Timeline for Bijective to algebraic space implies isomorphism
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 30, 2021 at 21:10 | answer | added | Ariyan Javanpeykar | timeline score: 1 | |
| May 29, 2021 at 16:30 | comment | added | R. van Dobben de Bruyn | @MatthieuRomagny the question is not whether $Y$ is integral (as opposed to just of finite type), but whether $Y$ is even a scheme. | |
| May 29, 2021 at 12:34 | comment | added | Jason Starr | The same proof from the answer that you link works for algebraic spaces as well as for schemes: Zariski's Main Theorem works for algebraic spaces. | |
| May 29, 2021 at 10:51 | comment | added | Matthieu Romagny | By miracle flatness (Matsumura, Comm. Ring Theory, Thm 23.1) the morphism $f$ is flat, hence open, hence a homeomorphism, so $Y$ is integral. | |
| May 29, 2021 at 10:02 | comment | added | Kim | @virkkunen Thanks! Is there any reference about this fact? I mean consider $Y$ as a complex manifold? | |
| May 29, 2021 at 9:15 | comment | added | user178279 | $Y$ can be considered a compact complex manifold by Artin and then $f$ is a holomorphic bijection. Holomorphic bijections have holomorphic inverses. | |
| May 29, 2021 at 9:10 | history | asked | Kim | CC BY-SA 4.0 |