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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 $N(IJ) = N(I)N(J)$.

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**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.

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**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.

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