Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $Aut(\bar{Q})$ be the automorphism group on the field of algebraic complex numbers. The order of an element $f \in Aut(\bar{Q})$ is the least natural number $n$ (if there exists one) such that $(f)^{n}$ is identity. What are the possible orders of elements of $Aut(\bar{Q})$?

share|improve this question

1 Answer 1

up vote 11 down vote accepted

This amounts to the Artin-Schreier theorem, which has come up several times already on MO (c.f. Examples of algebraic closures of finite index):

if $K/F$ is a field extension with $K$ algebraically closed and $[K:F] < \infty$, then $[K:F] = 1$ or $2$, and in the latter case, $F$ is real-closed.

Thus the answer here is that $n$ can be $1$, $2$ or $\infty$, and all possibilities occur: the field of real algebraic numbers gives an index $2$ subfield of $\overline{\mathbb{Q}}$.

(Also, just to be sure, there are elements of infinite order! E.g., if not then every element would have order $1$ or $2$, so the absolute Galois group would be abelian, and thus every finite Galois group over $\mathbb{Q}$ would be abelian, and this is certainly not the case.)

share|improve this answer
Every finite Galois group over $\mathbb{Q}$ would be abelian...:D –  Bombyx mori Feb 9 at 7:29

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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