6
$\begingroup$

Let $X$ and $Y$ be two varieties, and $f:X\rightarrow Y$ be a morphism. Suppose moreover there is a point $Q\in Y$ and $P=f^{-1}(Q)\in X$, such that the restriction $f:X\setminus P\rightarrow Y\setminus Q$ is an isomorphism. So in particular $f$ is a bijective birational morphism.

I was wondering under which conditions $f$ is an isomorphism. It is certainly not true in general. For example if $X$ is the affine line, $Y$ the cusp and $f:X\rightarrow Y: t\mapsto (t^3, t^2)$. Then $f: X\setminus 0\rightarrow Y\setminus (0,0)$ is an isomorphism, but of course $f:X\rightarrow Y$ is not an isomorphism.

Is it true if $X$ and $Y$ are smooth? Or if $X$ and $Y$ are projective?

$\endgroup$
1
  • 2
    $\begingroup$ Note that in positive characteristic, bijective doesn't imply birational (e.g. the Frobenius morphism). $\endgroup$
    – user18119
    Commented Apr 13, 2013 at 20:39

1 Answer 1

6
$\begingroup$

Projective will not be enough to make it work, because your example of the cusp can also arise for projective curves.

On the other hand, as long as $Y$ is normal and $f$ is proper, then it is true, by a version of Zariski's Main Theorem (Liu, 4.4.3).

Edit: As QiL'8 explains, the properness hypothesis is unnecessary. (Perhaps in future I should leave these things to the expert...)

$\endgroup$
6
  • $\begingroup$ in fact, this is true without assuming f proper. $\endgroup$
    – user18119
    Commented Apr 13, 2013 at 20:17
  • $\begingroup$ @QiL'8: Yes, I thought so, but couldn't quite put the pieces together in that case. Can you explain? $\endgroup$
    – user64687
    Commented Apr 13, 2013 at 20:22
  • $\begingroup$ because by ZMT, a quasi-finite morphism to a normal noetherian scheme is an open immersion. $\endgroup$
    – user18119
    Commented Apr 13, 2013 at 20:30
  • 1
    $\begingroup$ (Quasi-finite birational?) Of course! $\endgroup$
    – user64687
    Commented Apr 13, 2013 at 20:33
  • 5
    $\begingroup$ I disagree with the second part of edit. $\endgroup$
    – user18119
    Commented Apr 13, 2013 at 20:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .