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 a non-zero ideal of $R$. If $I + \mathfrak{f} = R$, we call $I$ regular. For the properties of regular ideals, see this. We call a non-zero ideal $J$ totally non-regular if every prime ideal containing $J$ is non-regular. I came up with the following proposition.

Proposition Let $I = [a, r + \omega]$ be a primitive ideal of $R$, where $a \gt 0$ and $r$ are rational integers. Then $I$ can be uniquely decomposed into $I = JM$, where $J$ is a regular ideal and $M$ is a totally non-regular ideal. Moreover there exist rational integers $g \gt 0, h \gt 0$ such that $a = gh$ and $J = [g, r + \omega], M = [h, r + \omega]$.

Outline of my proof I used the results of this question and mimicked my proof of this question.

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.


1 Answer 1


It is easy to see that there exist unique positive rational integers $g, h$ such that $a = gh$ and gcd$(g, f) = 1$ and every prime divisor of $h$ is a prime divisor of $f$.

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, $J = [g, r + \omega]$ and $M = [h, r + \omega]$ are ideals of $R$. By this question, $J$ is regular. We claim that $M$ is totally non-regular. Let $P$ be a prime ideal of $R$ containing $M$. It suffices to prove that $P$ is not regular. Let $p$ be the unique prime number contained in $P$. Since $h \in P$, $h$ is divisible by $p$. Since $f$ is divisible by $p$, $f \in P$. Hence $P$ is not regular by Lemma 4 of my answer to this question.

Next we prove that $I = JM$. Let $\theta = r + \omega$. Then $JM = (g, \theta)(h, \theta) = (gh, g\theta, h\theta, \theta^2) \subset I$. It is easy to see that $N(I) = a$. Similarly $N(J) = g, N(M) = h$. Hence $N(I) = N(J)N(M)$. Since $J$ is regular, $N(J)N(M) = N(JM)$ by this question. Hence $I = JM$.

It remains to prove the uniqueness of the decomposition $I = JM$. Note that, by this question, $J$ is invertible and uniquely decomposed as a product of regular prime ideals. Let $I = J'M'$ be another such decomposition. Since $J$ is invertible, it suffices to prove that $J = J'$. Suppose $J \subset P^e$, $e \ge 1$ where $P$ is a prime ideal. It suffices to prove that $J' \subset P^e$. Since $J \subset P$ and $J$ is regular, $P$ is also regular. Hence $P$ does not contain $M'$. Hence $M' + P^e = R$. Hence there exist $\mu \in M'$ and $\pi \in P^e$ such that $\mu + \pi = 1$. Let $x \in J'$. Since $x = \mu x + \pi x$ and $M'J' \subset P^e$, $x \in P^e$. Hence $J' \subset P^e$ as desired.


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