# The number of roots of unity in an imaginary quadratic number field

The book I am studying uses the notation $w_k$ to denote the number of roots of unity contained in $K$, an imaginary quadratic number field.

Supposedly, this is the same as the size of the unit group in $O$, the ring of integers in $K$. I am well familiar with the standard arguments for finding the units in $O$.

Im not really sure exactly what the phrase "the number of roots of unity in $K$" means precisely. Does anyone know?

Also, I was thinking that it is probibly the case that the only elements of $K$ with finite multiplicative order are the elements of the unit group. Is this correct? If so, then is this the correct line of reasoning to show that $w_k=|U(O)|$?

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A root of unity is a complex number $a$ such that $a^n = 1$ for some positive integer $n$. The "number of roots of unity" is the cardinality of the set of roots of unity.

For example, if $K=\mathbb{Q}(i)$, then the roots of unity in $K$ are $1$, $-1$, $i$, and $-i$, so the number of roots of unity contained in $K$ is $4$.

Yes: a complex number is a root of unity if and only if it has finite multiplicative order in $\mathbb{C}^*$. In fact, the group of roots of unity is the torsion subgroup of $\mathbb{C}^*$.

To prove that in the imaginary quadratic case the unit group of $\mathcal{O}_K$ coincides with the group of roots of unity in $K$ you could use Dirichlet's Unit Theorem (as Timothy Wagner suggests), but you don't have to. You can simply show that every unit in $\mathcal{O}_K$ must in fact be a root of unity. This is not hard, since you (probably) know exactly what $\mathcal{O}_K$ for $K=\mathbb{Q}(\sqrt{d})$ ($d$ square free, $d\neq 1$), and you can use the norm map. For example, to show it in the case of $K=\mathbb{Q}(i)$, note that an element in $\mathcal{O}_K = \mathbb{Z}[i]$ is a unit if and only if $N(a+bi) = a^2+b^2 = 1$ (since $N((a+bi)(c+di)) = N(a+bi)N(c+di)$). But since $a$ and $b$ are integers, that requires $a=\pm 1$ and $b=0$ or $a=0$ and $b=\pm 1$, giving you roots of unity.

(In fact, imaginary quadratic fields are the only number fields in which the group of units in the number field is finite and torsion; this follows from the aforementioned Dirichlet Unit Theorem)

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First off, thanks. This helps a lot. Is it easy to show that a complex number is a root of unity iff it has finite multiplicative order? Clearly, one direction is trivial. – Jason Smith Dec 7 '10 at 23:48
@Jason: both directions look trivial to me. Isn't this true by definition? – Qiaochu Yuan Dec 8 '10 at 0:03

You need Dirichlet's unit theorem.

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+1, You are fast – Bryan Yocks Dec 7 '10 at 23:08
You can use it, but to show that a unit in the ring of integers of an imaginary quadratic field must be a root of unity you don't need to use it. You can get it out of the norm map. – Arturo Magidin Dec 7 '10 at 23:09
@Arturo: Thanks, that is better. Note to self: If you hear hoofbeats think horses not zerbas. – Timothy Wagner Dec 7 '10 at 23:15
Oh, I'm a big fan of Dirichlet's Unit Theorem. Only a hair behind the Finiteness of the Class Number in my estimation. – Arturo Magidin Dec 7 '10 at 23:16
Yes, Timothy you are right, the Unit Theorem reduces this one to a triviality. Thanks. – Jason Smith Dec 7 '10 at 23:55