## Element in the absolute Galois group of the rationals

Usually when people talk on the absolute Galois group G of ℚ they have in mind two elements they can describe explicitly, namely the identity and complex conjugation (clearly, everything is up to conjugation), although the cardinality of the group is uncountable.

Can you describe other elements of G?

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See the last of this extended answer. I'm going to part company with everyone else and say that you can describe other elements of $\text{Gal}(\mathbb{Q})$. In other words, I claim that you can identity a specific element of $\text{Gal}(\mathbb{Q})$ in a wide range of ways, together with an algorithm to compute the values of that element as a function on $\mathbb{Q}$. You can use either a synthetic model of $\overline{\mathbb{Q}}$, or its model as a subfield of $\mathbb{C}$. Although this is all doable, what's not so clear is whether these explicit elements are interesting.

The other two parts of the answer raise interesting issues, but they are moot for the original question.

This is not exactly the question, but it is related. To begin with, it is difficult to "explicitly" describe $\overline{\mathbb{Q}}$ except as a subfield of $\mathbb{C}$. I found a paper, Algebraic consequences of the axiom of determinacy (in English translation of the title) that establishes that $\mathbb{C}$ does not have any automorphisms other than complex conjugation in ZF plus the axiom of determinacy (AD). So you need some part of the axiom of choice (AC) for this related question.

As for the smaller field $\overline{\mathbb{Q}}$, the Wikipedia page for the fundamental theorem of algebra suggests that you might not even be able to construct it in the first place without the axiom of countable choice. (I say "suggests" because I'm not entirely sure that that is a theorem. Note that AC and AD both imply countable choice even though they are enemy axioms.) Any construction with countable choice isn't truly "explicit". On the other hand, if you allow countable choice, then I suspect that you can build $\overline{\mathbb{Q}}$ synthetically by induction rather than as a subfield of $\mathbb{C}$, and that you can build many automorphisms of it as you go along.

So the questions for logicians is whether there is a universe over ZF in which $\overline{\mathbb{Q}}$ does not exist, or a universe in which it does exist but has no automorphisms.

I got email about this from Kevin Buzzard that made me look again at the paper referenced by Wikipedia, A weak countable choice principle by Bridges, Richman, and Schuster. According to this paper, life is pretty strange without countable choice. You want to make the real numbers as the metric completion of the rationals. However, there is a difference between general Cauchy sequences and what they called "modulated" sequences, which are sequences of rationals with a promised rate of convergence. They cite a result of Ruitenberg that the modulated complex numbers are algebraically closed in ZF. Hence $\mathbb{Q}$ has an algebraic closure in ZF.

But it still seems possible that without countable choice, algebraic closures of $\mathbb{Q}$ need not be unique up to isomorphism, and that the complex analysis model of $\overline{\mathbb{Q}}$ might not have automorphisms other than complex conjugation.

A better and hopefully final technical answer: As mentioned, $\overline{\mathbb{Q}}$ exists explicitly (in just ZF) as a subfield of $\mathbb{C}$. You can also construct it synthetically as follows: Consider the monic Galois polynomials over $\mathbb{Z}$. These are the polynomials such that the Galois group acts freely transitively on the roots; equivalently the splitting field is obtained by adjoining just one root. The Galois polynomials can be written in a finite notation and enumerated. Beginning with $\mathbb{Q}$, formally adjoin a root of $p_n(x)$, the $n$th monic Galois polynomial, for each $n$ in turn. If $p_n(x)$ factors over the field constructed so far, the factors can also be expressed in a finite notation; take the first irreducible factor. The result is an explicit, synthetically constructed $\overline{\mathbb{Q}}$.

For comparison, let $\widetilde{\mathbb{Q}}$ be the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$. Each element of it is computable: Its digits can be generated by an algorithm, even with an explicit bound on its running time. As we build $\overline{\mathbb{Q}}$, we can also build an isomorphism between $\widetilde{\mathbb{Q}}$. We can do this by sending the formal root of $p_n(x)$ to its first root in $\mathbb{C}$, using some convenient ordering on $\mathbb{C}$. Or we could just as well have used its last root, its second root if it has one, etc. Composing these many different isomorphisms between $\overline{\mathbb{Q}}$ and $\widetilde{\mathbb{Q}}$ gives you many field automorphisms.

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Leila Schneps claims that one cannot write down any other element.

See her paper in The Grothendieck theory of dessins d'enfants. Papers from the Conference on Dessins d'Enfant held in Luminy, April 19--24, 1993. Edited by Leila Schneps. London Mathematical Society Lecture Note Series, 200. Cambridge University Press, Cambridge, 1994. I think that http://www.aimath.org/WWN/motivesdessins/schneps1.pdf is more or less the same.

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what do you mean by write-down'? – Lior Bary-Soroker Nov 25 2009 at 13:23
I have no idea what exactly she meant. – Mariano Suárez-Alvarez Nov 25 2009 at 14:06
For Galois automorphisms of C, you actually need the axiom of choice for any to exist because (as an exercise) any others are Q-linear but not R-linear and hence are nonmeasurable functions. I am not sure if you actually need the axiom of choice to show that there are nontrivial Galois automorphisms of the closure of Q, but it may be similar - you need to make a choice of lift of the automorphism for every finite Galois extension F of Q in some compatible way. – Tyler Lawson Nov 25 2009 at 17:08

This is an analogous to the complex conjugation:

Let K be a local field, i.e. a finite extension of ℚp, for some p. Then its Galois group GK is finitely presented (See "Cohomology of number fields" of Jürgen Neukirch, Alexander Schmidt, and Kay Wingberg). Then its generators are in a sense explicit elements of G.

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I don't know of any other elements, but I just want to point out that it's hard to really explicitly describe "elements of $\text{Gal}~\bar{\mathbb{Q}}/\mathbb{Q}$"; see e.g. this answer for a much clearer explanation than I could provide.

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 I read this answer, and it's fine with me to find elements up to conjugation. Both the complex conjugation and the local elements' coming from local fields are defined up to conjugation. – Lior Bary-Soroker Nov 25 2009 at 13:25 Well, if you only care about them up to conjugacy, I suppose you could look at Galois representations of this group, about which some nontrivial things are in fact known, although I don't claim to understand it. – Harrison Brown Nov 25 2009 at 13:30

Dear "FC" Could you please give me a precise reference which I can find a proof of "Artin theorem" which is, up to conjugation non-trivial element of finite order in absolute Galois group of Q, is complex conjugation. I could not find it in the paper you have shared. you wrote "This explains Leila Schneps' remark" which is not clear where you mean.

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 the theorem appears, e.g., in Lang's "Algebra" – Lior Bary-Soroker Oct 16 2010 at 6:00