Given a prime $p$ and considering the finite field $\mathbb{F}_p$, I need to see that $\text{End}_{\mathbb{F}_p}$(E) is commutative using orders. It is known that $\text{End}_{\mathbb{F}_p} \subseteq \text{End}(E)$, and I have seen how $\text{End}(E)$ is one of the following:


An order in an imaginary quadratic field

An order in a quaternion algebra

The key is, if I show that $\text{End}_{\mathbb{F}_p}(E)$ is exactly an imaginary quadratic order, then it would be necessarily a commutative ring, but how to show that $\text{End}_{\mathbb{F}_p}(E)$ is indeed an imaginary quadratic order? This fact should be true regardless the curve is either ordinary or supersingular, or, in other words, regardless the inclusion $\text{End}_{\mathbb{F}_p}\subseteq \text{End}(E)$ is strict or not.

Any help on how to approach this will be appreciated, thanks.

  • $\begingroup$ In AEC $deg_i(\phi)=p$ and $p=\phi^*\phi$ so $deg_s(p)$ is 1 or p. If it's p then $E[p^r]=deg_s(p^r)=p^r$, $E[p^\infty]$ is infinite procyclic $\cong \mathbb{Z}[p^{-1}]/\mathbb{Z}$ and $End(E)$ injects in $End(E[p^\infty])$.The harder case $E[p^\infty]=O$ is obtained from considering the dual isogeny map $f\to f^*$ as an anti-involution giving a norm and trace from which it is shown $End(E)$ is a subring of a quaternion algebra, then $\phi^{2n}-t_n\phi^n +p^n=0$ lets us conclude about the subring commuting with $\phi^n$. $\endgroup$
    – reuns
    Jan 26, 2019 at 18:05

1 Answer 1


$\text{End}_{\Bbb F_p}(E)$ consists of the endomorphisms of $E$ that commute with the Frobenius automorphism $F$. As $p$ is prime, $F$ has norm $p$, and so $\Bbb Z[F]$ is a quadratic imaginary order.

If $E$ is ordinary, $\text{End}(E)$ is a quadratic imaginary order and so $\text{End}_{\Bbb F_p}(E)=\text{End}(E)$.

If $E$ is supersingular, then $\text{End}(E)$ is a non-commutative quaternion order, and the elements that commute with $F$ are just those in $\Bbb Q(F)\cap\text{End}(E)$, which form an order in the quadratic field $\Bbb Q(F)$.

In both cases, $\text{End}_{\Bbb F_p}(E)$ may be an order strictly containing $\Bbb Z [F]$.


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