TITLE: Computing a polynomial product over roots of unity
QUESTION [7 upvotes]: I'm trying to compute the coefficients of the following polynomial, where $\omega$ is a primitive $p$-th root of unity, for $p$ prime:
$$a(x) = \prod_{i=0}^{p-1} f(\omega^ix).$$
It turns out that the $i$-th coefficient is always an integer, and non-zero only when $i$ is a multiple of $p$. So it seems to me like there should be an elementary expression for $a$.
So far I've got this expression for the $i$-th coefficient:
$$a_i = x^i\sum_{k_0 + \ldots + k_{p-1} = i}f_{k_0} \cdots f_{k_{p-1}} \omega^{k_1 + 2k_2 + \ldots + (p-1)k_{p-1}}$$
where each $k_i$ is non-negative and bounded by the degree of $f$.
Clearly the roots of unity all cancel out somehow, but I can't figure out how to get a 'nice' expression out. Any suggestions?

REPLY [3 votes]: A slightly more general result is shown in Theorem A here
https://www.maths.tcd.ie/pub/ims/bull47/R4701.pdf
which states:
If $r$ is relatively prime to $n$ then the coefficient of $x^r$ in $\prod_{i=1}^n f(\omega^i x)$ is 0.
(Here $\omega$ is a primitive $n$-th root of unity, $f$ has rational coefficients.)