In a comment on [this](https://mathoverflow.net/q/37737/58001) 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?



The product of two supersingular elliptic curves is independent of which ones we pick
See Theorem 3.5 in "Supersingular K3 surfaces" by TetsuJi Shioda, or a recent paper "Abelian varieties isogenous to a power of an elliptic curve" at <https://arxiv.org/abs/1602.06237>. 


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. 


