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Jun 12, 2021 at 23:53 comment added Wojowu @AndréHenriques It is explained in detail in Neukirch, Schmidt, Wingberg Cohomology of Number Fields, section VII.5, specifically Theorem 7.5.14.
Jun 12, 2021 at 23:02 comment added André Henriques @Wojowu I would be interested to see the explicit presentation of this Galois group as a profinite group. Would you mind sharing?
Jun 12, 2021 at 22:51 comment added KConrad See mathoverflow.net/questions/195907/automorphisms-of-mathbb-c-p
Jun 12, 2021 at 21:53 comment added Wojowu Yes, any such automorphism is automatically a homeomorphism. To see this, note that the automorphisms preserve $\mathbb Z_p$, and hence its integral closure $\overline{\mathbb Z_p}$. The topology is generated by elements $p^n\overline{\mathbb Z_p}$ for $n\in\mathbb N$ hence the result.
Jun 12, 2021 at 21:49 comment added Very Forgetful Functor @Wojowu If I remember correctly, the absolute Galois group would be the group of field automorphisms of $\mathbb{Q}_p^\text{alg}$ fixing $\mathbb{Q}_p$. Unless there is some guarantee that any such automorphism is also a homeomorphism, then presumably what I am looking for is a subgroup of the absolute Galois group of $\mathbb{Q}_p$. I'm looking at topological fields here rather than just the algebraic structure.
Jun 12, 2021 at 21:41 comment added Wojowu Through the observations you have made, you are asking about the absolute Galois group of $\mathbb Q_p$. This term should give you plenty enough search results. For instance for $p>2$ an explicit presentation of this group as a profinite group is known.
Jun 12, 2021 at 21:40 review First posts
Jun 13, 2021 at 3:39
Jun 12, 2021 at 21:38 history asked Very Forgetful Functor CC BY-SA 4.0