Multiplying primitive roots of unity.

Artusartos
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Homework Statement

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.

Homework Equations


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?

Thank you in advance
 
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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}##.
 
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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.
 
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micromass said:
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}##.

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?
 
Artusartos said:
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?

Sure.
 
But note that it doesn't work for ##\zeta_{15}^5##, for example.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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