Proofs that every field has a unique (up to isomorphism) algebraic closure use some form of the axiom of choice. For uniqueness this is provably necessary: there are models of ZF in which $\mathbb{Q}$ has two non-isomorphic algebraic closures. Existence is trickier, because algebraic closures of many familiar fields can be constructed by hand.

For example, we can construct an algebraic closure of $\mathbb{Q}$ (or any number field) as the algebraic numbers inside $\mathbb{C}$. More generally, algebraic closures for any countable field can be constructed explicitly (enumerate the irreducible polynomials, adjoin a root to the first and enumerate the new field, factor the second over the new field and adjoin a root of the lexicographically first factor,...). As a further example, the function field of an algebraic curve over $\mathbb{C}$ embeds in the field of Laurent series (choose a local parameter at a point), so this field has an algebraic closure sitting inside the field of Puiseaux series.


Are there examples of specific, explicit fields for which no algebraic closure can be constructed in ZF?


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There is [a similar question on math.SE][1]  with no answers. I tried offering a bounty, to no avail.


  [1]: https://math.stackexchange.com/questions/803524/fields-whose-algebraic-closure-cannot-be-constructed-without-the-axiom-of-choice