# Multiplicative order of elements in an imaginary quadratic field

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.
@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