3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @LaurentMoret-Bailly Ofcourse. Sorry, I forgot to specify the flatness. I have edited the question. $\endgroup$ – Chen Sep 1 '17 at 13:05
  • 2
    $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.