# Question about Elliptic curves over finite field and Frobenius morphism

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?

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.
• @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 $p$.
• @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 $p \nmid a$.