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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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  • $\begingroup$ 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. math.umn.edu/~garrett/m/mfms/notes/07b_surjectivity.pdf, although this may be too trivial for the situation you're interested in... $\endgroup$ – M Turgeon Oct 18 '11 at 13:35
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    $\begingroup$ @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. $\endgroup$ – Ira L Oct 18 '11 at 14:25
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    $\begingroup$ If $R$ is a PID, then $q \neq \{0\}$ is just enough: $R/q$ is a zero-dimensional Noetherian ring, so that its Bass stable rank is $1$. (Alternatively, you could use the fact that $R/q$ is Artinian, hence semi-local). As a result $SL_n(R/q)$ is generated by elementary matrices. You may want to replace PID by Noetherian one-dimensional domain. $\endgroup$ – Luc Guyot Jun 6 '16 at 21:19
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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 $n>2$.

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    $\begingroup$ @ Tom: Really nice! $\endgroup$ – Alain Valette Oct 18 '11 at 3:56
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    $\begingroup$ This is wonderful. $\endgroup$ – Autumn Kent Oct 18 '11 at 4:15
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    $\begingroup$ 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! $\endgroup$ – Ira L Oct 18 '11 at 4:24
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    $\begingroup$ 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. $\endgroup$ – Tom Goodwillie Oct 18 '11 at 4:38
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    $\begingroup$ 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. $\endgroup$ – Torsten Ekedahl Oct 18 '11 at 6:06
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The following example relates to the second part of the question while echoing back the example $R = \mathbb{Z}[X, Y]$ and $q = (X^2 + Y^2 - 1)$ discussed above: Let $k$ be a finite field and let $q$ be any ideal of $R = k[X, Y]$, then the natural map $SL_n(R) \rightarrow SL_n(R/q)$ is surjective for all $n \ge 2$.

This is Theorem 1.7.(2) of "On the groups $SL_2(\mathbb{Z}[x])$ and $SL_2(k[x,y])$" by F. Grunewald, J. Mennicke and L. Vaserstein, 1994 (MR1276133).

In the spirit of the requirements of the first part of the question, I would ask whether $2$ is the smallest Krull dimension we can get for a ring $R$ generated by finitely many elements and for which surjectivity of the reduction of matrix coefficients modulo $q$ fails for some ideal $q \subset R$.

The case of $\mathbb{Z}[X]$ is somehow settled by Theorem 1.7.(1) of the same paper: for $R = \mathbb{Z}[X]$, the image of $SL_n(R)$ in $SL_n(R/q)$ is of finite index for every $n \ge 2$. In some sense, it is optimal since F. Grunewald et al. have a recipe to build quotients of $\mathbb{Z}[X]$ with non-trivial $SK_1$ (see Proposition 1.9 of the same paper and this MO post) whereas $SK_1(\mathbb{Z}[X]) = 1$.

As for the general part of the question, the group $SK_0(q)$ (see Definition II.2.6 and Exercise III.2.1 of C. Weible's K-book) is the natural obstruction to the surjectivity of coefficients reduction modulo $q$. You may argue that's kind of tautological though.

Addendum: T. Goodwillie's example originates from Example 13.5 of "Introduction to algebraic $K$-theory" by J. Milnor, 1971 (MR0349811). To some extent, it is also discussed in "Serre's problem on projective modules" by T. Y. Lam, 2006 (MR2235330), see in particular Proposition I.8.12 and Remark I.8.14.

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