Frobenius endorphism for elliptic curves

Question

In the Pairing for Beginners book, I read:

Frobenius endomorphism $\pi$ for $E$:

• $\pi : E \rightarrow E, (x, y) \mapsto (x^q,y^q)$
• Note: $\pi$ maps any point $E(\overline{\mathbb{F}}_q)$ to $E(\overline{\mathbb{F}}_q)$, but the set of points fixed by $\pi$ is exactly the group $E(\mathbb{F}_q)$.

Why is the set of points $\#E(\mathbb{F}_q)$ ?

Answer 1

You can try with $E : y^2 = x^3-x-2$ over $\Bbb{F}_5$ in http://magma.maths.usyd.edu.au/calc/

    F := FiniteField(5); A<x,y> := AffineSpace(F,2);
    E := Curve(A,y^2-x^3+x+2);
    Points(E);    

    F2<a> := FiniteField(5^2);
    E2 := BaseChange(E,F2);
    P2 := Points(E2); P2;
    #Points(E2)+1;

    Q := P2[2]; Q;
    [Q[1]^5,Q[2]^5] // Frobenius

Don't forget to add the point at $\infty$ to obtain a group, equivalently look at

     Points(ProjectiveClosure(E))