About the separable maps of elliptic curves

Question

We know that when $\alpha \neq 0$ is separable, $\deg(\alpha) = \#E(F_q)$. Also the Frobenius map $ \phi_q (x,y) = (x^q,y^q)$ is not separable and it has degree $q$ and the map $1-\phi_q = (x-x^q,y-y^q)$ is separable and has degree $q$. But this seems to enforce $E(F_q)$ to always have exactly $q$ points since $\deg(\alpha) = \#E(F_q)$, which is not the case. I am sure there is something wrong with my argument, I appreciate any help for me to understand this!

Answer 1

Your very first sentence isn't true. What we do know is that if $\alpha$ is a separable isogeny, then $\deg(\alpha) = \#\ker(\alpha)$. But we only know that $\ker(\alpha) = E(\mathbb{F}_q)$ in the case where $\alpha = 1 - \operatorname{Frob}_q$, since $P \in \ker(1 - \operatorname{Frob}_q) \iff P \in E(\mathbb{F}_q)$.

It's easy to construct explicit counterexamples to this mistaken claim. The multiplication-by-$m$ isogeny $[m]$ has degree $m^2$ and is separable as long as $p \nmid m$, where $p$ is the characteristic of the base field. So taking $\alpha = [m]$, we can make the lefthand side of the "equality" $\deg(\alpha) = \#E(\mathbb{F}_q)$ as large as we want while the righthand side stays the same.

It's also not true that $\deg(1 - \operatorname{Frob}_q) = q$; instead we have $\deg(\operatorname{Frob}_q) = q$. As far as I know, there is no simple expression for $\deg(1 - \operatorname{Frob}_q)$, but it can be bounded using the Cauchy-Schwarz inequality---see Silverman's The Arithmetic of Elliptic Curves Lemma V.1.2 (p. 138).