Timeline for Must the $j$-invariant of an elliptic curve with an isogeny be integral?
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| Nov 3, 2021 at 19:08 | comment | added | Maarten Derickx | The L-function part of this answer will not work. As is for example mentioned on page 1 of ams.org/journals/proc/2013-141-08/S0002-9939-2013-11532-7/…. If you twist a newform $f \in S_2(p)$ by a quadratic character $\chi$ of conductor then the sign changes by a factor of $\chi(p)$. In particular if $\chi(p)=-1$ then the functional equations of $f$ and $f \times \chi$ have opposing signs, and hence $L(f,1)L(f \times \chi,1)$ will be zero for all newfors in $S_2(p)$. Maybe that it will produce something if $\chi(p)=1$. The "But note" part however is great! | |
| Sep 7, 2012 at 8:50 | vote | accept | Barinder Banwait | ||
| Sep 7, 2012 at 8:50 | comment | added | Barinder Banwait | Thank you for your answer JSE, this approach is indeed very interesting. I've meanwhile realised that Mazur himself proved the existence of a rank zero quotient over imaginary quadratic fields at primes inert in the field. The approach you outline however would work for real quadratic fields also. Thanks again! | |
| Aug 31, 2012 at 16:34 | history | answered | JSE | CC BY-SA 3.0 |