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As far as I know there is many "suggestions" of what should be a "field with one element" $\mathbf{F}_{1}$.

My question is the following:

How we should think or what should be the "absolute Galois group" of the "hypothetical" $\mathbf{F}_{1}$ ?

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    $\begingroup$ See answers to this question: mathoverflow.net/q/6508/2024. $\endgroup$ – Dror Speiser Dec 28 '16 at 20:09
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    $\begingroup$ Class field theory suggests ${\mathbb R}$ (with addition). $\endgroup$ – Felipe Voloch Dec 28 '16 at 22:29
  • $\begingroup$ @FelipeVoloch Could you develop your suggestion please (comment or answer) $\endgroup$ – Muhammed Ali Dec 29 '16 at 9:03
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The Galois group of the maximal abelian extension of $\mathbb Q$ (or any number field) is given (class field theory) as the quotient of the idele class group by the connected component of the identity which is isomorphic to $\mathbb R$. If there is an ${\mathbb F}_1$ then its extensions provide "constant field extensions" of $\mathbb Q$. So, to be compatible with class field theory and keep the analogy with function fields, $\mathbb R$ must be at least the abelianization of the absolute Galois group of ${\mathbb F}_1$. But finite fields are abelian, so this suggests the answer. A slightly different perspective is that Weil constructed canonically and functorially an extension of the absolute Galois group of any number field by ${\mathbb R}$ (the Weil group) and that should be the extension one would get by allowing constant field extensions again.

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  • $\begingroup$ $\mathbb F_1$, whatever it is or isn't, is weird. $\endgroup$ – Wojowu Dec 29 '16 at 12:58
  • $\begingroup$ I think the “which is isomorphic to $\mathbb{R}$” part is wrong: it should be $\prod_p \mathbb{Z}_p^\times$ (at any rate, it should be profinite and there is torsion). $\endgroup$ – Gro-Tsen Aug 15 '19 at 12:50
  • $\begingroup$ @Gro-Tsen I was referring to the connected component of the identity and you seem to be talking about the quotient. $\endgroup$ – Felipe Voloch Aug 15 '19 at 15:57
  • $\begingroup$ OK, but then I don't understand “$\mathbb{R}$ must be at least the abelianization of the absolute Galois group of $\mathbb{F}_1$”… $\endgroup$ – Gro-Tsen Aug 15 '19 at 22:33
  • $\begingroup$ Imagine that the whole ideal class group somehow describes abelian extensions of $\mathbb{Q}$ in some extended sense which includes "constant field extensions" so we should see the absolute Galois group of $\mathbb{F}_1$ (or its abelianization if it's nonabelian) in the ideal class group. The connected component of the identity is a candidate. $\endgroup$ – Felipe Voloch Aug 15 '19 at 23:38

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