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Suppose we work in a model in which the Axiom of Choice does not hold, and in which $\mathbb{C}$ only has one nontrivial automorphism (such models exist).

Question: "how many" subfields of $\mathbb{C}$ are isomorphic to $\mathbb{R}$?

More precisely: in what way does the size of the automorphism group of $\mathbb{C}$ influence/control this question?

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    $\begingroup$ It is consistent that there are no such subfields other than $\mathbb R$ itself, but I don't know if it happens in every model in which there are no nontrivial automorphisms. $\endgroup$ – Wojowu Mar 21 at 22:22
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    $\begingroup$ Just for context, under AC, the answer is $2^{2^{\aleph_0}}=2^c$. (The reason is (a) that the automorphism group has this cardinal, and (b) acts on the set of such subfields with $\mathbf{R}$ having a stabilizer of cardinal 2, so its orbit has cardinal $2^c$.) $\endgroup$ – YCor Mar 21 at 22:22
  • $\begingroup$ I have a feeling that I recently (last few months?) saw this question somewhere online. Odd. $\endgroup$ – Asaf Karagila Mar 21 at 22:34
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    $\begingroup$ By the way, a related question is: how many subfields are isomorphic to $\mathbf{C}$. Under AC, the answer is again $2^c$. In general, the "algebraic closure" map from the set of subfields isomorphic to $\mathbf{R}$ to the set of subfields isomorphic to $\mathbf{C}$ is surjective. $\endgroup$ – YCor Mar 21 at 23:06
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    $\begingroup$ @LSpice All fibers of this map have the same cardinality, since all fields isomorphic to $\mathbb C$ will have the same number of copies of $\mathbb R$ which make it an algebraic closure. I believe all of these copies of $\mathbb R$ form one orbit under the automorphism group of $\mathbb C$, whose size is half the size of the automorphism group. $\endgroup$ – Wojowu Mar 22 at 1:39

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