We consider $G_{\mathbb Q} = Gal(\mathbb {\bar Q}/\mathbb Q)$. The Frobenius elements corresponding to each prime are well-studied. But these are really not elements; these are only defined as some conjugacy classes(upto inertia, etc..)

Question: Are these the only conjugacy classes in the absolute Galois group? If there are others, please give examples or methods to construct them.

The conjugacy classes are of course defined algebraically; this question is not asking for results of the form that the Frobenii form a dense set.


Firstly, Frobenius elements aren't even conjugacy classes, as you know. So you had better look at quotients $Gal(K/\mathbf{Q})$ of the Galois group which are unramified outside some set $S$. Now you have Frobenius conjugacy classes for all $p$ not in $S$. But now there's a dichotomy. If the quotient is finite, then we see every conj class infinitely often. If the quotient is infinite, then the Frobenius conj classes are dense, as you know, but again trivially there are conj classes that aren't Frobenius elements, because there are only countably many of them, and there will almost always (and perhaps always? not sure) be uncountably many conj classes in the Galois group---for example if $K$ is the maximal extension unramified outside $S$ and $S$ is non-empty then $K$ contains the $p$-cyclotomic extension for some prime $p\in S$ and so the Galois group has a quotient isomorphic to $\mathbf{Z}_p^\times$, and this group is uncountable and the Frobenius conj classes here are just the elements $\ell\in\mathbf{Z}_p^\times$ for $\ell$ running through the primes other than $p$. So now we can see that not only are there uncountably many conj classes that aren't Frobenius elements, we can see uncountably many that aren't in the group generated by the Frobenius elements. I have no idea how to construct these guys in any sort of natural way.

  • $\begingroup$ This is a side comment, but I think it's plausible that if G is any infinite profinite group, then there are uncountably many conjugacy classes in G. In fact, I believe that in such a G the Haar measure of any conjugacy class is zero (but please correct me if I am wrong). $\endgroup$ – senti_today Mar 10 '10 at 20:45
  • 2
    $\begingroup$ @senti_today : The first statement is true, because the number of conjugacy classes in a finite group cannot remain bounded as the size of the group increases. The second statement is false: Reflections inside a dihedral group are a conjugacy class of measure $\frac{1}{2}$, and you can convert this to an infinite example. $\endgroup$ – moonface Mar 10 '10 at 23:16
  • $\begingroup$ @moonface: I am not sure I follow your first comment. How does it imply that the number of conjugacy classes is not countable? In the second comment, you probably mean "dihedral group of order not divisible by 4" (otherwise there are two conjugacy classes of reflections, if I'm not mistaken). If I understand you correctly, the sort of example you have in mind is: consider $\mathbb{Z}/2\mathbb{Z}$ acting on $\mathbb{Z}_p$ via $x\mapsto-x$, where $p$ is an odd prime, and form the semidirect product. Then the nontrivial coset of $\mathbb{Z}_p$ is a single conjugacy class of Haar measure 1/2. $\endgroup$ – senti_today Mar 10 '10 at 23:49
  • $\begingroup$ Yes, dihedrals of 2xodd order, and that's exactly what I had in mind. About the first comment, I don't follow it either (I had hastily assumed that any inverse limit of finite sets with increasing size and surjective transition maps is uncountable...) Thanks for catching. $\endgroup$ – moonface Mar 11 '10 at 2:38

I agree with the answer of Kevin Buzzard -- you better look only at quotients $Gal(K/Q)$ unramified outside some (finite) set $S$ to make the question make sense as it is.

But regardless, you can ask about the conjugacy classes in the absolute Galois group, and what they "mean". An answer was given many years ago by Ax (I guess in one of his Annals papers, 1968 or 1969), and there are elaborations and new results in the thesis of James Gray, now published in the J. of Symbolic Logic as "Coding complete theories in Galois groups" (his thesis is also available freely online).

The reason why these "other conjugacy classes" (which are essentially all of them) are difficult to construct is that the associated fixed fields are the algebraic (over $Q$) subfields of pseudofinite fields of characteristic zero. In Theorem 1.27 of his thesis, Gray states that there is a [natural, homeomorphic in the appropriate topology] bijection between the set of conjugacy classes in $Gal(\bar Q / Q)$ and the Stone space of completions of $ACFA_0$ -- the theory of algebraically closed fields of characteristic zero with generic automorphism. Given such a completion of $ACFA_0$, realized by a model $(K, \sigma)$ where $K$ is algebraically closed, and $\sigma$ is a generic automorphism (see MacIntyre, "Generic automorphisms of fields" for the definition of generic used here), the fixed field $K^\sigma$ is a pseudofinite field of characteristic zero.

Now, while two conjugacy classes in $Gal(\bar Q / Q)$ are quite simple to describe -- the trivial conjugacy class and the conjugacy class of order 2 elements -- other conjugacy classes contain elements of infinite order. The associated fixed fields (in this infinite-order case) in $\bar Q$ are psuedofinite fields, which are difficult to get your hands on. Probably the best way is to take an ultraproduct (for a non-principal ultrafilter on the set of prime numbers) of finite fields, and take the algebraic elements within. This is certainly nonconstructive, relying heavily on Zorn's lemma.

Still, such fields have arithmetic significance. The model theory related to $ACFA_0$ has had a great deal of impact on number theory lately, and Fried-Jarden also touch on related matters in their "Field Arithmetic" book.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.