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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.

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  • $\begingroup$ 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 $\endgroup$
    – ali
    Sep 21 at 16:46
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    $\begingroup$ Note that $\mathbb{Q}_p(p^{1/p})$ (if $p \geq 3$) is never Galois, so its Galois closure cannot be abelian. $\endgroup$
    – Mindlack
    Sep 21 at 17:12
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    $\begingroup$ @Mindlack (+1) very nice counterexample... Why not make it an answer? $\endgroup$
    – peter a g
    Sep 21 at 17:35
  • $\begingroup$ If you are ok, I would like to know extension degree and it's Galois group. $\endgroup$
    – dandelion
    Sep 21 at 17:36
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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.

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  • $\begingroup$ If you are ok, could you give an proof of Qp(p^1/p)/Qp is not Galois ? $\endgroup$
    – dandelion
    Sep 22 at 4:43
  • $\begingroup$ And this is a tiny typo, but the first paragraph Q(p^1/p)/Qp is Qp(p^1/p)/Qp. $\endgroup$
    – dandelion
    Sep 22 at 4:44
  • $\begingroup$ @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. $\endgroup$
    – Mindlack
    Sep 22 at 6:34
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    $\begingroup$ 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 $G(K\vert \mathbb Q_p) \simeq \mathbb F_p \rtimes \mathbb F_p^\times, \tau \mapsto (1_+, 1_\times), \sigma \mapsto (0, v)$. $\endgroup$ Sep 22 at 19:41

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