0
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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
1
$\begingroup$

$$\Phi_p(X)= \sum_{m=0}^{p-1} X^m, \qquad \#\Bbb{Z}[\zeta_5]/(1-\zeta_5)=N_{Q(\zeta_5)/Q}(1-\zeta_5) = \prod_{k=1}^4(1-\zeta_5^k) = \Phi_5(1) = 5$$ 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 $p$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.