Let $K$ be a cubic field and let $\mathcal{O}\_K$ be its ring of integers. Does there always exist elements $\alpha, \beta \in \mathcal{O}\_K$ with $\text{Tr}(\alpha) = \text{Tr}(\beta) = 0$ such that $\{1, \alpha, \beta\}$ forms a $\mathbb{Z}$-basis for $\mathcal{O}\_K$? If the answer is generally no, then what about the case when $K$ is a cyclic cubic field?



Trace-free basis for $\mathcal{O}_K$, $K$ a cubic field
No. There do not always exist such $\alpha$ and $\beta$. If $K$ is a cubic field and such $\alpha$ and $\beta$ exist, then for all $x \in \mathcal{O}\_{K}$, ${\rm Tr}\left(\frac{1}{3} \cdot x\right) \in \mathbb{Z}$ and this implies that $1/3$ is in the inverse different of $\mathcal{O}\_{K}$, and hence that the different of $\mathcal{O}\_{K}$ is contained in $(3)$. Since the norm of the different is the discriminant, this implies that the discriminant of $\mathcal{O}\_{K}$ is a multiple of $27$. This need not be true, even for cyclic cubic fields.


