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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?

Asaf Karagila
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    Not fully sure about the tags, though. – Asaf Karagila Mar 19 '19 at 09:41
  • 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? – Alec Rhea Mar 19 '19 at 11:12
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    Alec, it appears in Hodges' Läuchli's algebraic closure of $Q$. Math. Proc. Cambridge Philos. Soc. 79 (1976), no. 2, 289–297. MR422022. – Asaf Karagila Mar 19 '19 at 11:25
  • Much appreciated Asaf; it seems I take choice for granted all too often, I thought these facts were more intrinsic to field theory. – Alec Rhea Mar 19 '19 at 11:30
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    Is this related? https://math.stackexchange.com/questions/114978/algebraic-closure-for-mathbbq-or-mathbbf-p-without-choice – Matt Cuffaro Mar 19 '19 at 17:43
  • @Matt: Not really. That tells us that for the rational numbers there is a canonical algebraically closed algebraic extension, so the spectrum is non-empty. But it's pretty far from anything I'm asking. – Asaf Karagila Mar 19 '19 at 17:45
  • Oh my misunderstanding. As an outsider to this, I'm excited to see what comes out of this question. Cheers. – Matt Cuffaro Mar 19 '19 at 17:48
  • @Matt: Not much, I expect. But maybe with time someone will find this interesting enough to pursue from a research point of view. – Asaf Karagila Mar 19 '19 at 17:49
  • It seems we can ask a similar question even with choice (but without global choice) about proper class sized fields like the field of fractions of the Grothendieck ring of the ordinals, since the poset/chains we want to Zorn are proper classes. It might be interesting to see how many algebraic closures a proper class sized field can have in the presence of choice without global choice. – Alec Rhea Mar 19 '19 at 18:05
  • Does the consistency of two closures follow because there are two sets for which no bijection exists and each set is contained by one of the closures? In which case, I suspect there are more than two closures. Gerhard "It's Not About The Algebra?" Paseman, 2019.03.19. – Gerhard Paseman Mar 19 '19 at 20:47
  • @Gerhard: In the case of the rationals, I guess? There is a canonically constructed closure which is countable. But it is consistent that there is an algebraically closed algebraic extension which is not countable. In fact, it does not have a real-closed subfield. – Asaf Karagila Mar 19 '19 at 20:51
  • @AlecRhea The answer seems pretty likely to be "proper class many", right? – Will Sawin Mar 19 '19 at 23:58
  • @WillSawin My intuition says yes, but I’m not sure — perhaps field properties like being formally real or having a particular valuation could have some impact on the answer. Hopefully looking at the arguments in the paper Asaf linked above can sharpen it somewhat. – Alec Rhea Mar 20 '19 at 00:35
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    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? – Noah Snyder Mar 20 '19 at 02:14
  • @Noah: That's a good question. I don't know. It is true that the standard axiomatization of ACF is complete. – Asaf Karagila Mar 20 '19 at 08:28
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    @Noah: Some months ago I was talking with another MO user about this question, and I realised that one of my recent papers implies that it is consistent that every field has a proper class of pairwise non-isomorphic algebraic closures, unless it is real-closed or algebraically closed. It's still not clear whether or not we can have some of them having the same cardinality. That's a great little idea. – Asaf Karagila Aug 30 '20 at 00:43
  • @AsafKaragila hi, may I ask which paper of yours implies this result? I would like to read it – Holo Jul 23 '21 at 09:11
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    @Holo: The one I am putting in arxiv later today. Check Monday's Replacements. – Asaf Karagila Jul 23 '21 at 09:13
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    @Holo: https://arxiv.org/abs/1911.09285 – Asaf Karagila Jul 26 '21 at 09:20
  • @AsafKaragila thanks! – Holo Jul 26 '21 at 19:15

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