Prime ideals lying above a prime number in $\mathbb{Q}(\zeta_n)$ If we assume we have a prime number $p$ such that $p\nmid n$, and $\mathscr{P}$ is a prime ideal lying above $p$ in the field extension $\mathbb{Q}(\zeta_n)$, is it always true that $\mathscr{P}\nmid n$?

Yes. If $\mathscr{P} \mid k$, then $\mathrm{Nm}(\mathscr{P}) \mid \mathrm{Nm}(k)$. Since $\mathrm{Nm}(\mathscr{P})$ is a power of $p$ and $\mathrm{Nm}(k)$ is a power of $k$, so this can only happen if $p \mid k$.
