Let $K$ be an algebraic number field of degree $n$. Let $\mathcal{O}_K$ be the ring of algebraic integers. An order of $K$ is a subring $R$ of $K$ such that $R$ is a free $\mathbb{Z}$-module of rank $n$. I am interested in the ideal theory on $R$ because when $n = 2$ it has a deep connection with the theory of binary quadratic forms as shown in this. Let $I$ be a non-zero ideal of $R$. It is easy to see that $R/I$ is a finite ring. The number of elements of $R/I$ is called the norm of $I$ and is denoted by $N(I)$. Let $\mathfrak{f} = \{x \in R | x\mathcal{O}_K \subset R\}$. If $I + \mathfrak{f} = R$, we call $I$ regular. Properties of regular ideals are stated in this question. Let $I, J$ be regular ideals of $R$. If $R = \mathcal{O}_K$, it is well-known that $N(IJ) = N(I)N(J)$. I wondered if this holds when $R \ne \mathcal{O}_K$. And I came up with the following proposition.
Proposition Let $I, J$ be regular ideals of $R$. Then $N(IJ) = N(I)N(J)$.
Outline of my proof I used the result of this question and reduced the problem to the case $R = \mathcal{O}_K$.
My question How do you prove the proposition? I would like to know other proofs based on different ideas from mine. I welcome you to provide as many different proofs as possible. I wish the proofs would be detailed enough for people who have basic knowledge of introductory algebraic number theory to be able to understand.
Related question A generalization of this question is asked here.
