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Let $K$ be a quadratic number field. Let $\mathcal{O}_K$ be the ring of algebraic integers in $K$. Let $R$ be an order of $K$, $D$ its discriminant. I am interested in the ideal theory on $R$ because it has a deep connection with the theory of binary quadratic forms as shown in this. By this question, $1, \omega = \frac{(D + \sqrt D)}{2}$ is a basis of $R$ as a $\mathbb{Z}$-module.

Let $x_1,\cdots, x_n$ be a sequence of elements of $R$. We denote by $[x_1,\cdots,x_n]$ the $\mathbb{Z}$-submodule of $R$ generated by $x_1,\cdots, x_n$.

Let $I$ be a non-zero ideal of $R$. I am interested in a decomposition of $I$ into a product of ideals. By this question, there exist unique rational integers $a, b, c$ such that $I = [a, b + c\omega], a \gt 0, c \gt 0, 0 \le b \lt a, a \equiv 0$ (mod $c$), $b \equiv 0$ (mod $c$). If $c = 1$, we say $I$ is a primitive ideal. Let $a = ca'$, $b = cb'$. Then $I = cJ$, where $J = [a', b' + \omega]$. Clearly $J$ is a primitive ideal. So the decomposition problem can be reduced to the case when $I$ is primitive.

Let $\frak{f}$ $= \{x \in R | x\mathcal{O}_K \subset R\}$. Let $I$ be an ideal of $R$. If $I + \mathfrak{f} = R$, we call $I$ regular.

My Question Is the following proposition correct? If yes, how do you prove it?

Proposition Let $I = [a, r + \omega]$ be a primitive regular ideal of $R$, where $a \gt 0$ and $r$ are rational integers. Suppose $a = gh, g \gt 0, h \gt 0$ Then $J_1 = [g, r + \omega], J_2 = [h, r + \omega]$ are primitive regular ideals and $I = J_1J_2$.

Remark I am not 100% 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 the proposition is correct. I also would like to know other proofs based on different ideas if the proposition is correct. I welcome you to provide as many different proofs of the result as possible. Please provide full proofs which can be understood by people who have basic knowledge of introductory algebraic number theory.

My method Use the results of this question and this question.

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Makoto Kato
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    So instead of opening new meta threads, you now incorporate them into your main site questions? (Also you should be well aware by now, I know that I told you that at least three times before, that the SE policies are not binding and every community has its own unwritten norms about self-answering a question. And as you were told, there is no issue with answering your own questions, the issues are with how you do it.) – Asaf Karagila Nov 08 '13 at 11:11
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    @AsafKaragila Could you tell me what exactly is wrong with my question? Otherwise I would not be able to improve it. Regards, – Makoto Kato Nov 09 '13 at 21:38

1 Answers1

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Let $\sigma$ be the unique non-identity automorphism of $K/\mathbb{Q}$. Since $\omega + \sigma(\omega) = D$, $\sigma(\omega) \in R$. Hence $N_{K/\mathbb{Q}}(r + \omega) = (r + \omega)(r + \sigma(\omega)) \in I$. Since $N_{K/\mathbb{Q}}(r + \omega) \in \mathbb{Z}$, it is divisible by $a$. Hence it is divisible by $g$ and $h$. Hence by this question, $J_1 = [g, r + \omega]$ and $J_2 = [h, r + \omega]$ are ideals of $R$. Let $\theta = r + \omega$. Then $J_1J_2 = (g, \theta)(h, \theta) = (gh, g\theta, h\theta, \theta^2) \subset I$. It is easy to see that $N(I) = a$. Similarly $N(J_1) = g, N(J_2) = h$. Hence $N(I) = N(J_1)N(J_2)$. By this question, $J_1$ and $J_2$ are regular. Hence $N(J_1)N(J_2) = N(J_1J_2)$ by this question. Hence $I = J_1J_2$.

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Makoto Kato
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  • I noticed that someone serially upvoted for my questions or answers including this one. While I appreciate them, I would like to point out that serial upvotes are automatically reversed by the system. – Makoto Kato Nov 27 '13 at 07:09