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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.

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    $\begingroup$ 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 $\Bbb{Z}$. $\endgroup$ Commented Dec 4, 2020 at 20:28
  • $\begingroup$ @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? $\endgroup$
    – user832207
    Commented Dec 4, 2020 at 23:42

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

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    $\begingroup$ 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 $\phi_3$. $\endgroup$
    – Lubin
    Commented Dec 4, 2020 at 21:48
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    $\begingroup$ Right $\phi_3$ is in $End(E/\Bbb{F}_9)$ $\endgroup$
    – reuns
    Commented Dec 4, 2020 at 21:50
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    $\begingroup$ @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. $\endgroup$
    – reuns
    Commented Dec 5, 2020 at 0:08
  • $\begingroup$ Thanks very much! I can only accept one answer, but your answer is very helpful too! $\endgroup$
    – user832207
    Commented Dec 5, 2020 at 0:13
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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.

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