# The discriminant of an integral binary quadratic form and the discriminant of a quadratic number field

Let $ax^2 + bxy + cy^2$ be a binary quadratic form over $\mathbb{Z}$. Let $D = b^2 - 4ac$ be its discriminant. It is easy to see that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$).

Conversely suppose $D$ is a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). Then there exists an integral binary quadratic form of discriminant $D$(see this question).

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 $4$). Then $D$ can be written uniquely as $D = f^2 d$, where $f$ is a positive integer and $d$ is the discriminant of a unique quadratic number field.

Lemma 1 Let $m \ne 0, 1$ be a square-free integers. Then $\mathbb{Q}(\sqrt m)$ is a quadratic number field.

Proof: It suffices to prove that the polynomial $x^2 - m$ is irreducible in $\mathbb{Q}[x]$. If $m \lt 0$, the polynomial has no rational roots. Hence the assertion is clear.

If $m \gt 1$, the assertion follows from Eisenstein's criterion.

Lemma 2 Let $m \ne 0, 1$ and $m' \ne 0, 1$ be square-free integers. Suppose $\mathbb{Q}(\sqrt m) = \mathbb{Q}(\sqrt m')$. Then $m = m'$.

Proof: By Lemma 1, $\mathbb{Q}(\sqrt m)$ and $\mathbb{Q}(\sqrt m')$ are quadratic number fields. Hence there exist rational integers $a, b$ such that $\sqrt m = a + b\sqrt m'$. Taking the traces of the both sides, we get $0 = 2a$. Hence $\sqrt m = b\sqrt m'$. Squaring the both sides, we get $m = b^2 m'$. Since $m$ is square-free, $m = m'$.

Lemma 3 Let $D \ne 0$ be a non-square integer. Then $D$ can be written as $D = f^2 c$, where $f \gt 0$ is an integer and $c \ne 1$ is a square-free integer.

Proof: We can assume $D \gt 0$ without loss of generality. Let $D = p_1^{n_1}\cdots p_r^{n_r}$, where $p_1,\cdots, p_r$ are distinct prime numbers. Let $n_k = 2m_k + e_k$ for $k = 1,\cdots, r$, where $m_k$ is an integer and $e_k = 0$ or $1$. Let $f = p_1^{m_1}\cdots p_r^{m_r}$ and $c = p_1^{e_1}\cdots p_r^{e_r}$ and we are done.

Lemma 4 Let $K$ be a quadratic number field. Then there exists a unique square-free integer $m$ such that $K = \mathbb{Q}(\sqrt m)$.

Proof: By Lemma 2, it suffices to prove the existence of such $m$. Let $\alpha$ be an irrational element of $K$. Then $K = \mathbb{Q}(\alpha)$. Since $(K \colon \mathbb{Q}) = 2$, there exist rational integers $a, b, c$ such that $a \ne 0$ and $\alpha$ is a root of the polynomial $ax^2 + bx + c$. Let $D = b^2 - 4ac$. Then $\alpha = (-b + \sqrt D)/2a$ or $(-b - \sqrt D)/2a$. Hence $K = \mathbb{Q}(\sqrt D)$. By Lemma 3, $D$ can be written as $D = f^2 m$, where $f \gt 0$ is an integer and $m \ne 1$ is a square-free integer. Clearly $K = \mathbb{Q}(\sqrt m)$.

Lemma 5 Let $m$ be an integer. Let $k, l$ be integers such that $k^2 \equiv l^2 m$ (mod $4$). If $m \equiv 1$ (mod $4$), then $k \equiv l$ (mod $2$). If $m \equiv 2, 3$ (mod $4$), then $k$ and $l$ are both even.

Proof: We first note the following facts.

If $k$ is even, then $k^2 \equiv 0$ (mod $4$).

If $k$ is odd, then $k^2 \equiv 1$ (mod $4$).

Case 1: $m \equiv 1$ (mod $4$)

If $k$ is even, then $0 \equiv l^2$ (mod $4$). Hence $l$ is even. If $k$ is odd, then $1 \equiv l^2$ (mod $4$). Hence $l$ is odd.

Case 2: $m \equiv 2$ (mod $4$)

If $k$ is even, then $0 \equiv l^2 2$ (mod $4$). Hence $l$ is even.

If $k$ is odd, then $1 \equiv l^2 2$ (mod $4$). This is impossible.

Case 3: $m \equiv 3$ (mod $4$)

If $k$ is even, then $0 \equiv l^2 3$ (mod $4$). Hence $l$ is even.

If $k$ is odd, then $1 \equiv l^2 3$ (mod $4$). This is impossible.

Lemma 6 Let $m \ne 0, 1$ be a square-free integer.

If $m \equiv 1$ (mod $4$), then let $\omega = (1 + \sqrt m)/2$.

If $m \equiv 2, 3$ (mod $4$), then let $\omega = \sqrt m$.

Then the ring of algebraic integers of $\mathbb{Q}(\sqrt m)$ is a free $\mathbb{Z}$-module with a basis $1, \omega$.

Proof: Let $K = \mathbb{Q}(\sqrt m)$. Let $\mathcal{O}_K$ be the ring of algebraic integers of $K$. Since it is easy to see that $\omega \in \mathcal{O}_K$, it suffices to prove that $\mathcal{O}_K \subset \mathbb{Z} + \mathbb{Z}\omega$. Let $\alpha = a + b\sqrt m$ be an element of $\mathcal{O}_K$, where $a$ and $b$ are rational numbers. $Tr_{K/\mathbb{Q}}(\alpha) = 2a$ is a rational integer. $N_{K/\mathbb{Q}}(\alpha) = a^2 - b^2 m$ is a rational integer. Hence $4(a^2 - m b^2) = (2a)^2 - (2b)^2 m$ is a rational integer. Hence $(2b)^2 m$ is a rational integer. Since $m$ is square-free, $2b$ is a rational integer. Let $k = 2a, l = 2b$. Then $k^2 \equiv l^2 m$ (mod $4$).

Case 1 : $m \equiv 1$ (mod $4$)

By Lemma 5, $k \equiv l$ (mod $2$). Hence there exists a rational integer $t$ such that $k = l + 2t$. Then $\alpha = a + b\sqrt m = (k + l \sqrt m)/2 = (l + 2t + l\sqrt m)/2 = t + l\omega$. Hence $\alpha \in \mathbb{Z} + \mathbb{Z}\omega$.

Case 2 : $m \equiv 2, 3$ (mod $4$)

By Lemma 5, $k$ and $l$ are both even. Hence $a$ and $b$ are both rational integers. Hence $\alpha \in \mathbb{Z} + \mathbb{Z}\omega$.

Lemma 7 Let $m \ne 0, 1$ be a square-free integers.

If $m \equiv 1$ (mod $4$), then the discriminant of $\mathbb{Q}(\sqrt m)$ is $m$.

If $m \equiv 2, 3$ (mod $4$), then the discriminant of $\mathbb{Q}(\sqrt m)$ is $4m$.

Proof: This follows immediately from Lemma 6.

Lemma 8 Let $K$ be a quadratic number field. Let $D$ be the discriminant of $K$. Then $K = \mathbb{Q}(\sqrt D)$.

Proof: By Lemma 4, there exists a unique square-free integer $m$ such that $K = \mathbb{Q}(\sqrt m)$. By Lemma 7, $D = m$ or $4m$. Hence $K = \mathbb{Q}(\sqrt D)$.

Proposition Let $D \ne 0$ be a non-square integer such that $D \equiv 0$ (mod $4$) or $D \equiv 1$ (mod $4$). Then $D$ can be written as $D = f^2 d$, where $f \gt 0$ is an integer and $d$ is the discriminant of a unique quadratic number field.

Proof: Since the uniqueness of the quadratic number field follows from Lemma 8, it suffices to prove the existence.

Case 1: $D \equiv 0$ (mod 4)

By Lemma 3, $D/4$ can be written as $D/4 = g^2m$, where $g \gt 0$ is an integer and $m \ne 1$ is a square-free integer. Hence, $D = 4g^2m$. If $m \equiv 1$ (mod $4$), then by Lemma 7, $m$ is the discriminant of $\mathbb{Q}(\sqrt m)$. Let $f = 2g, d = m$, and we are done.

If $m \equiv 2, 3$ (mod $4$), then by Lemma 7, $4m$ is the discriminant of $\mathbb{Q}(\sqrt m)$. Let $f = g, d = 4m$, and we are done.

Case 2: $D \equiv 1$ (mod $4$)

By Lemma 3, $D$ can be written as $D = f^2m$, where $f \gt 0$ is an integer and $m \ne 1$ is a square-free integer. Note that $f^2 \equiv 0, 1$ (mod $4$). However, if $f^2 \equiv 0$ (mod $4$), then $D \equiv 0$ (mod $4$). This is a contradiction. Hence $f^2 \equiv 1$ (mod $4$). Hence $D \equiv m$ (mod $4$). Hence $m \equiv 1$ (mod $4$). Hence $m$ is the discriminant of $\mathbb{Q}(\sqrt m)$.

• Is this something you came up by yourself ? If not, please give the source by respect of the authors. – ogerard Feb 5 '15 at 12:30

According to wikipedia, your proposition is true: http://en.wikipedia.org/wiki/Fundamental_discriminant

Probably one of the books in the reference list of the page above proves your proposition.

• I'm expecting that someone will post his original proof. – Makoto Kato Sep 15 '12 at 3:51