Ring of integers of $K=\mathbb{Q}(\alpha)$ where $\alpha$ is a root of $X^3-3X+1$

Question

I have to show that $\mathcal{O}_K=\mathbb{Z}[\alpha]$. The hint that the exercise gives is that I should first show that $\mathcal{O}_K = \mathbb{Z}[\alpha]+3\mathcal{O}_K$ and I know how to finish from there.

I have already proved that $(\alpha+1) \subset \mathcal{O}_K$ is a prime ideal of norm $3$ with $\frac{\mathcal{O}_K}{(\alpha+1)} \cong \mathbb{F}_3$ and I feel like the deduction should be obvious from here but I must be missing something.

Thank you for your help.

Answer 1

From the isomorphism $\mathcal{O}_K /(\alpha +1) \cong \mathbb{Z}/3$ we deduce $\mathcal{O}_K = \mathbb{Z} + (\alpha +1)\mathcal{O}_K$. In particular, we have $$\mathcal{O}_K = \mathbb{Z}[\alpha+1] +(\alpha +1)\mathcal{O}_K.$$ Multiplying this equation by $\alpha+1$ on both sides, we obtain $$(\alpha +1)\mathcal{O}_K =(\alpha +1) \mathbb{Z}[\alpha+1] +(\alpha +1)^2\mathcal{O}_K$$ Combining the two displayed equations, we get $\mathcal{O}_K = \mathbb{Z}[\alpha +1] +(\alpha+1)^2\mathcal{O}_K$. Repeating this last step, we arrive at $\mathcal{O}_K =\mathbb{Z}[\alpha +1] + (\alpha+1)^3\mathcal{O}_K$. Now using the fact that $\mathbb{Z}[\alpha+1] = \mathbb{Z}[\alpha]$ and $(\alpha+1)^3\mathcal{O}_K =3\mathcal{O}_K$, we have the desired result.