"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 still understand your third question, but it looks like Scott Carnahan's first comment might answer it. 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 $\alpha$.

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

"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 still understand your third question, but it looks like Scott Carnahan's first comment might answer it.

"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 still understand your third question, but it looks like Scott Carnahan's first comment might answer it. 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 $\alpha$.