Timeline for Are morphism from $\mathbb{P}^{1}$ to itself often liftable to a morphism from a curve to an elliptic curve with bounded degree?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 25, 2023 at 18:08 | vote | accept | question-asker | ||
| Apr 15, 2023 at 9:32 | history | edited | Peter Mueller | CC BY-SA 4.0 |
revised the answer
|
| Apr 15, 2023 at 9:27 | comment | added | Peter Mueller | @question-asker I revised the answer. The point is that you can always reduce to the case $d=2$. | |
| Apr 15, 2023 at 9:23 | history | edited | Peter Mueller | CC BY-SA 4.0 |
revised the answer
|
| Apr 14, 2023 at 22:23 | comment | added | question-asker | Do you might expanding your answer? Because I still don't understand it. I guess I'm having trouble seeing how $\pi_{E}$ came into play. Because, for example, I could make $C$ has genus 1 without having the branch point of $\pi_{E}$ being among the branch point of $f$ (of course here the condition is that $f$ must correspond to $n$-torsion points which is still special). | |
| Apr 14, 2023 at 22:05 | comment | added | Peter Mueller | @question-asker If you replace $E$ by $\mathbb P^1$, and keep the assumption that the degree of the map $\pi_E$ (so the $E$ now is $\mathbb P^1$) is bigger than $1$, you still will have the genus of $C$ go to infinity. | |
| Apr 14, 2023 at 21:54 | comment | added | question-asker | Sorry, I think maybe I misunderstood this answer, so let me ask a clarification questions. It seems to me like your argument about the genus of $C$ involves only looking at the ramifications of $f\circ\pi_{C}$. Don't we need to make use of the fact that $E$ is an elliptic curve somewhere? Because if we allows $E$ to be a non-elliptic curve, such as $\mathbb{P}^{1}$ then it's definitely always possible. | |
| Apr 14, 2023 at 20:03 | history | answered | Peter Mueller | CC BY-SA 4.0 |