0

Is the product of all primitive nth roots of unity equal to 1? Equivalently, is $\Phi_n(0) = 1$ for $n>2$?

More exactly, I'm trying to prove that $$x^{\phi(n)} \Phi_n(x^{-1}) = \Phi_n(x).$$

Since $$\Phi_n(x) = \prod_{\substack{k = 1 \\ (k,n) = 1}}^{n} \left(x - e^{\frac{2k\pi i}{n}} \right),$$ and since the inverses of all primitive nth roots is still the set of primitive nth roots, we have $$\Phi_n(x) = \prod_{\substack{k = 1 \\ (k,n) = 1}}^{n} \left(x - e^{-\frac{2k\pi i}{n}} \right)$$ and so $$\Phi_n(x^{-1}) = \prod_{\substack{k = 1 \\ (k,n) = 1}}^{n} \left(x^{-1} - e^{-\frac{2k\pi i}{n}} \right) = \Phi_n(x) = \prod_{\substack{k = 1 \\ (k,n) = 1}}^{n} \left( \dfrac{e^{\frac{2k\pi i}{n}} - x }{xe^{\frac{2k\pi i}{n}}}\right). $$

I have basically finished, because $\phi(n)$ is even so I can switch around the terms in the parantheses, but I still need the products of primitive nth roots of unity to be equal to 1.

Cois Monis
  • 123

1 Answers1

2

If $n \neq 1,2$, then if $\zeta$ is a primitive $n$'th root of unity, then $\zeta^{-1}$ is a different $n$'th root of unity, so in this case the answer is yes. If $n = 2$, then $-1$ is the only primitive square root of unity.

primes.against.humanity
  • 2,301