Let $K$ be a finite extension of $\mathbb{Q}_p$.
Is the centraliser of $\operatorname{Gal}(\overline{K}/K)$ in $\operatorname{Gal}(\overline{\mathbb{Q}_p} / \mathbb{Q}_p)$ trivial ? If yes, how can I show it ?
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Sign up to join this communityLet $K$ be a finite extension of $\mathbb{Q}_p$.
Is the centraliser of $\operatorname{Gal}(\overline{K}/K)$ in $\operatorname{Gal}(\overline{\mathbb{Q}_p} / \mathbb{Q}_p)$ trivial ? If yes, how can I show it ?
For an extension $L/\mathbf{Q}_p$, let $G_L$ denote the absolute Galois group $\mathrm{Gal}(\overline{L}/L)$.
If $\sigma \in G_{\mathbf{Q}_p}$ acts centrally on $G_K$, then it also acts centrally on the subgroup $G_L \subset G_K$ for any finite $L/K$. But then it also acts trivially on the abelianization of $G_L$. By class field theory, the abelianization of $G_L$ may be identified with the pro-finite completion of $L^{\times}$, and moreover this isomorphism is compatible with the action of $G_{\mathbf{Q}_p}$. Hence $\sigma$ must also act trivially on $L^{\times}$ for any finite $L/\mathbf{Q}_p$, and thus $\sigma$ is trivial.