For primes $p, n$, let $\zeta$ be a $n$-th root of unity, and $p \equiv 1$ mod $n$. Then $\mathcal{O} = \mathbb{Z}[\zeta]$ is the ring of integers of the cyclotomic field $\mathbb{Q}(\zeta)$.
What are the prime divisors of $p\mathcal{O}$?
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See this. In $\mathbb{Z}[\zeta]$, with $f(X)$ the minimal polynomial of $\zeta$ : $$(q) = \prod_{j=1}^{g} (q,f_j(\zeta))^{e_j}$$ where $\prod_{j=1}^{g} f_j(X)^{e_j}$ is the factorization of $f(X)$ in $\mathbb{F}_q[X]$.
Now if $p \equiv 1 \bmod n$ then $X^n-1$ splits completely in $\mathbb{F}_p[X]$ and $$\Phi_n(X) \equiv \displaystyle\prod_{a \in \mathbb{F}_p, ord(a) = n} (X-a) \equiv\prod_{j=1,gcd(j,n)=1}^n (X-b^j) \bmod p$$ where $b$ is any integer of order $n$ in $\mathbb{F}_p$.
so in $\mathbb{Z}[\zeta]$ : $$(p) = \prod_{a \in \mathbb{F}_p, ord(a) = n} (p,a-\zeta) = \prod_{j=1,gcd(j,n)=1}^n (p,b^j-\zeta)$$
