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?

  • $\begingroup$ Quite easy to see that totally real fields are so generated. It is also easy to see that the field generated can be strictly larger than a maximal totally real subfield. $\endgroup$ – Charles Matthews Jun 7 '10 at 20:08
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    $\begingroup$ See <a href="mathoverflow.net/questions/15260/… question</a> $\endgroup$ – Franz Lemmermeyer Jun 8 '10 at 21:56

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 $K$.

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    $\begingroup$ Try $K$ an imaginary quadratic field other than Q(i) and Q(sqrt(-3)). More generally, the units of $k$ and the roots of unity in $K$ generate a subgroup of the units of $K$ with index 1 or 2 (see Washington's book on cycl. fields Theorem 4.12 on page 40 of the 2nd edition), so as long as K has no roots of unity other than 1 and -1 you could expect to find interesting examples after searching a bit. $\endgroup$ – KConrad Jun 7 '10 at 20:39
  • $\begingroup$ Also, something should be said about the relation between multiplicative independence and linear independence over Q for units in order for the rank formula to be compared with the degree formula. $\endgroup$ – KConrad Jun 7 '10 at 20:41
  • $\begingroup$ Thanks, Keith for the comments. I almost convinced myself that if $k$ is a totally real field, then all but finitely many CM quadratic extensions of $k$ have the same unit group. If I finally do convince myself, I'll add the details to the reply. $\endgroup$ – Robin Chapman Jun 8 '10 at 6:20
  • $\begingroup$ @Robin: See MR0063403, Remak, Robert, Über algebraische Zahlkörper mit schwachem Einheitsdefekt. Compositio Math. 12, (1954). 35--80 $\endgroup$ – Victor Protsak Jun 8 '10 at 6:52
  • $\begingroup$ Keith, I'm not sure what you're getting at with your remark above multiplicative independece. The rank of a group of units in a number field doesn't change in a larger number field. $\endgroup$ – Robin Chapman Jun 8 '10 at 18:20

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.

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  • $\begingroup$ Although a Kummer extension can be generated by roots of binomial polynomials, it can also be generated in other ways too. So why do you say the problem reduces to asking when the extension can be realized by adjoining n-th roots of units in the base field? There is no reason I see that the extension being generated by units requires them to be roots of units in the base field. In fact, Q(sqrt(2)) shows it doesn't have to happen like that. This field is generated over Q by 1 + sqrt(2) which is not the square root of a unit in Z. $\endgroup$ – KConrad Jun 7 '10 at 20:47
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    $\begingroup$ Please edit this display of irrelevant considerations. $\endgroup$ – Charles Matthews Jun 8 '10 at 7:24

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