Timeline for Supersingular points on the modular curve $X_0(p)$ over $\mathbb{F}_p$ and their Frobenius action
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Jun 7, 2023 at 8:42 | comment | added | Chris Wuthrich | As a group scheme $E[p]$ and the kernel of $\pi$ are non-trivial. You can no longer just look at $\bar{\mathbb{F}}_p$-rational points. See Katz-Mazur or chapter 8 in Saito's second book on Fermat's Last Theorem. The fibre is a formed of two projective lines (swapped by $W_p$) meeting precisely at the supersingular points. [I now I start abstaining from commenting for a while.] | |
| Jun 6, 2023 at 20:35 | comment | added | did | @WillSawin, thanks for your reply. I am a bit confused why this is true. My reasoning was: if $E$ is supersingular, then $\pi^2=[-p]$ and $E[-p]=E[p]=0$. This implies $[-p]$ is injective, which implies $\pi$ is injective. Would you mind pointing out where this argument goes wrong? | |
| Jun 6, 2023 at 16:08 | comment | added | Will Sawin | The kernel of Frobenius is not trivial but instead is a group scheme of degree $p$. So for Question 1 you would just take C to be the kernel. | |
| Jun 6, 2023 at 15:41 | history | asked | did | CC BY-SA 4.0 |