If $K$ is a number field, $\Alg_K$ is the ring of algebraic integers in $K$ , and $\UK$ is the (multiplicative) group of units in $\Alg_K$ . Dirichlet's unit theorem gives the structure of the unit group. We can use that theorem to characterize CM-fields:

We use the notation of the article on Dirichlet's unit theorem, where $r$ (and $r_F, r_K$ ) is used to count real embeddings and $s$ (as well as $s_F, s_K$ ) to count complex embeddings, and we write $\mu(F)$ or $\mu(K)$ for the group of roots of unity in $\UF$ or $\UK$ .

Proof.
Write $n=[F:\Rats],\ m=[K:F]>1$ .

($\Rightarrow$ ): If $K/F$ is CM, then since $F$ is totally real, $r_F = n,\ s_F = 0$ . Hence by Dirichlet's unit theorem, $\UF\cong \mu(F)\times \Ints^{n-1}$ . Since $K/F$ is a complex quadratic extension, $[K:\Rats]=2n$ and all its embeddings are complex. Thus $r_K=0,\ 2s_K = 2n$ . Hence $\UK \cong \mu(K) \times \Ints^{n-1}$ as well. Clearly $\UF\subset \UK$ , and since they have the same rank, their quotient is torsion and thus finite.

($\Leftarrow$ ): Since $\UK/\UF$ is finite, the ranks of these groups are equal and thus $r_F+s_F=r_K+s_K$ again by Dirichlet's unit theorem.

Now,

		(1)
		(2)

Finally, by (3), we then have $s_K = r_F+3s_F$ ; (2) says that $s_K = r_F+s_F$ , and thus $s_F=0$ . It follows that $F$ is totally real and, since $r_K=0$ , $K$ must be an imaginary quadratic extension of $F$ .