Orders of field automorphisms of algebraic complex numbers - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T19:50:00Z http://mathoverflow.net/feeds/question/13769 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/13769/orders-of-field-automorphisms-of-algebraic-complex-numbers Orders of field automorphisms of algebraic complex numbers Ashutosh 2010-02-02T01:44:28Z 2010-02-02T01:47:32Z <p>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 $Aut(\bar{Q})$?</p> http://mathoverflow.net/questions/13769/orders-of-field-automorphisms-of-algebraic-complex-numbers/13771#13771 Answer by Pete L. Clark for Orders of field automorphisms of algebraic complex numbers Pete L. Clark 2010-02-02T01:47:32Z 2010-02-02T01:47:32Z <p>This amounts to the Artin-Schreier theorem, which has come up several times already on MO (c.f. <a href="http://mathoverflow.net/questions/8756/examples-of-algebraic-closures-of-finite-index" rel="nofollow">http://mathoverflow.net/questions/8756/examples-of-algebraic-closures-of-finite-index</a>):</p> <p>if $K/F$ is a field extension with $K$ algebraically closed and $[K:F] &lt; \infty$, then $[K:F] = 1$ or $2$, and in the latter case, $F$ is real-closed.</p> <p>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 $\overline{\mathbb{Q}}$.</p> <p>(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.) </p>