4
$\begingroup$

Is there an algebraic field extension $K / \Bbb Q$ such that $\text{Aut}_{\Bbb Q}(K) \cong \Bbb Z$?

Here I mean the field automorphisms (which are necessarily $\Bbb Q$-algebras automorphisms) of course.

According to this answer, one can find some extension of $\Bbb Q$ whose automorphism group is $\Bbb Z$. But I've not seen that one can expect this extension to be algebraic.

At least such an extension can't be normal, otherwise $\Bbb Z$ would be endowed with a topology turning it into a profinite group, which can't be countably infinite. (So typically, if we replace $\Bbb Q$ by $\Bbb F_p$, then the answer to the above question is no, because any algebraic extension of a finite field is Galois).

Thank you!

$\endgroup$
  • $\begingroup$ Another reason for which an infinite countable group can't be profinite was given here $\endgroup$ – Watson Jun 24 '18 at 10:12
  • $\begingroup$ Interestingly enough, Fried and Kollár showed that any finite group is the automorphism group of some finite extension of $\Bbb Q$ (which might not be Galois in general, so it doesn't solve the inverse Galois problem). $\endgroup$ – Watson Jun 24 '18 at 12:02
2
$\begingroup$

Let $L$ be the fixed field of $\text{Aut}_{\Bbb Q}(K)$, so $\Bbb Q \subsetneq L \subset K$, and $K/L$ is a normal extension with Galois group $\Bbb Z$, which is impossible.

$\endgroup$
  • 1
    $\begingroup$ I see (details are written here). The main point is that $\mathrm{Aut}(K / K^G)$ has a natural structure of profinite group, since we deal with algebraic extensions. $\endgroup$ – Watson Jun 23 '18 at 14:51
  • $\begingroup$ @Watson Maybe this is a dumb question. Why $Q$ is a proper field of $L$ here? What is being contradicted here? $\endgroup$ – user45765 Jun 23 '18 at 15:14
  • $\begingroup$ @user45765 because $K/L$ is a Galois extension. $\endgroup$ – Kenny Lau Jun 23 '18 at 15:15
  • $\begingroup$ @KennyLau Sorry for dumbness. I could not see the full argument. Assume the proof goes by contradiction. Assume $K/Q$ is algebraic.(i.e. separable.) $K/L$ is galois and by galois group 1-1 correspondence, $L=Q$. Why $L\neq Q$ to start with here? Or I am heading towards the wrong thought? Thanks. $\endgroup$ – user45765 Jun 23 '18 at 15:21
  • 1
    $\begingroup$ Anyway we don't really need a strict inclusion $\Bbb Q \subsetneq L$ (in my reasoning, I did not use it…) ? We just use that in our case, the automorphism group of $K / K^G$ is $G$ (since $G$ is the full automorphism group of $K$, i.e. $G= \Bbb Z$ by assumption) — see part (d) in the theorem of the document I cited. $\endgroup$ – Watson Jun 23 '18 at 15:22

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.