Let $K$ be an abelian number field. Let $p$, $q$ be rational primes. Is there some condition like $p\equiv q$ modulo some integer which depends on conductor of $K$ or $\operatorname{disc}(K)$ that implies $\genfrac(){}{}{K/\Bbb Q}{p}=\genfrac(){}{}{K/\Bbb Q}{q}$?
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7$\begingroup$ When $K\subset L=\Bbb{Q}(\zeta_n)$ with $n$ the conductor and $p\nmid n$ then $Frob_{p,L/\Bbb{Q}}= (\zeta_n \to \zeta_n^{p\bmod n})$ from which you deduce $Frob_{p,K/\Bbb{Q}}$. $\endgroup$– reunsCommented Oct 3, 2020 at 5:28
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