Let $R$ be a commutative ring with unit and let $q$ be an ideal of $R$. There is thus a natural map $SL(n,R) \rightarrow SL(n,R/q)$ for all $n$. This map is surjective if $SL(n,R/q)$ is generated by elementary matrices, but I very much doubt that it is surjective in general (though I don't know any examples).


My questions are as follows. 


1. Can someone give me an example of a ring $R$ and an ideal $q$ of $R$ such that the map $SL(n,R) \rightarrow SL(n,R/q)$ is not surjective for any $n$? I'd like the examples to be as nice as possible. For instance, it would be great to have an example where $R$ is Noetherian and has finite Krull dimension.
2. What conditions can I put on $R$ and $q$ to assure that this map is surjective, at least for large $n$?



When is $SL(n,R) \rightarrow SL(n,R/q)$ surjective?
A sort of universal example: Let $R$ be the polynomial ring $\mathbb Z[x\_{11},x\_{12},x\_{21},x\_{22}]$ and let $q$ be the ideal generated by $x\_{11}x\_{22}-x\_{12}x\_{21}-1$. The obvious element of $SL\_2(R/q)$ does not come from $SL\_2(R)$. You can see this by comparing with the example of the ring $\mathbb R[u,v]$ and the ideal generated by $u^2+v^2-1$, using the ring map taking $(x\_{11},x\_{12},x\_{21},x\_{22})$ to $(u,v,-v,u)$. If the resulting matrix came from an element of $SL\_2(\mathbb R[u,v]$), then topologically the corresponding map from the circle in $\mathbb R^2$ defined by $u^2+v^2=1$ to $SL\_2(\mathbb R)$ would extend to a continuous map $\mathbb R^2\to SL\_2(\mathbb R)$, which it doesn't. This example persists to $SL\_n$ for $n>2$.


