My question refers to following comment in this thread: About cyclotomic extensions of $p$-adic fields
@Ted wrote:
For n not divisible by p, the Galois group of $\mathbb{Q}_p(\zeta_n) $ over $\mathbb{Q}_p$ is the same as the Galois group of $\mathbb{F}_p(\zeta_n) $ over $\mathbb{F}_p $, basically because of Hensel's Lemma. But finite extensions of finite fields are cyclic, so it's easy to calculate the latter Galois group: It's cyclic of size equal to the multiplicative order of p modulo n. The extension is also unramified...
My problem is that I don't see the how he does conclude it by Hensel's Lemma:
My try: I consider the cyclotomic polynomial $\Phi_n$ with coefficients in $\mathbb{Z}_p \subset \mathbb{Q}_p$.
$\Phi_n$ is as polynomial over it's residue field $k= \mathbb{F}_p$ is obvioulsy irreducible. But Hensel's Lemma can help me only if the polynomial splits over residue field in some factors. These using Hensel's Lemma can be "lifted" to a factorisation of $\Phi_n$ in $\mathbb{Z}_p[X]$, right?
But I don't see why this property helps me to see that Galois groups of $\mathbb{Q}_p(\zeta_n) $ over $\mathbb{Q}_p$ is the same as the Galois group of $\mathbb{F}_p(\zeta_n) $ over $\mathbb{F}_p $.
$\def\Gal{\operatorname{Gal}}\def\FF{\mathbb{F}}\def\QQ{\mathbb{Q}}\def\NN{\mathbb{N}}$A basic fact of finite fields is that $F=\FF_p(\zeta_n)$ (with $p \not\mid n$) is the splitting field over $\FF_p$ of $f(x) := x^{p^d-1}-1$ where $d \in \NN$ is minimal such that $n \mid p^d-1$. Also every non-zero element of $F$ is a root of $f$, and $(F:\FF_p)=d$, and $\Gal(F/\FF_p) \cong C_d$ is generated by $\zeta \mapsto \zeta^p$.
Let $E=\QQ_p(\zeta_n)$, then clearly its residue class field is $F$. Take any non-zero $\bar\alpha \in F$, it is a root of $f(x)$, and lift it to an element $\alpha \in E$, i.e. such that $\alpha$ reduces to $\bar\alpha$ in the residue class field. Now $f'(\alpha) = (p^d-1)\alpha^{p^d-2}$ and since $v_p(p^d-1)=0$ and $v_p(\alpha)=0$ we deduce that $v_p(f'(\alpha))=0$. Therefore we can apply Hensel's lemma to deduce that there is $\beta \in E$ such that $f(\beta)=0$ and $\bar\alpha = \bar\beta$. From this we deduce that $E$ is in fact the splitting field over $\QQ_p$ of $f(x)$, and so in particular $E = \QQ_p(\zeta_{p^d-1})$.
If $\sigma \in \Gal(E/\QQ_p)$ then it restricts to $\bar\sigma \in \Gal(F/\FF_p)$. Further, such a $\sigma$ must map $\zeta_{p^d-1} \mapsto \zeta_{p^d-1}^k$ for some $k$. Putting this together, then $\bar\sigma(\zeta_{p^d-1}) = \zeta_{p^d-1}^k$, but the only elements of $\Gal(F/\FF_p)$ can have $k=p^j$. We deduce that $\Gal(E/\QQ_p)$ is cyclic and generated by the automorphism $\zeta_{p^d-1} \mapsto \zeta_{p^d-1}^p$, just like $\Gal(F/\QQ_p)$ and so they are naturally isomorphic, the isomorphism being $\sigma \mapsto \bar\sigma$.