Let $\alpha$ be an algebraic integer on the unit circle in $\mathbb{C}$ such that all the conjugates of $\alpha$ lie on the unit circle. Does it follow that $\alpha$ is a root of unity?
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asked Apr 7, 2015 at 9:01
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9$\begingroup$ Yes; see Mark Sapir's answer to this question: mathoverflow.net/questions/38680/… $\endgroup$– R.P.Commented Apr 7, 2015 at 9:09
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9$\begingroup$ It is called Kronecker's theorem. $\endgroup$– Andreas ThomCommented Apr 7, 2015 at 9:29
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3$\begingroup$ The original reference is [L.Kronecker, Zwei Sätze über Gleichungen mit ganzzahligen Coefficienten, J. Reine u. Angew. Math. 53, (1857), 173-175.] $\endgroup$– Andreas ThomCommented Apr 7, 2015 at 11:13
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3$\begingroup$ Since the product of norms of $\alpha$ over various completions is 1, it has to be a algebraic unit. Then it follows that the cyclic subgroup $C$ generated by $\alpha$ is bounded for every norm. Since $\mathbf{Z}[\alpha^{\pm 1}]$ is discretely embedded as a subring in some finite product of completions, we deduce that $C$ is finite. $\endgroup$– YCorCommented Apr 7, 2015 at 11:15
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3$\begingroup$ See Theorem 8 in Section IV.4 of Weil: Basic number theory. I mention this as the proof here is very transparent: it relies on the fact that the intersection of a compact set and a discrete set is finite. $\endgroup$– GH from MOCommented Apr 7, 2015 at 18:48
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