Let $K = \mathbb{Q}\left(\frac{-1+\sqrt{-3}}{2}\right)$. Show that $\mathcal{O}_K = \mathbb{Z}\left[\frac{-1+\sqrt{-3}}{2}\right]$.

Question

Let $K = \mathbb{Q}\left(\frac{-1+\sqrt{-3}}{2}\right)$. Can someone explain me how to show that

$\mathcal{O}_K = \mathbb{Z}\left[\frac{-1+\sqrt{-3}}{2}\right]$ ?

I know that the ring of integers $\mathcal{O}_K$ is the integral closure of $\mathbb{Z}$ in $K$, i.e. it contains the elements of $K$ which are integral over $\mathbb{Z}$.

Thanks for your help.

Answer 1

This is proved in Dummit and Foote's book abstract algebra, 3rd edition, page 698, the first example. Just notice $\mathbb{Q}\left(\frac{-1+\sqrt{-3}}{2}\right)=\mathbb{Q}(\sqrt{-3})$ and $\mathbb{Z}\left[\frac{-1+\sqrt{-3}}{2}\right]=\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$.