Let $K$ be a quadratic number field. Let $R$ be an order of $K$, $D$ its discriminant. By this question, $1, \omega = \frac{(D + \sqrt D)}{2}$ is a basis of $R$ as a $\mathbb{Z}$-module. Let $I$ be a non-zero ideal of $R$. By this question, there exist unique integers $a, b, c$ such that $I = \mathbb{Z}a + \mathbb{Z}(b + c\omega), a \gt 0, c \gt 0, 0 \le b \lt a, a \equiv 0$ (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 $J = \mathbb{Z}a' + \mathbb{Z}(b' + \omega)$. Clearly $J$ is a primitive ideal.

Let $I = \mathbb{Z}a + \mathbb{Z}(b + \omega)$ be a primitive ideal, where $a$ and $b$ are integers such that $a \gt 0, 0 \le b \lt a$. Let $\omega'$ be the conjugate of $\omega$, i.e. $\omega' = \sigma(\omega)$, where $\sigma$ is the unique non-identity automorphism $K/\mathbb{Q}$. Since $\omega + \omega' = D$, $\omega' \in R$. Hence $N_{K/\mathbb{Q}}(b + \omega) = (b + \omega)(b + \omega') \in I$. Hence $N_{K/\mathbb{Q}}(b + \omega)$ is divisible by $a$.

I came up with the following proposition.

Proposition Let $K, R, \omega$ be as above. Let $a \ne 0$ and $b$ be integers. Suppose $N_{K/\mathbb{Q}}(b + \omega)$ is divisible by $a$. Then $I = \mathbb{Z}a + \mathbb{Z}(b + \omega)$ is an ideal of $R$.

Outline of my proof I showed $(b + \omega)\omega \in I$ using $\omega + \omega' = D$.

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.


It suffices to prove that $(b + \omega)\omega \in I$. Let $N_{K/\mathbb{Q}}(b + \omega) = ac$. Then $(b + \omega)(b + \omega') = (b + \omega)(b + D - \omega) = ac$. Hence $(b + \omega)(b + D) - (b + \omega)\omega = ac$. Hence $(b + \omega)\omega = (b + \omega)(b + D) - ac \in I$ as desired.

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