Finite galois extension over $\Bbb Q_p$ is always abelian? Is finite galois extension over $\Bbb Q_p$ always abelian ?
I often counts number of give degree extension of $\Bbb Q_p$ using local class field theory, but I'm worrying there are counter example of titled statement.
If the titled statement is not true,could you give me
counterexample of the titled question?
Thank you in advance.
 A: As explained in my comment, for an odd $p$, the extension $\mathbb{Q}_p(p^{1/p})/\mathbb{Q}_p$ is not Galois, thus its Galois closure $K$ cannot be abelian.
As for the Galois group $G$ of $K$: note that $K$ is generated by $p^{1/p}$ of degree $p$ and $u$ ($p$-th root of unity) of degree $p-1$, so that $[K:\mathbb{Q}]$ has degree $p(p-1)$.
I am reasonably confident that this group is the semi-direct product of $\mathbb{F}_p$ and $\mathbb{F}_p^{\times}$ where $x \in \mathbb{F}_p^{\times}$ acts by $z \in \mathbb{F}_p \longmapsto x^rz$ – and I think that we can take $r=1$. This is true, for instance, as long as $G$ has a trivial center.
Edit: this is true indeed, as Torsten Schoeneberg's argument shows.
Here's the one I managed to come up with: let $\sigma \in Gal(K/\mathbb{Q}_p)$ act on $\mu_pp^{1/p}$. It's easy to see that this action is of the form $\sigma(\zeta p^{1/p}) = \zeta^{a(\sigma)}b(\sigma)p^{1/p}$, where $b(\sigma) \in \mu_p, a(\sigma) \in \mathbb{F}_p^{\times}$, so under a fixed bijection $\mu_pp^{1/p} \cong \mu_p \cong \mathbb{F}_p$ (where the second map is an isomorphism), $\sigma$ acts on $\mathbb{F}_p$ by the affine bijection $z \longmapsto a(\sigma)z+b(\sigma)$.
It is easy to check that this induces a morphism $G \rightarrow AffBij(\mathbb{F}_p)$ which is injective between groups of cardinality $p(p-1)$, hence is an isomorphism. And it's clear that $AffBij(\mathbb{F}_p)$ is the semi-direct product of $\mathbb{F}_p$ and $\mathbb{F}_p^{\times}$ where the latter acts on the former by multiplication.
