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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?

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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

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.

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