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

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

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Don't you also need to add $n\geq 3$? –  Gjergji Zaimi Feb 29 '12 at 0:47
Yes I think so. –  Sean Eberhard Mar 1 '12 at 11:18

As Gjergji Zaimi notes in his comment, Suslin's theorem does require the assumption $n \geq 3$. –  Jim Humphreys Mar 1 '12 at 0:02