Richard Taylor once explained his research (in number theory) in a very simple way: understanding the absolute Galois group $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$.

It is known that in Zermelo-Fraenkel theory without the Axiom of Choice, it is consistent to say that $\vert \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \vert = 2$.

What are the consquences for number theory in such a model ?

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    $\begingroup$ Why would you need the axiom of choice to show that $\mathbb{Q}(\sqrt[3]{7},\zeta_3)$ is a Galois extension of degree larger than $2$? $\endgroup$ Jun 2, 2022 at 13:26
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    $\begingroup$ @ChrisWuthrich Is there a way to lift those automorphisms to $\overline{\mathbb Q}$ without AC? The usual argument uses Zorn's Lemma. $\endgroup$ Jun 2, 2022 at 13:37
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    $\begingroup$ @ArielWeiss In the absence of AC you have to be careful what you mean by the symbol $\overline{\mathbb{Q}}$ since one cannot prove the uniqueness of the algebraic closure. If you are referring to the usual algebraic closure, then you don't need Zorn's lemma because you can use the natural ordering on $\mathbb{N}$ to list all polynomials with integer coefficients in a canonical way. $\endgroup$ Jun 2, 2022 at 13:47
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    $\begingroup$ AC has no relevant consequence on arithmetic. In fact, defining things carefully it can even be a theorem that arithmetical statement are independant of AC. When people say they want to understand the absolue Galois group, what they mean is that they want to understand the category of finite extention of $\mathbb{Q}$. If you assume AC, then that's indeed the same as understanding the absolute Galois group, but if you don't the absolute Galois group is no longer relevant. We can rephrase most of what we know about it in terms of finite extension and make these independent of AC. $\endgroup$ Jun 2, 2022 at 14:17
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    $\begingroup$ @SamHopkins There is a more charitable reading of the question, which is that it asks if there exists an interesting exotic object that is similar to a known object but that has been undeservedly neglected. In the case at hand, I think the answer is no (if we define "interesting" as "number-theoretically interesting"), but I don't see anything wrong with asking the question. $\endgroup$ Jun 2, 2022 at 14:42

2 Answers 2


In the absence of the axiom of choice, it is still possible to define the "usual" algebraic closure of $\mathbb{Q}$ because you can just explicitly enumerate all polynomials with integer coefficients. Richard Taylor's remark should be interpreted as referring to this particular algebraic closure.

What you can't prove without choice is the uniqueness of the algebraic closure. Exotic algebraic closures could exist. So in the absence of choice, it's a bit misleading to use the notation $\overline{\mathbb{Q}}$ as if it referred to some unique thing. Wilfrid Hodges, in his paper, Läuchli's algebraic closure of $Q$, shows that an exotic algebraic closure of $\mathbb{Q}$ can have all kinds of unexpected properties. But what this shows, in my opinion, is that what we usually think of as "algebraic number theory" doesn't make too much sense in that context. When people talk about "algebraic number theory" they are tacitly referring to the usual algebraic closure.

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    $\begingroup$ I think it is more correct to say that when people talk about "algebraic number theory" they only talk about finite extention of $\mathbb{Q}$. The "usual" algebraic closure can have very few automorphisms. $\endgroup$ Jun 2, 2022 at 14:17
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    $\begingroup$ @SimonHenry Is it that the usual algebraic closure can have few automorphisms? Specifically, using $\overline{\mathbb{Q}}$ for the algebraic closure of $\mathbb{Q} \subset \mathbb{C}$, is it possible that there is a Galois extension $K/\mathbb{Q}$ for which $\text{Aut}(\overline{\mathbb{Q}}/\mathbb{Q}) \to \text{Gal}(K/\mathbb{Q})$ is not surjective? I used to think this could happen, but Timothy Chow's comment on the original question suggests not, and now I think I can prove that it can't happen. $\endgroup$ Jun 2, 2022 at 14:29
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    $\begingroup$ @SimonHenry Every automorphism of a finite extension of $\mathbb Q$ lifts to an automorphism of the “usual” algebraic closure, as the latter is countable. $\endgroup$ Jun 2, 2022 at 14:30
  • $\begingroup$ Interesting. So there is only one countable algebraic closure. Thanks ! $\endgroup$ Jun 2, 2022 at 16:03
  • $\begingroup$ I'm a bit confused, how can one prove that the algebraic closure is countable ? The "order the polynomials canonically" only seems to give that the algebraic closure is a countable union of finite sets. Must that be countable without AC ? (once you have a well-order on $\overline {\mathbb Q}$ clearly there is no issue) $\endgroup$ Jun 9, 2022 at 13:34

There would be no consequences, for two reasons:

  1. As Timothy Chow points out, if we define $\overline{\mathbb Q}$ as the set of complex numbers that are roots of a nonzero polynomial with rational coefficients, it is easy to prove the Galois group maps surjectively to the automorphism group of each finite Galois extension of $\mathbb Q$.

  2. If we wanted to consider the Galois group of another field where there there really are set-theoretic issues with finding enough automorphisms (e.g. studying the automorphism group of $\mathbb C$) we could simply redefine our notion of "group" so it doesn't necessarily have a set of elements (e.g. as a group object in the category of locales) and there would be no difficulty with redoing the standard arguments in this setting.

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    $\begingroup$ On the other hand, we could say that number theory is about studying the absolute Galois groupoid of $\mathbb{Q}$, and the statement that there is more than one algebraic closure of $\mathbb{Q}$ means its absolute Galois groupoid is not connected, which seems like its number-theoretically meaningful. I think it makes sense to ask more about these other, nonstandard, components, and I'm not convinced there's no interesting number-theoretic question there. $\endgroup$
    – Gro-Tsen
    Jun 9, 2022 at 14:52
  • $\begingroup$ (This reminds me of the question of whether we can define right-derived functors for $M \mapsto M^I$ (where $I$ is a set and $M$ a module or even a vector space) in ZF, and, if so, what they look like. I can't decide whether it's a dumb question.) $\endgroup$
    – Gro-Tsen
    Jun 9, 2022 at 14:57

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