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Let $K$ be a quadratic number field.

Let $R$ be an order of $K$, $D$ its discriminant.

I am interested in the ideal theory on $R$ because it is closely related to 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 $I$ be a non-zero ideal of $R$.

It is easy to see that there exist integers $a \gt 0, c \gt 0, b$ such that $I = \mathbb{Z}a + \mathbb{Z}(b + c\omega)$. $a$ is the smallest positive integer contained in $I$.

$c$ is the smallest postive integer $y$ such that there exist integer $x$ such that $x + y\omega \in I$.

Since $a\omega \in I$, $a \equiv 0$ (mod $c$).

It is easy to see that we can choose $b$ such that $0 \le b \lt a$.

I came up with the following proposition(see my answer below).

**Proposition**

Let $K, R, \omega$ be as above.

Let $I$ be a non-zero ideal of $R$.

Then 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 $c$).

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**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.

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Since $I$ is an ideal, $(b+c\omega)\omega \in I$.Since $\omega^2 = d\omega + \frac{d(1-d)}{4}$, $(b+c\omega)\omega = (b + cd)\omega + c\frac{d(1-d)}{4}$ Hence $b + cd \equiv 0$ (mod $c$). Hence $b \equiv 0$ (mod $c$)

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