characterizing CM-fields using Dirichlet’s unit theorem

If $K$ is a number field, ${𝒪}_K$ is the ring of algebraic integers in $K$, and ${𝒪}_K^{\star }$ is the (multiplicative) group of units in ${𝒪}_K$. Dirichlet’s unit theorem gives the structure of the unit group. We can use that theorem to characterize CM-fields:

Theorem 1.

Let $\mathrm{Q}\mathrm{\subset }F\mathrm{\subset }K$ be nontrivial extensions of number fields. Then $K$ is a CM-field, with $F$ its totally real subfield, if and only if ${\mathrm{O}}_K^{\mathrm{\star }}\mathrm{/}{\mathrm{O}}_F^{\mathrm{\star }}$ is finite.

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 ${𝒪}_F^{\star }$ or ${𝒪}_K^{\star }$.

Proof.
Write $n=[F:\mathbb{Q}],m=[K:F]>1$.

($\Rightarrow$): If $K/F$ is CM, then since $F$ is totally real, $r_F=n,s_F=0$. Hence by Dirichlet’s unit theorem, ${𝒪}_F^{\star }\cong \mu (F)\times {\mathbb{Z}}^{n-1}$. Since $K/F$ is a complex quadratic extension, $[K:\mathbb{Q}]=2n$ and all its embeddings are complex. Thus $r_K=0,\mathrm{\hspace{0.25em}2}{s}_K=2n$. Hence ${𝒪}_K^{\star }\cong \mu (K)\times {\mathbb{Z}}^{n-1}$ as well. Clearly ${𝒪}_F^{\star }\subset {𝒪}_K^{\star }$, and since they have the same rank (http://planetmath.org/FreeModule), their quotient is torsion and thus finite.

($\Leftarrow$): Since ${𝒪}_K^{\star }/{𝒪}_F^{\star }$ 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,

$r_K+2s_K$$=mn=m(r_F+2s_F)$(1)

$r_K+s_K$$=r_F+s_F;$(2)

subtracting (2) from (1), we get

$$s_K=(m-1)(r_F+2s_F)+s_F\ge (m-1)n$$(3)

and thus $mn=r_K+2s_K\ge r_K+2(m-1)n$ so that $0\le r_K\le n(2-m)$. Thus $m\le 2$, and since $K$ is a nontrivial extension, we must have $m=2$ so that $K/F$ is quadratic and $r_K=0$ (since $n(2-m)=0$).

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 $F$.