This question began as Why are procyclic subgroups of Galois groups of number fields free profinite?, which fizzled out, but which garnered some helpful comments from YCor.

Let $K$ be a field, take $\omega \in \mathrm{Gal}(\bar{\mathbb{Q}}/K)$, and let $L_{\omega}$ be the fixed field of $\omega$. Then $\Gamma = \mathrm{Gal}(\bar{\mathbb{Q}}/L_{\omega})$ is a procyclic group, and (using Artin-Schreier) must be isomorphic to $$1, \mathbb{Z}/2\mathbb{Z}, \prod_{p\in S} \mathbb{Z}_p\textrm{, or }(\mathbb{Z}/2\mathbb{Z}) \times \prod_{2 \neq p \in S} \mathbb{Z}_p$$ for a subset $S$ of primes.

Question: Which of these groups can occur? Using the normalized Haar measure on $\mathrm{Gal}(\bar{K}/K)$, can one calculate the probability that an element will generate a given procyclic group?

An element $\omega \in \hat{\mathbb{Z}}$ will generate a subgroup isomorphic to $\hat{\mathbb{Z}}$ with probability 1. So this answers the case for finite fields, and also leads me to believe that $\hat{\mathbb{Z}}$ is likely to occur in general, but I don't know whether any $\hat{\mathbb{Z}}$ occurs in general or whether they would be outnumbered by many $\mathbb{Z}_p$ that do not extend.

I would be most interested in the field $K$ being $\mathbb{Q}$ or a number field, but also in $\mathbb{Q}^{ab}$ and function fields over $\mathbb{F}_q$ or $\bar{\mathbb{F}_q}$. The case of number fields may be easier if we assume totally imaginary, to dispense with the $\mathbb{Z}/2\mathbb{Z}$.

It would be interesting to know whether number fields can be distinguished by these probabilities.

  • $\begingroup$ The fourth type of groups, when $S$ is noempty, cannot occur as an absolute Galois group. More generally, it is a consequence of Artin-Schreier theory that the normalizer of an involution in an absolute Galois group is the sugroup of order $2$ generated by this involution. $\endgroup$
    – user05811
    Apr 14 '18 at 15:21

$\hat{\mathbb Z}$ occurs with probability one.

It is easy to define a homomorphism from the Galois group of a number field to $\hat{\mathbb Z}$. To do this, just define a homomorphism to $\mathbb Z_p$ for each $p$. It is sufficient to define such homomorphisms for $\mathbb Q$, as the restriction to the Galois group of a number field will be an open subgroup, also isomorphic to $\mathbb Z_p$. Over $\mathbb Q$, this can be done using the $p$-power cyclotomic tower and the $p$-adic logarithm.

Then by the argument you noted, with probability $1$ the image of your element inside this generates a $\hat{\mathbb Z}$. It is easy to see from this that your element generates $\hat{\mathbb Z}$ in the original Galois group.


For $K=\mathbb{Q}$, this was proved byJ. Ax (Solving diophantine problems modulo every prime, Ann. Math. 85 (1967), 161-183). M. Jarden extended this to arbitrary Hilbertian fields (Algebraic extensions of finite corank of Hibertian fields, Israel J. Math 18 (1974), 279-307). Moreover, he showed that a randomly chosen list of $e$ elements in the absolute Galois group generates a free profinite group on $e$ elements. Global fields, as well as $\mathbb{Q}^{\rm ab}$, are Hilbertian. An excellent source for all this is the book "Field Arithmetic" by M. Fried and M. Jarden.

Regarding the last question, there are non-Hilbertian fields $K$ such that almost all $\omega$ in the absolute Galois group generate $\hat{\mathbb{Z}}$ - see Example 26.1.11 in the Fried-Jarden book (second edition).


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