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
Canonical name	CharacterizingCMfieldsUsingDirichletsUnitTheorem
Date of creation	2013-03-22 17:57:26
Last modified on	2013-03-22 17:57:26
Owner	rm50 (10146)
Last modified by	rm50 (10146)
Numerical id	4
Author	rm50 (10146)
Entry type	Theorem
Classification	msc 11R04
Classification	msc 11R27
Classification	msc 12D99