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

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.

• In this case, for every element of the field you can easily find a polynomial with intger coefficients which has it as a root. Can you use that? – Mariano Suárez-Álvarez Jan 22 '18 at 16:25
• Ok. I know how to find a monic polynomial with integer coefficients annihilating $\frac{-1+\sqrt{-3}}{2}$. Is this sufficient or do I have to show something more ? – Crystal Jan 22 '18 at 16:38
• Well, what does it mean for an element to be integral over $Z$? – Mariano Suárez-Álvarez Jan 22 '18 at 16:42
• An element $a$ is integral over $\mathbb{Z}$ if there is a monic polynomial $f \in \mathbb{Z}[X]$ such that $f(a) = 0$. The definition is clear to me. So $\frac{-1+\sqrt{-3}}{2}$ is integral over $\mathbb{Z}$. – Crystal Jan 22 '18 at 16:54
• Yes, and that means that the whole ring generated by that is contained in the ring of integer elements, but you also need to prove the other inclusion. – Mariano Suárez-Álvarez Jan 22 '18 at 18:08

## 1 Answer

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 $\mathbb{Z}\left[\frac{-1+\sqrt{-3}}{2}\right]=\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$.