# How to prove that $\text{End}_{\mathbb{F}_p}(E)$ is commutative for a given elliptic curve E?

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:

$$\mathbb{Z}$$

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.

• 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 $\phi^n$. Jan 26, 2019 at 18:05

$$\text{End}_{\Bbb F_p}(E)$$ 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 $$\Bbb Z [F]$$.