Let $D$ be a division ring, and $R=D[t_1,\ldots,t_n]$. If $GL_m(R)$ is the usual group of invertible matrices over $R$, then by $E_m(R)$ I mean the subgroup of $GL_m(R)$ generated by the elementary matrices of the form $I_m+r\epsilon(i,j)$, $r\in U(R)$. It is known that the whitehead group $K_1(R)$ of $R$ is isomorphic to the commutator quotient group of $U(D)$. Is there any way we can extract from this, anything about $GL_m(R)/E_m(R)$?
1 Answer
In general what you are asking about is the problem of $K_1$stabilization, that is the study of maps $GL(n, R)/E(n, R)\to GL(n+1, R)/E(n+1, R)$. For rings of finite dimension all such maps are bijective starting with some $n$ depending on the dimension.
The first relevant link is "On the stabilization of general linear group" by L. Vaserstein. I believe that any paper on the subject cites this one, so you can explore them by following the link.
For the particular problem about polynomials over a division ring, take a look at "The general linear group of polynomial rings over regular rings" by T. Vorst. Here's the abstract:
In this note we shall prove for two types of regular rings $A$ that every element of $GL_r(A[X_1,\ldots,X_n])$ is a product of an element of $E_r(A[X_1,\ldots,X_n])$ (the group of elementary matrices) and an element of $GL_r(A)$, for $r\geq3$ and $n$ arbitrary. This is a kind of $GL_r$analogue of results of Lindel and MohanKumar and is an extension of a result of Suslin.

$\begingroup$ @ Andrei Smolensky Thanks, this is exactly what I needed. I am having a bit of trouble accessing the paper by Vorst, is there any chance you know what the two types of regular ring mentioned in the abstract actually are? $\endgroup$ Dec 30, 2016 at 15:19

1$\begingroup$ i) complete regular equicharacteristic local ring; ii) regular and of essentially finite type over a perfect field; And now I start to wonder whether division algebras fall into one of these two types... $\endgroup$ Dec 30, 2016 at 15:29

$\begingroup$ Well I am mostly interested in rings that are finite dimensional over their center (which in my case would be a perfect field). That said, I don't think it is of essentially finite type $\endgroup$ Dec 30, 2016 at 15:39

$\begingroup$ such as $\left(\frac{1,1}{\mathbb{Q}}\right)$. $\endgroup$ Dec 30, 2016 at 15:46