The definitions of a divisible group that I have seen all seem to assume abelian is an a priori property of the group. My question is as to whether or not it is known that--given a non-torsion element $\tau\in\mathbf{G}$* and $n\in\mathbb{N}$, do we know if $\exists \tau'\in\mathbf{G}_\mathbb{Q}$ such that $\tau=\tau'^n$?

The motivation is that--if this never happens--then the fields $\overline{\mathbb{Q}}^\tau$ should be sort of "maximal" subfields of $\overline{\mathbb{Q}}$ since fixing by the generator would be equivalent to fixed by any power of it.

Edit: anon has noted that quotients of divisible groups are divisible, so we can lay to rest that $\mathbf{G}_\mathbb{Q}$ is not divisible. Can we tell if the opposite is true? I.e. the answer to "can I find such a $\tau '$ given $n$" is "no", but how about the somewhat interesting and related question: If I have a cyclic subgroup of the absolute Galois group, $C=\langle\tau\rangle$ can we find a maximal (with respect to inclusion), cyclic group containing $C$? Of course this is equivalent to a minimal field in a chain, so it seems like if that is so something interesting must be going on.

The other related question which deals with the original spirit of the problem is: are there $\tau$ such that $\langle \tau^n\rangle$ fixes $k\subseteq\overline{\mathbb{Q}}$ then $\langle\tau\rangle$ fixes $k$? i.e. is $k=\overline{\mathbb{Q}}^\tau$ equivalent to $k=\overline{\mathbb{Q}}^{\tau^n}$ possible for some $\tau$ non-torsion? If so can such elements be characterized?

The immediate observation is that, by the FTGT, if we identify $\langle\tau\rangle\cong\mathbb{Z}$ and write $[n]$ for the index $n$ subgroup, that--in the topology of $\mathbf{G}_\mathbb{Q}$--necessarily $\overline{[n]}=\overline{[1]}$. I.e. the monothetic group $\overline{[1]}$ has every element is a generator. If $\overline{[1]}$ were connected, of course this would be trivial to check since the set of generators has Haar measure 1 in this case, and a finite index subgroup has positive Haar measure, and so contains a generator, but the topology here is totally disconnected so it is not easy to see one way or the other.

*My TeX was not rendering properly when I wrote $\mathbf{G}_\mathbb{Q}$ here, so I just included it as a sidenote rather than have it look like a mess.

  • 9
    $\begingroup$ Quotients of divisible groups are divisible, which is certainly not true of the absolute Galois group of Q. $\endgroup$ – anon Nov 28 '12 at 18:39
  • $\begingroup$ That's a fantastically simple reason. Thank you, anon. I've edited the original question to be a bit stronger now that the easy part is revealed. $\endgroup$ – Adam Hughes Nov 28 '12 at 18:48

There are the following results:

  • Let $K = \bar{K}$ be algebraically closed and $\sigma \in \mathrm{Aut}(K)$. Then every finite extension of $K^\sigma$ is cyclic.
  • Let $x \in K^{sep} \setminus K$. Let $L/K$ be a subfield of $K^{sep}$ maximal with respect to the property of not containing $x$. Then $G_L = \mathbf{Z}/2$ or $\mathbf{Z}_p$.

And $1 \in \mathbf{Z}_p$ is not $p$-divisible.


Here is a result of Haran that might be relevant : For almost all $\sigma$ in the absolute Galois group of $\mathbb{Q}$ the fixed field $\bar{\mathbb{Q}}^\sigma$ of $\sigma$ in $\bar{\mathbb{Q}}$ has no proper cofinite subextensions. (I.e. $L\subsetneq \bar{\mathbb{Q}}^\sigma$ implies $[\bar{\mathbb{Q}}^\sigma:L]=\infty$.)

Here almost all is in the sense of the Haar measure on the absolute Galois group.

  • 1
    $\begingroup$ For reference purposes, this goes by the name The Bottom Theorem, and it holds more generally for finitely (topologically) generated subgroups of the absolute Galois group of any Hilbertian field. $\endgroup$ – PrimeRibeyeDeal Jul 20 '17 at 15:20

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.