Endomorphism ring of elliptic curve over $\overline{\mathbb{F}_p}$ I want to prove that the endomorphism ring of elliptic curve $E$ over $\overline{\mathbb{F}_p}$ is not isomorphism to $\mathbb{Z}$.
We can find a $q=p^n$ such that $E$ is defined over $\mathbb{F}_q$. I have proved that the Frobenius endmorphism $(x,y)\rightarrow (x^q,y^q)$ is not equal to $[m]$ for any $m\in\mathbb{Z}$, thus the injective map $\mathbb{Z}\rightarrow\operatorname{End}E$ given by $m\rightarrow[m]$ is not surjective, thus this map is not an isomorphism, but can there exist other map such that $\mathbb{Z}\rightarrow\operatorname{End}E$ is an isomorphism?
For example, the map $\mathbb{Z}\rightarrow\mathbb{Z}$ given by $x\rightarrow 5x$ is injective and not surjective, but $\mathbb{Z}\cong\mathbb{Z}$.
Thanks.
 A: $E/\Bbb{F}_3 : y^2=x^3+x$ then $\# E(\Bbb{F}_3)= 4$ thus $\phi_3$ is a root of $X^2+3$ so that $\phi_9 = [-3]$.
$End(E/\Bbb{F}_9)=\Bbb{Z}[i,\phi_3]$ (or maybe $\Bbb{Z}[i,\phi_3]$ just has finite index in $End(E/\Bbb{F}_9)$ ?) it is still not $\Bbb{Z}$ but this is not due to the Frobenius $\phi_9$.
A: Depends on whether you want to use deep results. The way I see it depends on work of Deuring from 1941, and other general facts not elementary but still not as deep as Deuring:
In case the Frobenius endomorphism $(x,y)\mapsto(x^q,y^q)$ is not an integer, you’re done. If it is an integer, it will be $[q^{1/2}]$, and since a power of $p$ is purely inseparable, your elliptic curve is supersingular. But Deuring showed that a supersingular elliptic curve in characteristic $p$ has an endomorphism ring that is an order in (i.e. free over $\Bbb Z$ of same dimension as) the central division algebra over $\Bbb Q$ of dimension four, ramified only at $p$ and infinity.
So in this interesting case, the endomorphism ring is much larger than $\Bbb Z$.
I hope that someone else can give a more elementary way of seeing what you want — I don’t consider this answer to be at all satisfactory.
