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Jul 15, 2013 at 21:25 vote accept Barinder Banwait
Jul 10, 2013 at 0:04 comment added v08ltu Or if you want to think in terms of $L$-series equality, here $a_p(E)=\chi_d(p) a_p(E')$ for all primes $p$ with the (nonsquare) twisting factor $d$, and $a_p=0$ for primes that are inert in $K$, and over $K$ one has $a_{\frak p}(E)=\chi_d(p)a_{\overline{\frak p}}(E)$ and similarly for $E'$, so $\{a_{\frak p}(E),a_{\overline{\frak p}}(E)\}=\{a_{\frak p}(E'),a_{\overline{\frak p}}(E')\}$.
Jul 9, 2013 at 15:54 history answered Felipe Voloch CC BY-SA 3.0