Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Dirichlet's unit theorem computes the group of units of the algebraic numbers of a number field. There are a few generalisations for orders available.

Assume the order has an involution. For example, the number field $\mathbb{Q}(\xi_1,\dots,\xi_n)$ can be equipped with an involution $\xi_k \mapsto \overline{\xi_k} = \pm \xi_k$ and the ring of algebraic numbers will (hopefully) inherit this involution.

It is natural to ask for the "unitary units" of the order i.e. all $x$ such that $x\overline{x}=1$.

Is there a theorem that tells us the structure of the group of unitary units? Do we know at least when this group is (in-)finite?

share|improve this question
1  
What kind of involution do you want; any automorphism of order 2? Can you tell us what examples you have checked already? –  KConrad May 18 '12 at 16:28
    
If your field is CM or totally real and the involution is complex conjugation, the "unitary units" are exactly the roots of unity. Otherwise, if the involution is non-trivial there will be infinitely many such units. –  Kevin Ventullo May 18 '12 at 20:32
add comment

1 Answer 1

Yes. Let $K \subset \mathbb{C}$ be a number field that is closed under complex conjugation (not necessarily Galois). Modulo the roots of unity, the group of unitary units in $K$ is free of rank equal to the number of infinite primes of $K$ minus the number of infinite primes of the real subfield $K \cap \mathbb{R}$. Recall that the number of infinite primes of a field is the number of real embeddings + half the number of complex embeddings. This is proved nicely in this paper by Daileda.

Note that $K$ and its real subfield have the same number of infinite places iff either $K$ is totally real or $K$ is CM (meaning a totally imaginary quadratic extension of a totally real field). So for precisely these cases, the only unitary units in $K$ are roots of unity.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.