Let $K$ be an imaginary quadratic field and $U$ denote the unit group in the ring of integers in $K$. Are there $\alpha \in K-U$ with finite multiplicative order? That is, is there $n \in N$ such that $\alpha^n=1$?
Multiplicative order of elements in an imaginary quadratic field
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number-theory
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1Dear Jason, Maybe you know this, but just in case: if $K$ is an imag. quad. field, then $K^{\times}$ actually contains very few elements of finite order. Unless $K = \mathbb Q(i)$ or $\mathbb Q(\sqrt{-3}),$ the only elements of finite order that $K^{\times}$ contains are $\pm 1$. In the latter two cases, the elements of finite order are $\{\pm 1, \pm i\}$ and $\{\pm 1, \pm \zeta_3, \pm \zeta_3^{-1}\}$ resp. (where $\zeta_3 = (-1 + \sqrt{-3})/2$). Of course your question makes sense for any number field $K$, and the answer by Qiaochu and Arturo applies just as well in that general context. – 2010-12-08
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If $\alpha^n=1$, then $\alpha$ satisfies the polynomial $x^n-1$, hence is an algebraic integer. Thus, $\alpha\in \mathcal{O}_K$; once you have that, you can make the easy observation Qiaochu did in the comments that $\alpha^{-1}=\alpha^{n-1}$ is also in $\mathcal{O}_K$ to show $\alpha\in U$ (or you can use the fact that if an algebraic integer satisfies a monic polynomial with integer coefficients and constant term $1$, then it must be a unit). So the answer is "no."
Of course, every root of unity is integral (satisfying $x^n-1$ for some $n$, or the appropriate cyclotomic polynomial if you insist on getting the minimal polynomial), so roots of unity in a number field are always in the ring of integers.
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1@Alex Bartel: The original poster thought Qiaochu had answered the question, and that the question was dumb given that answer. But the answer, as Qiaochu notes, was at best incomplete. The OP's belief that his question was dumb, based on that answer, was itself unfounded. – 2010-12-08