1
$\begingroup$

I am trying to understand certain points on the modular curve $X_0(p)$ over $\mathbb{F}_p$ (where $p$ is a rational prime) via their moduli description.

A point $P \in X_0(p)$ can be viewed as $P:=(E, C)$ where $E$ is an elliptic curve and $C$ is a cyclic subgroup of order $p$, corresponding to the kernel of an isogeny $\phi$ of degree $p$.

Question 1: If $E$ is a supersingular curve defined over $\mathbb{F}_{p^2}$, the Frobenius map $\pi$ is a purely inseparable isogeny of degree $p$. However, this means its kernel is trivial. I am thus wondering if $E$ together with $\pi$ give rise to a point on $X_0(p)(\mathbb{F_{p^2}})$ and if so, how does this point look like in the moduli description (what is $C$ ?).

Question 2: Let $P:=(E,C) \in X_0(p)(\bar{\mathbb{F}_p})$, where $E$ be a supersingular curve defined over $\mathbb{F}_{p^2}$. What can we say about $\pi(P)$? What can we say about $w_p(P)$, where $w_p$ is the Artin-Lehner involution? Are they the same action on supersingular points?

Improve this question
$\endgroup$
3
  • 2
    $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Commented Jun 6, 2023 at 16:08
  • $\begingroup$ @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? $\endgroup$
    – did
    Commented Jun 6, 2023 at 20:35
  • 2
    $\begingroup$ 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.] $\endgroup$
    – Chris Wuthrich
    Commented Jun 7, 2023 at 8:42

0

Reset to default

You must log in to answer this question.