Timeline for Deduce kernel of isogeny from action on torsion points
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 9, 2023 at 9:59 | comment | added | Ben Smith | Since $\gcd(N_1',N_2) = 1$, the isogeny $\psi_{N_1'}$ restricts to an isomorphism on $N_2$-torsion; so $\psi(R) = 0$ for some $R$ in $E[N_2]$ iff $\psi_{N_2}(R) = 0$. | |
| Feb 8, 2023 at 20:31 | vote | accept | Manuel Bravi | ||
| Feb 8, 2023 at 14:44 | comment | added | Manuel Bravi | I think that i understood the second part: let $\psi|_{E[N_2]} : E[N_2] \to E[N_2](?)$ and $\{P,Q\}$ one basis of $E[N_2].$ Then i can write $\psi(P) = a_{11}P+a_{12}Q$ and $\psi(Q)=a_{21}P + a_{22}Q$ with $a_{ij} \in \mathbb{Z}/N_2\mathbb{Z}$ and I calculate the kernel as usual in such cases. I also understand why $\ker \psi_{N_2}\subseteq \ker \psi|_{E[N_2]},$ but not why $\ker \psi|_{E[N_2]}\subseteq \ker \psi_{N_2}$ | |
| Feb 7, 2023 at 18:38 | history | answered | Ben Smith | CC BY-SA 4.0 |