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Let be $E/\mathbb{F}_p$ an elliptic curve and $\phi:E\longrightarrow E:(x,y)\longrightarrow(x^q,y^q)$ the Frobenius morphism.

I read at "Arithmetic of elliptic curves" at page 138:

$$P\in E(\mathbb{F}_p) \Leftrightarrow \phi(P)=P$$

I don't understand it, does someone know why?

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The reason is Frobenius map fixes $E(\mathbb{F}_p)$. As the finite field $\mathbb{F}_p$ has $p$ elements, so the $p^{th}$-power map on $\mathbb{F}_p$ is identity. Hence $E^{(p)}=E$. That is, $\phi(P)=P$, $ \ \ \forall P \in E(\mathbb{F}_p)$.

For example, take the map $f(x)=x^3$ in $\mathbb{F}_3=\{0,1,2 \}$. Then $f(0)=0^3=0, \ f(1)=1^3=1, \ f(2)=2^3=2 \mod 3$. So in this case $f$ is identity.

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    $\begingroup$ so we could say that the reason is the "Fermat’s little theorem" ? $\endgroup$ May 24, 2020 at 13:28
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    $\begingroup$ @danihelovick, I am not sure whether we should say like that, it may be. But one thing if you take $E(\mathbb{F})$ over the infinite field $\mathbb{F}$, then the $p^{th}$ -power map is not identity. It is only true for reduction of $\mathbb{F}$ modulo $p$. $\endgroup$
    – MAS
    May 24, 2020 at 13:40
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    $\begingroup$ @danihelovick,I think you are right. Because every finite group Fermat's little theorem holds. Here $\mathbb{F}_p$ is a finite group also. So $a^p \equiv a \mod p$ provided $ p \nmid a$. $\endgroup$
    – MAS
    May 24, 2020 at 13:54

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