TITLE: number fields generated by units of number fields
QUESTION [8 upvotes]: Which number fields are generated by the units of some number field? That is, if $K$ is a number field and $U(K)$ its group of units, the field $k = \mathbb{Q}(U(K))$ is a subfield of $K$. But which number fields $k$ occur in this way as $K$ varies over all number fields?
This is an unmotivated question that struck me while making up exercises. It does not seem usable for an exercise, but who knows?

REPLY [9 votes]: The rank of the unit group of $K$ is $r+s-1$ where $r$ and $2s$
are the numbers of real and complex embeddings of $K$. As $r+2s=n$
this rank is less than $n=r+2s$, the degree of $K$. It follows that the
degree of $k$ is at least $r+s\ge n/2$. Thus if $K\ne k$ then the degree
of $k$ is $n/2$, $r=0$ and $s=n/2$. Hence $K$ is a totally complex
quadratic extension of the totally real field $k$, and so is a so-called
CM field.
When we have a CM field, the unit groups of $K$ and $k$ have the same
rank (where $k$ is the real subfield of $K$) but the unit group of $K$
may still be strictly larger (as an example the $p$-th cyclotomic field
where $p$ is prime). I suspect that there must be examples where the
unit groups are the same, but I don't have any to hand.
Added (8/6/2010) I claim that each totally real number field $k$
has infinitely many quadratic extensions $K$ with the same unit
group. Such an extension must be a CM field: $K=k(\sqrt a)$ where $a\in k$
is totally negative.
First I claim that $k$ has infinitely many CM quadratic extensions.
Given a prime ideal $P$ in $k$, by weak approximation there is
a totally negative $a\in K$ whose valuation at $P$ is $1$. Then
$k(\sqrt{a})$ is a CM field ramified over $P$. So there must be infinitely
many such fields.
I now claim that only finitely many CM quadratic extensions of $k$ have
larger unit group. Such an extension $K$ may be a cyclotomic extension,
got by adjoining a root of unity. But the number of roots of unity in $K$
is bounded by a number depending on the absolute degree of $K$, so there
are only finitely many cyclotomic quadratic extensions of $k$.
Assume that $K/k$ is a CM quadratic extension, not cyclotomic and
with larger unit group. Then $K=k(\epsilon)$ where $\epsilon$ is a unit.
As the ranks of the unit groups coincide, then some power of $\epsilon$
lies in $k$. We may assume that $\epsilon^p=\eta\in k$ where $p$
is prime. If $\epsilon^\ast$ is the Galois conjugate of $\epsilon$ over
$k$ then $\epsilon^\ast{}^p=\eta$ and so $(\epsilon^\ast/\epsilon)^p=1$.
Therefore $\epsilon^\ast/\epsilon$ is a nontrivial $p$-th root of unity.
If $\epsilon^\ast/\epsilon\notin k$ then $K/k$ is cyclotomic. So
$\epsilon^\ast/\epsilon\in k$, and $k$ contains all $p$-th roots of unity.
This is only possible if $p=2$ since $k$ is totally real.
Hence $K=k(\sqrt{\eta})$ where $\eta$ is a unit. As the unit group is finitely
generated, there are only finitely many such extensions $K$.