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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

PS

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.

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It should be pretty clear that such a group must be infinitely generated, since it is divisible. In other words, there is no polynomial whose roots generate the group. It seems pretty likely that the structure of the abstract group is $\mathbb{Q}/\mathbb{Z} \oplus \mathbb{Q}^{\oplus \infty}$ –  S. Carnahan Oct 1 '10 at 10:15
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@Scott: I'm not sure what's going on. Isn't ${1,-1}$ a set of algebraic integers which lie on the unit circle that generate a multiplicative subgroup of the circle and consists of the roots of a quadratic polynomial? –  Dror Speiser Oct 1 '10 at 11:54
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I think there's more confusion than question here, so I'm voting to close. The third paragraph is particularly confusing, as when one talks about units in a number field one is automatically talking about algebraic integers - but maybe "extension field" doesn't mean "number field" - but we shouldn't have to guess what the question means. –  Gerry Myerson Oct 1 '10 at 12:27
    
@Dror: I mentally replaced the article "a" with "the" in the first sentence. You are quite correct. –  S. Carnahan Oct 1 '10 at 13:31
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2 Answers 2

"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 $n\in\mathbb{Z}$.

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Thank you for your answer. I will try to elaborate on my question, hope it will clear some of the confusion. The reason why I was asking this question is the following. I have a monic-polynomial with integer coefficient, it may not be irreducible. I know that some of its root lies on the unit circle and my interest lie in finding some arithmetic/geometric relation between these zeros which lies in the unit circle. Now as you said by Kronecker theorem if all the roots lie on the unit circle then I can conclude that the roots are roots of unity. But I have no way to conclude that. –  Vagabond Oct 1 '10 at 14:17
    
The argument that the the roots lie on the unit circle is much simpler than Kronecker's Theorem. I'm sure it's in the first section or two of Washington's Cyclotomic Fields. Restricting to the irreducible case seems prudent -- otherwise you could take a cyclotomic polynomial and multiply it by any other monic polynomial and be unable to conclude much of anything. –  Cam McLeman Oct 1 '10 at 14:20
    
So, I was thinking may be I should look into the group generated by these zeros which lie on the unit circle. Its quite possible that even if they are not roots of unity i.e. they d not satisfy $\theta^n=1$ for any n, a linear combination of them might do, for example if $\theta_1/\theta_2 = e^{2 \pi i k/n}$ then we get $(\theta_1 * \overline{ \theta_2})^n$ =1 –  Vagabond Oct 1 '10 at 14:26
    
more generally its possible that such a relation between the roots holds $\theta_1^{m_1} * \theta_2^{m_2} *\dots *\theta_k^{m_k} =1$. In that case certainly we wont get something like a free group as there are non trivial relation involving the roots ? –  Vagabond Oct 1 '10 at 14:32
    
Right, though these non-trivial relations are themselves pretty trivial -- I'd spend some time with Dirichlet's Unit Theorem. –  Cam McLeman Oct 1 '10 at 14:40
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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 $v \in S$ ?

  1. 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$\cap O_K^\times$"

  2. 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.

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For a quadratic field, if $\alpha\ne\pm1$, and $\alpha$ is on the unit circle, then so are all of its conjugates (since it only has one, its complex conjugate), so by a theorem of Kronecker it's a root of unity. –  Gerry Myerson Oct 1 '10 at 22:47
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