@Alessandro Let $\alpha=\frac{1+\sqrt{-31}}{2}$. Then $\mathfrak{p}_1^3=(2,\alpha)^3=(8,4\alpha,2\alpha^2,\alpha^3)$. The minimal polynomial of $\alpha$ is $x^2-x+8$, so $\alpha^2-\alpha+8=0$, so $\alpha^2=\alpha-8$, so $\alpha^3=\alpha^2-8\alpha=-7\alpha-8$, so $(8,4\alpha,2\alpha^2,\alpha^3)=(8,4\alpha,2\alpha-16,-7\alpha-8)=(\alpha)$ (skipping doing the whole double inclusions), note that $\alpha$ is contained in the LHS and $\alpha$ has norm $8$ which is the norm of the ideal