Let $p$ be an add prime greater than 3, $\omega$ is the $p$-th root of unity. Show that if $$\alpha = a_0+ a_1\omega+a_2\omega^2+\cdots +a_{p-2}\omega^{p-2} \in \mathbb{Z}[\omega], a_i \in \mathbb{Z}$$ and $p \mid \alpha$, then $p \mid a_i$ for all $i$.
Any ideas? Hint is every element of $\mathbb{Q}[\omega]$ is uniquely representable in the form
$$a_0+ a_1\omega+a_2\omega^2+\cdots +a_{p-2}\omega^{p-2}, a_i \in \mathbb{Q}.$$
Because of $p\mid \alpha$ we have $\alpha\in p\Bbb{Z}[\omega]$. So $\alpha$ is representable as a sum of terms $pk\omega^j$. Since $\alpha=a_0+\cdots +a_{p-2}\omega^{p-2}$, and the representation is unique, these two representations must be equal. So every $a_i$ is divisible by $p$.