Timeline for The group of $k$-automorphisms of $k[x_1,\ldots,x_n,x_1^{-1}]$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 12, 2020 at 18:10 | comment | added | Johannes Hahn | @LSpice: It's not the same. It's only a subgroup of it. For example $x_2\mapsto x_2+a$ is an automorphism of the polynomial algebra for every $a$ from the ring of scalars. If $a\in k(x_1) \setminus S$, this is an automorphism of $k(x_1)[x_2]$, but not of $S[x_2]$, because it doesn't even map $S[x_2]$ to itself. | |
| Mar 12, 2020 at 17:57 | history | edited | Luc Guyot | CC BY-SA 4.0 |
Fixes two typos.
|
| Mar 12, 2020 at 17:13 | comment | added | LSpice | It's clear that $\operatorname{Aut}_S(S[x_2, \dotsc, x_n])$ is a subgroup of $\operatorname{Aut}_k(k[x_1^{\pm1}, x_2, \dotsc, x_n])$, but is it obvious that it's the same as $\operatorname{Aut}_{k(x_1)}(k(x_1)[x_2, \dotsc, x_n])$? | |
| Mar 12, 2020 at 17:08 | history | edited | Johannes Hahn | CC BY-SA 4.0 |
Aut(S) is dihedral
|
| Mar 11, 2020 at 17:37 | vote | accept | user237522 | ||
| Mar 11, 2020 at 0:13 | history | answered | Johannes Hahn | CC BY-SA 4.0 |