For a commutative ring $R$ define $\mathrm{SK}_1(n, R)=\mathrm{SL}(n, R)/\mathrm{E}(n, R)$, the quotient of the special linear group by its subgroup generated by the elementary matrices. When $n\geqslant3$, $\mathrm{E}(n, R)$ is normal in $\mathrm{SL}(n, R)$, and $\mathrm{SK}_1(n, R)$ is a group.

When $n$ is large compared to the stable rank of $R$, the group $\mathrm{SK}_1(n, R)$ is abelian. In general it is not. However, if the Bass—Serre dimension $\delta(R)$ is finite, $\mathrm{SK}_1(n, R)$ is nilpotent of degree at most $\delta(R)+1$ (BS-dimension is something very similar to the dimension the maximal spectrum).

It was shown by A. Bak (PDF) that the nilpotency degree can indeed be arbitrary large, and similar examples were constructed by W. van der Kallen (link).

Question: How bad can $\mathrm{SK}_1(n, R)$ be for a commutative ring $R$?

I'm interested in the examples of two kinds:

  1. If $R$ is so and so, then $\mathrm{SK}_1(n, R)$ is not very bad;
  2. Here's an example of $R$ such that $\mathrm{SK}_1(n, R)$ is terrible, e.g., perfect.
  • 2
    $\begingroup$ The examples of Bak and van der Kallen are commutative. So, what kind of example would you like for question 2? $\endgroup$ – Matthias Wendt Apr 9 '16 at 18:33
  • 1
    $\begingroup$ @MatthiasWendt Explicit examples of commutative rings such that $\mathrm{SK}_1$ is far from being abelian, say, is perfect. $\endgroup$ – Andrei Smolensky Apr 9 '16 at 19:10

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