Let $\pi: X \rightarrow Y$ be a birational surjective morphism. Let us also suppose that $Y$ is normal and $X$ is smooth. Is it true that $\pi$ becomes the isomorphism after restricting on $\pi^{-1}(U)$ for some $U \subset Y$ such that $\operatorname{codim}(Y \backslash U) \geq 2$?
2
-
7$\begingroup$ Yes. Since $\operatorname{Sing}(Y) $ has codimension $\geq 2$, you can take it out and assume that $Y$ is smooth. Then you can apply Van der Waerden purity theorem (EGA IV.21.12.12); the proof is much easier in your case but I have no time to look for a referene. $\endgroup$– abxCommented Sep 9, 2017 at 7:38
-
4$\begingroup$ Or you can use Zariski's Main Theorem. That implies that if $\pi$ is not an isomorphism over $y\in Y$, then $\dim \pi^{-1}(y)>0$, so the locus where it is not an isomorphism can have dimension at most $\dim Y-2$. $\endgroup$– Sándor KovácsCommented Sep 9, 2017 at 12:43
Add a comment
|