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?


1 Answer 1


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.

  • $\begingroup$ @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/…. $\endgroup$
    – sawdada
    Oct 8, 2019 at 3:46
  • $\begingroup$ Your link doesn't help much, I'm just asking for the idea (and it seems to be the key point of this question) $\endgroup$
    – reuns
    Oct 8, 2019 at 3:49
  • $\begingroup$ @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. $\endgroup$
    – sawdada
    Oct 8, 2019 at 3:57
  • $\begingroup$ Thanks for the answer, I'll have to think about it for a bit $\endgroup$
    – Asvin
    Oct 8, 2019 at 14:36

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