Timeline for A variation on Abhyankar–Moh–Suzuki theorem
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 29, 2021 at 10:55 | comment | added | user237522 | Thank you. It is too strong for me to require that $t \mapsto (f(t),g(t)$ is injective.. I wish to consider $f$ and $g$ of degree $\geq 4$ satisfying (1)-(4). | |
| Dec 29, 2021 at 10:40 | comment | added | Jérémy Blanc | Asking that $f',g'$ do not have a common zero implies to have an immersion. If $t\mapsto (f(t),g(t))$ is moreover injective, then you are done. But maybe this is not enough mild for you? | |
| Dec 29, 2021 at 1:15 | comment | added | user237522 | I remember a paper with condition $f',g' \in \mathbb{C}[f,g]$, but it does not seem 'mild' to me. | |
| Dec 29, 2021 at 1:10 | vote | accept | user237522 | ||
| Dec 29, 2021 at 1:08 | comment | added | user237522 | Thank you very much! I was curious about the case $\deg(h) \geq 2$, which also has a counterexample math.stackexchange.com/questions/4340517/… Please, do you think there exists a 'mild' condition (in addition to (1)-(4)) that would imply $\mathbb{C}[f,g]=\mathbb{C}[t]$? | |
| Dec 28, 2021 at 21:39 | history | answered | Jérémy Blanc | CC BY-SA 4.0 |