Let R be a "nice" ring with 1 (e.g. Euclidean domain). Then the subgroup E(n,R) generated by the elements $I+te\_{i,j}$ is equal to $SL(n,R)$. 


My question is as follows: Instead of $SL(n,R)$ I look at left and right multiplication by elements of E(n,R) on elements of $M(n,R)$, set of all matrices, (i.e. it need not be even invertible). So equivalently, I want the coset representatives of double coset-decomposition. That is, I would like to determine :


$E(n,R)\backslash M(n,R)/ E(n,R)$.


In the case of $R$ a field, the coset representatives are diagonal matrices with either 1 or 0 on the diagonals.


I believe that one can show the coset representatives are diagonals. It will be useful to have finer result that what appears on the diagonal, e.g., how many 1s and how many 0s etc. 



generalization of result on K_1 of $SL(n,R)$
This is an expansion of my comment.
The [Smith normal form](https://en.wikipedia.org/wiki/Smith_normal_form) is a normal form of a matrix with entries in any given PID (but this probably works for non-domains and for Bezout rings in general). It goes as follows: given an $m\times n$ matrix $A$, there exist invertible $m\times m$ and $n\times n$ matrices $B$ and $C$ such that $BAC=\operatorname{diag}(a\_1,\ldots,a\_r,0,\ldots,0)$. Moreover, the entries $a\_i$ satisfy $a\_i\mid a\_{i+1}$ and are unique up to the multiplication by a unit. This decribes the representatives of $GL(m,R) \backslash M(m,n,R) / GL(n,R)$. From here you can obtain the decription for $SL(m,R) \backslash M(m,n,R) / SL(n,R)$ — no independent multiplication by a unit anymore, so the difference is the same as between $K\_1$ and $SK\_1$.


When $R$ is a Euclidean ring, one has an algorith for computing the Smith normal form. This works for any PID, in fact, but the resulting matrices $B$ and $C$ are not necessary elementary in this case. For a Euclidean ring the algorithm gives you the chain of elementary transfromation for obtaining the SNF.


As you mentioned, $E(n,R)=SL(n,R)$ when $R$ is Euclidean, but this is not the case for a PID. This equality fails, for example, for the ring $S^{-1}\mathbb{Z}[x]$, where $S$ is the multiplicative system generated by all cyclotomic polynomials. This is a result of
[D. R. Grayson — $SK\_1$ of an interesting principal ideal domain](http://www.sciencedirect.com/science/article/pii/002240498190089X).


Here are two other links with examples:


* [H.W. Lenstra Jr. — Grothendieck groups of abelian group rings](http://www.sciencedirect.com/science/article/pii/0022404981900918)
* [F. Ischebeck — Hauptidealringe mit nichttrivialer $SK\_1$-Gruppe](http://link.springer.com/article/10.1007%2FBF01235330)


The complete answer to your question would contain the computation of $K\_1$ for an arbitrary PID, so basically there are no chances. Howerer, it is easy to show that using the elementary transformation you can *almost* get the SNF. Namely, the usual procedure for Euclidian rings will fail in the setting of PID only in the last step, when you are facing the problem of simplifying a $2\times2$ matrix. So each coset contains a block diagonal matrix with only one block of size >1 (which is of size 2). The determinants of the blocks satisfy the same relation as $a\_i$'s.


