# Thread: Prove, that an ideal is prime in O_K

1. ## Prove, that an ideal is prime in O_K

Hello,

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

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

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

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

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

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

Bye,
Alexander

2. Originally Posted by AlexanderW
Hello,

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

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

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

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

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

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

Bye,
Alexander
Hint: $\mathcal O_K = \mathbb{Z}[\zeta_5]$.

3. Hello

thanks for the hint.

So i have to proof at first, that $\mathcal O_K = \mathbb Z[\zeta_5]$ for $K=\mathbb Q(\zeta_5)$

The minimal polynomial of $\zeta_5$ is (since 5 is a prime number) the cyclotomic polynomial:

$f_{\zeta_5}(x)=x^4+x^3+x^2+x+1$

So, i have to add the zeros of $f_{\zeta_5}$ to $\mathbb Q$ to get $\mathbb Q(\zeta_5)$.

But how does one find out the zeros of a polynomial of degree 5??

Bye,
Alexander

4. Originally Posted by AlexanderW
Hello

thanks for the hint.

So i have to proof at first, that $\mathcal O_K = \mathbb Z[\zeta_5]$ for $K=\mathbb Q(\zeta_5)$

The minimal polynomial of $\zeta_5$ is (since 5 is a prime number) the cyclotomic polynomial:

$f_{\zeta_5}(x)=x^4+x^3+x^2+x+1$

So, i have to add the zeros of $f_{\zeta_5}$ to $\mathbb Q$ to get $\mathbb Q(\zeta_5)$.

But how does one find out the zeros of a polynomial of degree 5??

Bye,
Alexander
Recall that the roots of $f_{\zeta_5}(x)$ satisfy the equation $x^5 = 1$. However, if your textbook doesn't contain a proof that $\mathcal O_K = \mathbb Z[\zeta_5]$, I doubt that you should be proving that fact to solve the problem. Suppose that P is a prime ideal in $\mathcal O_K$ dividing the ideal generated by 3. Do you know anything about the structure of the residue field $\mathcal O_K/P$?