Let $M_2(\Ints)$ be the ring of $2{x}2$ matrices with integer entries, and define $GL_2(\Ints)$ to be the subring of matrices invertible over $\Ints$ . Thus for $M\in M_2(\Ints)$ , $$ M\in GL_2(\Ints)\iff \det M=\pm $$

Let $Aut_{\Ints}(\Ints\oplus\Ints)$ be the ring of automorphisms of $\Ints\oplus\Ints$ as a $\Ints$ -module. Then $GL_2(\Ints)\cong Aut_{\Ints}(\Ints\oplus\Ints)$ as rings, under the obvious operations.

To see this, we demonstrate a natural correspondence between endomorphisms of $\Ints\oplus\Ints$ and $M_2(\Ints)$ and show that invertible endomorphisms correspond to invertible matrices. Let $\varphi: \Ints\oplus\Ints\to \Ints\oplus\Ints$ be any ring homomorphism. It is clear that $\varphi$ is determined by its action on $(1,0)$ and $(0,1)$ , since $$ \varphi(x,y)=\varphi(x(1,0)+y(0,1))=x\varphi(1,0)+y\varphi(0,1 $$ Suppose then that $\varphi(1,0)=(a,b)$ and $\varphi(0,1)=(c,d)$ . Then $$ \varphi(x,y)=(ax,bx)+(cy,dy)=(ax+cy,bx+dy)=\begin{pmatrix}a&c\\b&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix $$ Now, $\varphi$ is surjective if both $(1,0)$ and $(0,1)$ are in its image. But $(1,0)\in{im}\ \varphi$ if and only if there is some $(x,y)$ such that

$A=Aut_{\Ints}(\Ints\oplus\Ints)$ has a simple and well-known set of generators as a group:

Define the subgroup $A'\subset A$ by $A'=<r,s>$ , the subgroup of $A$ generated by $r$ and $s$ . If $\varphi_1,\varphi_2\in A$ , define $\varphi_1\sim\varphi_2$ if $\varphi_1$ and $\varphi_2$ are in the same $A'$ -coset.

Our objective is to show that $A'=A$ , which we can do by showing that each $\varphi\sim e$ , where $e$ is the identity transformation of $A$ . This demonstration is essentially an application of the Euclidean algorithm. For suppose $$ \varphi(x,y)=(ax+cy,bx+dy $$ Assume, by applying $r$ if necessary, that $a\leq b$ , and choose $m$ such that $b=am+q, 0\leq q<a$ . Then $rs^{-m}\varphi(x,y)=r(ax+cy,(b-am)x+(d-cm)y)=r(ax+cy,qx+dy)=(qx+dy,ax+cy)$ , so that $$ \varphi\sim (x,y)\mapsto(qx+dy,ax+cy $$ Continuing this process, we eventually see that $$ \varphi\sim (x,y)\mapsto(cy,bx+dy $$ But $ad-bc=\pm 1$ , so we have $bc=\pm 1$ . Applying either $s^d$ or $s^{-d}$ as appropriate, we get $$ \varphi\sim (x,y)\mapsto(cy,bx)\sim(x,y)\mapsto(bx,cy $$

Thus, we are done if we show that all such forms $(bx,cy)$ with $b,c=\pm 1$ are in the same $A'$ -coset as $e$ . The case where $b=c=1$ is obvious. For the other cases, note that

This result is often phrased by saying that the matrices $$ \begin{pmatrix}1&0\\1&1\end{pmatrix},\quad\begin{pmatrix}0&1\\1&0\end{pmatrix $$ generate $GL_2(\Ints)$ as a multiplicative group.