Is this simple proof that the Frobenius endomorphism of an elliptic curve defined over $\mathbb F_q$ is surjective valid?

Question

I am quite sure that the following "proof" is flawed, but I don't see why:

Let $E$ be an elliptic curve defined over $\mathbb F_q$. Since $E$'s ideal is generated by a polynomial in $\mathbb F_q[X]$, it is left invariant by the Frobenius endomorphism $$\varphi\colon\;E\to E,\;(x,y)\mapsto(x^q,y^q) \text.$$ Hence, the image of $E$ under $\varphi$ is the curve whose ideal is $E$'s ideal, that is $\varphi(E)=E$, thus $\varphi$ is a surjection.

Is this reasoning valid?

Answer 1

What about the curve $X+Y=0$ defined over $k = \Bbb F_2(Z)$. The point $(Z,Z)$ of the curve is not in the image of $\phi$ even though the ideals of the curve and of its image by $\phi$ are the same.