generalization of result on K_1 of $SL(n,R)$ 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.  
 A: Over any PID $R$, using the classification of submodules of f.g. free modules, you can get that orbits of the left-right action of $\mathrm{GL}_n(R)\times \mathrm{GL}_n(R)$ on $\mathrm{M}_n(R)$ have representatives given by diagonal matrices, and the question boils down to understand which matrices are in the same orbit. Two diagonal matrices are in the same orbit if they are equal up to multiplication by an invertible diagonal matrix, and up to conjugation by permutation matrices, but this is not the only condition: for instance, the matrices $A=\begin{pmatrix}1 & 0\\ 0 & 6\end{pmatrix}$ and $B=\begin{pmatrix}2 & 0\\ 0 & 3\end{pmatrix}$ are in the same orbit, because the endomorphism $B$ can be written as $A$ in the pair of bases $((1,1),(3,2))$, $((2,3),(1,1))$. Maybe the only good invariant is the "local" one, that is, to check for each prime how many times this prime appears in the diagonal entries, and with which multiplicity.  
For the left-right action of $\mathrm{SL}_n(R)\times \mathrm{SL}_n(R)$, the determinant also has to be fixed, and this is the only other invariant: if two diagonal matrices with the same determinant are in the same orbit for the left-right action of $\mathrm{GL}_n(R)\times \mathrm{GL}_n(R)$, say $D_1=UD_2V$, then letting $S$ be the diagonal matrix with diagonal entries $(\det(U),1,\dots,1)$, we have $D_1=(US^{-1})D_2(SV)$, and both $US^{-1}$ and $SV$ have determinant 1. So everything boils down the the left-right action of $\mathrm{GL}_n(R)\times \mathrm{GL}_n(R)$.
If the ring is Euclidean, $\mathrm{EL}_n(R)=\mathrm{SL}_n(R)$ so we are in the previous case. So the only issue is to classify precisely diagonal matrices are in the same orbit for the left-right action of $\mathrm{GL}_n(R)\times \mathrm{GL}_n(R)$ on $\mathrm{M}_n(R)$.
A: This is an expansion of my comment.
The 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.
Here are two other links with examples:


*

*H.W. Lenstra Jr. — Grothendieck groups of abelian group rings

*F. Ischebeck — Hauptidealringe mit nichttrivialer $SK_1$-Gruppe 
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.
