Algebraic number field with non trivial integral basis

Question

So far I have only seen extensions of $\mathbb{Q}$ with "trivial" integral basis. Meaning that the integral basis is the most natural one e.g. the integral basis for $\mathbb{Q}(\sqrt[3]{2})$ is just $\{1, \sqrt[3]{2}, \sqrt[3]{2}^2 \}$ (I know this is not easily proven). So is there a extension which integral basis is really hard to intuitively see?

Answer 1

Considering pure cubic number fields shows already that we do not always have such a nice integral basis. Suppose that $K=\mathbb{Q}(\sqrt[3]{d})$ with $d=ab^2$ cubefree with coprime squarefree integers $a$ and $b$.

Theorem If $a^2\not\equiv b^2(9)$, then $K$ is monogenic if and only if the Diophantine equation $ax^3+by^3=1$ has integer solutions. If $a^2\equiv b^2(9)$, then $K$ is monogenic if and only if the Diophantine equation $ax^3+by^3=9$ has integer solutions.

In particular, take any prime $p\equiv 1(9)$, which can be represented by $7x^2+3xy+9y^2$ with integers $x,y$. Then $K=\mathbb{Q}(\sqrt[3]{p})$ is not monogenic, i.e., the ring of integers $\mathcal{O}_K$ is not of the easy form $\mathbb{Z}[\alpha]$ for some $\alpha\in \mathcal{O}_K$. For example, $p=19$ can be represented with $x=y=1$ in this form, so that $K=\mathbb{Q}(\sqrt[3]{19})$ is not monogenic. So can you "intuively see" the integral basis here ?

P.S. As to the question, why this is hard in general, see the discussion here.