I'm doing an exercise whose main purpose is to show that $\mathscr{O}_K=\mathbb{Z}[\xi]$, where $K=\mathbb{Q}[\xi]$ and $\xi$ is a primitive $p$-th root of $1$, $p$ prime.
So, let $\alpha\in\mathscr{O}_K$. Certainly, $\xi,\xi^2,\ldots,\xi^{p-1}$ is a $\mathbb{Q}$-basis for $K$, thus we can write $\alpha=\sum_{i=1}^{p-1}a_i\xi^i$, for some coefficients $a_i\in\mathbb{Q}$. Now i want to prove that $\alpha$ can also be written as $$\alpha=\sum_{i=0}^{p-2}b_i\xi^i$$ where $b_i$ are rational numbers such that $pb_i\in\mathbb{Z}$ for every $i$.
To do this i've proved that $$\operatorname{tr}_{K|\mathbb{Q}}(\alpha(\xi^{-l}-1))=pa_l$$ for every $l=1,\ldots,p-1$. Substituing in the previous formula for $\alpha$, i get $$\alpha=\sum_{i=1}^{p-1}\frac{\operatorname{tr}(\alpha(\xi^{-i}-1))}{p}\xi^i$$
By setting $b_i:=\frac{\operatorname{tr}(\alpha(\xi^{-i}-1))}{p}$, i find $$\alpha=\sum_{i=1}^{p-1}b_i\xi^i$$ and since for $i=0$ the $i$-th summand is zero, I can also write $$\alpha=\sum_{i=0}^{p-1}b_i\xi^i$$ but this is not what I was looking for, since there is $p-1$ and not $p-2$ in my formula. How can $I$ adjust that?
Expanding on my comment:
$\sum_{i=0}^{p-1}\xi^i=0$, so $\sum_{i=0}^{p-1}b_{p-1}\xi^i=0$, so $$\alpha=\sum_{i=0}^{p-1}b_i\xi^i=\sum_{i=0}^{p-1}b_i\xi^i-\sum_{i=0}^{p-1}b_{p-1}\xi^i=\sum_{i=0}^{p-1}(b_i-b_{p-1})\xi^i=\sum_{i=0}^{p-2}(b_i-b_{p-1})\xi^i$$ as desired.