Timeline for Does a small contraction occur between smooth varieties?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 18, 2012 at 0:33 | comment | added | tarosano | @ roy smith, thank you very much for the references. | |
| Jun 15, 2012 at 16:25 | comment | added | roy smith | I think this is Zariski's "main theorem", factorial or smooth case, as in Shafarevich BAG vol. 1, p.120, or Mumford's red book, SLN 1358, 2nd ed. p.210. | |
| Jun 14, 2012 at 12:55 | comment | added | tarosano | @ Karl Schwede, thank you very much for teaching me the related question. | |
| Jun 14, 2012 at 12:54 | comment | added | tarosano | @ Jason Starr, Thank you very much for the answer. I think that this answers my question. | |
| Jun 14, 2012 at 12:02 | comment | added | Karl Schwede | Also see Sándor's answer to this question. mathoverflow.net/questions/31696/… | |
| Jun 14, 2012 at 11:59 | answer | added | J.C. Ottem | timeline score: 5 | |
| Jun 14, 2012 at 10:47 | comment | added | Jason Starr | Yes, such an $f$ is an isomorphism. Consider the pullback map on relative differentials, $f^*:f^*\Omega^1_V \to \Omega^1_{\tilde{V}}$. This is a map of locally free sheaves of the same rank. It is an isomorphism if and only if the associated determinant is an isomorphism, i.e., it is everywhere nonzero considered as a section of the associated Hom sheaf. This Hom sheaf is invertible, so this section is zero on a Cartier divisor. Your hypotheses imply this Cartier diviser is empty. Hence $f^*$ is everywhere an isomorphism. | |
| Jun 14, 2012 at 10:36 | history | asked | tarosano | CC BY-SA 3.0 |