5
$\begingroup$

Let $k$ be a field, $X$ be a complete variety over $k$, $V$ be an open subvariety of $X$, $Y$ be a scheme over $k$. Suppose $L$ is a line bundle on $V\times Y$. If $L|_{V\times\lbrace y\rbrace}$ extends to a line bundle on $X\times\lbrace y\rbrace$ for every closed point $y$ of $Y$, does the line bundle $L$ extend to $X\times Y$?

What if a stronger condition is assumed ,namely for any functor $\phi\colon\operatorname{Pic}(V\times Y) \to \operatorname{Pic}(V)$ (here $\operatorname{Pic}$ denotes the Picard functors), the line bundle $\phi(L)$ on $V$ extends to $X$. Does $L$ extends to $X\times Y$?

Edit: $X$ is assumed to be smooth i.e. a smooth complete variety.

Improve this question
$\endgroup$
2
  • 1
    $\begingroup$ Welcome new contributor. That is not true. Let $X$ be a nodal plane quartic curve, let $Y$ be the normalization of $X$, let $V$ be the open complement of the node, i.e., the maximal open subscheme over which $Y\to X$ is an isomorphism, and let $L$ be the invertible sheaf of the effective Cartier divisor in $V\times Y$ that is the image of the graph of the open inclusion of $V$ in $Y$. $\endgroup$
    – Jason Starr
    Aug 18, 2020 at 23:15
  • $\begingroup$ Hi @JasonStarr , thank you for your answer. I should've said $X$ is smooth. Would my statements be true? $\endgroup$
    – Parkey
    Aug 19, 2020 at 2:50

1 Answer 1

Reset to default
1
$\begingroup$

Welcome new contributor. This is not true, even if $X$ is smooth. One example permutes the role of $X$ and $Y$ in my previous example.

Let $X$ be a smooth, geometrically connected, projective curve of genus $g>0$. Let $f:X\to Y$ be the normalization of a nodal curve with a single node $p$ that is a $k$-rational point. For instance, $Y$ could be a nodal plane quartic, and $X$ could be the normalization (a genus $3$ curve). Assume that the preimage of $\{p\}$ in $X$ is split, i.e., $\{r',r''\}$ for $k$-rational points $r',r''$ of $X$.

Let $V$ be the open complement of $\{r',r''\}$ in $X$. Denote the graph morphism of the restriction of $f$ to $V$ as follows, $$\Gamma:V\to V\times Y.$$ The image of this graph morphism is a prime Cartier divisor in $V\times Y$. Denote by $L$ the invertible sheaf on $V\times Y$ associated to this Cartier divisor.

The pullback of this Cartier divisor to $V\times X$ does extend to a Cartier divisor on $X\times X$. Every such extension is of the form $$D_{c',c''} = \underline{\Delta} + \text{pr}_1^*\left(c' \underline{r'} + c''\underline{r''}\right).$$

For each of these extended Cartier divisors, the restrictions over $X\times \{r'\}$ and over $X\times \{r''\}$ are not rationally equivalent. Indeed, if they were, then $\underline{r'}$ and $\underline{r''}$ would be rationally equivalent, so that the genus $g$ equals $0$. (This was my reason for working with smooth curves of positive genus.) Since $X\times X$ is smooth, the homomorphism from the group of rational equivalence classes of Cartier divisors to the Picard group is an isomorphism. Thus, every invertible sheaf on $X\times X$ that extends the pullback of $L$ has non-isomorphic restrictions over $X\times\{r'\}$ and over $X\times\{r''\}$. Therefore each extended invertible sheaf on $X\times X$ is not isomorphic to the pullback of an invertible sheaf on $X\times Y$.

Edit. In the example above, for every Zariski cover $Y'\to Y$, the same result holds. However, there is an étale cover $Y'\to Y$ such that the invertible sheaf extends to $X\times Y'$. For an example where there is no such extension even after an étale cover, instead of letting $X\to Y$ be the normalization of a nodal curve, let is be the normalization of a cuspidal curve. Then the same construction gives an invertible sheaf $L$ on $V\times Y$ such that for every étale cover $Y'\to Y$, there is no extension of the invertible sheaf to $X\times Y'$.

Improve this answer
$\endgroup$
15
  • $\begingroup$ Thank you very much for the example and detailed elaboration! I realize I only want to consider the extension result locally on $Y$. So I have a further question: Can the line bundle $L$ (in your example and in general) extends at least locally on $Y$ i.e. there exists a (Zarisiki, etale or fppf) cover $\lbrace Y_i \to Y \rbrace$ such that $L|_{V \times Y_i}$ extends to $L|_{X \times Y_i}$ for each $i$ ? $\endgroup$
    – Parkey
    Aug 19, 2020 at 11:36
  • 1
    $\begingroup$ That still does not work. Let me modify the example above to explain this. $\endgroup$
    – Jason Starr
    Aug 19, 2020 at 13:34
  • 1
    $\begingroup$ Thx @JasonStarr. I don't see your modification. But I guess you were trying to say that there is no neighborhood $U_p$ around $p$ such that $L|_{V \times U_p} $extends to $L|_{X \times U_p}$ for similar reason? I'm not entirely sure though... $\endgroup$
    – Parkey
    Aug 20, 2020 at 11:13
  • 1
    $\begingroup$ Yes, that is precisely what I expect. I will try to write a proof soon. $\endgroup$
    – Jason Starr
    Aug 22, 2020 at 13:11
  • 1
    $\begingroup$ I am probably not going to have time to write a proof soon. There are three key steps. First, by Zariski's Main Theorem, for the normalization $\widetilde{Y}$ of $Y$, every extension of the Cartier divisor on $X\times \text{Spec}\ k(Y)$ extends uniquely to $X\times \widetilde{Y}$. Next, by Artin approximation (or something simpler), after base change by an etale extension, the irreducible containing a specified closed point of $Y$ are geometrically irreducible. Finally, a sufficiently ramified fppf cover factors through the normalization. $\endgroup$
    – Jason Starr
    Aug 23, 2020 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.