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

2012-02-04T09:33:20Z

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 $\gcd (disc.(\alpha ), a ) \not= 1$ ?

Answer by Franz Lemmermeyer for 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

2012-02-04T19:17:57Z

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.

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 - MathOverflow

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

Answer by Franz Lemmermeyer for 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