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.
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.