Is the field of real numbers $\mathbb{R}$ a finite extension of some subfield $k\subset \mathbb{R}$?
closed as offtopic by Mark Sapir, Daniel Moskovich, Emil Jeřábek, Andrey Rekalo, Theo JohnsonFreyd Jul 16 '13 at 17:20This question appears to be offtopic. The users who voted to close gave this specific reason:



there is no such subfield. [Edited] It is a theorem of Emil Artin that the only automorphism of the field of complex numbers of finite order is of order two. If such a subfield $F$ existed, then $Aut ({\mathbb C}/F)$ would have only elements of order two and hence abelian. In particular, ${\mathbb R}/F$ would be abelian (and Galois). But $\mathbb R$ has no field automorphisms and hence $F=\mathbb R$. I thank Peter clark for pointing out that I was attributing a wrong result to Emil Artin(!). 

