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Let $\alpha \in \mathbb{C}$ be an algebraic number. If the minimal polynomial of $\alpha$ over $\mathbb{Q}$ has degree $2$, we say $\alpha$ is a quadratic number. Let $\theta$ be a quadratic number. $\theta$ is a root of the unique polynomial $ax^2 + bx + c \in \mathbb{Z}[x]$ such that $a > 0$ and gcd$(a, b, c) = 1$. $D = b^2 - 4ac$ is called the discriminant of $\theta$.

Let $\sigma = \left( \begin{array}{ccc} p & q \\ r & s \end{array} \right)$ be an element of $GL_2(\mathbb{Z})$.

We claim that $\omega = \frac{p\theta + q}{r\theta + s}$ is a quadratic number. Since $\omega \in \mathbb{Q}(\theta)$, it suffices to prove that $\omega$ is not a rational number. Suppose otherwise. Since $p\theta + q = \omega r\theta + \omega s$, and $1, \theta$ are linearly independent over $\mathbb{Q}$, $p = \omega r$ and $q = \omega s$. Hence det $\sigma = ps - qr = \omega rs - \omega rs = 0$. This is a contradiction.

My question Is the following proposition correct? If yes, how do you prove it?

Proposition Let $\theta, D, \sigma, \omega$ be as above. Then the discriminant of $\omega$ is $D$.

Makoto Kato
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$\theta$ is a root of the unique polynomial $ax^2 + bx + c \in \mathbb{Z}[x]$ such that $a > 0$ and gcd$(a, b, c) = 1$.

Let $\sigma^{-1} = \left( \begin{array}{ccc} p' & q' \\ r' & s' \end{array} \right)$.

Then $\theta = \frac{p'\omega + q'}{r'\omega + s'}$. Hence $a(p'\omega + q')^2 + b(p'\omega + q')(r'\omega + s') + c(r'\omega + s')^2 = 0$. Let $f = ax^2 + bxy + cy^2$. Let $f^{\sigma^{-1}} = kx^2 + lxy + my^2$, where $f^{\sigma^{-1}}$ is defined in this question. Then $\omega$ is a root of the polynomial $kx^2 + lx + m$. By this question, $kx^2 + lxy + my^2$ is primitive, i.e. gcd$(k, l, m) = 1$ and its discriminant is $D$. Note that $k\ne 0$, because $D$ is not a square integer. If $k \gt 0$, we are done. If $k \lt 0$, $\omega$ is a root of polynomial $-kx^2 -lx - m$ and we are done.

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Makoto Kato
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