Algebraic integers on the unit circle - MathOverflow

Algebraic integers on the unit circle

2010-10-01T09:41:35Z

Consider a set of algebraic integers which lie on the unit circle, they will generate a multiplicative subgroup of $\mathbb S^1$. Do these objects have a name?

I would guess they contain useful arithmetic/number theoretic information, for example if the generating set is the set of roots of an irreducible polynomial, what kind of information would they contain?

Has the group structure of the elements of a number field which lie on the unit circle $\mathbb S^1$ and the subgroup of it which is made up of algebraic integers been studied.

Would greatly appreciate if you could suggest a reference.

Regards Vagabond

It would be real nice if you could answer keeping in mind that I do not know much Algebra/ commutative algebra/ algebraic geometry. But I hope that does not stop you from answering.

Answer by Cam McLeman for Algebraic integers on the unit circle

2010-10-01T13:35:04Z

"Do these objects have a name"

Probably not. "Multiplicative subgroup of $S^1$ generated by algebraic integers" is pretty descriptive, and not of such fundamental importance that it's worth shortening.

"If the generating set is the set of roots of an irreducible polynomial, what kind of information would they contain?"

Almost none. Assuming you're still talking about algebraic integers, if all of the roots of a monic irreducible polynomial have absolute value 1, then the polynomial is cylcotomic and the roots are roots of unity. Your multiplicative group is then finite and well-understood.

I'm not sure I completely understand your third question, but it looks like Scott Carnahan's first comment points you in the right direction. To elaborate very slightly, note that if $\alpha$ is an algebraic integer on the unit circle, then so is $\alpha^r$ for any $r\in\mathbb{Q}$, so you get a copy of $\mathbb{Q}$ in your multiplicative subgroup for each "independent" such $\alpha$. Of course, if you're inside a fixed number field (which re-reading seems to be the focus of this question), you at least get the subgroup $\alpha^n$ for $n\in\mathbb{Z}$.

Answer by Dror Speiser for Algebraic integers on the unit circle

2010-10-01T14:42:53Z

The question is a bit ambiguous because "lie on the unit circle" is ambiguous. A closely worded question is:

Let $S$ be a set of archimedean places of a number field $K$. What is the subgroup $K_S \subset K^{\times}$ of nonzero numbers with $|x|_v = 1$ for all $v \in S$ ?

If we call the complement of $S$ (in the set of places) $\bar{S}$, then the above group is called $\bar{S}$-units. The notion is usually applied to finite $\bar{S}$ containing all archimedean places, so $S$ contains none. But we don't have to be so restrictive. So to answer the first question: the name should probably be "$\bar{S}$-units$\cap O_K^\times$"
Assume $S$ is non empty. For a non-cyclotomic quadratic field the group seems to be empty. For the two exceptional cyclotomic fields the group is of course the roots of unity. I would guess that for most fields the group is finite, and in fact consists of the roots of unity in the field (for most notions of "most fields" that most of us have). Of course this doesn't have to be the case - for example if the field has a Salem number.