Let $K=\mathbb{Q}(\alpha)$ be an imaginary quadratic field. Suppose $\mathcal{O}=\mathbb{Z}[\alpha]$ is a non-maximal order such that a prime $p$ divides its conductor. We might take $\alpha=\frac{d+\sqrt{d}}{2}$ where $d$ is the discriminant of $\mathcal{O}$.
Show that $\mathfrak{a}=(p, \alpha)$ is a non-invertible ideal in $\mathcal{O}$ and compute the absolute norm $N(\alpha)$.
I've heard about that in number fields if the ideal is relatively prime to the the conductor then it is invertible. I'm not sure if the converse is also true. For computing the absolute norm, if $\mathcal{O}$ was maximal we can consider the prime factorization of $p$ in $\mathcal{O}$ but what can we do when $\mathcal{O}$ is not maximal?