Primacy of ideals in a cyclotomic field

Let $$p,q$$ be odd rational primes and $$n=pq$$, $$\zeta_{n}$$ a primitive $$n$$th root of unity, $$K = \mathbb{Q}(\zeta_n)$$ the $$n$$th cyclotomic field and $$\mathcal{O}_{K} = \mathbb{Z}[\zeta_n]$$ its ring of integers.

Are the ideals $$I = (\zeta^{p}_{n} -1), J = (\zeta^{q}_{n} -1)$$ prime in $$\mathcal{O}_{K}$$? How would you prove this?

• Have you computed its norm? Sep 23 '19 at 16:23
• $N_{Q(\zeta_5)/Q}(\zeta_5-1)=5$ thus $(\zeta_5-1)$ is prime of norm $5$ in $\Bbb{Z}[\zeta_5]$ thus for $p$ prime $\ne 5$, $(\zeta_5-1)$ is prime in $\Bbb{Z}[\zeta_5,\zeta_p]$ iff $(5)$ is prime in $\Bbb{Z}[\zeta_p]$ iff the order of $5$ modulo $p$ is $p-1$. Sep 23 '19 at 18:56
• @DietrichBurde Yes, I think I so. I get $N(I) = N_{\mathbb{Q}(\zeta_n) / \mathbb{Q}} (\zeta_n^p - 1) = \prod_{k \in \mathbb{Z}_n^{\times}}(\zeta_n^{kp} - 1)$ which with a few manipulations I think boils down to $q^{p-1}$. But while that doesn't prove $I$ is not prime, I don't think it proves that it is either.
– JMP
Sep 25 '19 at 15:31
• @reuns that seems very elegant. I don't follow on the penultimate iff: $(\zeta_5 - 1)$ is prime in $\mathbb{Z}[\zeta_5, \zeta_p] \iff (5)$ is prime in $\mathbb{Z} [\zeta_p]$. Could you elaborate or suggest a reference?
– JMP
Sep 25 '19 at 15:38

$$\Phi_p(X)= \sum_{m=0}^{p-1} X^m, \qquad \#\Bbb{Z}[\zeta_5]/(1-\zeta_5)=N_{Q(\zeta_5)/Q}(1-\zeta_5) = \prod_{k=1}^4(1-\zeta_5^k) = \Phi_5(1) = 5$$ Thus $$1-\zeta_5$$ is prime in $$\Bbb{Z}[\zeta_5]$$.
Then for $$p \nmid 5$$ because $$\Phi_p(X)$$ is irreducible over $$\Bbb{Q}(\zeta_5)$$
$$\Bbb{Z}[\zeta_5,\zeta_p]/(1-\zeta_5)\cong\Bbb{Z}[\zeta_5][X]/(\Phi_p(X))/(1-\zeta_5)\cong\Bbb{Z}[\zeta_5]/(1-\zeta_5)[X]/(\Phi_p(X))$$ $$\cong \Bbb{Z}/(5)[X]/(\Phi_p(X))\cong \Bbb{Z}[X]/(\Phi_p(X))/(5)\cong \Bbb{Z}[\zeta_p]/(5)$$ and hence $$(1-\zeta_5)$$ is prime in $$\Bbb{Z}[\zeta_5,\zeta_p]$$ iff $$(5)$$ is prime in $$\Bbb{Z}[\zeta_p]$$ iff $$5$$ is of order $$p-1$$ modulo $$p$$.