Kummer ring - special monic polynomial with zero at root of unity

Let $$\zeta_n = e^{2 \pi i / n}$$ be the n-th root of unity.

Let

$$P(z) = \sum_{k = 0}^{n-1} s_k z^k$$

be a monic polynomial over $$z \in \mathbb{C}$$, specified by integer coefficients $$s_k \in \mathbb{Z}$$.

I am looking for a way to show if (or if not) a given polynomial $$P$$ has a zero $$P(\zeta) = 0$$.

In particular I am interested in the special case $$s_k \in \{0,1\}$$. So given the coefficients $$s_k$$, does $$P$$ has the zero $$\zeta$$?

A polynomial $$P\in\Bbb{Z}[X]$$ has the primitive $$n$$-th root of unity $$\zeta_n$$ as its root if and only if it is divisible by the $$n$$-th cyclotomic polynomial $$\Phi_n$$. So polynomial long division of $$\sum_{k=0}^{n-1}s_kX^k$$ by $$\Phi_k$$ has remainder $$0$$ if and only if $$P$$ has $$\zeta$$ as a root.
• Thanks! So in general $P(z)$ has root $\zeta_n$ iff $P(z)$ is divisible by $\Phi_n(z)$. Any ideas regarding the special case $s_k \in \{0,1\}$? Note that this is related to the problem of balancing less than n glasses on a round tray with n equidistant holders for glasses. – TomS Feb 12 at 22:16
• Yes, if and ony if $P$ is divisible by $\Phi_n$. – Servaes Feb 12 at 22:21