How can we prove the existence of algebraic closures from first-order compactness?

Question

A devilish remark in „Aluffi: Algebra Chapter 0“[1, Remark VII.2.7] claims that

$^7$ Allegedly, the existence of algebraic closures is a consequence of the compactness theorem for first-odrer logic, which is known to be weaker than the axiom of choice.

I then found out the following things:

• The Compactness theorem is equivalent to the Ultrafilter lemma (equivalently, the boolean prime ideal theorem)
• We can modify Artins iterative approach by extending a certain ideal to a prime ideal (see [2]) and taking the fraction field, which thus constructs an algebraic closure only using the boolean prime ideal theorem.

However, I have not seen a model-theoretic proof, i.e., a proof using the compactness theorem for first-order logic directly.

Is such a proof known?

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[1] Aluffi, Paolo. Algebra: Chapter 0: Chapter 0. Vol. 104. American Mathematical Soc., 2009.

[2] Banaschewski, Bernhard. "Algebraic closure without choice." Mathematical Logic Quarterly 38.1 (1992): 383-385.

Answer 1

Sure. Fix a field $K$, and consider the theory over the language of commutative rings together with a constant symbol for each element of $K$ with the following axioms:

• The field axioms
• Axioms specifying the ring operations on all the elements of $K$
• For each nonzero polynomial $p\in K[x]$, an axiom saying that $p$ splits.

Any finite subset of these axioms has a model: just take a finite field extension of $K$ in which all the polynomials mentioned in axioms of the third type split. So by compactness, there is a model $L$ of all the axioms. Taking $\overline{K}$ to be the subfield of $L$ consisting of elements which are algebraic over $K$, we see that $\overline{K}$ is an algebraic closure of $K$.