# $\omega$ a $p$th root of unity then $\prod_{k=0}^{p-1}P(\omega^k)$ is an integer

Let $p$ be prime and let $P=a_0+a_1x+...+a_{p-1}x^{p-1}$ be a polynomial in $\mathbb{Z}[x]$. Let $\omega$ be a $p$th root of unity and define

$D=\prod_{k=0}^{p-1}P(\omega^k)$

How can I show that $D$ is an integer and furthermore that $D \equiv (a_0+...+a_{p-1}) \mod p$?

Note: the above relation is obvious in the context of the calculation of a ceartain determinant, but I'm looking for a purely number theory/polynomials solution.

The $\omega^k, 1 \leq k \leq p-1$ are conjugate over $\mathbb{Q}$ and integral, so $D$ is a rational integer. Furthermore, $(\omega^k - 1)$ has $p$-adic valuation $1/(p-1)$ so $\prod_{k=0}^{p-1}P(\omega^k) \equiv \prod_{k=0}^{p-1}P(1) \equiv (a_0+...+a_{p-1})^p \equiv (a_0+...+a_{p-1})$ modulo $(\omega - 1)$ and also modulo $p$ because all are rational integers.
• The $\omega^k, 1 \leq k \leq p-1$ are the set of all conjugates of $\omega$ over $\mathbb{Q}$, the Galois group over $\mathbb{Q}$ permutes them and so fixes $D$, therefore $D$ is rational. It is integral because $\omega$ and all $a_i$ are integral. An integral rational number is an integer. Oct 24, 2017 at 11:28
• An integral number is a root of a monic (i.e. leading coefficient=1) polynomial with integer coefficients. The sum or product of integral numbers is integral. The $\omega^k, 1 \leq k \leq p-1$ are integral roots of the irreducible polynomial $x^{p-1} + x^{p-2} + ... + 1 = 0$. An integral rational number is an integer. If $P(x_1,..,x_{p-1}) \in \mathbb{Q}[x_1,...,x_{p-1}]$ is a symmetric polynomial then $P(\omega,...,\omega^{p-1}) \in \mathbb{Q}$; this is the implication of Galois theory that I wrote about; it may benefit you to work out a few specific examples with a small prime $p$. Oct 24, 2017 at 12:57