I've posted this question on math.stackexchange, but haven't gotten any responses so I'm trying here instead.


Let $A = F\_q[T]$ be the ring of polynomials in one variable with coefficients in a finite field, and let $r>1$ be an integer.
I'm currently looking for the abelianisation of the congruence subgroup $Γ(N)$ of the special linear group $SL(r,A)$, i.e. the kernel of the 'modulo $N$' map $SL(r,A) \to SL(r,A/N)$, where $N \in A$ is a nonconstant polynomial; more particularly, I'm looking for the torsion-free part of the abelianisation, but I wouldn't complain about knowing the abelianisation itself if that's possible.


So, here are my questions, in reverse order of importance:


1. What are the commutator subgroups of $SL(r,A)$ and $Γ(N)$?
2. What are the abelianisations of these groups?
3. What are the torsion-free abelianisations of these groups?



Non-torsion part of the abelianisation of congruence subgroups
In a more general setting, let $A$ be a commutative ring, $I, J$ its ideals, $n\geqslant3$, then
$$[E(n,A,I),E(n,A,J)]\geqslant E(n,R,IJ),$$
where $E(n,A,I)$ is the normal closure in $E(n,A)$ of the subgroup $E(n,I)$ generated by the elementary generators $x\_{ij}(\xi)=1+\xi e\_{ij}$ of level $I$, that is, with $\xi\in I$. In particular, $SL(n,A)$ is perfect once it coincides with $E(n,R)$ (which is the case for polynomials over a field).


This is well-known since Bass' 1964 paper. Since $\mathbb{F}\_q[T]$ is Euclidean (and thus of stable rank $2$), one has $E(n,R,I)=SL(n,R,I)$ by the $K\_1$-stability [result of Vaserstein](http://mi.mathnet.ru/eng/msb3595) (Theorem 3.2).


It is also known that $E(n,R,I)=\langle z\_{ij}(\xi,\eta)\colon \xi\in I,\ \eta\in A \rangle$, where $z\_{ij}(\xi,\eta)=x\_{ji}(\eta)x\_{ij}(\xi)x\_{ji}(-\eta)$.


Now the projection $SL(n,A,I)\to SL(n,A,I)\_{\mathrm{ab}}$ factors through $S=SL(n,R,I)/SL(n,r,I^2)$, and it is not hard to show for $A=\mathbb{F}\_q[T]$ and $I=fA$ that the image of $z\_{ij}(\xi,\eta)$ is $S$ has finite order, hence this abelianization is also finite.


Alternatively, one can do it with an even more straightforward computation. A reasonably good set of generators is known for $[E(n,A,I),E(n,A,J)]$ (see [this paper](https://doi.org/10.1017/S0013091515000188)). Namely, it is generated (as a normal subgroup) by the elements of the following three types:


1. $[x\_{ij}(\xi),z\_{ij}(\zeta,\eta)]$,
2. $[x\_{ij}(\xi),x\_{ji}(\zeta)]$,
3. $x\_{ij}(\xi\zeta)$,


where $\xi,\zeta\in I$, $\eta\in A$.


Now $z\_{ij}(\xi)^k=x\_{ji}(\eta)x\_{ij}(k\xi)x\_{ji}(-\eta) \equiv x\_{ij}(k\xi)$, so taking $k=\operatorname{char}(\mathbb{F}\_q)$ gives you the finiteness of orders of the generators, while the third relation shows that it suffices to take only finitely many of them and allows to calculate the abelianization explicitly (similar to [this answer](https://mathoverflow.net/a/76366/)).


