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Let $p$ be a rational prime. Consider the ring of integers $\mathbb{Z}[\zeta_p] $ of the $p$-th cyclotomic field $\mathbb{Q}(\zeta_p)$. If the norm $N(\alpha)$ of $\alpha \in \mathbb{Z}[\zeta_p]$ is a rational prime, must $\alpha$ be a prime element of $\mathbb{Z}[\zeta_p] $?

If it helps, I only need the case where $N(\alpha) \equiv 1$ mod $p$.

Solveit
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    This is true in any number ring; it follows painlessly from the unique decomposition in prime ideals, but a direct proof without ideals is a bit tough. –  Dec 28 '16 at 08:00
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    Yes it means it is a prime element. And for showing $(\alpha)$ is a prime ideal, you need to show that $|N(x)|$ extends to an integer-valued (multiplicative) norm on ideals – reuns Dec 28 '16 at 08:01
  • Thank you! If either of you would write it up as an answer, I will accept it. – Solveit Dec 28 '16 at 08:14

1 Answers1

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As pointed out by user1952009, we can look at the norm of the ideal $(\alpha) \subset \Bbb Z[\zeta_n]$.

Definition. Let $K$ be a number field and let $I$ be a non-zero ideal of $\mathcal O_K$. Then the (absolute) norm of $I$ is defined as the cardinality of the quotient ring $\mathcal O_K / I$, which is finite.

Proposition. Let $x \in \mathcal O_K$ be non-zero. Then $$N( x\mathcal O_K) = |N_{K/\Bbb Q}(x)|$$

Finally, if $\alpha$ has a prime norm, then the ideal $(\alpha)$ has a prime absolute norm, say $p$, which means that the ring $\mathcal O_K/(\alpha)$ has $p$ elements. It implies that this ring is the field $\Bbb F_p$ (see also here), and $(\alpha)$ is a maximal ideal of $\mathcal O_K$ and $\alpha$ is prime in $\mathcal O_K$.

Watson
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