Let $D$ be a division ring and $R=D[t_1,\ldots,t_n]$ the polynomial ring in $n$ variables. Now let $GL_m(R),\,E_m(R)$ be the usual general linear group and its subgroup generated by the elementary matrices of the form $I_m+r\epsilon(i,j)$ ($r\in R$). Further, by $U(R)$ I mean the units of $R$. This may be confused with $GL_1(R)$ in a natural way. My question is this, when is $GL_m(R)=U(R)\cdot E_m(R)$, i.e. when can any $X\in GL_m(R)$ be written as a product of elementary matrices (up to a unit)?

I know that, by a result of Cohn, that $GL_2(R)\neq U(R)\cdot E_2(R)$, but what about the case when $m>2$? If $D$ was commutative then we could obviously apply Suslin's stability theorem, but as far as I know this is not true for the case of division rings.

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    $\begingroup$ If $D$ is commutative you don't need any theorem: computing the determinant trivially implies $E_n(R)\neq GL_n(R)$. Or do you mean something else? Suslin's theorem indeed says that $E_n(R)=SL_n(R)$. $\endgroup$ – YCor Dec 27 '16 at 18:48
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    $\begingroup$ By the way you should use two different letters for the size of matrices and the number of indeterminates, unless you really want them to match for any precise reason. $\endgroup$ – YCor Dec 27 '16 at 20:15
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    $\begingroup$ OK; this lengthy discussion would have been avoided if you had asked whether $GL_m(R)$ is generated by elementary and diagonal (invertible) matrices. $\endgroup$ – YCor Dec 28 '16 at 15:11
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    $\begingroup$ @Sam I agree it seems it was obvious to you, which is a good start. It wasn't to me, as you might have noticed. As regards your second question, the answer is clear from my previous comment :) $\endgroup$ – YCor Dec 28 '16 at 16:11
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    $\begingroup$ The question would look much better if the embedding of $U(R)$ was there, and it is not too late to fix this. $\endgroup$ – Ilya Bogdanov Dec 29 '16 at 22:12

If $D$ is a field then every matrix of $\mathrm{SL}_m(R)$ can be written as product of elementary matrices for $m$ not equal to 2, because for $m=2$ Cohn gives an example which contradict this result. For $\mathrm{GL}_m(R)$ how you can say?

  • $\begingroup$ You could have read the comments before posting. You were also misled by the unclear wording of the question. $\endgroup$ – YCor Dec 28 '16 at 11:54
  • $\begingroup$ @Ram Krishna Verma you just repeated what I asked in the question $\endgroup$ – Sam Williams Dec 28 '16 at 13:46

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