# characterizing CM-fields using Dirichlet’s unit theorem

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

###### Theorem 1.

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

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, $\mathcal{O}_{F}^{\star}\cong\mu(F)\times\mathbb{Z}^{n-1}$. Since $K/F$ is a complex quadratic extension, $[K:\mathbb{Q}]=2n$ and all its embeddings  are complex. Thus $r_{K}=0,\ 2s_{K}=2n$. Hence $\mathcal{O}_{K}^{\star}\cong\mu(K)\times\mathbb{Z}^{n-1}$ as well. Clearly $\mathcal{O}_{F}^{\star}\subset\mathcal{O}_{K}^{\star}$, and since they have the same rank (http://planetmath.org/FreeModule), their quotient is torsion and thus finite.

($\Leftarrow$): Since $\mathcal{O}_{K}^{\star}/\mathcal{O}_{F}^{\star}$ 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,

 $\displaystyle r_{K}+2s_{K}$ $\displaystyle=mn=m(r_{F}+2s_{F})$ (1) $\displaystyle r_{K}+s_{K}$ $\displaystyle\phantom{=mn\ }=r_{F}+s_{F}\ ;$ (2)

subtracting (2) from (1), we get

 $s_{K}=(m-1)(r_{F}+2s_{F})+s_{F}\geq(m-1)n$ (3)

and thus $mn=r_{K}+2s_{K}\geq r_{K}+2(m-1)n$ so that $0\leq r_{K}\leq n(2-m)$. Thus $m\leq 2$, 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 $F$.

Title characterizing CM-fields using Dirichlet’s unit theorem CharacterizingCMfieldsUsingDirichletsUnitTheorem 2013-03-22 17:57:26 2013-03-22 17:57:26 rm50 (10146) rm50 (10146) 4 rm50 (10146) Theorem msc 11R04 msc 11R27 msc 12D99