# Ring of integers

• Jun 13th 2013, 12:13 AM
AlexanderW
Ring of integers
Hello,

let $K=\mathbb Q(\sqrt d)$ be a number field.

It is well known, that the ring of integers $\mathcal O_K$ of $K$ is given by $\mathbb Z[\xi]$ with

$\xi = \frac{1}{2}+\frac{1}{2}\sqrt d \ \ \quad \text{ if } \quad d \equiv 1 \text{ mod }4$
$\xi = \sqrt d \quad \quad \ \ \quad \text{ if } \quad d \equiv 2,3 \text{ mod }4.$

How to prove that this is the same as the ring $\mathbb Z[ \theta]$ with $\theta = \frac{1}{2}d + \frac{1}{2} \sqrt{d}$ ?

Bye,
Alexander
• Jun 17th 2013, 08:32 PM
Drexel28
Re: Ring of integers
Quote:

Originally Posted by AlexanderW
Hello,

let $K=\mathbb Q(\sqrt d)$ be a number field.

It is well known, that the ring of integers $\mathcal O_K$ of $K$ is given by $\mathbb Z[\xi]$ with

$\xi = \frac{1}{2}+\frac{1}{2}\sqrt d \ \ \quad \text{ if } \quad d \equiv 1 \text{ mod }4$
$\xi = \sqrt d \quad \quad \ \ \quad \text{ if } \quad d \equiv 2,3 \text{ mod }4.$

How to prove that this is the same as the ring $\mathbb Z[ \theta]$ with $\theta = \frac{1}{2}d + \frac{1}{2} \sqrt{d}$ ?

Bye,
Alexander

Alexander,

Just check that $\xi\in\mathbb{Z}[\theta]$ and $\theta\in\mathbb{Z}[\xi]$. Where are you having trouble?

Best,
Alex
• Jun 18th 2013, 02:00 AM
AlexanderW
Re: Ring of integers
Maybe the exercise is incorrect, because:

I would like to check that $\theta \in \mathbb Z[\xi]$ ... then the case $d \equiv 2,3 \text{ mod 4}$ yields a contradiction:

$\frac{1}{2}d+\frac{1}{2}\sqrt{d} = a+b \cdot \sqrt{d} \in \mathbb Z[\xi]$ is not possible for $a,b \in \mathbb Z$.

Quote:

Originally Posted by Drexel28
Alexander,

Just check that $\xi\in\mathbb{Z}[\theta]$ and $\theta\in\mathbb{Z}[\xi]$. Where are you having trouble?

Best,
Alex

• Jun 18th 2013, 12:31 PM
Drexel28
Re: Ring of integers
Quote:

Originally Posted by AlexanderW
Maybe the exercise is incorrect, because:

I would like to check that $\theta \in \mathbb Z[\xi]$ ... then the case $d \equiv 2,3 \text{ mod 4}$ yields a contradiction:

$\frac{1}{2}d+\frac{1}{2}\sqrt{d} = a+b \cdot \sqrt{d} \in \mathbb Z[\xi]$ is not possible for $a,b \in \mathbb Z$.

You seem to be correct. The minimal polynomial for $\frac{1}{2}d+\frac{1}{2}\sqrt{d}$ when $d=2$ is $2x^2-4x+1$, and thus this element isn't even integral over $\mathbb{Z}$. Are you sure the exercise is written down correctly? Can you give more context (book its in, etc.)?

Best,
Alex