# How many algebraic closures can a field have?

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?

• Not fully sure about the tags, though. Mar 19, 2019 at 9:41
• Alec, it appears in Hodges' Läuchli's algebraic closure of $Q$. Math. Proc. Cambridge Philos. Soc. 79 (1976), no. 2, 289–297. MR422022. Mar 19, 2019 at 11:25
• Is this related? math.stackexchange.com/questions/114978/… Mar 19, 2019 at 17:43
• 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? Mar 20, 2019 at 2:14
• @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. Aug 30, 2020 at 0:43