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Sep 9, 2017 at 12:43 comment added Sándor Kovács 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 $\dim Y-2$.
Sep 9, 2017 at 7:38 comment added abx 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.
Sep 9, 2017 at 3:57 history asked Din CC BY-SA 3.0