Let $E$ be a supersingular elliptic curve over $\overline{\mathbb{F}_q}$ where $q=p^n$ and $p$ is prime. Then $B:=\text{End}(E) \otimes \mathbb{Q}$ is a unique quaternion algebra over $\mathbb{Q}$ ramified exactly at $p$ and $\infty$, and $\text{End}(E)$ is a maximal order in $B$.

I want the proof of the following statement:

For every maximal order $O' \subseteq B$, there exists $E'$ such that $O' \simeq \text{End}(E')$.

This is given in Voight 42.2.21 (p.790): For every isogeny $\phi: E \rightarrow E'$, there exists a left $O$-ideal $I$ and an isomorphism $\rho:E_I \rightarrow E'$ such that $\phi=\rho \phi_I$. Moreover, for every maximal order $O' \subseteq B$, there exists $E'$ such that $O' \simeq \text{End}(E')$.

Here $E$ is supersingular and $O:=\text{End}(E)$. Definition of $E_I=E/E[I]$ is given in 42.2.1.

The proof just says use a connecting ideal between orders to prove the second statement, but I don't get how we can introduce the connecting ideal here. I suppose he means connecting ideal between $O:=\text{End}(E)$ and $O'$, but how do we know that they are connected in the first place? According to his definition two orders are connected if and only if they are locally isomorphic, i.e. $O_\mathfrak{p} \simeq O'_\mathfrak{p}$ for all primes $\mathfrak{p}$, and I don't think this can happen if we just choose a random maximal order $O' \subseteq B$. Can anyone give me a full proof of this?

  • $\begingroup$ Where should your link lead? $\endgroup$ Commented Jul 24, 2020 at 8:11
  • $\begingroup$ @AnginaSeng Sorry I fixed the link $\endgroup$
    – Andy
    Commented Jul 24, 2020 at 8:20

1 Answer 1


Apologies if this was too brief. We are given a maximal order $\mathcal{O}' \subset B$. Since $\mathcal{O}$ is also a maximal ideal, we may apply Lemma 17.4.7 (in the most recent version, v.0.9.23) to get $I' := \mathcal{O}\mathcal{O}'$ a connecting fractional ideal (indeed, there is a unique genus of maximal orders, 17.4.10), and in particular $\mathcal{O}_{\mathsf{R}}(I') = \mathcal{O}'$. Clearing denominators (section 9.3), there exists $a \in \mathbb{Z}$ such that $a\mathcal{O}' \subseteq \mathcal{O}$, so letting $I := a I' \subseteq \mathcal{O}$ we still have $\mathcal{O}_{\mathsf{R}}(I) = \mathcal{O'}$. Now apply Lemma 42.2.9.


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