3
$\begingroup$

I am trying to understand when a morphism defined in an open dense subset of a variety can be extended to the whole variety.

For curves, it is known that if f:C->C’ is a rational morphism from the curve C to the curve C’ then f can be uniquely extended to the whole curve C.

On the other hand, for higher dimensional varieties the situation is more complicated, for instance,one can not extend a morphism from $\mathbb{A}^2\setminus(0,0) \to \mathbb{P}^2$.

I would like to know if is there any “nice” condition on a morphism $f:U \subset X \to Y$ where U is a open dense subset of a (smooth, if necessary) projective variety X, and Y is a projective variety, that make possible to extend $f$ to the whole X.

Thank you in advance.

$\endgroup$
  • 2
    $\begingroup$ Even for curves, what you say is true only if you assume that $C$ is smooth and $C'$ projective in general. $\endgroup$ – Mohan Apr 2 '18 at 1:33
  • 3
    $\begingroup$ There are many cases in which extending is possible, but there is probably no general framework that covers them all. Some sufficient conditions are given in chapter 3 of Daniel Litt's thesis. $\endgroup$ – R. van Dobben de Bruyn Apr 2 '18 at 1:40
  • 1
    $\begingroup$ You might search for the following keywords, "Abhyankar's theorem", "Lang-Nishimura", "Weil extension", and "N'eron extension property". $\endgroup$ – Jason Starr Apr 2 '18 at 10:33
4
$\begingroup$

Since $Y$ is projective, the question reduces to the case $Y = \mathbb{P}^n$. In this case, a morphism to $Y$ is given by an epimorpism $\mathcal{O}^{\oplus n+1} \to L$ for a line bundle $L$. So, if you want to extend a morphism, you need to extend the line bundle and the epimorphism.

A line bundle $L$ always extends to scheme points of codimension 1, so we may assume that $\mathrm{codim}(X \setminus U) \ge 2$. In this case there is a unique extension of $L$ as a reflexive sheaf, this sheaf is just the pushforward of $L$ from the open subset. In particular, an extension as a line bundle exists if and only if this reflexive sheaf is locally free (if $X$ is smooth this is always true). Note, however, that in general an extension is only defined modulo codimension 1 components of $X \setminus U$.

For a discussion of extension of the epimorphism, let me assume that $\mathrm{codim}(X \setminus U) \ge 2$, so that $L$ extends uniquely. Then the morphism $\mathcal{O}^{\oplus n+1} \to L$ also extends uniquely (just by taking the pushforward), and the only question is whether the extension is surjective. Again, the image of the extension is an ideal on $X$ (twisted by $L$), and this ideal is the obstruction for the extension of the morphism.

$\endgroup$
  • $\begingroup$ Dear @Sasha, thank you very much for your answer. Maybe I would like to add a question, you say in the second paragraph that a line bundle always extend to scheme points of codimension 1, and then we may assume $codim(X \setminus U) \geq 2$. But, this means that also for the case that $codim(X \setminus U) = 0$ the extension problem can be solved ? $\endgroup$ – User43029 Apr 2 '18 at 14:21
  • $\begingroup$ Maybe I should add, that trying to follow your answer for the case where the codimension is zero, from the smoothness of $X$ the line bundle can always be extende, but the problem should be the extension of the epimorphism. right? $\endgroup$ – User43029 Apr 2 '18 at 14:22
  • $\begingroup$ @user123456: I don't quite understand what do you mean by codimension 0 case. If X is irreducible and U is nonempty, then the codimension of the complement is positive. $\endgroup$ – Sasha Apr 2 '18 at 14:43
  • $\begingroup$ Dear @Sasha, I am really sorry, I got confused with the codimension of U and its complement. Now I get it. Thank you very much. $\endgroup$ – User43029 Apr 2 '18 at 14:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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