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Jul 15, 2014 at 3:29 comment added Will Sawin Yes, you can see this very easily by multiplying the Grossencharacter with a quadratic character $\mathbb A_L^\times \to \mu_2$.
Jul 15, 2014 at 3:27 comment added Hugo Chapdelaine So the point is as @Cesnavicius wrote, the support of the conductor won't be bounded.
Jul 15, 2014 at 3:24 vote accept Hugo Chapdelaine
Jul 15, 2014 at 3:12 comment added Hugo Chapdelaine Yes sure, but it seems to me that it should be possible to bound the support of the conductor of $\psi$ just in terms of the number field $L$ (the discriminant of $L$, its degree etc).
Jul 14, 2014 at 23:24 history answered Will Sawin CC BY-SA 3.0