For $K$ a field, is it known what the abelianization of $GL\_2(K[X])$ is?



Abelianization of general linear group of a polynomial ring
There is a discussion on whether $K$ has two elements or is larger, which strongly affects the conclusion.


One has the determinant map $\mathrm{GL}\_2(K[X])\to K^\*$. To show that it's the abelianization, it's enough to show that the kernel $\mathrm{SL}\_2(K[X])$ is contained in the derived subgroup of $\mathrm{GL}\_2(K[X])$. 


Since $K[X]$ is a Euclidean domain, $\mathrm{SL}\_2(K[X])$ is generated by elementary matrices $e\_{12}(y)$ and $e\_{21}(y)$ where $y$ ranges over $K[X]$, $e\_{ij}(y)=I\_2+yE\_{ij}$, and $(E\_{ij})\_{1\le i,j\le 2}$ is the canonical basis of the space of matrices.


For $t\in K^\*$, write $d\_1(t)=tE\_{11}+E\_{22}$. Then $d\_1(t)e\_{12}(y)d\_1(t)^{-1}=e\_{12}(ty)$, and in particular, the commutator (with suitable conventions) is $e\_{12}((t-1)y)$.


If $K$ is not reduced to the field on 2 elements, one concludes that $e\_{12}(y)$ (and similarly $e\_{21}(y)$) is a commutator for each $y$, since one can choose $t\in K^\*$ such that $t-1$ is a nonzero scalar. In this case, we deduce that:



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> For $|K|\ge 3$ the abelianization of $\mathrm{GL}\_2(K[X])$ is the determinant map onto $K^\*$.
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[Side note: the argument also shows that if $|K|\ge 4$ then $\mathrm{SL}\_2(K[X])$ is a perfect group: indeed one then chooses $t\in K^\*$ with $t^2\neq 1$ and use $c\_{12}(t)=tE\_{11}+t^{-1}E\_{22}$ instead of $d\_t$, which satisfies $c\_{12}(t)e\_{12}(y)c\_{12}(t)^{-1}=e\_{12}(t^2y)$.]




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It remains to deal with the case when $K$ is the 2-element field $\mathbf{F}\_2$, for which we have $\mathrm{GL}\_2(\mathbf{F}\_2[X])=\mathrm{SL}\_2(\mathbf{F}\_2[X])$. In this case the determinant map is trivial. But yet the abelianization map is non-trivial (in contrast to $n\ge 3$ where $\mathrm{SL}\_n(K[X])$ is a perfect group for every field $K$).


Indeed, it is known (for $K$ any field) that $\mathrm{GL}\_2(K[X])$ is the amalgamated product of its subgroups $\mathrm{GL}\_2(K)$ and $B(K[X])$ over their intersection $B(K)$. Here $B$ is the upper triangular group. A reference for this nontrivial fact is Serre's Astérisque book *"Arbres, amalgames, $\mathrm{SL}\_2$"*, or its English translation *"Trees"*.


Let us specify the latter to $K=\mathbf{F}\_2$: in this case, $\mathrm{GL}\_2(\mathbf{F}\_2)$ is a non-abelian group of order $6$, while $B(\mathbf{F}\_2[X])$ is the upper unipotent (abelian) group consisting of those matrices $e\_{12}(y)$ where $y$ ranges over $K[X]$. Let me write presentations for these two groups:
$$\mathrm{GL}\_2(\mathbf{F}\_2)=\langle z,x\_0\mid z^2=x\_0^2=(zx\_0)^3=1\rangle$$
$$B(\mathbf{F}\_2[X])=\langle x\_i:i\ge 0\mid x\_i^2=[x\_i,x\_j]=1:i,j\ge 0\rangle$$
These presentations are arranged so that $x\_0$ represents the element $e\_{12}(1)$, which generates the amalgamated subgroup (of order two). Hence a presentation of the amalgamated product is obtained by amalgamating the presentations:
$$\mathrm{GL}\_2(\mathbf{F}\_2[X])=\langle z, x\_i:i\ge 0\mid z^2=(zx\_0)^3=x\_i^2=[x\_i,x\_j]=1:i,j\ge 0\rangle.$$
To abelianize we just have to read this as presentation of abelian group. We know that $\mathrm{GL}\_2(\mathbf{F}\_2)$ abelianizes to a cyclic group of order $2$ in which $x\_0$ and $z$ are identified, and we get 
$$\mathrm{GL}\_2(\mathbf{F}\_2[X])\_{\mathrm{ab}}=\langle x\_i:i\ge 0\mid x\_i^2: i\ge 0\rangle\_{\mathrm{AbGrp}}:$$
this freely generated by $(x\_i)\_{i\ge 0}$ as 2-elementary abelian group, and $x\_i$ is the image of $e\_{12}(X^i)$ in the abelianization map.


[Side note: given that the derived subgroup of the $24$-element group $\mathrm{SL}\_2(\mathbf{F}\_3)$ has index $3$, similarly using an amalgam decomposition yields that the abelianization of $\mathrm{SL}\_2(\mathbf{F}\_3[X])$ is the free 3-elementary abelian group on the images of $e\_{12}(X^i)$, for $i\ge 0$.]


