Timeline for Are Chow groups invariant under universal homeomorphisms?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 19, 2020 at 20:05 | history | edited | Alon Amit | CC BY-SA 4.0 |
typo
|
| Jun 19, 2020 at 19:01 | vote | accept | Mikhail Bondarko | ||
| Jun 19, 2020 at 18:35 | answer | added | R. van Dobben de Bruyn | timeline score: 1 | |
| Jun 19, 2020 at 17:21 | comment | added | R. van Dobben de Bruyn | @MikhailBondarko how about the resolution of a cusp? (Tag 0BRC) | |
| Jun 19, 2020 at 11:42 | answer | added | D.-C. Cisinski | timeline score: 2 | |
| Jun 19, 2020 at 8:50 | comment | added | Mikhail Bondarko | Well, in characteristic $0$ only "trivial" universal homeomorphisms exist, yes.:) | |
| Jun 19, 2020 at 8:47 | history | edited | Mikhail Bondarko | CC BY-SA 4.0 |
added 64 characters in body
|
| Jun 19, 2020 at 8:28 | comment | added | abx | I am afraid this will happen only in trivial cases. For instance if $Y$ is a smooth projective curve over $\mathbb{Q}$, with Jacobian variety $J$, $CH^1(Y)=\operatorname{Pic}(Y) $ is finitely generated, while $\operatorname{Pic}(X) =\mathbb{Z}\oplus J(\bar{\mathbb{Q}})$ is not. | |
| Jun 19, 2020 at 8:21 | history | asked | Mikhail Bondarko | CC BY-SA 4.0 |