Exercise 2.9.4

Question

Show that for any pair $(x,y)$ of integers with $(x,y) 
e (0,0)$ there is a matrix $U \in \mathbb{Z}^{(2,2)}$ with first column $(x,y)$ and determinant $\mathrm{gcd}(x,y)$. Also show that the set of all such matrices is the $\Gamma$-orbit of any such matrix $U$.

Richard Ganaye · Accepted Answer

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\begin{proof}
Write $d = \mathrm{gcd}(x,y)$. By B\'ezout's theorem, $d \Z = x \Z + y \Z$, there is a pair of integers $(u,v)$ such that $d = xu + yv$, and $d 
e 0$, because $(x,y)
e (0,0)$.

Then the matrix $A = \begin{pmatrix} x & -v\ y & u\end{pmatrix}$, whose first column is $(x,y)$, satisfies $\det(A) = xu +yv = d$.

If $M = \begin{pmatrix} x & s\ y & t\end{pmatrix}$ is any matrix in $\Z^{(2,2)}$ with first column $(x,y)$ such that 
$\det(M) = d$, then $xt - ys = d$.
Using $ xu +yv = d$, we obtain 
\begin{align}
x(t-u) = y(s+v),
\end{align} that is 
$
\frac{x}{d}(t-u) = \frac{y}{d}(s+v),
$
 where $\frac{x}{d} \wedge \frac{y}{d} = 1$. Therefore $\frac{x}{d} \mid s+ v$, thus there is some $k \in \Z$ such that $s+ v = k \frac{x}{d}$. If we substitute this value in (1), we obtain $x(t- u) = x k\frac{y}{d}.$

\bigskip
$\bullet$ If $x 
e 0$, then 
\begin{align*}
s &= -v + k \, \frac{x}{d},\
t &= {\phantom -}u + k \, \frac{y}{d}.
\end{align*}
Therefore
\begin{align*}
M &= \begin{pmatrix} x &  -v + k \, \frac{x}{d}\ y & {\phantom -}u + k \, \frac{y}{d}\end{pmatrix}\
&=\begin{pmatrix} x & -v\ y & u\end{pmatrix} \begin{pmatrix} 1 & \frac{k}{d} \ 0& 1\end{pmatrix}.
\end{align*}
But there is no necessity that $k/d \in \Z$. For instance, if $A = \begin{pmatrix} 3 & -1\ 6 & -1\end{pmatrix}$ and $B = \begin{pmatrix} 3 & 1\ 6 & -3\end{pmatrix}$, with determinant $3$, are not in the same $\Gamma$-orbit, since 
$$\begin{pmatrix} 3 & 1\ 6 & 3\end{pmatrix} = \begin{pmatrix} 3 & -1\ 6 & -1\end{pmatrix} \begin{pmatrix} 1& \frac{2}{3}\ 0 & 1\end{pmatrix},$$
thus 
$$A^{-1} B = \begin{pmatrix} 1& \frac{2}{3}\ 0 & 1\end{pmatrix} 
ot \in \langle T angle.$$

To obtain the result of the sentence, we must assume that $d = x \wedge y = 1$. This is sufficient to achieve the proof of Proposition 2.6.2.

Then, in the case $x 
e 0$,
\begin{align*}
s &= -v + k x,\
t &= {\phantom -}u + k y.
\end{align*}
and 
\begin{align*}
M &=\begin{pmatrix} x & -v\ y & u\end{pmatrix} \begin{pmatrix} 1 & k \ 0& 1\end{pmatrix}\
&= A T^k, 
\end{align*}
so that $M$ is in the $\Gamma$-orbit of $A$ (where $\Gamma \subset \mathrm{SL}_2(\Z)$ acts on $ \mathrm{SL}_2(\Z)$ by right multiplication).

$\bullet$ If $x = 0$, since $x\wedge y = 1$, $y = \varepsilon = \pm 1$. Then there is some $t \in \Z$ such that
$$M = \begin{pmatrix} 0 & -\varepsilon\ \varepsilon & t\end{pmatrix} = \begin{pmatrix} 0 & -\varepsilon\ \varepsilon & 0\end{pmatrix}\begin{pmatrix} 1& \varepsilon t\ 0 & 1\end{pmatrix} = A T^{\varepsilon t},$$
thus $M$ is in the $\Gamma$-orbit of $A = \begin{pmatrix} 0 & -\varepsilon\ \varepsilon & 0\end{pmatrix}$.

\bigskip
Conversely, if $M$ is in the $\Gamma$-orbit of $A$, that is $M = A T^k$ for some $k \in \Z$, then the first column of $M$ is $(x,y)$, and $\det(M) = \det(A) = 1$.

The set of matrixes $U \in \Z^{(2,2)}$ with determinant $1$, whose first column is $(x,y)$,  is the $\Gamma$-orbit of any such matrix.
\end{proof}

Exercise 2.9.4

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Exercise 2.9.4

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