# Proof of Deuring's Correspondence

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?

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.