Abelianization of general linear group of a polynomial ring For $K$ a field, is it known what the abelianization of $GL_2(K[X])$ is?
 A: As an alternative to YCor's beautiful answer, one can use the following theorem of P. M Cohn [1, Theorem 9.5].

Theorem. Let $R$ be a ring which is quasi-free for $\text{GE}_2$ and denote by $N$ the ideal generated by all $\alpha - 1$  with $\alpha \in U(R)$. Then there is a split exact sequence
  $$0 \rightarrow R/N \rightarrow \text{GE}_2(R)^\text{ab} \rightarrow U(R)^\text{ab} \rightarrow 0.$$
  and the mapping $\alpha \mapsto \begin{pmatrix} \alpha^\text{ab} & 0 \\ 0 & 1 \end{pmatrix}$ induces a splitting.

In the above statement


*

*$U(R)$ denotes the unit group of $R$, 

*$G^\text{ab}$ denotes the abelianization of a group $G$ and $g \mapsto g^\text{ab}$ the abelianization homomorphism (in particular in the case of  $U(R)\to U(R)^\text{ab}$),

*$\text{GE}_2(R)$ is the subgroup of $\text{GL}_2(R)$ generated by the elementary matrices $\begin{pmatrix} 1 & r \\ 0 & 1 \end{pmatrix}$, $\begin{pmatrix} 1 & 0 \\ s & 1 \end{pmatrix}$ with $r,s \in R$ and the diagonal matrices $\begin{pmatrix} \alpha & 0 \\ 0 & \beta \end{pmatrix}$ with $\alpha, \beta \in U(R)$.


I will not expand the definition of a quasi-free ring for $\text{GE}_2$ but just mention that a discretely normed ring is quasi-free for $\text{GE}_2$. In particular $K[X]$ is quasi-free for $\text{GE}_2$ if $K$ is a field. 
Note that if $\text{SL}_2(R)$ is generated by the elementary matrices, e.g. $R$ is Euclidean, then $\text{GE}_2(R) = \text{GL}_2(R)$. This holds in particular for $R = K[X]$ with $K$ a field since $K[X]$ is Euclidean (only) in this case.
If $K$ is the field with two elements, then the above theorem yields an isomorphism of the additive group of $K[X]$ with $\text{GL}_2(K[X])^\text{ab}$. If $K$ is a field with more than two elements, the same theorem yields an isomorphism of $\text{GL}_2(K[X])^\text{ab}$ with $U(K)$.

[1] P. M. Cohn, "On the structure of the $\mathrm{GL}_2$ of a ring", 1966 (MSN).
A: 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:

For $|K|\ge 3$ the abelianization of $\mathrm{GL}_2(K[X])$ is the determinant map onto $K^*$.

[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 $c_{12}(t)e_{12}(y)c_{12}(t)^{-1}=e_{12}(t^2y)$.]

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 $i\ge 0$.]
