# 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?

• No they are not always abelian. In fact local class field theory only study subfield of maximal abelian extension. For the whole galios group you have to study local langlands
– ali
Sep 21 at 16:46
• Note that $\mathbb{Q}_p(p^{1/p})$ (if $p \geq 3$) is never Galois, so its Galois closure cannot be abelian. Sep 21 at 17:12
• @Mindlack (+1) very nice counterexample... Why not make it an answer? Sep 21 at 17:35
• If you are ok, I would like to know extension degree and it's Galois group. Sep 21 at 17:36

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.

• If you are ok, could you give an proof of Qp(p^1/p)/Qp is not Galois ? Sep 22 at 4:43
• And this is a tiny typo, but the first paragraph Q(p^1/p)/Qp is Qp(p^1/p)/Qp. Sep 22 at 4:44
• @dandelion: I corrected the typo. $\mathbb{Q}_p(p^{1/p})$ is not Galois because it doesn’t have the same degree as $K$ – and I computed the degree of $K$ directly. Sep 22 at 6:34
• For $\zeta$ a primitive $p$-th unit root, $E:=\mathbb Q_p(\zeta)$, we have $\tau: p^{1/p} \mapsto \zeta p^{1/p} \in G(K\vert E) \subset G(K\vert \mathbb Q_p)$; for $v$ a generator of $\mathbb F_p^\times$, lift $s: \zeta \mapsto \zeta^v \in G(E\vert\mathbb Q_p)$ to $\sigma \in G(K\vert \mathbb Q_p)$ and via twisting with a power of $\tau$ ensure $\sigma$ fixes $p^{1/p}$. Then $\sigma \circ \tau \circ \sigma^{-1} = \tau^v$ and $G(K\vert \mathbb Q_p) \simeq \mathbb F_p \rtimes \mathbb F_p^\times, \tau \mapsto (1_+, 1_\times), \sigma \mapsto (0, v)$. Sep 22 at 19:41