# 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.

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Some off-topic discussion of meta-matters (downvoting, etc) has been purged. Please refrain from discussing meta matters on the main site. Instead, please use the meta site. – Bill Dubuque Oct 16 '12 at 3:46
How is this elementary-number-theory? Per chance a retag is in order? – awllower Feb 22 '13 at 11:56
@awllower The theory of integral binary quadratic forms belongs to elementary number theory. – Makoto Kato Feb 22 '13 at 14:33
@MakotoKato The introduction to this tag does not specify the point, and I think this question, quite involved with algebraic notions such as number fields, or ideals..., should pertain to algebraic number theory. Of course, this could be said to be a problem of taste, while tastes differ. – awllower Feb 23 '13 at 2:22
@awllower I have a book titled "elementary number theory" written by Teiji Takagi in Japanese. The author is one of the main pioneers of class field theory. The book uses only elementary methods as the title shows. It contains the theory of integral binary quadratic forms and the theory of quadratic number fields. I would like to point out that quadratic number fields can be treated quite elementarily without the knowledge of algebraic number theory. The book, however, does not provide an answer to my question. – Makoto Kato Feb 23 '13 at 3:12

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$(a, b, c), \frac{(-b + \sqrt D)}{2}] = [1, \frac{(-b + \sqrt D)}{2}] = [1, \frac{(-b - D)}{2} + \frac{(D + \sqrt D)}{2}] = [1, \frac{(D + √D)}{2}] = R$

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