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.

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    $\begingroup$ I think you're supposed to put the answer in the answer box, and save the meta for meta. $\endgroup$ – dfeuer Nov 10 '13 at 10:13
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    $\begingroup$ The spoiler shows how I tried to solve the problem. When you post a question, You need to show your effort to solve it. $\endgroup$ – Makoto Kato Nov 10 '13 at 12:58
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    $\begingroup$ I opened the following meta thread asking why this question was put on hold. meta.math.stackexchange.com/questions/11617/… $\endgroup$ – Makoto Kato Nov 10 '13 at 15:59

By this question, $N(I) = N(I\mathcal{O}_K), N(J) = N(J\mathcal{O}_K), N(IJ) = N(IJ\mathcal{O}_K)$. Hence $N(IJ) = N(I\mathcal{O}_K)N(J\mathcal{O}_K) = N(I)N(J)$.

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