# divisibility of isogenies by constant map

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.

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.