Originally Posted by

**AlexanderW** Hello,

let $\displaystyle \zeta_5$ be a primitive fifth root of unity and $\displaystyle \mathcal O_K$ the ring of integers in $\displaystyle K=\mathbb Q(\zeta_5)$.

How to prove, that $\displaystyle 3 \cdot \mathcal O_K$ is a prime ideal in $\displaystyle \mathcal O_K$?

I know that $\displaystyle \mathcal O_K$ is a dedekind ring and therefore each ideal has a factorization in prime ideals and furthermore i know that the ring of integers of $\displaystyle \mathbb Q(\sqrt m)$ is $\displaystyle \mathbb Z \cdot 1 \oplus \mathbb Z \cdot \sqrt{m}$ if $\displaystyle m \equiv 2,3 \text{ mod }4$ and $\displaystyle \mathbb Z\cdot 1 \oplus \mathbb Z\cdot \left ( \frac{1}{2}+\frac{1}{2}\sqrt{m} \right ) $ if $\displaystyle m \equiv 1 \text{ mod }4$.

In the case of $\displaystyle \zeta_3$ it is easy because of

$\displaystyle \mathbb Q(\zeta_3) = \mathbb Q \left ( \frac{-1}{2}+\frac{1}{2}i\sqrt{3} \right )=\mathbb Q(i\sqrt{3}) = \mathbb Q(\sqrt{-3})$ and $\displaystyle -3 \equiv 1 \text{ mod }4$.

But $\displaystyle \zeta_5={{\rm e}^{ \frac{2 \pi i}{5} }} = \cos \left ( \frac{2 \pi }{5} \right ) + i \sin \left ( \frac{2 \pi }{5} \right )$ and now i can't simplify the term and i don't know how the ring $\displaystyle \mathcal O_K$ looks like if $\displaystyle K=\mathbb Q(\zeta_5)$.

Please help me.

Bye,

Alexander