# Isogeny between j-invariants

I am studying $$l$$-isogeny graphs (volcanoes). As I understand these graphs have $$j$$-invariants as vertices but I am having a hard time understanding the edges. The following is not clear to me:

Suppose $$E_1/\mathbb{F}_q, E_2/\mathbb{F}_q$$ are elliptic curves with the same $$j$$-invariant $$j$$. If $$\phi_1: E_1 \to E_1'$$ is an $$l$$-isogeny, then there exist an $$l$$-isogeny $$\phi_2: E_2 \to E_2'$$.

I've searched for the proof but everyone is using modular polynomial. I would like to avoid modular polynomial as it seems to me that it could be done in more elementar way. Am I wrong?

This was my (maybe naive) approach: Let $$\rho$$ be an isomorphism over $$\overline{\mathbb{F}_q}$$ between $$E_1$$ and $$E_2$$. Every separable isogeny is determined by its kernel, let $$G=ker \phi_1$$. Since $$G$$ is a cyclic subgroup of $$E_1[l]$$ then $$\rho(G)$$ is a cyclic subgroup of $$E_2[l]$$. For $$\rho(G)$$ to be a kernel of $$l$$-isogeny it would have to be defined over $$\mathbb{F}_q$$ (invariant under the elements of $$Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q)$$). We know that $$G$$ is defined over $$\mathbb{F}_q$$. This is where my idea ends.

It would be marvelous If I could show that $$E_2, E_2'$$ have the same $$j$$-invariant along the way. I will be grateful for any tips, ideas, links, anything. Thanks a lot.

EDIT: I realized that if $$j\neq 0,1728$$ (so let's assume that) then $$\rho$$ is defined over quadratic extension $$L$$ of $$\mathbb{F}_q$$. The isomorphism can be then written as $$\rho(x,y)=(c_1x+c_2, (d_1x+d_2)y)$$ where $$c_1,c_2,d_1,d_2 \in L$$. Also it suffices to show that $$\rho(G)$$ is invariant under Frobenius automorphism (not every automorphism from $$Gal(\overline{\mathbb{F}}_q/\mathbb{F}_q)$$). So it breaks down to prove the following: For each $$P=(x,y) \in G$$: $$\pi(\rho(P))=\pi(c_1x+c_2, (d_1x+d_2)y)=(c_1^qx^q+c_2^q, (d_1^qx^q+d_2^q)y^q) \in \rho(G)$$ where we know that $$(x^q,y^q) \in G$$.

Not sure what to do next, everything I tried was too messy.

I think I figured it out. We want to show that $$\rho(G)$$ is invariant under Frobenius automorphism. The key fact is that every element in the finite field $$L$$ has a $$p$$th root. So there is an isogeny $$\rho'$$ such that $$\rho\pi_p = \pi_p \rho'$$ where $$\pi_p$$ is the map $$(x,y)\mapsto (x^p,y^p)$$. Comparing degrees we get that $$\rho'$$ is also an isomorphism. Now, the automorphism group of $$E_1$$ is just $$\mathbb{Z}_2$$ because $$j\neq 0,1728$$. And since $$\rho^{-1}\rho'$$ is an automorphism, it is necessary that $$\rho^{-1}\rho'=-id$$ or $$id$$. The rest is easy, as well as the fact that $$E_2$$, $$E_2'$$ have the same $$j$$-invariant (construct an isomorphism using the two isogenies and $$\rho$$)