In an answer to this question I was led to show the trick proving that $\mathbb R$ is the fraction field of some strict subring $A\subsetneq \mathbb R=\operatorname{Frac}(A)$.
A crucial point in the trick is the existence of a transcendence basis of $\mathbb R$ over $\mathbb Q$.
So this necessitates the axiom of choice and I would be grateful to a savvy user for confirming (or infirming) my feeling that the existence of such an $A$ is indeed only possible under acceptance of the axiom of choice.
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asked Apr 7, 2014 at 23:12
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3$\begingroup$ 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. $\endgroup$– Asaf Karagila ♦Commented Apr 7, 2014 at 23:24
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$\begingroup$ 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 $\Bbb R$? $\endgroup$– Asaf Karagila ♦Commented Apr 10, 2014 at 18:05
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$\begingroup$ 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 $A$ . $\endgroup$– Georges ElencwajgCommented Apr 15, 2014 at 7:07
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$\begingroup$ 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. $\endgroup$– Asaf Karagila ♦Commented Apr 15, 2014 at 7:14
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2$\begingroup$ 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? $\endgroup$– fedjaCommented Nov 12, 2017 at 21:28
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