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Exercise 12, Chapter 4 of Marcus' Number Fields asks the following: Let $p$ be a prime not dividing $m$. Determine how $p$ splits in $\mathbb{Q}(\zeta_m +\zeta_m^{-1})$.

Certainly $p$ is unramified, but I'm having trouble finding what the inertial degree is. I think it must divide $f$, where $f$ is the order of $p$ modulo $m$, since $\mathbb{Q}(\zeta_m +\zeta_m^{-1})\subset\mathbb{Q}(\zeta_m)$. How does one find the inertial degree here?

Beyond this, what can we say about ramification in composita? For example, I have the following problem: Let $q\in\mathbb{Z}$ be an odd prime, $2$ divide $n$, and gcd$(q,n)=1$. Suppose we have the extension $\mathbb{Q}\subset\mathbb{Q}(\zeta_q +\zeta_q^{-1}, \zeta_{2n})$. How does an unramified prime $p\in\mathbb{Z}$ split in $\mathbb{Q}(\zeta_q +\zeta_q^{-1}, \zeta_{2n})$? Again, I think one can use the fact that $\mathbb{Q}(\zeta_q +\zeta_q^{-1}, \zeta_{2n})\subset\mathbb{Q}(\zeta_{2qn})$ combined with multiplicativity of the inertial degree, but I'm struggling to arrive at a conclusion.

a196884
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    The splitting behavior is determined by the Frobenius element in the Galois group. If the Frobenius element has order $f$, then the inertial degree is $f$. The Frobenius element in $\mathrm{Gal}(\mathbf{Q}(\zeta_m)/\mathbf{Q}) = (\mathbf{Z}/m \mathbf{Z})^{\times}$ at $p$ is $[p]$, and similarly the Frobenius element in $\mathrm{Gal}(\mathbf{Q}(\zeta_m + \zeta^{-1}_m)/\mathbf{Q}) = (\mathbf{Z}/m \mathbf{Z})^{\times}/\pm 1$ at $p$ is also $[p]$. So literally $f$ is the smallest integer such that $p^f \equiv \pm 1 \mod m$. – user760870 Apr 17 '20 at 17:01
  • Similarly, the Galois group of the next extension is $(\mathbf{Z}/q \mathbf{Z})^{\times}/\pm 1 \oplus (\mathbf{Z}/2n \mathbf{Z})^{\times}$ and Frobenius is $([p],[p])$, so it's the smallest power such that $p^f \equiv \pm 1 \mod 2nq$. – user760870 Apr 17 '20 at 17:02
  • Thanks so much! – a196884 Apr 20 '20 at 10:01

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