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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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.

  • $\begingroup$ 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

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