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Jul 9, 2023 at 7:51 comment added The Thin Whistler @FelipeVoloch could you explain again in a little more détail that sentence "$\mathbb{R}$ must be at least the Abelianization of the absolute Galois group of $\mathbb{F}_{1}$"?
Dec 11, 2021 at 17:16 comment added Asvin If we think about a curve over a finite field, then the Galois group of the finite field shows up as the quotient of the total fundamental group, right? So your argument really suggests that the galois group of F_1 should be the units in the profinite completion. In fact the Cyclotomic extensions should be the "constant" extensions as suggested by many things including Iwasawa theory.
Aug 15, 2019 at 23:38 comment added Felipe Voloch 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.
Aug 15, 2019 at 22:33 comment added Gro-Tsen OK, but then I don't understand “$\mathbb{R}$ must be at least the abelianization of the absolute Galois group of $\mathbb{F}_1$”…
Aug 15, 2019 at 15:57 comment added Felipe Voloch @Gro-Tsen I was referring to the connected component of the identity and you seem to be talking about the quotient.
Aug 15, 2019 at 12:50 comment added Gro-Tsen 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).
Dec 29, 2016 at 12:58 comment added Wojowu $\mathbb F_1$, whatever it is or isn't, is weird.
Dec 29, 2016 at 9:28 history answered Felipe Voloch CC BY-SA 3.0