I am thinking about why the cyclotomic polynomial $\Phi_n = \prod_{i \in \mathbb{Z}/n\mathbb{Z}^\times} (x - \zeta^i)$ is irreducible, where $\zeta$ is the $n$th root of unity over $\mathbb{Q}$. Specifically, I am trying to figure out how it is related to ramification and lifting the Frobenius element.
Let $f(x)$ be the minimal polynomial of $\zeta$. $f(x) | \Phi_{n}(x)$, and we want to show that we have an equality. It suffices to show that, for any root $\zeta$ of $f(x)$, $\zeta^p$ is a root of $f(x)$. Take a prime $p$ relatively prime to $n$, and a primitive root of unity $\zeta$ which is a root of $f(x)$.
Write $K = \mathbb{Q}(\zeta_n)$. Write $p \mathcal{O}_K = \prod_{i = 1}^n \mathfrak{q}_i^{e_i}$, where $e_i$ are the ramification indices. Write $f_i = [\mathcal{O}_K / \mathfrak{q}_i: \mathbb{F}_p]$. The automorphism $\sigma \in \text{Aut}_{\mathbb{Q}}(K)$ sending $\zeta$ to $\zeta^p$ is a lift for the Frobenius element in $\text{Aut}_{\mathbb{F}_p}(\mathcal{O}_K / \mathfrak{q}_i)$. It is also the unique one. See here.
Can someone explain the below paragraph in terms of ramification and lifting factorizations? The proof is standard as it is though.
$\sigma(\zeta) = \zeta^p$ is a root of $x^n - 1$. Write $x^n - 1= f(x) g(x)$ in $\mathbb{Z}[x]$. If $\zeta^p$ is a root of $g(x)$, then $g(x^p)$ has $f(x)$ as a factor, i.e. $g(x^p) = f(x) h(x)$. In $\mathbb{F}_p[x]$, we get $g(x)^p = f(x) h(x)$. This shows that $g(x)$ and $f(x)$ have a common factor in $\mathbb{F}_p[x]$, so that $x^n - 1$ has a multiple root in $\mathbb{F}_p[x]$, whereas it is separable by the standard derivation test.
I think something is going on in this last paragraph in terms of lifts of frobenius maps and ramification at $p$. Can someone explain this to me?