7
$\begingroup$

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)

$\endgroup$
1
  • 1
    $\begingroup$ 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, 2017 at 18:33

1 Answer 1

3
$\begingroup$

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 $q\equiv1\pmod n$.

$\endgroup$
1
  • 1
    $\begingroup$ I agree with your first sentence, but I do not see any justification for the second (?). $\endgroup$
    – Speedy
    Oct 21, 2017 at 10:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.