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?
1 Answer
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.
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1$\begingroup$ This is very nice, as it gives a simple necessary condition for the existence of $\alpha$ and $\beta$. As an answer to the question of whether there always exist such $\alpha$ and $\beta$, bringing in the discriminant and different and inverse different is a bit of overkill, isn't it? All one needs is a monic, irreducible cubic with quadratic term not a multiple of three, such as $x^3-x^2-1$, isn't that right? Or, $x^3+x^2-2x-1$, if you want a cyclic example. $\endgroup$ Commented Jul 7, 2022 at 3:14
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1$\begingroup$ To add some detail to Gerry's comment, if such $\alpha$ and $\beta$ exist then ${\rm Tr}_{K/\mathbf Q}(a + b\alpha + c\beta) = 3a$ for all integers $a$, $b$, and $c$, so ${\rm Tr}_{K./\mathbf Q}(\mathcal O_K) = 3\mathbf Z$ and thus we just need a cubic field containing an algebraic integer whose trace is not divisible by $3$. $\endgroup$– KConradCommented Jul 10, 2022 at 18:46