Let $l$ be an odd prime number and $\zeta$ be a primitive $l$-th root of unity in $\mathbb{C}$. Let $\mathbb{Q}(\zeta)$ be the cyclotomic field. Let $A$ be the ring of algebraic integers of $\mathbb{Q}(\zeta)$. Let $\alpha \in A$. Let $N(\alpha)$ be the norm of $\alpha$.

My question: How can we prove that $N(\alpha) \equiv 0$ or $\equiv 1$ (mod $l$)?

  • $\begingroup$ Is $A = \Bbb{Z}[\zeta_l]$ where $l$ is the odd prime in the assumption? $\endgroup$ – user38268 Jul 23 '12 at 7:11
  • $\begingroup$ @BenjaLim It's a well known fact. So you can take it for granted. $\endgroup$ – Makoto Kato Jul 23 '12 at 7:41

Observe that $\zeta^n-\zeta\in (1-\zeta)$ if $l$ does not divide $n$. This implies that all the conjugates of $\alpha$ are congruent mod $(1-\zeta)$, hence $N\alpha\equiv \alpha^{l-1}$ mod $(1-\zeta)$. Since $A/(1-\zeta)\cong\mathbb{F}_l$, we have $N\alpha\equiv0$ or $\equiv1$ mod $l$.

| cite | improve this answer | |
  • $\begingroup$ I'd like to add a bit of explanation just in case. Since $\zeta \equiv 1$ (mod $1 - \zeta$), $\zeta^n \equiv 1$ (mod $1 - \zeta$). Hence $\zeta \equiv \zeta^n$ (mod $1 - \zeta$). $\endgroup$ – Makoto Kato Jul 23 '12 at 10:45

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.