TITLE: An abelian number field is either totally real or CM-field
QUESTION [6 upvotes]: The wikipedia article of totally real number fields says:

The totally real number fields play a significant special role in
  algebraic number theory. An abelian extension of Q is either totally
  real, or contains a totally real subfield over which it has degree
  two.

Can someone give me a reference for this fact, or a sketch of a proof?
Thanks in advance!

REPLY [8 votes]: Assume that $K\subset \Bbb{C}$ is an abelian extension of $\Bbb{Q}$ that is not totally real. Because it is Galois, it is stable under complex conjugation, and equal to its image in all embeddings $K\to \Bbb{C}$. As it is not totally real, it thus cannot be real either. Therefore the restriction of the usual complex conjugation, $\sigma$, is an element of order two in the Galois group $G=Gal(K/\Bbb{Q})$. 
Let $L$ be the fixed field of the cyclic subgroup $N=\langle\sigma\rangle$. Then $L=K\cap\Bbb{R}$. Furthermore, because $G$ is abelian, $N\unlhd G$. Therefore $L$ is also Galois, and hence equal to its own image under all embeddings to $\Bbb{C}$. This means that $L$ is totally real. Clearly $[K:L]=2$.