Let $K$ be a cubic extension of the rational numbers of discriminant $D$ and $\{ 1, \omega\_2, \omega\_3 \}$ be an integral basis for the ring of integers $\mathcal{O}\_K$ of $K$. Let $\alpha \in \mathcal{O}\_K$ be primitive so that no rational prime divides $\alpha$, let the norm of $\alpha$ be equal to $a^3$, with $a \in \mathbb{Z}$, and assume that $a$ is prime to $D$.


**Question: Do there always exist rational integers $\beta\_2$, $\beta\_3$ such that $$\{ a, \beta\_2 + \omega\_2, \beta\_3 + \omega\_3 \}$$ is an integral basis for the ideal $(a, \alpha )$ ?** Even when $\gcd (disc.(\alpha ), a ) \not= 1$ ?



Do there always exist integers $\beta_2$, $\beta_3$ such that $\{a, \beta_2 + \omega_2, \beta_3 + \omega_3 \}$ is an integral basis for the ideal $(a, \alpha )$ of a cubic field
If I understand this correctly, then the answer is no. If an ideal $I$ has a basis of the desired form, then clearly any element of the ring of integers is congruent to a rational integer modulo $I$. But this implies that $I$ is a product of ideals of inertia degree $1$.


Thus if $\alpha$ is the cube of an ideal of degree $2$, such a basis cannot exist.


