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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1$\begingroup$ See answers to this question: mathoverflow.net/q/6508/2024. $\endgroup$– Dror SpeiserCommented Dec 28, 2016 at 20:09
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1$\begingroup$ Class field theory suggests ${\mathbb R}$ (with addition). $\endgroup$– Felipe VolochCommented Dec 28, 2016 at 22:29
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$\begingroup$ @FelipeVoloch Could you develop your suggestion please (comment or answer) $\endgroup$– Muhammed AliCommented Dec 29, 2016 at 9:03
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1 Answer
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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.
answered Dec 29, 2016 at 9:28
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$\begingroup$ $\mathbb F_1$, whatever it is or isn't, is weird. $\endgroup$– WojowuCommented Dec 29, 2016 at 12:58
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$\begingroup$ 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-TsenCommented Aug 15, 2019 at 12:50
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$\begingroup$ @Gro-Tsen I was referring to the connected component of the identity and you seem to be talking about the quotient. $\endgroup$ Commented Aug 15, 2019 at 15:57
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$\begingroup$ 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-TsenCommented Aug 15, 2019 at 22:33
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$\begingroup$ 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. $\endgroup$ Commented Aug 15, 2019 at 23:38