# The product of two supersingular elliptic curves is independent of which ones we pick

In a comment on this MO question, Qing Liu says "In positive characteristic p, if you take two supersingular elliptic curves $$E_1,E_2$$, then $$E_i×E_j$$ is isomorphic to $$E^2_1$$ for any pair $$i,j$$."

Why is this true?

Let $$C_0$$ be a supersingular elliptic curve over an algebraically closed field $$k$$ of char $$p>0$$, and $$R:= \operatorname{End}(C)$$ which is a maximal order in the quaternion algebra $$D_{p,\infty}=\operatorname{End}(C)\otimes \mathbb Q$$.
Note all supersingular elliptic curves are isogenus, and there is a bijection between supersingular elliptic curves over $$k$$ and rank one projective right $$R$$ modules (both up to isomorphism) given by $$C \mapsto \operatorname{Hom}(C,C_0)$$. The key point for us is that if the natural right $$R$$ module $$\operatorname{Hom}(C,C_0)$$ is free i.e $$\operatorname{Hom}(C,C_0) \cong R$$, then $$C \cong C_0$$. Similar results hold for product of supersingular elliptic curves.
Now the proof is finished by an old fact that any projective module of rank $$g \geq 2$$ over $$R$$ is free, see "M. Eichler, Über die Idealklassenzahl hyperkomplexer Systeme, Math. Z. 43 (1938), 481–494", which is written in old language and it seems only a few people know the proof.
• @reuns Here we work over an algebraically closed field $k$, and you can choose a specific descent to $\mathbb F_{p^2}$ then the result is standard, for example see mathoverflow.net/questions/18982/…. – sawdada Oct 8 '19 at 3:46
• @reuns Two elliptic curves over a finite field $\mathbb F_q$ are isogenus iff they have same number of $\mathbb F_q$ points, which is determined by the Frobenius action. This is quite standard, and is not the key point. – sawdada Oct 8 '19 at 3:57