Timeline for Is $\mathbb{A}_k^n(k)$ dense in the Berkovich analytification of $\mathbb{A}_k^n$?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Sep 28, 2020 at 20:39 | vote | accept | Jakob Werner | ||
| Sep 28, 2020 at 20:30 | comment | added | Wojowu | @JakobWerner You are right (and indeed $\mathbb A^1_k(K)$ is not even literally speaking a subspace of $\mathbb A^1_k$, so one does have to be careful). I have realized the construction of the neighbourhood is not quite as straightforward as it at first seemed to me, so I've spelled out some details. | |
| Sep 28, 2020 at 20:29 | history | edited | Wojowu | CC BY-SA 4.0 |
added 663 characters in body
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| Sep 28, 2020 at 20:01 | comment | added | Jakob Werner | I don't see how you get the open neighbourhood in $\mathbb{A}^1$. It is clear to me that $\mathbb{A}^1(k)$ can't be dense in $\mathbb{A}^1(K)$, but the metric topologies on these sets are not the subspace topologies from $\mathbb{A}^1$. Maybe you can explain where your argument fails if $k$ is algebraically closed and $K$ is an infinite complete non-archimedean extension? | |
| Sep 28, 2020 at 17:46 | history | answered | Wojowu | CC BY-SA 4.0 |