Timeline for Set isomorphism of smooth varieties implies isomorphism as varieties
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 3, 2014 at 0:36 | vote | accept | Li Yutong | ||
| Apr 2, 2014 at 13:53 | answer | added | Puzzled | timeline score: 1 | |
| Mar 17, 2013 at 9:58 | comment | added | Jérémy Blanc | Since it is injective, it is birational with its image (compute the degree of field extension: it is $1$). Because it is surjective, it is birational. | |
| Mar 16, 2013 at 21:54 | comment | added | Li Yutong | @Ray Hoobler, could you give more details on how the map is birational? | |
| Mar 16, 2013 at 21:22 | comment | added | Jérémy Blanc | It is in EGA IV, 3ème partie, Théorème 8.12.6, page 45, it is called " ,,Main Theorem'' de Zariski." It says that your map decomposes into an open immersion, followed by a finite morphism. In your case, you should be able to see that the both are isomorphisms. | |
| Mar 16, 2013 at 21:15 | comment | added | Ray Hoobler | The map must be birational. Since Y is normal, it has to be an isomorphism. | |
| Mar 16, 2013 at 19:46 | comment | added | anon | Use (a special case of) Zariski's Main Theorem. | |
| Mar 16, 2013 at 19:31 | history | asked | Li Yutong | CC BY-SA 3.0 |