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Let $K$ be a number field Which is Galios over $\Bbb Q$.The group $Gal(K/\Bbb Q)$ is not neccesarly abelian. Let $p_1$, $p_2$ be rational primes. In this link they show if $p_1\equiv p_2$ modulo conductor of $K$ then $\genfrac(){}{}{K/\Bbb Q}{p}=\genfrac(){}{}{K/\Bbb Q}{q}$ if The group $Gal(K/\Bbb Q)$ is abelian.

I am looking for such a condition in general number fields. Precisely, is there some condition like $p_1\equiv p_2$ modulo some integer which depends on conductor of $K$ or $\operatorname{disc}(K)$ that implies $\genfrac(){}{}{K/\Bbb Q}{p}=\genfrac(){}{}{K/\Bbb Q}{q}$?

Thanks for your help.

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  • $\begingroup$ How do you define $\left(\frac{K/\mathbf{Q}}{p}\right)$ if $K$ is not abelian? You can define $\left(\frac{K/\mathbf{Q}}{\mathfrak{P}}\right)$ for $\mathfrak{P} \mid p$ an unramified prime, but if $K$ is nonabelian then it will depend on the choice of $\mathfrak{P}$. $\endgroup$
    – David Loeffler
    Commented Oct 30, 2020 at 8:32
  • $\begingroup$ For $K/\mathbb{Q}$ non-abelian the symbol $\genfrac(){}{}{K/\Bbb Q}{p}$ should be viewed as a conjugacy class, namely the class of the frobenius element. In any case, no this is not determined by congruence conditions for $K/\mathbb{Q}$ non-abelian, this is the whole point of modern algebraic number theory (modular forms, Langlands etc..) See here: mathoverflow.net/questions/11688/… $\endgroup$
    – Daniel Loughran
    Commented Oct 30, 2020 at 9:08
  • $\begingroup$ Does this answer your question? Why do congruence conditions not suffice to determine which primes split in non-abelian extensions? $\endgroup$
    – Daniel Loughran
    Commented Oct 30, 2020 at 9:10
  • $\begingroup$ Not able to see how it answer my question. $\endgroup$
    – SUNIL PASUPULATI
    Commented Oct 31, 2020 at 5:12
  • $\begingroup$ Can you please clarify what you mean by "Artin Symbol" for a non-abelian extension? It is just that it is usually only defined in the abelian case and there are differing conventions for non-abelian extensions. I understand "Artin symbol" to mean the associated frobenius conjugacy class. In which case, this symbol is trivial at an unramified prime $p$ if and only if $p$ is completely split in $K$, and the cited question provides the answer "no" to this special case of your question, hence "no" to the general case. $\endgroup$
    – Daniel Loughran
    Commented Oct 31, 2020 at 20:02

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There is no uniform, arithmetically "simple" condition (like lying in a residue class) that classifies the Artin symbol in nonabelian extensions. Roughly speaking, if you write $K=\mathbb{Q}(\alpha)$ and $f$ is the minimal polynomial of $\alpha$ (over $\mathbb{Q}$), then the Artin symbol at $p$ (defined up to conjugacy) is determined by the factorization of $f\pmod{p}$. This is pleasantly described here.

Special cases do arise. For example, if $n\equiv 1,2\pmod{4}$ is a positive squarefree integer, then $p\nmid 4n$ splits completely in the Hilbert class field of $\mathbb{Q}(\sqrt{-n})$ if and only if there exists $(x,y)\in\mathbb{Z}^2$ such that $p=x^2+ny^2$.

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