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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1If a^n = 1, then a^{n-1} is a multiplicative inverse to a, so a is in the unit group. – 2010-12-07
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0This is kind of a silly question. How does one delete a question? – 2010-12-07
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0There should be a link that says "delete" under the tags. – 2010-12-08
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1@Qiaochu, @Jason Smith: I don't think Qiaochu's argument by itself suffices: he is showing that $\alpha$ has an inverse, but then every nonzero element of $K$ has an inverse. The unit group $U$ is the group of invertible *algebraic integers* in $K$, so you also need to mention that $\alpha$ must be in $\mathcal{O}_K$ before you conclude that $\alpha\in U$: it *is*, because $\alpha$ satisfies the monic polynomial $x^n-1$. (I had put this as an answer, but deleted it after comments by Alex Bartel to allow Jason to delete the question if he wants to). – 2010-12-08
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0@Arturo: sorry, I misread the question. Not sure what I was thinking. As you say, you also need to use the fact that O_K is integrally closed. – 2010-12-08
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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