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

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!

• Another reason for which an infinite countable group can't be profinite was given here – Watson Jun 24 '18 at 10:12
• 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). – Watson Jun 24 '18 at 12:02

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.
• 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. – Watson Jun 23 '18 at 14:51
• @Watson Maybe this is a dumb question. Why $Q$ is a proper field of $L$ here? What is being contradicted here? – user45765 Jun 23 '18 at 15:14
• @user45765 because $K/L$ is a Galois extension. – Kenny Lau Jun 23 '18 at 15:15
• @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. – user45765 Jun 23 '18 at 15:21
• 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. – Watson Jun 23 '18 at 15:22