Timeline for When are the Artin symbols of two rational primes equal?
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Oct 31, 2020 at 20:02 | comment | added | Daniel Loughran | 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. | |
Oct 31, 2020 at 19:15 | answer | added | 2734364041 | timeline score: 1 | |
Oct 31, 2020 at 5:12 | comment | added | SUNIL PASUPULATI | Not able to see how it answer my question. | |
Oct 30, 2020 at 9:34 | review | Close votes | |||
Nov 4, 2020 at 3:02 | |||||
Oct 30, 2020 at 9:10 | comment | added | Daniel Loughran | Does this answer your question? Why do congruence conditions not suffice to determine which primes split in non-abelian extensions? | |
Oct 30, 2020 at 9:08 | comment | added | Daniel Loughran | 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/… | |
Oct 30, 2020 at 8:32 | comment | added | David Loeffler | 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 $\mathfrak{P}$. | |
Oct 30, 2020 at 6:43 | history | asked | SUNIL PASUPULATI | CC BY-SA 4.0 |