 Solution

Put $u(t)=\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)$ and $l(s)=\left(\begin{array}{cc}1& 0\\ s& 1\end{array}\right)$. These elementary matrices belong to the special linear group. Let $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ have determinant one. If $a\ne 0$, elementary row elimination givesThe remaining diagonal matrix is also a product of the given generators. Indeed, for $a\ne 0$, direct multiplication givesHence $g=l(c/a)w(a)w(-1)u(b/a)$. If $a=0$, its determinant forces $c\ne 0$, and $u(1)g$ has upper-left entry $c$. The established case applies to it, and $g=u(-1)(u(1)g)$ is again generated. No division by two was used, so the proof works over every field. The upper and lower elementary unipotent matrices generate $S{L}_2(K)$.