# Norm of the product of two regular ideals of an order of an algebraic number field

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.

#### Solutions Collecting From Web of "Norm of the product of two regular ideals of an order of an algebraic number field"

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 $N(IJ) = N(I\mathcal{O}_K)N(J\mathcal{O}_K) = N(I)N(J)$.