A doubt about ramification in cyclotomic fields I'm studying the Algebraic Number Theory Notes of Robert B. Ash. I really like his notes, but I don't understand a suggestion he gives at page 7 of chapter 8 (Factoring of prime ideals in Galois Extensions).
He's using all the theory of that chapter to discover new properties of cyclotomic fields. So pick $\zeta$ a primitive $m^{th}$ root of unity and let $L=\mathbb Q(\zeta)$, $ A=\mathbb Z$ and $K=\mathbb Q$.
Consider $p$ rational prime that does not divide $m$. Say $B$ the integral closure of $A$ in $L$ and that $(p)$ factors in $B$ as $Q_1....Q_g$. We know that the relative degree $f$ is the same for all $Q_i$. 

He wants to find the Frobenius automorphism $\sigma$ explicitly. We know that $\sigma$ has the property that $\sigma(x)\equiv x^p\pmod {Q_i}$ for all $i$ and for all $x\in B$. From this, why do we deduce that $\sigma(\zeta)=\zeta^p$? 


All finite extensions of $\Bbb F_p$ are separable (in fact Galois), so all polynomials are separable--note some authors demand separable to mean "distinct roots" I use the "irreducible factors have distinct roots" definition, if you are using the former, then clearly this is not always true as $x^{pk}-1$ has repeated roots modulo $p$.
Now, the point of the suggestion is that $\zeta\mapsto\zeta^p$ generates the Galois group of $x^m-1$ over $\Bbb F_p$ because it generates all finite extension Galois groups over $\Bbb F_p$. We know that it permutes the roots of $x^m-1$ in the big field as well, so it lifts to an element of $\text{Gal}(L/\Bbb Q)$. What Ash is referencing is his theorem 8.1.8 which says that the map from the decomposition group is surjective in the case that the residue field extension is separable on page 3 of his notes. This is used to life the mod $Q$ Frobenius to the Frobenius element in the big Galois group.
