Timeline for Finding $K$-rational points on $X_0(35)$
Current License: CC BY-SA 4.0
3 events
| when toggle format | what | by | license | comment | |
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| Sep 12, 2020 at 10:44 | comment | added | Nulhomologous | So, the argument shows that for any quadratic field $K=\mathbb{Q}(\sqrt{d})$, where $d\in \mathbb{Z}$ is square free, $X_0(35)(K)=X_0(35)(\mathbb Q)$ if and only if the quadratic twist of $X_0(35)$ by $d$ does not have any rational point. Even more,$$X_0(35)(K)=X_0(35)(\mathbb Q)\cup X_0^{\langle d\rangle }(35)(\mathbb Q)$$ where the union is disjoint and $X_0^{\langle d\rangle }(35)(\mathbb Q)$ denotes the quadratic twist by $d$ of $X_0(35)(\mathbb Q)$ | |
| Aug 3, 2020 at 20:49 | comment | added | 5W1H | I wasn't sure how to deal with the case where x-coordinate is not in $\mathbb{Q}$. Thank you so much for clarifying my confusion. | |
| Aug 2, 2020 at 12:14 | history | answered | Michael Stoll | CC BY-SA 4.0 |