# On the norm formula $N(IJ) = N(I)N(J)$ in 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 of $K$.
Let $R$ be an order of $K$, i.e. a subring of $K$ which is a free $\mathbb{Z}$-module of rank $n$.
Let $\mathfrak{f} = \{x \in R | x\mathcal{O}_K \subset R\}$.
Let $I$ be a non-zero ideal of $R$.
If $I + \mathfrak{f} = R$, we call $I$ regular.
For the properties of regular ideals, see this.
We denote by $N(I)$ the number of elements of the finite ring $R/I$.

If $I$ and $J$ are regular ideals of $R$, we have $N(IJ) = N(I)N(J)$ as proved in my answer to this question.
I wonder if the formula still holds when one of them is not regular.
And I came up with the following proposition.

Proposition
Let $R$ be an order of an an algebraic number field.
Let $I$ be a regular ideal, $J$ be a non-zero ideal of $R$.
Then $N(IJ) = N(I)N(J)$.

Outline of my proof
I generalized the result of this and proved $R/J$ is $R$-isomorophic to $I/IJ$.
See my answer below for the detail.

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.

#### Solutions Collecting From Web of "On the norm formula $N(IJ) = N(I)N(J)$ in an order of an algebraic number field"

Lemma
Let $I$ be a regular ideal, $J$ be a non-zero ideal of $R$.
Then there exists non-zero $\alpha \in I$ and an ideal $M$ such that $\alpha R = IM, M + J = A$.

Proof:
By this question, $I$ is uniquely decomposed as a product of regular prime ideals.
Let $I = P_1^{e_1}…P_n^{e_n}$ be the prime decomposition, where $P_1,\cdots,P_n$ are distinct regular prime ideals and $e_i \ge 1$ for all $i$.
Let $Q_1, …, Q_m$ be all the distinct prime ideals which contain $J$, but not contain $I$.
For each $i$ choose $\alpha_i\in\ P_i^{e_i}\setminus P_i^{e_i+1}$.
For each $j$ choose $\beta_j\in\ A\setminus Q_j$.
Note that $P_1,\cdots,P_n$ and $Q_1, …, Q_m$ are maximal ideals.
Hence, by the Chinese Remainder Theorem, there exists $\alpha\in A$ such that
$\alpha\equiv\alpha_i\pmod{P_i^{e_i+1}}$ for all $i$ and
$\alpha\equiv\beta_j\pmod{Q_j}$ for all $j$.
Since $\alpha \in P_i^{e_i}$ for all $i$, $\alpha \in I$.
Since $I$ is invertible by this question, there exists an ideal $M$ such that $\alpha R = IM$.
Clearly $M + J = A$

Proof of the proposition
It suffices to prove that $R/J$ is isomorphic to $I/IJ$.
By the lemma, there exist non-zero $\alpha \in I$ and an ideal $M$ such that $\alpha R = IM, M + J = R$.
Then $IM + IJ = I$.
Let $\psi\colon R \rightarrow I/IJ$ be the $R$-homomorphism sending $x$ to $\alpha x$ (mod $IJ$).
Since $\alpha R + IJ = I$, $\psi$ is surjective.
It remains to prove that $\psi$ is injective.
Suppose $\alpha x \in IJ$ for $x \in R$.
It suffices to prove that $x \in J$.
Since $\alpha R = IM$, $IMx \subset IJ$.
Since $I$ is invertible, $Mx \subset J$.
Since $M + J = R$, there exist $\mu \in M$ and $\beta \in J$ such that $\mu + \beta = 1$.
Since $\mu x + \beta x = x$, $x \in J$ as desired.