number fields generated by units of number fields 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?
 A: 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 $K$.
A: Very interesting matter maybe.
If we restrict this question for a moment to abelian extensions of a number field (not just Q) in the presence of enough roots of unity, this question reduces to a question in Kummer theory: Can we realize this extension by taking a root/radical of a unit.
So when is this possible? This relies actually on the ramification/branching behaviour of our extension. For the radical equation X^n - a one can compute its discriminant and I've forgotten the precise result, but this shows us something like ramification can only come from divisors of n or prime divisors of a if I remember correctly; so if a is required to be a unit, this restricts the possibilities for ramification quite heavily.
Now this argument doesn't actually answer anything for you since over Q this Kummer argument is nearly void.
On the other hand, by Kronecker-Weber, over Q, even abelian ext lies in a cyclotomic one, and that's roots of unity, so units only.
Class field theory should help for a more general answer, but I fear I haven't paid attention enough to classes to say how. Problematically, it does not tell us too much about explicit generators of fields.
Over local fields we may still touch the question: Indeed, every extension can be obtained by adjoining a unit, this is not so hard to see. On the other hand, maybe this is irrelevant since local fields really have too many units to serve as a model case.
Finally, it occurs to me, going back to the Kummer case: Normally we use H^1(G_m) =1 (Hilbert 90) and basically you ask whether H^1(O*) = 1. This is somehow related to Iwasawa theory, but I forgot how, but I remember for sure that the latter statement can fail to be true.
Sorry, this is not an answer for sure and rather a wordy mess; but maybe helpful (certainly helpless :-) anyway.
