I know the usual proof of the existence of an algebraic closure for any field using Zorn's Lemma. The answer to this previous question makes it clear that in general, some nonconstructive axiom (not necessarily the full AC) is needed to guarantee an algebraic closure. My question is if we can avoid any of this in the cases of $\mathbb{Q}$ and $\mathbb{F}_p$.
Can algebraic closures for $\mathbb{Q}$ and $\mathbb{F}_p$ be constructed in ZF?
Intuitively, it seems plausible to me that they can. There are two places I see a need for an AC-typed axiom in the construction of an algebraic closure: one is to create some order (i.e. bijection with $\mathbb{N}$) for the set of polynomials whose roots I need to adjoin; two is to handle what happens when I start adjoining roots and the polynomials start factoring into smaller factors. (Which factor do I approach first?) It seems to me that $\mathbb{Q}$ and $\mathbb{F}_p$ both have structure that could be used cleverly to resolve both of these points without recourse to AC. However, I don't see the path clearly.
Best Answer
For any finite or countable field $K$, you can well-order $K$ even without AC, and then you don't actually need any further choice to construct a closure for it.
Namely, since $K$ can be well-ordered, you can well-order all monic polynomials over $K$ and adjoin roots for the irreducible ones one by one by transfinite induction up to $\omega_1$. Each time we adjoin elements, we can well-order the new elements and stick them at the end of the well-ordering of the ones we already have. If we order the polynomials primarily by "maximal coefficient" rather than by degree, the new polynomials that become possible after each extension will always come after the ones we already know.
By the time we reach $\omega_1$, there cannot be any more polynomials that need to have roots adjoined. Namely, every polynomial we can form at that point will have had each of its coefficients added at a time when there were only countably many polynomials in our list of polynomials to process, so this polynomial will have been processed at some step before $\omega_1$.
Edit to add: In fact, as Zhen Lin points out, one only has to adjoin roots for polynomials with coefficients in $K$. For $K=\mathbb Q$ that is shown in this question, but the arguments there appear to work in general. This is easy enough to do for any well-orderable $K$, not just countable ones.