Apparently this should be a straightforward / standard homework problem, but I'm having trouble figuring it out.
Let $D$ be a square-free integer not divisible by $3$. Let $\theta = \sqrt[3]{D}$, $K = \mathbb{Q}(\theta)$. Let $\mathcal{O}_K$ be the ring of algebraic integers inside $K$. I need to find explicitly elements generating $\mathcal{O}_K$ as a $\mathbb{Z}$-module.
It is reasonably clear that $\theta$ is itself an algebraic integer and that $\mathbb{Z}[\theta] \le \mathcal{O}_K$, but I strongly suspect it isn't the whole ring. I'm not sure where the hypotheses on $D$ come in at all… any hints would be much appreciated.
Best Answer
A very belated answer: This is (part) of the content of Exercise 1.41 of Marcus' Number Fields (a great source of exercises in basic number theory). In it, one proves that, for $K = \mathbf{Q}(m^{1/3})$, $m$ squarefree, an integral basis is given by
$\begin{cases} 1, m^{1/3}, m^{2/3},& m \not \equiv \pm 1 \pmod 9 \\ 1, m^{1/3}, \frac{m^{2/3} \pm m^{1/3} + 1}{3},& m \equiv \pm 1 \pmod 9 \end{cases}$
This is leveraged out of his Theorem 13 (among other things).