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Let $R$ be a ring. An elementary matrix over $R$ is a matrix with $1$s along the diagonal and at most one other nonzero entry. Let $\text{EL}_n(R)$ denote the subgroup of $\text{GL}_n(R)$ generated by the elementary matrices.

I understand that $\text{EL}_n(R) = \text{SL}_n(R)$ provided that $R$ is a Euclidean domain.

Why does $\text{EL}_n(\mathbf{Z}[X_1,\ldots,X_m]) = \text{SL}_n(\mathbf{Z}[X_1,\ldots,X_m])$?

flag	asked Feb 29 2012 at 0:15 Sean Eberhard 1,257●3●10

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Don't you also need to add $n\geq 3$? – Gjergji Zaimi Feb 29 2012 at 0:47

Yes I think so. – Sean Eberhard Mar 1 2012 at 11:18

Igor Rivin · Accepted Answer · 2012-02-29 03:32:25Z UTC

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This is the "Suslin stability theorem". There is an algorithmic proof in Park+ Woodburn, "An algorithmic proof of the Suslin stability theorem for polynomial rings."

answered Feb 29 2012 at 3:32

Igor Rivin
31.2k●3●37●71

As Gjergji Zaimi notes in his comment, Suslin's theorem does require the assumption $n \geq 3$ . – Jim Humphreys Mar 1 2012 at 0:02

Why do elementary matrices generate the special linear group over polynomial rings?

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