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?

  • $\begingroup$ Have you computed its norm? $\endgroup$ Sep 23 '19 at 16:23
  • $\begingroup$ $N_{Q(\zeta_5)/Q}(\zeta_5-1)=5$ 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 $p-1$. $\endgroup$
    – reuns
    Sep 23 '19 at 18:56
  • $\begingroup$ @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. $\endgroup$
    – JMP
    Sep 25 '19 at 15:31
  • $\begingroup$ @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? $\endgroup$
    – 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 $p$.


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