Originally Posted by

**AlexanderW** Hello,

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

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

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

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

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

Thanks for your help in advance.

Bye,

Alexander