Question. Let $p$ be a prime and $\omega$ be a primitive $p$-th root of unity. Let $a$ be a nonzero element of $\mathbb Z[\omega]$. Can it happen that for each $n\in \mathbb N$, $(1-\omega)^n$ divides $a$?
This can be reframed as follows. Let $I$ be the ideal generated by $1-\omega$ in $\mathbb Z[\omega]$. Is $I\cap I^2\cap I^3\cap \cdots$ the zero ideal?
The motivation for the above question comes from the proof of Chebatarev's Theorem as given in this arXiv paper. On page 3, the last line of the proof mentions an infinite descent. But what exactly is descending is not clear to me. If the answer to the question posed above is indeed 'no' then I'll be done.
If $\alpha$ is a non-zero element of $\mathbb{Z}[\omega]$, then since $\mathbb{Z}[\omega]$ is a Dedekind domain there is a unique factorization $$(\alpha)=\mathfrak{p}_1^{n_1}\cdots\mathfrak{p}_k^{n_k}$$ where the $\mathfrak{p}_i$ are (non-zero) prime ideals. This implies that for each prime $\mathfrak{p}$, there is a unique non-negative integer $n$ such that $\alpha\in \mathfrak{p}^n$ but $\alpha\not\in \mathfrak{p}^{n+1}$.
Moreover, $(1-\omega)$ is a prime ideal (it's the prime lying over $p$), so it follows that $$\bigcap_{n=0}^{\infty}(1-\omega)^n=(0)$$