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?



Computing a polynomial product over roots of unity
Notice that
$$
y^p-x^p=\prod\_{i=0}^{p-1} (y-\omega^i x)\ .
$$
Therefore
$$
\prod\_{i=0}^{p-1} f(x\omega^i)= Res\_y (y^p-x^p, f(y))\ .
$$
Where $Res\_y$ is the resultant of the polynomials in $y$. The above is just a particular case of the so called Poisson product formula. You can then compute the resultant using for instance Sylvester's determinant formula. 
See the review
<http://mate.dm.uba.ar/~alidick/papers/chapter1cd.pdf>
by Cattani and Dickenstein
for a nice introduction to resultants and their properties.


