TITLE: Algebraic closure for $\mathbb{Q}$ or $\mathbb{F}_p$ without Choice?
QUESTION [30 upvotes]: 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.

REPLY [3 votes]: I believe we can generalize a bit: every well-orderable field $K$ has a well-orderable algebraic closure. (and any two are isomorphic)
First, choose a well-ordering $<$ on $K$. Also, choose a well-ordering of the polynomials over $K$ (e.g. a lexicographic order). The $\alpha$-th polynomial is $f_\alpha$.
Using transfinite recursion, construct a family $L_\alpha$ of algebraic extension fields of $K$, and well-orderings $<_\alpha$ on $L_\alpha$ such that, for any $\alpha < \beta$:

$L_\alpha$ is a subfield of $L_\beta$
If $x \in L_\alpha$ and $y \in L_\beta \setminus L_\alpha$, then $x < y$.

The base case will be $L_0 = K$ and $<_0 = <$. At each $\alpha > 0$, perform the following:

$E = \cup_{\beta < \alpha} L_\beta$ is an algebraic extension of $K$
$<_E = \cup_{\beta < \alpha} <_\beta$ is a well-ordering on $E$
Construct $L_\alpha$, a splitting field of $f_\alpha$ over $E$
Modify $L_\alpha$ so that $L \subseteq L_\alpha$.
Construct a well-ordering $<_\alpha$ on $L_\alpha$ (e.g. lexicographical, as a finite-dimensional $E$-vector space)
Modify $<_\alpha$ so that $E$ is an initial segment of $L_\alpha$

The main thing to make this work, I think, is to verify making the splitting field is constructive -- e.g. note that we have a canonical choice of well-ordering on polynomials over $E$, and thus a canonical way to choose factor of $f_\alpha$ each time we need to do so.