Easy algebraic construction of $\overline{\mathbb Q}$?

Question

Let $\overline{\mathbb Q}$ be the algebraic closure of $\mathbb Q$. One usually viewes $\overline{\mathbb Q}$ as a subfield of $\mathbb C$. But $\mathbb R$, and hence $\mathbb C$, is defined analytically (e.g., as the completion of $\mathbb Q$). Is there a simple, purely algebraic way to construct $\overline{\mathbb Q}$, and prove that it is algebraically closed? Of course one can just follow the proof that all fields have an algebraic closure, but that seems quite complicated. Could the construction of $\overline{\mathbb Q}$ from $\mathbb Q$ be as easy as the construction of $\mathbb C$ from $\mathbb R$?

Answer 1

I'm not aware of a "simple" or "easy" way to do this. It's worth knowing that, in general, both the existence and uniqueness of algebraic closures are independent of ZF and require some use of choice. This MO discussion is quite nice and shows that the ultrafilter lemma suffices, but it's not known whether the two are equivalent. There is also a discussion at this math.SE question.

To be clear, proving that the algebraic closure of $\mathbb{Q}$ exists can be done in ZF without any choice; you can just explicitly describe an order in which to adjoin the roots of all irreducible polynomials. However, it's independent of ZF whether the algebraic closure is unique, and it's consistent with ZF that fields exist which have no algebraic closure.

This is all to give some indication that some "real work" is required to prove that algebraic closures exist. For $\mathbb{Q}$ I don't know a way to bypass this work without constructing $\mathbb{C}$, which as you say you want to avoid.

The nicest proof I know that algebraic closures exist in general is explained in this note by Keith Conrad, who attributes it to Brian Conrad and indirectly to Zorn. It's discussed in this $n$-category cafe post by Tom Leinster. The idea is that for a field $K$ we can (canonically, with no choice) construct a "splitting ring" $R$ (not field) of all polynomials over $K$, by writing down generators and relations corresponding to formally adjoining all roots of all (nonzero) polynomials. Then we need either the axiom of choice to find a maximal ideal $M$ of $R$, which gives the algebraic closure as $R/M$, or the ultrafilter lemma to find a prime ideal $P$ of $R$, which gives the algebraic closure as $\text{Frac}(R/P)$.

This argument by Joel David Hamkins proving both existence and uniqueness from the compactness theorem for first-order logic (equivalent to the ultrafilter lemma) is also quite nice and conceptual but requires some familiarity with first-order logic.