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 most recent 30 from http://mathoverflow.net 2013-05-26T00:02:00Z http://mathoverflow.net/feeds/question/87514 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/87514/do-there-always-exist-integers-beta-2-beta-3-such-that-a-beta-2-om 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 Samuel Hambleton 2012-02-04T09:33:20Z 2012-02-04T19:17:57Z <p>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 $D$.</p> <p><strong>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 $(a, \alpha )$ ?</strong> Even when $\gcd (disc.(\alpha ), a ) \not= 1$ ?</p> http://mathoverflow.net/questions/87514/do-there-always-exist-integers-beta-2-beta-3-such-that-a-beta-2-om/87544#87544 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 Franz Lemmermeyer 2012-02-04T19:17:57Z 2012-02-04T19:17:57Z <p>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 $1$.</p> <p>Thus if $\alpha$ is the cube of an ideal of degree $2$, such a basis cannot exist.</p>