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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?

[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.

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  • $\begingroup$ Might be that model-theory is not really a suitable tag, feel free to modify if you have strong opinions on that. $\endgroup$
    – Lukas Juhrich
    Jul 23, 2019 at 19:27
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    $\begingroup$ The model-theory tag is definitely appropriate here. $\endgroup$
    – Alex Kruckman
    Jul 23, 2019 at 22:42

1 Answer 1

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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 $K$.

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  • $\begingroup$ Wait a second. Don't we sneakily use choice here (or in most proofs using compactness) by not spexifying the models explicitly? $\endgroup$
    – Lukas Juhrich
    Jul 27, 2019 at 18:41
  • $\begingroup$ The hypotheses of the compactness theorem are that for each finite subset there exists a model, not that there exists a family of models simultaneously for all of them. So you don't need to choose them in order to invoke compactness. $\endgroup$
    – Eric Wofsey
    Jul 27, 2019 at 19:52
  • $\begingroup$ Note that in particular you can prove the compactness theorem without making a choice of such models: the mere existence of such models tells you that the theory is consistent (syntactically), and then you can construct a model of it using the completeness theorem. $\endgroup$
    – Eric Wofsey
    Jul 27, 2019 at 19:55
  • $\begingroup$ (Or, working semantically, you can imitate a proof of the completeness theorem using "finitely satisfiable" in place of "consistent", which gives you Henkin's proof of compactness which never uses the specific models which witness finite satisfiability as a family, just the fact that each one exists.) $\endgroup$
    – Eric Wofsey
    Jul 27, 2019 at 20:25
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    $\begingroup$ Note that even if you did need a family of models (say, if you were using the ultraproduct proof of compactness), that still doesn't require AC, because you can just make all choices at once (nothing requires you to have exactly one model for each finite subset of the theory). Or more precisely, in order to have a set and not a proper class, you could use all models of minimal rank. $\endgroup$
    – Eric Wofsey
    Jul 27, 2019 at 20:56

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