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Let $E_1, E_2$ be elliptic curves and $\phi:E_1 \rightarrow E_2$ be a (separable) isogeny such that $\phi(E_1[m])=\{O_{E_2}\}$. Then $\phi$ is divisible by the multiplication by $m$-map $[m]$, i.e., there is an isogeny such that $\phi = [m]\circ\psi$.

I was reviewing the arithmetic of elliptic curves and I'm wondering this is true. I remember there was a similar statement regarding factoring an isogeny given another isogeny whose kernel is contained in the other's, but that was for non-constant isogenies.

Jason Dil
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1 Answers1

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Well, it think the problem is not with the "non-constantness". All $[m]$ are non-constant if $m \neq 0$. Silverman $(4.2a)$. The problem I see in your statement is that you need $[m]$ to be separable (not $\phi$).

Silverman Corollary 4.11:

Let $\phi:E_1 \rightarrow E_2$, $\psi:E_1 \rightarrow E_3$ be non-constant isogenies, and assume $\phi$ is separable. If $\text{Ker}(\phi) \subseteq \text{Ker}(\psi)$ then there exists a unique isogeny $\lambda$ s.t. $\psi = \lambda \circ \phi$.

In your case, $\phi(E_1[m]) = \mathcal{O}$ means that $E_1[m] = \text{Ker}([m])\subseteq \text{Ker}(\phi)$. Which is different from the theorem where the $\phi$ kernel is the subset and $\phi$ is separable.

$[m]$ is separable always if char$(K)=0$. If $K$ is a finite field of characteristic $p$ ($K$ is the field over which the isogenies/curves are defined) then by Silverman Corollary $5.5$ $[m]$ is separable if and only if $p$ does not divide $m$.

Therefore, if $[m]$ is separable, then by the theorem we get $\phi = \lambda \circ [m] = [m] \circ \lambda$ ($[m]$ "commutes" because $\lambda$ is an isogeny but technically those are different endomorphisms over diff. curves).

Not sure what to do if $p$ divides $m$. I guess then $\phi$ (separable) cannot be equal to $[m] \circ \lambda$ since $[m]$ is not separable.

honzaik
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