SK1

Remarks

Equivalently, $SK_1(R) = SL(R)/E(R)$, where $SL(R)$ is the union of the special linear groups $SL_n(R)$ and $E(R)$ is the elementary subgroup. The map $u \mapsto [\mathrm{diag}(u, 1, 1, \dots)]$ splits $\det$, so $K_1(R) \cong R^\times \oplus SK_1(R)$. Example: for a field $F$, row reduction writes each matrix of determinant $1$ as a product of elementary matrices, so $SK_1(F) = 0$. Non-example: the definition needs $R$ commutative. For a noncommutative ring such as a division ring there is no determinant with values in $R^\times$; one uses the Dieudonné determinant or the reduced norm instead. For a central simple algebra $A$ over a field $F$, one sets $SK_1(A) = \ker(\mathrm{Nrd} \colon K_1(A) \to F^\times)$, where $\mathrm{Nrd}$ is the reduced norm.