4
$\begingroup$

Let $X,Y$ be two normal algebraic surfaces (for instance projective) and let $\varphi\colon X\dashrightarrow Y$ be a birational map which restricts to an isomorphism $(X\setminus F)\to (Y\setminus G)$ where $F\subset X,G\subset Y$ are finite subsets. Does it follow that $\varphi$ is an isomorphism?

(This is true at least when $X$ and $Y$ are smooth).

$\endgroup$
1
  • 3
    $\begingroup$ There are trivial (smooth) counterexamples when "projective" is dropped. I guess you mean this is true in the smooth projective (or complete) case; probably the affine smooth case follows by some kind of Hartogs phenomenon. $\endgroup$
    – YCor
    Commented Nov 10, 2016 at 7:58

1 Answer 1

5
$\begingroup$

I think so (even in dimension higher than 2, assuming that $F$ and $G$ are still finite sets, and not codimension two subvarieties of course). Let $H$ be an ample divisor in $X$ avoiding $F$ and let $H'$ be its strict transform in $Y$. Then $H'$ is ample in $Y$. In particular $Y={\rm Proj}(Y,H')={\rm Proj}(X,H)=X$.

$\endgroup$

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.