Timeline for Extending automorphisms from $\mathbb{A}^1$ to $\mathbb{P}^1$ over general rings
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 1, 2023 at 23:55 | comment | added | E. KOW | The paper seems to deal only with k points, no? The conclusion there for n=1 is it's just affine transformations | |
| Jan 1, 2023 at 20:35 | comment | added | abx | Yes, this has been worked out (more generally for $\operatorname{Aut}(\mathbb{A}^n) $) by Shafarevitch, On some infinite-dimensional groups II, Math. USSR-Izvestyia, Vol. 18, 1982, pp. 185-194. | |
| Jan 1, 2023 at 20:14 | comment | added | E. KOW | Thanks @abx! Is there a description of $Aut \mathbb{A}^1$ as an ind group scheme? | |
| Jan 1, 2023 at 12:30 | comment | added | Martin Brandenburg | Answers can be posted as answers :) | |
| Jan 1, 2023 at 5:05 | comment | added | abx | @Tom Goodwillie: Oops, you are right of course! It should be $x\mapsto x+\varepsilon x^n$ with $n\geq 3$. | |
| Dec 31, 2022 at 21:54 | comment | added | Tom Goodwillie | @abx Are you sure? It seems to me that when $\varepsilon^2=0$ then this is the fractional linear transformation $x\mapsto \frac{x}{1-\varepsilon x}$. (In the other affine line with coordinate $y=1/x$ it is $y\mapsto y-\varepsilon$.) | |
| Dec 31, 2022 at 20:15 | comment | added | abx | No, it's not. Take a field $K$ and put $R:=K[\varepsilon ]/(\varepsilon ^2)$. Then $x\mapsto x+\varepsilon x^2$ is an automorphism of $\mathbb{A}^1_R$ which does not extend to an automorphism of $\mathbb{P}^1_R$. | |
| Dec 31, 2022 at 18:53 | history | undeleted | E. KOW | ||
| Dec 31, 2022 at 18:33 | history | deleted | E. KOW | via Vote | |
| Dec 31, 2022 at 18:25 | history | asked | E. KOW | CC BY-SA 4.0 |