K$ and $\overline{R}$ the integral closure of $R$ in $\overline{K}$. Since $k$ is algebraically closed, the residue field of $\overline{R}$ is $k$.

Let $\pi:\mathcal{X} \to \mathrm{Spec}(R)$ be a flat family of projective varieties and smooth generic fiber $\mathcal{X}_{K}$. Denote by $\overline{\pi}:\mathcal{X}_{\overline{R}} \to \mathrm{Spec}(\overline{R})$ the base change of $\pi$ under the natural inclusion $R \hookrightarrow \overline{R}$. Note that the generic fiber of $\overline{\pi}$ is again smooth and the special fiber is the same as $\mathcal{X}_k$ (as $k$ is algebraically closed). Let $E$ be a coherent sheaf on $\mathcal{X}_k$. My question is:

If $E$ lifts to a coherent sheaf $E_{\overline{R}}$ on $\mathcal{X}_{\overline{R}}$ (in the sense, $E_{\overline{R}}$ is flat over $\overline{R}$ and $E_{\overline{R}} \otimes_{\overline{R}} k \cong E$) does it also lift to a coherent sheaf on $\mathcal{X}$ i.e., does there exists a coherent sheaf $E_R$ on $\mathcal{X}$, flat over $R$ such that $E_{{R}} \otimes_{{R}} k \cong E$? If not true in general, is there any known condition on $R$ or $K$ under which this holds true?

EDIT Does the above question have a positive answer, if we substitute coherent by locally free?

  • $\begingroup$ @LaurentMoret-Bailly Ofcourse. Sorry, I forgot to specify the flatness. I have edited the question. $\endgroup$ – Chen Sep 1 '17 at 13:05
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    $\begingroup$ Here is a possible counterexample: Let $t$ be a uniformizer of $R$ and let $\mathcal{X}=\{xy = t\} \subseteq \mathbf{A}^2_R$. Let $0$ be the node in the special fiber. One checks easily that there is no section of $\pi$ through $0$. On the other hand, the point $(\sqrt{t}, \sqrt{t})\in \mathcal{X}(\bar R)$ is such a section $s:\bar R\to \mathcal{X}_{\bar R}$ over $\bar R$. Now take $E$ to be the structure sheaf of this section i.e. $E = s_* \mathcal{O}_{{\rm Spec}\, R}$. $\endgroup$ – Piotr Achinger Sep 1 '17 at 14:54
  • $\begingroup$ Sorry, I meant $E$ to be the skyscraper sheaf of $0$, and above I show that it extends to a flat $E_{\bar R}$ but it doesn't extend to an $E_R$. $\endgroup$ – Piotr Achinger Sep 1 '17 at 14:57
  • $\begingroup$ @PiotrAchinger What if I substitute coherent by locally free in the above question? $\endgroup$ – Chen Sep 1 '17 at 15:02
  • $\begingroup$ In general you can imagine that there is a relative moduli scheme (or stack) $\mathcal{M}/R$ parametrizing coherent sheaves on $\mathcal{X}$ flat over the base. Then the question becomes: suppose $\mathcal{M}(\bar R)\to \mathcal{M}(k)$ is surjective (which is close to saying that $\mathcal{M}$ itself is flat), then is $\mathcal{M}(R)\to \mathcal{M}(k)$ surjective as well? The answer is yes if $R$ is complete (or henselian) and $\mathcal{M}$ is smooth over $R$, and you might be able to detect this using the deformation theory of $E$. For example, the vanishing of ${\rm Ext}^2(E, E)$ is enough. $\endgroup$ – Piotr Achinger Sep 1 '17 at 18:15

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