A binary quadratic form and an ideal of an order of a quadratic number field

Let $F = ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{Z}$. We say $D = b^2 - 4ac$ is the discriminant of $F$. If $D$ is not a square integer and gcd($a, b, c) = 1$, we say $ax^2 + bxy + cy^2$ is primitive.

Is the following proposition true? If yes, how do we prove it?

Proposition Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). By this question, there exists an order $R$ of a quadratic number field $K$ such that the discriminant of $R$ is $D$. Let $ax^2 + bxy + cy^2$ be a binary quadratic form whose discriminant is $D$. Then $I = \mathbb{Z}a + \mathbb{Z}\frac{(-b + \sqrt{D})}{2}$ is an ideal of $R$. Moreover, $I$ is invertible if and only if $ax^2 + bxy + cy^2$ is primitive.

Proof that I is an ideal of $R$

By this question, $R = \mathbb{Z} + \mathbb{Z}\frac{(D + \sqrt D)}{2}$. Hence it suffices to show that $a\frac{(D + \sqrt D)}{2} \in I$ and $\frac{(-b + \sqrt D)}{2}\frac{(D + \sqrt D)}{2} \in I$.

$a\frac{(D + \sqrt D)}{2} = \frac{(aD + a\sqrt D)}{2} = \frac{(aD + ab + a(-b + \sqrt D))}{2} = a\frac{(D + b)}{2} + a\frac{(-b + \sqrt D))}{2}$

Since $D \equiv b^2$ (mod $4$), $D \equiv b^2 \equiv b$ (mod $2$). Hence $\frac{(D + b)}{2}$ is an integer. Hence $a\frac{(D + \sqrt D)}{2} \in I$.

$(-b + \sqrt D)(D + \sqrt D) = -bD - b\sqrt D + D\sqrt D + D = -bD + D + (D - b)\sqrt D = D - b^2 + (D - b)(-b + \sqrt D)$.

Hence $\frac{(-b + \sqrt D)}{2}\frac{(D + \sqrt D)}{2} = \frac{(D - b^2)}{4} + \frac{(D - b)}{2}\frac{(-b + \sqrt D)}{2}$.

Since $D \equiv b^2$ (mod $4$) and $D \equiv b$ (mod $2$), $\frac{(D - b^2)}{4}$ and $\frac{(D - b)}{2}$ are integers. Hence $\frac{(-b + \sqrt D)}{2}\frac{(D + \sqrt D)}{2} \in I$. QED

Lemma 1 Let $K$ be a quadratic number field. Let $R$ be an order of $K$, $D$ its discriminant. Then $R = \{\frac{(x + y\sqrt D)}{2} |\ x \in \mathbb{Z}, y \in \mathbb{Z}, x \equiv yD$ (mod $2)\}$.

Proof: By this question, every element $\alpha \in R$ can be writen as $\alpha = u + y\frac{(D + \sqrt D)}{2}, u, y \in \mathbb{Z}$. Hence $\alpha = \frac{(2u + yD + y\sqrt D)}{2} = \frac{(x + y\sqrt D)}{2}$, where $x = 2u + yD$. QED

Lemma 2 Let $D$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). By this question, there exists an order $R$ of a quadratic number field $K$ such that the discriminant of $R$ is $D$. Let $ax^2 + bxy + cy^2$ be an integral binary quadratic form whose discriminant is $D$. By what we have proved in the above, $I = \mathbb{Z}a + \mathbb{Z}\frac{(-b + \sqrt{D})}{2}$ is an ideal of $R$. Let $(R : I) = \{ z \in K |\ zI \subset R \}$. Then $(R : I) = \mathbb{Z} + \mathbb{Z}\frac{(b + \sqrt D)}{2a}$.

Proof(based on Cohen's A course in computational algebraic number theory): Since $D$ is non-square, $a \ne 0$. Let $z \in (R : I)$ Then $za \in R$. By Lemma 1, there exist integers such that $za = \frac{(x + y\sqrt D)}{2}$ and $x \equiv yD$ (mod $2$). Then $z = \frac{(x + y\sqrt D)}{2a}$ $z\frac{(-b + \sqrt{D})}{2} = \frac{(x + y\sqrt D)}{2a}\frac{(-b + \sqrt{D})}{2} = \frac{(-bx + x\sqrt D - by\sqrt D + yD)}{4a} = \frac{(-bx + yD + (x - by)\sqrt D)}{4a} \in R$. By Lemma 1, $x \equiv by$ (mod $2a$). Hence there exists an integer $u$ such that $x = by + 2au$. Then $z = \frac{(x + y\sqrt D)}{2a} = \frac{(2au + y(b + √D))}{2a} = u + y\frac{(b + \sqrt D)}{2a}$. Hence $z \in \mathbb{Z} + \mathbb{Z}\frac{(b + \sqrt D)}{2a}$. Hence $(R : I) \subset \mathbb{Z} + \mathbb{Z}\frac{(b + \sqrt D)}{2a}$.

It remains to prove the opposite inclusion. Let $\gamma = \frac{(b + \sqrt D)}{2a}$. Since $\mathbb{Z} \subset (R : I)$, it suffices to prove that $\gamma \in (R : I)$. Since $D \equiv b$ (mod $2$), $\gamma a = \frac{(b + \sqrt D)}{2} \in R$ by Lemma 1. On the other hand, $\gamma\frac{(-b + \sqrt{D})}{2a} = \frac{(D - b^2)}{4a} = -c \in R$. Hence $\gamma I \subset R$. Hence $\gamma \in (R : I)$. QED

Proof that $I$ is invertible if and only if $ax^2 + bxy + cy^2$ is primitive The idea of the following proof is borrowed from Cohen's A course in computational algebraic number theory.

We use the following notation. Let $x_1, \cdots, x_n$ be a finite sequence of elements of $K$. We denote by $[x_1, \cdots, x_n]$ the subgroup of $K$ generated by $x_1, \cdots, x_n$.

By this question, $I$ is invertible if and only if $I(R : I) = R$.

By Lemma 2, $(R : I) = [1, \frac{(b + \sqrt D)}{2a}]$. Hence $I(R : I) = [a, \frac{(-b + \sqrt D)}{2}][1, \frac{(b + \sqrt D)}{2a}]= [a, \frac{(b + \sqrt D)}{2}, \frac{(-b + \sqrt D)}{2}, \frac{(D - b^2)}{4a}]=[a, \frac{(b + \sqrt D)}{2}, \frac{(-b + \sqrt D)}{2}, c]=[a, b, c, \frac{(-b + \sqrt D)}{2}]=$ [gcd$(a, b, c), \frac{(-b + \sqrt D)}{2}]$.

Hence if $I(R : I) = R$, gcd$(a, b, c) = 1$.

Conversely suppose gcd$(a, b, c) = 1$. Since $D \equiv b$ (mod $2$), $\frac{(-b - D)}{2}$ is an integer. Hence $I(R : I) = [$gcd$(a, b, c), \frac{(-b + \sqrt D)}{2}] = [1, \frac{(-b + \sqrt D)}{2}] = [1, \frac{(-b - D)}{2} + \frac{(D + \sqrt D)}{2}] = [1, \frac{(D + √D)}{2}] = R$