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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 $\alpha^n=1$?

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If a^n = 1, then a^{n-1} is a multiplicative inverse to a, so a is in the unit group. – Qiaochu Yuan Dec 7 '10 at 23:44
This is kind of a silly question. How does one delete a question? – Jason Smith Dec 7 '10 at 23:57
There should be a link that says "delete" under the tags. – Qiaochu Yuan Dec 8 '10 at 0:01
@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). – Arturo Magidin Dec 8 '10 at 3:15
Dear 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. – Matt E Dec 8 '10 at 5:16

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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I am curious: why did you post the answer if Qiaochu already answered the question in the comments and the poster expressed the wish to delete the question altogether? If your answer gets upvoted, he will lose the ability to delete. – Alex B. Dec 8 '10 at 1:57
@Alex Bartel: I don't think Qiaochu's comment is sufficient: Qiaochu noted that the element is invertible, but he didn't say why it would necessarily be an algebraic integer as well. He just noted that $a$ has an inverse; every nonzero element of $K$ satisfies that. – Arturo Magidin Dec 8 '10 at 3:06
@Alex Bartel: But you are right, I should have raised my point in comments first; I deleted my answer, raised the issue, and only later undeleted. But, again, you are right, I should have gone that route first. – Arturo Magidin Dec 8 '10 at 4:42
I still don't understand why you are putting this answer here, thereby preventing the OP from deleting his question (now that your answer has 2 votes), after he said that that's what he wanted to do. The OP hasn't even come back here within that tiny time window that you allowed him. Ah well, you two will work it out. – Alex B. Dec 8 '10 at 4:53
@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. – Arturo Magidin Dec 8 '10 at 4:57

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