Timeline for Is $\operatorname{Spec}( \mathbb Q[x_1\ldots,x_n])$ simply connected space? [closed]
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Sep 18, 2018 at 13:00 | review | Reopen votes | |||
| Sep 18, 2018 at 14:59 | |||||
| Sep 13, 2018 at 14:15 | history | closed |
abx Dan Petersen YCor Steven Landsburg user1073 |
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| Sep 13, 2018 at 9:00 | answer | added | nGroupoid | timeline score: 11 | |
| Sep 13, 2018 at 8:43 | comment | added | Dan Petersen | The point is that $\pi_1(\mathrm{Spec}(\mathbb Q)) = \mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$ sits inside $\pi_1(\mathrm{Spec}(\mathbb Q[x_1,\ldots,x_n]))$. Konstantin, you would be better off reading something like Milne's notes on étale cohomology rather than trying to ask questions here at this stage of your learning. | |
| Sep 13, 2018 at 8:41 | comment | added | Najib Idrissi | I'm no algebraic geometer, but I guess @abx is saying that $\mathbb{Q}$ isn't even connected, much let alone simply connected. | |
| Sep 13, 2018 at 8:07 | comment | added | Konstantin | Simply connected means any covering of Spec has ramification. | |
| Sep 13, 2018 at 7:56 | comment | added | abx | You consider $\Bbb{Q}$, which is very far from being simply connected. | |
| Sep 13, 2018 at 7:40 | history | edited | Konstantin | CC BY-SA 4.0 |
added 32 characters in body
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| Sep 13, 2018 at 7:25 | comment | added | Konstantin | I consider zero characteristic fields only. | |
| Sep 13, 2018 at 7:13 | comment | added | Dan Petersen | No, it's not simply connected. | |
| Sep 13, 2018 at 7:10 | review | Close votes | |||
| Sep 13, 2018 at 14:20 | |||||
| Sep 13, 2018 at 6:50 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added MathJax
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| Sep 13, 2018 at 6:50 | review | First posts | |||
| Sep 13, 2018 at 6:51 | |||||
| Sep 13, 2018 at 6:46 | history | asked | Konstantin | CC BY-SA 4.0 |