Cubic field of discriminant minus 307 ($K=\mathbb{Q}(\alpha ),{\alpha }^3-5\alpha +8=0$)

The polynomial $X^3-5X+8$ is irreducible over $\mathbb{Q}$. Forone hasThe field has ring of integers $\mathbb{Z}[\beta ]$ and field discriminant $-307$.

The elementsform an integral basis. Relative to this basis, multiplication by $\beta$ has integer matrixThe basis discriminant is one quarter of the discriminant $-4\cdot 307$ of $(1,\alpha ,{\alpha }^2)$, hence is the square-free integer $-307$.

For $g(X)=X^3-5X^2+11X-12$, the reductions areandBoth quadratic factors are irreducible, soand