Prove that the Frobenius map is a homomorphism

Question

I want to prove that the Frobenius map $\phi$ is a homomorphism from the group of points on an elliptic curve $E(F_{2^k})$ to itself (endomorphism). It is trivial to check that if a point $P \in E$ then so is $\phi(P)$. However, i cannot find a simple way to show that $\phi$ respects the group structure i.e. $\phi(P+Q)=\phi(P)+\phi(Q)$.

Answer 1

Think about the group operation in terms of chords and tangents. So if the chord $L$ connecting $P$ and $Q$ has equation $ax+by+c=0$, then the chord $\phi(L)$ connecting $\phi(P)$ and $\phi(Q)$ has equation $\phi(a)x+\phi(b)y+\phi(c)=0$. If $L$ intersects $E$ at a third point $R$, then $\phi(L)$ intersects $E$ at the point $\phi(R)$ (here it is essential that the defining equation of $E$ has coefficients invariant under $\phi$.

Do the same with tangents.