# Multiplying primitive roots of unity.

1. Jul 30, 2013

### Artusartos

1. The problem statement, all variables and given/known data

Let $ζ_3$ and $ζ_5$ denote the 3rd and 5th primitive roots of unity respectively. I was wondering if I could write the product of these in the form $ζ_n^k$ for some n and k.

2. Relevant equations

3. The attempt at a solution

We know that $ζ_3$ is a root of $x^3=1$, and $ζ_5$ is a root of $x^5=1$, so $ζ_3ζ_5$ must be a root of $x^{15}=1$, right?...so $ζ_3ζ_5 = ζ_{15}$. And in general $ζ_nζ_k = ζ_{[n,k]}$, where [n,k] denotes the lcm of n and k. Is that right?

Also is it true that $ζ_{15}^8 = ζ_{15}$? If so, then why/how?

2. Jul 30, 2013

### micromass

Staff Emeritus
You got to be careful to specify which root you work with. It is certainly true that $\zeta_{15}^8$ is a primitive 15th root of unity, but it might not be the same one!

For example, in $\mathbb{C}$, we have $\zeta_{15} = e^{2\pi i /15}$ is a possible choice. But then $\zeta_{15}^8 = e^{2\pi i (8/15)}$ is a 15th root of unity but not the same one.

It is indeed true that $\zeta_3\zeta_5$ is a primitive 15th root of unity. And depending on how you defined $\zeta_{15}$, equality holds. For example:

$$e^{2\pi i/3} e^{2\pi i/5} = e^{ 2\pi i (8/15) }$$

So this is a primitive 15th root of unity. But it is not $\zeta_{15}$ if you defined $\zeta_{15} = e^{2\pi i /15}$.

3. Jul 30, 2013

### Boorglar

Perhaps instead of using $ζ_{n}$ to represent a particular primitive n-th root, you might use it to denote the whole set of primitive n-th roots. Then change "=" with $\in$, and define the product of two sets as the set containing all possible products of elements from either set. Then what you say will be true.

Last edited: Jul 30, 2013
4. Jul 30, 2013

### Artusartos

Thanks. We can also conclude that $\Bbb{Q}(\zeta_{15}^8) = \Bbb{Q}(\zeta_{15})$ since both are primitive roots of unity and $\zeta_{15}^{8k}=\zeta_{15}$ and $\zeta_{15}^m = \zeta_{15}^8$ for some $k$ and $m$, right?

5. Jul 30, 2013

### micromass

Staff Emeritus
Sure.

6. Jul 30, 2013

### CompuChip

But note that it doesn't work for $\zeta_{15}^5$, for example.