Assuming the axiom of choice given a field $F$, there is an algebraic extension $\overline F$ of $F$ which is algebraically closed. Moreover, if $K$ is a different algebraic extension of $F$ which is algebraically closed, then $K\cong\overline F$ via an isomorphism which fixes $F$. We can therefore say that $\overline F$ is the algebraic closure of $F$.

(In fact, much less than the axiom of choice is necessary.)

Without the axiom of choice, it is consistent that some fields do not have an algebraic closure. It is consistent that $\Bbb Q$ has two non-isomorphic algebraically closed algebraic extensions.

It therefore makes sense to ask: Suppose there are two non-isomorphic algebraically closed algebraic extensions. Is there a third? Are there infinitely many? Are there Dedekind-infinitely many?

What is provable from $\sf ZF$ about the spectrum of algebraically closed algebraic extensions of an arbitrary field? What about the rational numbers?

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    $\begingroup$ Not fully sure about the tags, though. $\endgroup$ – Asaf Karagila Mar 19 '19 at 9:41
  • $\begingroup$ Very cool question, could you provide a reference/proof hint for the comment about the consistency $\mathbb{Q}$ having more than one algebraic closure without choice? $\endgroup$ – Alec Rhea Mar 19 '19 at 11:12
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    $\begingroup$ Alec, it appears in Hodges' Läuchli's algebraic closure of $Q$. Math. Proc. Cambridge Philos. Soc. 79 (1976), no. 2, 289–297. MR422022. $\endgroup$ – Asaf Karagila Mar 19 '19 at 11:25
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    $\begingroup$ Is this related? math.stackexchange.com/questions/114978/… $\endgroup$ – Matt Cuffaro Mar 19 '19 at 17:43
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    $\begingroup$ What happens to the model theoretic results about algebraically closed fields without choice? Can you have uncountable nonisomorphic ACFs with the same characteristic and cardinality without choice? $\endgroup$ – Noah Snyder Mar 20 '19 at 2:14

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