Let $\zeta = e^\frac{2 \pi i}{p}$, with $p$ prime. Show that the elements of the form $1+\zeta +\zeta^2 + \dots + \zeta^m$, with $1 \leq m \leq p-2$, are invertible in $\mathbb{Z}[\zeta]$.
I know that $f(\zeta) \in \mathbb{Z}[\zeta]$ is invertible if and only if $N(f(\zeta)) \in \{1, -1\}$.
Definition : $N(f\zeta)) = \prod_{i=1}^{p-1} \sigma_i(f(\zeta)) = \prod_{i=1}^{p-1} f(\zeta^i) $ and $\sigma_m(f(\zeta)) = f(\zeta^m) $
This doesn't make sense for me because $|\prod_{i=1}^{p-1} f(\zeta^i)| > 1 $. I already know that $1+\zeta +\zeta^2 + \dots + \zeta^m = \frac{\zeta^{m-1}-1}{\zeta-1}$, but it not clear this expression could help me.
Is anyone could give me a hint how to solve this problem?
Let’s see whether I can get this right.
The key is, indeed, your observation that your quantity is equal to $(\zeta_p^{m+1}-1)\big/(\zeta_p-1)$. Now, in accordance with your observation about the norms, all you need to do is show that the norm here is equal to $1$. But $\zeta_p^{m+1}-1$ and $\zeta_p-1$ have the same norm (down to $\Bbb Q$) as long as $m+1<p$, because $\zeta_p$ and $\zeta_p^{m+1}$ are conjugate over $\Bbb Q$.