Given that $\alpha$ is the root of the polynomial, $x^3 - x - 1$ is $\alpha$ and $K=\mathbb{Q}(\alpha)$, show that the norm of the ideal $\langle 5, \alpha-2\rangle$ is $5$ and the norm of the ideal $\langle 5, \alpha^2+2\alpha +3\rangle$ is $5^2$.
The discriminant of the field is $\triangle_k=-19$, so since $5\not|\triangle_k$ and we can apply Dedekind-Kummer to find the ideals above $\langle 5\rangle$. So it turns out that $\langle 5\rangle=\langle 5,\alpha - 2\rangle\langle 5, \alpha^2 + 2\alpha + 3\rangle$ is the prime factorization for $\langle 5\rangle$ over $O_K$. Since $Nm_{K/\mathbb{Q}}(\langle 5\rangle)=Nm_{K/\mathbb{Q}}(5)=5^3$, and since $Nm_{K/\mathbb{Q}}(\langle 5\rangle)=Nm_{K/\mathbb{Q}}(\langle 5,\alpha - 2\rangle)Nm_{K/\mathbb{Q}}(\langle 5,\alpha^2 + 2\alpha + 3\rangle)=5^3$.
Therefore, $Nm_{k/\mathbb{Q}}(\langle 5, \alpha-2\rangle) = 5$ or $5^2$, but how do I show that it must be $5$, once I show that, then it's clear on the norm of the second ideal.