Is $\operatorname{SL}_2(\mathcal O)$ finitely generated? Let $F$ be a number field and $\mathcal O$ its integer ring: is $\operatorname{SL}_2(\mathcal O)$ finitely generated?
It's well known that $\operatorname{SL}_2(\mathbf Z)$ is generated by $T=\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and $S=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$.
It seems easy enough when $\mathcal O$ is a Euclidean domain, because then we can replace $T$ with $T_i:=\begin{pmatrix}1&x_i\\0&1\end{pmatrix}$ where $\mathcal O=\bigoplus x_i\mathbf Z$ and use the algebraic proof here.
But what happens when $\mathcal O$ is not a Euclidean domain?
 A: Yes, Vaserstein proved, that if $R$ is a Dedekind ring of arithmetic type with infinitely many units then  $SL_2(R) = E_2(R)$ holds, where
$E_2(R)$ denotes the subgroup of $SL_2(R)$ generated by the elementary matrices of the form $\begin{pmatrix} 1 & r \\ 0 & 1\end{pmatrix}$ and $\begin{pmatrix} 1 & 0 \\ r & 1\end{pmatrix}$ with $r \in R$. We can apply this for rings of integers, provided 
that the unit group is infinite. By Dirichlet’s unit theorem, a ring of integers $R$ of an algebraic number field $K$ has finitely many units if and only if $K$ is either the field of rationals $\mathbb{Q}$ or an imaginary quadratic number field. Under this assumption P. M. Cohn proved that $SL_2(R) = E_2(R)$ if and only if $R$ is Euclidean with respect to complex modulus. This leaves only the cases $F = \Bbb Q(\sqrt{-D})$ for $D$ any squarefree integer different from $1,2,3,7,11 $. Then $SL_2(\mathcal O)$ is not generated by the elementary matrices.  However, the index of the subgroup generated by elementary matrices in $SL_2(\mathcal O)$ is finite, so we can add a finite number of non-elementary matrices to the generating set, to ensure $SL_2(\mathcal O)$ will be generated by this set.
So it is finitely generated in all cases.
Reference (with many examples and explicit calculations): Generators and relations for certain special linear groups by R. G. Swan.
