Timeline for Do we have $G(\mathbb A_S) G(k) = G(\mathbb A)$ for sufficiently large $S$?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 22, 2019 at 2:47 | vote | accept | D_S | ||
| May 20, 2019 at 7:39 | answer | added | GH from MO | timeline score: 7 | |
| May 20, 2019 at 2:45 | history | edited | D_S | CC BY-SA 4.0 |
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| May 19, 2019 at 22:50 | comment | added | D_S | It just means $\prod\limits_{v \in S} G(k_v) \prod\limits_{v \not\in S} G(\mathcal O_v)$ | |
| May 19, 2019 at 22:39 | comment | added | Kevin Buzzard | What does $G(\mathbb{A}_S)$ mean in this generality? It seems to me that $\mathbb{A}_S$ is not a $k$-algebra in general. | |
| May 19, 2019 at 21:29 | history | edited | KConrad | CC BY-SA 4.0 |
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| May 19, 2019 at 19:44 | history | asked | D_S | CC BY-SA 4.0 |