Exercise 3.4.9 (Forms whose discriminant is a perfect square)

Proof.

• Suppose first that $a\ne 0$. Consider here $x,y$ as (indeterminate) variables in the field $\mathbb{Q}(x,y)$ of rational functions of two variables $x,y$. Since $d$ is a perfect square, Problem 7 shows that

  $$a{u}^2+\mathit{bu}+c=a(u-r_1)(u-r_2),$$

  where $r_1,r_2$ are rational numbers. For $u=x∕y$, we obtain

  $$\begin{array}{llll}\hfill f(x,y)& =y^2\left(a{\left(\frac{x}{y}\right)}^2+b\frac{x}{y}+c)\right)& \hfill & \\ \hfill & =a{y}^2\left(\frac{x}{y}-r_1\right)\left(\frac{x}{y}-r_2\right)& \hfill & \\ \hfill & =a(x-r_{1}y)(x-r_{2}y).& \hfill & \end{array}$$

  Then

  $$\begin{array}{llll}\hfill a{x}^2+\mathit{bxy}+c& =a(x-r_{1}y)(x-r_{2}y)& \hfill & \\ \hfill & =a{x}^2-a(r_1+r_2)\mathit{xy}+a{r}_1{r}_2{y}^2.& \hfill & \end{array}$$

  Therefore $-a(r_1+r_2)=b\in \mathbb{Z},a{r}_1{r}_2=c\in \mathbb{Z}$.

  Write $r_1=-\frac{s_1}{t_1},r_2=-\frac{s_2}{t_2}$, where $s_1\wedge t_1=1,s_2\wedge t_2=1$. Put

  $${\alpha }_1=a{r}_1,{\alpha }_2=a{r}_2.$$

  Then ${\alpha }_1,{\alpha }_2$ are the roots of the polynomial

  $$p(u)=(u-a{r}_1)(u-a{r}_2)=u^2+\mathit{bu}+\mathit{ac}.$$

  Moreover ${\alpha }_1,{\alpha }_2$ are rational numbers. If we write ${\alpha }_1=p∕q,p\wedge q=1,q>0$, then $p^2+\mathit{bpq}+q^2=0$, thus $q\mid p^2$, where $p\wedge q=1$, therefore $q\mid 1$ (and $q>0$), thus $q=1$. Hence ${\alpha }_1$, and similarly ${\alpha }_2$ are integers (more generally, in algebraic terms, an algebraic integer which is in $\mathbb{Q}$ is a rational integer). This shows that $t_1\mid a{s}_1$, where $t_1\wedge s_1=1$, thus $t_1\mid a$, and similarly $t_2\mid a$, so $a=q_1{t}_1=q_2{t}_2$, where $q_1,q_2$ are integers.

  Since $c=a{r}_1{r}_2=a\frac{s_1{s}_2}{t_1{t}_2}$, then

  $$\begin{array}{llll}\hfill f(x,y)& =a\left(x+\frac{s_1}{t_1}y\right)\left(x+\frac{s_2}{t_2}y\right)& \hfill & \\ \hfill & =\frac{a}{t_1{t}_2}(t_{1}x+s_{1}y)(t_{2}x+s_{2}y)& \hfill & \\ \hfill & =\left(q_{2}x+\frac{c}{s_2}y\right)(t_{2}x+s_{2}y),& \hfill & \end{array}$$

  where all coefficients are integers, except perhaps $c∕s_2$. But $b=q_2{s}_2+c\frac{t_2}{s_2}$ (coefficient of $\mathit{xy}$), thus $c\frac{t_2}{s_2}$ is an integer, so $s_2\mid c{t}_2$, where $s_2\wedge t_2=1$, therefore $s_2\mid c$.

  Put

  $$h_1=q_2=\frac{a}{t_2},k_1=\frac{c}{s_2},h_2=t_2,k_2=s_2.$$

  Then $h_1,k_1,h_2,k_2$ are integers, and

  $$f(x,y)=(h_{1}x+k_{1}y)(h_{2}x+k_{2}y).$$

• Suppose now that $a=0$. Then

  $$\begin{array}{llll}\hfill f(x,y)& =y(\mathit{bx}+\mathit{cy})& \hfill & \\ \hfill & =(h_{1}x+k_{1}y)(h_{2}x+k_{2}y),& \hfill & \end{array}$$

  where $h_1=0,k_1=1,h_2=b,k_2=c$ are integers.

If $f(x,y)$ is a quadratic form with integral coefficients whose discriminant $d$ is a perfect square, there are integers $h_1,h_2,k_1,k_2$ such that $f(x,y)=(h_{1}x+k_{1}y)(h_{2}x+k_{2}y).$ □

Proof. Put $a_1=a\wedge b$, and $a_2=\frac{a}{a_1}=\frac{a}{a\wedge b}$. Then $a=a_1{a}_2$ and $a_1\mid a$. Moreover $a\mid \mathit{bc}$, therefore

where $\frac{a}{a\wedge b}$ and $\frac{b}{a\wedge b}$ are relatively prime. Hence $a_2=\frac{a}{a\wedge b}\mid c$. □

Proof. Suppose first that $a=0$. Then

where $h_1=0,k_1=1,h_2=b,k_2=c$ are integers.

Now we suppose that $a\ne 0$. Since the discriminant $d$ is a perfect square, there is an integer $\delta$ such that $d={\delta }^2$. By equation (3.3),

Therefore

where $\frac{b-\delta }{2},\frac{b+\delta }{2}$ are integers. Indeed, $b$ and $d=b^2-4\mathit{ac}$ have same parity. If $b$ is even, $4\mid d$, so $\delta$ is even, and if $b$ is odd, $d$ and $\delta$ are odd.

Moreover

Thus $a\mid \frac{b-\delta }{2}\frac{b+\delta }{2}$. By the Lemma, there are integers $a_1,a_2$ such that

By equation (1),

where

are integers.

If $f(x,y)$ is a quadratic form with integral coefficients whose discriminant $d$ is a perfect square, there are integers $h_1,h_2,k_1,k_2$ such that $f(x,y)=(h_{1}x+k_{1}y)(h_{2}x+k_{2}y).$   □