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My understand (from this question) is that we can construct a countable algebraic closure of $\mathbb{Q}$ without using the axiom of choice (i.e. we can construct $\overline{\mathbb{Q}} \subseteq \mathbb{C}$), although perhaps this oversimplifying thing.

What can we say about $\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ without using the axiom of choice? I think the infinite Galois correspondence requires some form of choice, so maybe this the automorphism group of $\bar{\mathbb{Q}}$ doesn't deserve to be called a Galois group without choice. Can it be shown that it is uncountable? Or that it has more than $2$ elements?

Motivation: people often say that it is impossible to write down any non-trivial element of $\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. If it is consistent with ZF that $\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ has size $2$, then this would be a way to formalize that statement.

vukov
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  • See this very similar question, which went unanswered. – Alex Kruckman Dec 16 '18 at 23:39
  • To be clear, do you mean for $\bar{\mathbb{Q}}$ to refer specifically to a countable algebraic closure of $\mathbb{Q}$, or to an arbitrary algebraic closure? Not all algebraic closures are necessarily countable without choice. – Eric Wofsey Dec 17 '18 at 00:11
  • @EricWofsey I was thinking about the "standard algebraic closure" of $\mathbb{Q}$, i.e. the set of algebraic numbers in $\mathbb{C}$. If it isn't possible to make sense of that without choice, then I would be interested if it was possible to show that there exists an algebraic closure of $\mathbb{Q}$ with more than $2$ elements in its automorphism group. – vukov Dec 17 '18 at 00:17
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    I don't think people usually say it's impossible to write down a non-trivial element of $\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$ (there's just not any simple formula for one). Rather, it's $\text{Gal}(\mathbb{C}/\mathbb{Q})$ which you can't write down any nontrivial elements of. – Eric Wofsey Dec 17 '18 at 00:39
  • @EricWofsey I'm sure you understand this piece of theory better than I do, but I would think that in $Gal(\Bbb{C}/\Bbb{Q})$ we have at least the complex conjugation, but other non-trivial automorphisms exist only by virtue of Zorn's lemma. On the other hand, depending on how an element of $\overline{\Bbb{Q}}$ is described, it may be taxing to decide whether it is real or not. Making it difficult to describe its complex conjugate, no? – Jyrki Lahtonen Dec 17 '18 at 06:28
  • @JyrkiLahtonen: By "nontrivial" I meant "besides the obvious two", as I gathered OP meant in their original statement about $\text{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$. – Eric Wofsey Dec 17 '18 at 06:29
  • @EricWofsey Understood. Sorry about the nitpick. +1 for the explanation of the use of an enumeration (=well order). – Jyrki Lahtonen Dec 17 '18 at 06:30
  • IOW, I'm still wondering whether it is possible to describe complex conjugation in $\overline{\Bbb{Q}}$ without embedding it into $\Bbb{C}$? I would think not. – Jyrki Lahtonen Dec 17 '18 at 06:40
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    @JyrkiLahtonen: Well, every automorphism of $\overline{\mathbb{Q}}$ of order $2$ is conjugate to complex conjugation (and of course, if you don't have a chosen embedding in $\mathbb{C}$, you can only hope to describe it up to conjugation). See https://math.stackexchange.com/questions/2563758/are-all-elements-of-order-2-in-mathrmgal-overline-bbb-q-bbb-q-conj. – Eric Wofsey Dec 17 '18 at 18:47
  • This of course does not tell you how to construct such an automorphism, but I don't know how you would do that without essentially using an embedding into $\mathbb{C}$, since as soon as you have an automorphism of order $2$ you get an associated embedding into $\mathbb{C}$ (the embedding of the fixed field of an automorphism of order $2$ into $\mathbb{R}$ can easily be defined in terms of which elements are squares). – Eric Wofsey Dec 17 '18 at 18:47

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If $\overline{\mathbb{Q}}$ is a countable algebraic closure of $\mathbb{Q}$, the axiom of choice is hardly if ever needed to study $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. All of the usual arguments using the axiom of choice work without it, by just explicitly using an enumeration of $\overline{\mathbb{Q}}$ in place of the well-ordering that would be used via the axiom of choice. For instance, you can construct elements of $\text{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ by one-by-one extending automorphisms of subfields of $\overline{\mathbb{Q}}$ which are finite and Galois over $\mathbb{Q}$ (no choice is ever needed since you have a well-ordering of $\overline{\mathbb{Q}}$ which you can use to make all your choices). The usual diagonalization argument goes through to show there are uncountably many such elements. Similarly, you can prove that any two countable algebraic closures of $\mathbb{Q}$ are isomorphic.

More generally, pretty much the same can be said for any algebraically closed field $K$ that is well-orderable. The only way the usual theory of algebraically closed fields (and more generally of infinite field extensions) uses the axiom of choice is to have a well-ordering on the field so that you can construct homomorphisms out of (and/or into) it by transfinite recursion. So, if you already know that such a well-ordering exists, you don't need the axiom of choice.

Eric Wofsey
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  • As far as I can see, the set of algebraic numbers in $\mathbb C$ is countable, even without choice. The point is that (1) we can enumerate all the nonconstant polynomials over $\mathbb Q$, (2) each polynomial has only finitely many roots, and (3) each finite subset of $\mathbb C$ has a canonical ordering, namely the lexicographic ordering according to real parts and then imaginary parts. – Andreas Blass Dec 17 '18 at 03:32
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    Perhaps a better formulation of my preceding comment is that, if an algebraic extension of a countable field is linearly orderable as a set, then it is countable. – Andreas Blass Dec 17 '18 at 03:33
  • The "...to show there are uncountable many such elements" part seems to be, at least, confusing if not completely wrong. There are only countable many algebraic elements over the rationals, and thus $;|\overline{\Bbb Q}|=|\Bbb Q|=\aleph_0;$ . Nevertheless, it seems to be the "uncountable elements" there refers to the elements of the full Galois group, though I can't see how the diagonal argument of Cantor can be applied in a meaningful way there. – DonAntonio Dec 17 '18 at 08:11