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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 $n$?

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Well, an obvious condition you can put on R and q is that R is a PID and q is a maximal ideal. See e.g., although this may be too trivial for the situation you're interested in... – M Turgeon Oct 18 '11 at 13:35
@M Turgeon : In fact, you don't need $R$ to be a PID. If $q$ is a maximal ideal, then $R/q$ is a field, and in this case the usual proof shows that $SL(n,R/q)$ is generated by elementary matrices and the map is surjective. – Ira L Oct 18 '11 at 14:25
up vote 16 down vote accepted

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 $\mathbb R^2\mapsto SL_2(\mathbb R)$, which it doesn't. This example persists to $SL_n$ for $n>2$.

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@ Tom: Really nice! – Alain Valette Oct 18 '11 at 3:56
This is wonderful. – Richard Kent Oct 18 '11 at 4:15
This is a beautiful example. I'm going to hold off on accepting it for a while to see if any other interesting answers trickle in, but it's great! – Ira L Oct 18 '11 at 4:24
I think I learned this from Milnor's Algebraic $K$-Theory book. It gives an element of $K_1$ of a Dedekind domain that is not detected by the determinant. – Tom Goodwillie Oct 18 '11 at 4:38
Add to this that this loop is homotopically non-trivial even when passing from $SL_2$ to $SL$ which seems to be the OP's question. – Torsten Ekedahl Oct 18 '11 at 6:06

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