I'm stuck with the following problem:


In Petit's work "[Faster Algorithms for Isogeny Problems using Torsion Point Images](https://eprint.iacr.org/2017/571)", p. 8, he says that we can deduce $\ker \psi\_{N\_2}$ knowing the action of $\psi = \psi\_{N\_1'}\circ\psi\_{N\_2}$ on $E[N\_2],$ where $\psi\_{N\_1'}$ and $\psi\_{N\_2}$ are isogenies of degree respectively $N\_1'$ and $N\_2$ and $\psi = \psi\_{N\_1'}\circ\psi\_{N\_2}$ an endomorphism on the elliptic curve $E$.


Well, I don't understand this statement. I've tried writing $[N\_2] = \psi\_{N\_2}\circ \hat{\psi}\_{N\_2}$ and then fiddling with the expressions, but to no avail.


I have a feeling this problem is easy, but I'm stuck and would appreciate any help/ideas.



Deduce kernel of isogeny from action on torsion points
The key details here are


1. $N\_2$ is "smooth by assumption", so solving discrete logarithms in $E[N\_2]$ is supposed to be easy (using Pohlig-Hellman; how easy this is depends on how smooth $N\_2$ is).
2. $\gcd(N\_1,N\_2) = 1$, so the kernel of $\psi\_2$ is equal to the kernel of the restriction of $\psi$ to $E[N\_2]$.


Now, if you evaluate $\psi$ on a basis of $E[N\_2]$, then by solving
discrete logarithms you can view the restriction of $\psi$ to
$E[N\_2]$ as a matrix over $\mathbb{Z}/N\_2\mathbb{Z}$, and then
computing the kernel (and hence the kernel of $\psi\_2$) is a matter of linear algebra.


