Question about Elliptic curves over finite field and Frobenius morphism

Question

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?

Answer 1

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.