Show that product of integer polynomial values for arguments being complex roots of unity is integer.

Question

$P(x)$ is polynomial with integer coefficients and $w$ is complex (primitive) root of unity of a degree $n$. Show that $P(1)*P(w)*P(w^2)*...*P(w^{n-1})$ is integer number for every natural number $n$.

Answer 1

Let $w_k=w^k$. Then $Q(w_0, w_1, \dots,w_{n-1})=P(w_0) \cdot P(w_1) \cdot P(w_2) \cdots P(w_{n-1})$ is a polynomial symmetric in $w_0, w_1, \dots,w_{n-1}$. Therefore $Q$ can be expressed as a polynomial in the elementary symmetric polynomials in $w_k$ with integer coefficients. But, since $w$ is a primitive $n^{th}$ root of unity, $w_k=w^k \;|\; k=0 \dots n-1$ are the $n$ roots of $x^n - 1=0$ which is a monic polynomial with integer coefficients, so by Vieta's formulas all the elementary symmetric polynomials in $w_k$ are integers (in fact, all of them $0$ except the full product which is $1$), thus $Q$ is an integer as well.