Recently I have been fascinated by properties of fields in ZF-models in which the Axiom of Choice (AC) does not hold.

For instance, in some such models it is consistent to say that the field of complex numbers has only two automorphisms!

My question: let $\overline{\mathbb{Q}}$ be an algebraic closure of $\mathbb{Q}$. What can be said about $\mathrm{Aut}(\overline{\mathbb{Q}})$ in ZF-models without AC?

(I know one has to be very careful here, because algebraic closures of the rationals are not necessarily uniquely defined anymore in some models, if I am not mistaken.)

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    $\begingroup$ Have you checked out the various 'related' questions in the sidebar? There's not a direct answer to this question but there's a decent amount of discussion about similar topics. (e.g., 'The' countable $\overline{\mathbb{Q}}$ must have many automorphisms because most of the usual study of $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ doesn't need choice.) $\endgroup$ Apr 5, 2021 at 22:24
  • $\begingroup$ @Steven: But $\Bbb Q$ can have a proper class of pairwise non-isomorphic algebraic closures. Nevertheless, it seems that this is a question whose heart is model theoretic, in which case choice can probably be avoided. $\endgroup$
    – Asaf Karagila
    Apr 5, 2021 at 23:23
  • $\begingroup$ I would guess that there are models of ZF that have algebraic closures of $\mathbb{Q}$ with trivial automorphism group. $\endgroup$ Apr 6, 2021 at 2:11
  • $\begingroup$ Related: math.stackexchange.com/questions/1631918/… $\endgroup$
    – Harry West
    Jun 1, 2021 at 18:49


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