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 it is closely related to the theory of binary quadratic forms when $n = 2$ as shown in this. Let $I$ be a non-zero ideal of $R$. We denote by $N(I)$ the number of elements of the finite ring $R/I$. Let $\mathfrak{f} = \{x \in R | x\mathcal{O}_K \subset R\}$. If $I + \mathfrak{f} = R$, we call $I$ regular. For the properties of regular ideals, see this.

My question Is the following Proposition correct? If yes, how do you prove it?

Proposition Let $K$ be an algebraic number field of degree $n$ such that $K/\mathbb{Q}$ is Galois. Let $G$ be the Galois group of $K/\mathbb{Q}$. Let $R$ be an order of $K$ such that $\sigma(R) = R$ for every $\sigma \in G$. Let $I$ be a regular ideal of $R$. Then $\prod_{\sigma\in G} \sigma(I) = N(I)R$.

Remark I am not sure of the correctness of the proposition, though I think it is likely to be true(see my method below). I would like to know if I am on the right track. I also would like to know as many different proofs of the proposition as possible if it is correct. When you post an answer, please provide a full proof which can be understood by people who have basic knowledge of introductory algebraic number theory.

My method Use the results of this and reduce the problem to the case $R = \mathcal{O}_K$.

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    $\begingroup$ This seems likely. It is fairly well-known, for example, that $\displaystyle \prod_{\sigma \in G}\sigma(I)\cap\mathbb{Z}=N_{K/\mathbb{Q}}(I)$. Thus, we have the obvious containment $N(I)R\subseteq\prod_{\sigma}\sigma(I)$, and I think by checking indices that cardinalities force equality. $\endgroup$ – Alex Youcis Nov 11 '13 at 10:05

We first show that $\mathfrak{f}$ is $G$-invariant, i.e. $\sigma(\mathfrak{f}) = \mathfrak{f}$ for every $\sigma \in G$. It suffices to prove that $\sigma(\mathfrak{f}) \subset \mathfrak{f}$ for every $\sigma \in G$. Let $x \in \mathfrak{f}$. Since $x\mathcal{O}_K \subset R, \sigma(x)\mathcal{O}_K \subset \sigma(R) = R$. Hence $\sigma(x) \in \mathfrak{f}$ as desired.

Let $a = N(I)$. Let $J = \prod_{\sigma\in G} \sigma(I)$. Since $I + \mathfrak{f} = R$, $\sigma(I) + \sigma(\mathfrak{f}) = \sigma(R)$ for every $\sigma \in G$. Since $\sigma(\mathfrak{f}) = \mathfrak{f}$ and $\sigma(R) = R$, $\sigma(I)$ is regular. Hence $J$ is regular.

By this question, $a = N(I\mathcal{O}_K)$. It is well-known that $\prod_{\sigma\in G} \sigma(I\mathcal{O}_K) = a\mathcal{O}_K$. Since $J\mathcal{O}_K = \prod_{\sigma\in G} \sigma(I\mathcal{O}_K)$, $J\mathcal{O}_K = a\mathcal{O}_K$. By this question, $J\mathcal{O}_K \cap R = J$.

Hence it suffices to prove that $a\mathcal{O}_K \cap R = aR$. Since $\sigma(I)\mathcal{O}_K + \mathfrak{f} = \mathcal{O}_K$ for every $\sigma \in G$, $a\mathcal{O}_K + \mathfrak{f} = \mathcal{O}_K$. Hence by this question, $a\mathcal{O}_K \cap R$ is regular and $(a\mathcal{O}_K \cap R)\mathcal{O}_K = a\mathcal{O}_K$. By this question, $N(a\mathcal{O}_K \cap R) = N(a\mathcal{O}_K) = a^n$. Since $aR \subset a\mathcal{O}_K \cap R$ and $N(aR) = a^n$, $a\mathcal{O}_K \cap R = aR$ as desired.QED

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