Timeline for Are the reals really a fraction field?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 18, 2018 at 14:42 | comment | added | Franka Waaldijk | @fedja could you explain how this subgroup A is constructed? | |
| Nov 12, 2017 at 21:28 | comment | added | fedja | The funny thing is that one can easily construct a proper additive subgroup $A$ of $\mathbb R$ that generates $\mathbb R$ as a ring without AC. I surmise you know that already, but, once you observe that, the belief that the existence theorem in your question really requires AC gets somewhat shaken, doesn't it? | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Apr 15, 2014 at 7:14 | comment | added | Asaf Karagila♦ | I see. Thank you for the answer. If we can show that there are $2^\frak c$ automorphisms of $A$ then it certainly can't be doable without the axiom of choice (it is consistent there are only $\frak c$ automorphisms of $\Bbb R$ as an additive group). Of course it doesn't mean the entire result fails, just this particular approach, but it would be a first step, I suppose. | |
| Apr 15, 2014 at 7:07 | comment | added | Georges Elencwajg | Dear @Asaf, every automorphism of $A$ can be extended to an automorphism of $\mathbb R$ (apply the automorphism to the numerator and the denominator of a fraction), but I'm far for sure that every permutation of the $r_i$'s can be extended to $A$ . | |
| Apr 10, 2014 at 18:05 | comment | added | Asaf Karagila♦ | Georges, being not quite fluent in algebra; in your answer on MSE, when you consider the $A$ to be the integral closure of $\Bbb Q[r_i\mid i\in I]$, does every permutation of the $r_i$'s induce an automorphism of $A$, and can that automorphism be extended uniquely to an automorphism of $\Bbb R$? | |
| Apr 7, 2014 at 23:24 | comment | added | Asaf Karagila♦ | I have a feeling that you might be able to show that $A$ cannot have the Baire property; or maybe Lebesgue measurability; or that it cannot be Borel; or something like that. In either of these cases it seems that the axiom of choice is needed. | |
| Apr 7, 2014 at 23:12 | history | asked | Georges Elencwajg | CC BY-SA 3.0 |