# 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.

• You want a homomorphism of rings. $x\mapsto 5x$ does not respect products. A homomorphism of rings $\Bbb{Z}\to R$ must map $1$ to $1_R$. In your case the neutral element of the endomorphism ring is $[1]$. In other words $m\mapsto [m]$ is the only ring homomorphism from $\Bbb{Z}$. Commented Dec 4, 2020 at 20:28
• @JyrkiLahtonen Yes, you are right. I was silly just now. If I only consider homomorphism of groups, then $\operatorname{End}E\neq\mathbb{Z}$ may be not obvious？
– user832207
Commented Dec 4, 2020 at 23:42

$$E/\Bbb{F}_3 : y^2=x^3+x$$ 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 $$\phi_9$$.

• Actually, you have shown that the endomorphism ring is larger than $\Bbb Z[i]$, since $\phi_3$ behaves like square root of $-3$. I presume that by “$i\in End(E)$” you mean $(x,y)\mapsto(-x,iy)$. You’ll see that this does not commute with $\phi_3$. Commented Dec 4, 2020 at 21:48
• Right $\phi_3$ is in $End(E/\Bbb{F}_9)$ Commented Dec 4, 2020 at 21:50
• @Sate You can compute the rational map $(x,y)\to [-3](x,y)= (x,-y)+(x,-y)+(x,-y)$ which is given by some rational functions of $x,y$ and check that it is $(x,y)\to (x^9,y^9)$. But I used that $\phi_q$ is a root of $(X-\phi_q)(X-\phi_q^*) = X^2-t_q X+q$ where $\phi_q^*$ is the dual endomorphism, and $\#E(\Bbb{F}_q)= \deg(\phi_q-1)=(\phi^*-1)(\phi-1)= q+1-t_q$ so $\#E(\Bbb{F}_3)= 4$ gives $t_3=0$ and $\phi_3^2+3=0$. Those things are detailed in Silverman's AEC book. Commented Dec 5, 2020 at 0:08
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 $$\Bbb Z$$.