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The question is from Marcus' book, "Number Fields" (exercise 12, Chapter 4)

Let $\omega= e^{\frac{2\pi i}{m}}$ and $p$ be a rational prime not dividing $m$. Then how does $p$ split in $\mathbb{Q}[\omega + \omega^{-1}]$?

I know that $\mathbb{Q}[\omega + \omega^{-1}]$ is subfield of $\mathbb{Q}[\omega]$ and that $\text{Gal}(\mathbb{Q}[\omega]/\mathbb{Q})$ is isomorphic to $\mathbb{Z}_m^{\times}$.

The following result is known:

Lemma: Let $H$ be the subgroup of $\mathbb{Z}_m^{\times}$ fixing a subfield $K$ of $\mathbb{Q}[\omega]$ pointwise and $p$ be a rational prime not dividing $m$. Then $f$ is the inertial degree of any prime $\mathfrak{P}$ of $K$ lying over $p\mathbb{Z}$, where $f$ is the least positive integer such that $p^f \equiv x \pmod m$ with $x\in H$.

Since $\mathbb{Q}[\omega]$ has degree 2 over $ \mathbb{Q}[\omega + \omega^{-1}]$, I get $H = \mathbb{Z}_3^{\times}$ and have to find least possible values of $f$ such that $p^f \equiv 1,2 \pmod m$.

What can I conclude from all this, about the splitting of primes?

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    $\begingroup$ The real subfield $\Bbb{Q}[\omega+\omega^{-1}]$ is the fixed field of complex conjugation. In the identification of the Galois group with $\Bbb{Z}_m^*$ the complex conjugation corresponds to the residue class of $-1$. Therefore $H=\{\overline{1},\overline{-1}\}\le\Bbb{Z}_m^*$. I don't understand where your view that $H=\Bbb{Z}_3^*$ would come from? For one that is not a subgroup of $\Bbb{Z}_m^*$ in any natural way. $\endgroup$
    – Jyrki Lahtonen
    Jul 7, 2016 at 18:29
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    $\begingroup$ And because this extension is Galois, the equation $efg=n$ holds. You know $n$, the Lemma gives you a method for finding $f$. Do you know $e$ or $g$? $\endgroup$
    – Jyrki Lahtonen
    Jul 7, 2016 at 21:12
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    $\begingroup$ @JyrkiLahtonen I think, $\{\overline{1}, \overline{-1}\}=\{\overline{1}, \overline{2}\} = \mathbb{Z}_3^*$. Also, what I am asking in question is basically that how to figure out $e$. I know that for $\mathbb{Z}[\omega]$, ramification index is $\phi{(p^k)}$ where $k$ is the highest exponent of $p$ dividing $m$. Also, that $n=\frac{\phi{(m)}}{2}$. I don't know how to put the pieces together! $\endgroup$
    – rationalbeing
    Jul 8, 2016 at 4:06
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    $\begingroup$ No, $\overline{2}\neq\overline{-1}$. After all, you are looking at these modulo $m$, and $2\not\equiv-1\pmod m$. You are correct about $n$ and $e$. Did you forget the assumption $p\nmid m$? $\endgroup$
    – Jyrki Lahtonen
    Jul 8, 2016 at 5:55
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    $\begingroup$ Not really. You can describe the prime ideals using a factorization of the minimal polynomial of $\omega+\overline{\omega}$ over $\Bbb{F}_p$, but that is less explicit (and was probably covered in the proof the lemma already). $\endgroup$
    – Jyrki Lahtonen
    Jul 8, 2016 at 6:10

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Based on hints by Jyrki Lahtonen :

First of all, $H=\{\overline{1}, \overline{-1}\}$ (which is not same as $\mathbb{Z}_3^{\times}$). So, using the Lemma stated in question I can calculate $f$ to be the least positive integer such that $p^f \equiv \pm 1 \pmod m$.

Since $p\not|m$, by a standard theorem for $\mathbb{Q}[\omega]$, we know that its ramification index is 1. Moreover, $\mathbb{Q}[\omega + \omega^{-1}]$ is a subfield of $\mathbb{Q}[\omega]$, hence we conclude that its ramification index is also 1.

We know that $\mathbb{Q}[\omega]$ has degree 2 over $\mathbb{Q}[\omega + \omega^{-1}]$ and degree $\phi(m)$ over $\mathbb{Q}$ hence we conclude that $\mathbb{Q}[\omega + \omega^{-1}]$ and degree $\frac{\phi(m)}{2}$ over $\mathbb{Q}$.

Let $g$ be the number of prime factors in $\mathbb{Q}[\omega + \omega^{-1}]$, then by another standard theorem $fg = \frac{\phi(m)}{2}$ and we get $g = \frac{\phi(m)}{2f}$.

Thus, if $K=\mathbb{Q}[\omega + \omega^{-1}]$, then $$ p\mathcal{O}_K = \prod_{\ell = 1}^{\frac{\phi(m)}{2f}} \mathfrak{P}_{\ell}$$ where $f$ is the smallest positive integer satisfying $p^f \equiv \pm 1 \pmod m$.

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