Canonical basis of an ideal of a quadratic order

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 $c$).

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.

Solution 1:

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 $c$)