It is a theorem of Selberg that a lattice $\Gamma$ in a linear group has a torsion-free subgroup of finite index. Page 64 in 'Introduction to Arithmetic Groups' by Dave Morris asserts these can be realized as follows. Since $\Gamma$ is finitely generated, $\Gamma \subseteq \mathrm{SL}\_k(R)$ where $R = \mathbb{Z}[a\_1, \ldots,a\_n]$ for some $a\_1, \ldots, a\_n \in \mathbb{C}$. If $\mathfrak{m}$ is a maximal ideal in $R$ then the kernel of reduction modulo $\mathfrak{m}$ is torsion-free and has finite index in $\Gamma$ since $\mathrm{SL}\_k(R/\mathfrak{m})$ is finite. For the purposes of this question let $k=2$.


In the case $\Gamma = \mathrm{SL}\_2(\mathbb{Z})$ doing this gives the principal congruence subgroups, which are actually free for levels $\geq 3$. So my question is, when do lattices in $\mathrm{SL}\_2(\mathbb{C})$ have a free subgroup of finite index? Relatedly, for which commutative rings $R$ does $\mathrm{SL}\_2(R)$ have a free subgroup of finite index?



Finite-index free subgroups in lattices and matrix rings
A co-compact lattice in $SL\_2({\mathbb C})$ cannot contain a finite index free subgroup because of cohomological reasons. The cohomological dimension of the finite index torsion free subgroup is 3. \vskip 5mm


A non-cocompact lattice $\Gamma $ in $SL\_2({\mathbb C})$, after a conjugation, intersects the upper triangular unipotent matrices in a lattice; in particular, it contains a free abelian subgroup of rank two; therefore, $\Gamma $ cannot contain a finite index free subgroup. 


For the same reason, a finite index subgroup of $SL\_2(R)$ , for a commutative ring $R\subset {\mathbb C}$, contains the upper triangular unipotent matrices of the form 
$$\begin{pmatrix} 1 & Nx \cr 0 & 1\end{pmatrix}$$ where $N$ is a non-zero integer and $x\in R$. So, unless $R={\mathbb Z}$, the group $SL\_2(R)$ cannot contain a finite index free subgroup.


