It is the first part of an exercise in the book “Number Fields” written by Marcus. Let $K$ be a sub field of $\mathbb{Q}(w)$ with $w=exp(2i\pi/m).$ Identify $\mathbb{Z}_m^\star$ with the Galois group of $\mathbb{Q}(w)$ over $\mathbb{Q}$ in the usual way, and let $H$ be its subgroup fixing $K$ point wise. For a prime $p\in\mathbb{Z}$ not dividing $m$, let $f$ denote the least positive integer such that $p^f\equiv\pm1\mod{m}.$ Show that $f$ is the inertial degree $f(P|p)$ for any prime $P$ of $K$ lying over $p.$
I know that $f(P|p)$ is the order of the frobenius automorphism $\phi(P|p)=\phi$. Since the Galois group is abelian it depends only on $p.$ I also know that $p$ does not ramify in the bigger field, since it does not divide the discriminant, thus it does not ramify in $K$ too. I think I have to prove that $f$ is the order of $\phi$, but I’m not visualising how to do it, in particular I don’t see how to use the hypothesis on $f.$
The Galois morphism determined by $w\to w^p$ (so also $w^k\to (w^k)^p$) is a lift of $\text{Frob}_p$, acting on a corresponding residual field, so the Frobenius $\phi=\phi_p$ corresponds to $p\in(\Bbb Z/m)^\times$.
The order $f'$ of $p$ (modulo $m$) in this ring is the minimal $f'$ such that $p^{f'}$ is plus one modulo $m$.
This is also the order of $\phi_p$ in the Galois group of the given cyclotomic extension.
$\square$
Example using sage: Let $m=13$, and $p=5$. Then $p^2=25=26-1=-1$ modulo $m$, and $p^4=1$ modulo $m$.
With above notations, $f=2$, $f'=4$.
The cyclotomic field extension is done with respect to the cyclotomic polynomial $\Phi_{13}$ of order $13$, which splits as follows, when considered over $\Bbb F_p$:
In sage, one can also require the residual field:
(The last factor in the decomposition,
(x^4 + 3*x^3 + 3*x + 1), corresponds to the chosen second generatora^4 - 2*a^3 - 2*a + 1.)