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.

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
• And once again, @MakotoKato , one question that pops up is: why? Why are you "expecting" someone to post his original proof and why didn't you explicitly ask that in your OP? If you don't add insights, ideas, background (!), etc. to your questions, how can we guess what you want? Perhaps there's something in some book's proof that you don't like/understand or what? You've already asked lots of questions and you've already been adviced a lot about how to ask them. Haven't you learned anything from all those comments, ideas, etc.? – DonAntonio Sep 15 '12 at 8:55
• @DonAntonio I just want someone to write his proof here. Citing a book can help, but not everybody can have an access to it. – Makoto Kato Sep 15 '12 at 14:31

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