Timeline for Question about Zariski cancelation problem
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 16, 2020 at 15:25 | comment | added | A.Skutin | Is this argument valid for any other field which isn't algebraically closed but with zero characteristic? | |
| Sep 16, 2020 at 14:02 | comment | added | naf | $\mathrm{Spec}(A)$ is obviously a curve which is unirational over $\mathbb{Q}$. It is also clear that the space of complex points of $\mathrm{Spec}(A)$ is contractible (in the analytic topology). There is only one curve over $\mathbb{Q}$ which has these two properties. | |
| Sep 15, 2020 at 20:56 | comment | added | Steven Landsburg | @ulrich : what easy argument do you have in mind? | |
| Sep 15, 2020 at 11:10 | comment | added | A.Skutin | I can only prove in case of algebraically closed field. Any hints to not closed field case? | |
| Sep 15, 2020 at 10:48 | comment | added | naf | This is an easy exercise. | |
| Sep 15, 2020 at 10:41 | history | edited | YCor | CC BY-SA 4.0 |
removed capitals from title
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| Sep 15, 2020 at 10:17 | history | asked | A.Skutin | CC BY-SA 4.0 |