# 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. – Asaf Karagila Mar 19 '19 at 9:41
• 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
• 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
• Is this related? math.stackexchange.com/questions/114978/… – Matt Cuffaro Mar 19 '19 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? – Noah Snyder Mar 20 '19 at 2:14