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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 $R = \mathcal{O}_K$.

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**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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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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