This is a part of proof showing that the ring of integer for $K=\mathbb{Q}(\zeta)$ is $\mathbb{Z}[\zeta]$, where $\zeta$ is a $p$-th root of unity.
Show that if $x=a_0+a_1\zeta+\cdots+a_{p-2}\zeta^2 \in \mathcal{O}_K$, $a_i \in \mathbb{Q}$, then $\text{tr}(x(1-\zeta)) \in (1-\zeta)\mathcal{O}_K$.
I know that $\text{tr}(x(1-\zeta))=a_0p$, but I'm not sure why it should be in $(1-\zeta)\mathcal{O}_K$.
If $\sigma\in Gal(K/\Bbb{Q})$ then $\sigma(\zeta)=\zeta^k, p\nmid k$ so that $$\sigma(1-\zeta)= 1-\zeta^k=(1-\zeta)(\sum_{m<k} \zeta^m) \subset (1-\zeta)$$
$$Tr(x(1-\zeta))=\sum_{\sigma\in Gal(K/\Bbb{Q})} \sigma(x(1-\zeta))$$