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