Let $\theta$ be a root of the polynomial $p=x^3+6x-1$, and $K=\mathbb{Q}\left[\theta\right]$ the field generated by it, and $\mathcal{O}_K$ its ring of integers with integral basis $\left(1,\theta,\theta^2\right)$. Let $P$ be the principal ideal $\left(3\right)$. Since $p \mod 3 = \left(x-1\right)^3$ we know that $P$ splits as $I^3$, where $I$ is the ideal $\left(3,\theta-1\right)$. This is an indication that the class $\left[I\right]$ has order $3$, but in order to prove this one has to show that $I$ is not a principal ideal. Practically this results in proving that there doesn't exist an element in $\mathcal{O}_K$ that has norm $3$. In likewise situations in quadratic extensions this is proven by the non-existing solution of a Diophantine equation involving the expression of the norm in function of the coefficients of the elements. But unlike the quadratic case the cubic case doesn't yield a quadratic form. Now what I would like to do is the following:
The norm of the ideal $J=\left(\theta-1\right)$ is $N\left(J\right)=6$,and $N\left(P\right)=3^3$. If the ideal $I$ would be principal it would be generated by an algebraic integer that is $\gcd\left(3,\theta-1\right)$. But is there a way to obtain this value and show that it is not an algebraic integer?
Marc Bogaerts