TITLE: The product of two supersingular elliptic curves is independent of which ones we pick
QUESTION [5 upvotes]: 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?

REPLY [5 votes]: 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.