Exercise 3.5.3 (There exist integers $u$ and $v$ such that $\begin{bmatrix} x & y \\u & v \end{bmatrix} \in \Gamma$ if and only if $(x,y) = 1$)

Proof.

• If $A=\left(\begin{array}{cc}\hfill x\hfill & \hfill y\hfill \\ \hfill u\hfill & \hfill v\hfill \end{array}\right)\in \Gamma$ then $\det <!--\; FUNCTION\; APPLICATION\; -->(A)=1$, so $\mathit{xv}-\mathit{yu}=1$. Therefore $x\wedge y=1$.

• Conversely, if $x\wedge y=1$, there are integers $u,v$ such that $\mathit{xv}-\mathit{yu}=1$ by Bézout’s theorem. Then

  $$\left|\begin{array}{cc}\hfill x\hfill & \hfill y\hfill \\ \hfill u\hfill & \hfill v\hfill \end{array}\right|=1,$$

  so $\left(\begin{array}{cc}\hfill x\hfill & \hfill y\hfill \\ \hfill u\hfill & \hfill v\hfill \end{array}\right)\in \Gamma$.

There exist integers $u$ and $v$ such that $\left(\begin{array}{cc}\hfill x\hfill & \hfill y\hfill \\ \hfill u\hfill & \hfill v\hfill \end{array}\right)\in \Gamma$ if and only if $x\wedge y=1$. □