Let $n$ be a positive integer. It is well known that a rational prime $p$ splits in the $n$th cyclotomic field $\mathbb{Q}(\zeta_n)$ if and only if $p \equiv 1 \bmod n$. I am trying to understand what happens if we replace $\mathbb{Q}$ with a quadratic field, precisely:

Given a quadratic field $F = \mathbb{Q}(\sqrt{d})$, with $d$ squarefree integer, which prime ideals $P$ of the ring of integers $O_F$ split completely in $F(\zeta_n)$ ?

I guess that it is crucial to understand the minimal polynomial $f$ of $\zeta_n$ over $F$. Of course $f$ is a divisor of the $n$th cyclotomic polynomial $\Phi_n$, but the characterization of when $\Phi_n$ is reducible in $F$ does not seem trivial (see [1]). Maybe I am missing some easy way...

Thank you for any suggestion.

[1] Weisner, Quadratic Fields in Which Cyclotomic Polynomials are Reducible, Annals of Mathematics (1928) (http://www.jstor.org/stable/1968008?seq=1#page_scan_tab_contents)

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    $\begingroup$ For $(n,q)=1 $, since $\mathbf{F}_q^\times$ is cyclic with $q-1$ elements, $\mathbf{F}_q(\zeta_n)=\mathbf{F}_{q^m}$ where $m$ is the order of $q \bmod n$, thus you can deduce the factorization of $\Phi_n$ just from the cardinality of $\mathcal{O}_F/\mathfrak{p} \cong \mathbf{F}_q $ $\endgroup$ – reuns Oct 20 '17 at 18:33

Unless $F$ is a subfield of $\Bbb Q(\zeta_n)$ then $\Phi_n$ remains irreducible over $F$.

If $\frak p$ is a prime ideal of $\frak o$, the ring of integers of $F$ and $|\mathfrak{o/p}|=q=p^r$ then as long as $p\nmid n$, the ideal $\mathfrak p$ splits completely over $F(\zeta_n)$ iff $q\equiv1\pmod n$.

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    $\begingroup$ I agree with your first sentence, but I do not see any justification for the second (?). $\endgroup$ – Speedy Oct 21 '17 at 10:01

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