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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 $l$)?

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  • $\begingroup$ 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
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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 $l$.

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  • $\begingroup$ 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

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