Exercise 2.9.6

Question

Let $f$ be an integral form of discriminant $\Delta$ and let $(x,y)$ be integers such that $x^2 - \Delta y^2 = \pm 4$. Prove that $U(f,x,y)$ is an automorphism of $f$}

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

\begin{proof}
The matrix $U = U(f,x,y)$ is given by (2.20):
$$U(f,x,y) = 
\begin{pmatrix} \frac{x - yb}{2} & -cy \ay & \frac{x+yb}{2} \end{pmatrix}.
$$
Since $x^2 - (b^2-4ac) y^2 = \pm 4$, $x^2 - b^2 y^2 \equiv 0 \pmod 4$, therefore
\begin{align}
(x-by)(x+by) \equiv 0 \pmod 4.
\end{align}
Moreover $(x-by) - (x + by) = -2by \equiv 0 \pmod 2$, thus $x-by$ and $x +by$ have same parity. The relation (1) shows that they are not both odd, thus they are both even. This shows that the coefficients of $U$ are integers.

Since $\det(U) = \frac{x^2 - \Delta y^2}{4} = \pm1$, we conclude that $U \in \mathrm{GL}(2,\Z)$.

By definition, $U$ is an automorphism of $f$ if and only if $M(f) = (\det U) U^T M(f) U$.

\begin{align*}
M(f) U &= \begin{pmatrix} a &\frac{b}{2} \ \frac{b}{2} & c \end{pmatrix}\begin{pmatrix} \frac{x - yb}{2} & -cy \ay & \frac{x+yb}{2} \end{pmatrix}\
&=\frac{1}{4} \begin{pmatrix} 2ax & bx + \Delta y\ bx - \Delta y & 2cx \end{pmatrix},\
(U^T)^{-1} M(f) &= \frac{1}{\det(U)} \begin{pmatrix}  \frac{x + yb}{2} & -ay\ cy &\frac{x - yb}{2}\end{pmatrix}
\begin{pmatrix} a &\frac{b}{2} \ \frac{b}{2} & c \end{pmatrix},\
(\det U)(U^T)^{-1} M(f) &= \frac{1}{4} \begin{pmatrix} 2ax & bx + \Delta y \ bx - \Delta y & 2c x \end{pmatrix}.
\end{align*}
This shows that $M(f) = (\det U) U^T M(f) U$ and that $U$ is an automorphism of $f$.

Note that if $x^2 - \Delta y^2 = 1$, then $\det(U) = 1$ and $U(f,x,y)$ is a proper automorphism.
\end{proof}

Exercise 2.9.6

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Exercise 2.9.6

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