Relative K-groups

Remarks

The relative Whitehead lemma says that $E(R, I)$ is a normal subgroup of $GL(R)$ and contains the commutator subgroup of $GL(I)$, so $K_1(R, I)$ is an abelian group. There is an exact sequence $$K_1(R, I) \to K_1(R) \to K_1(R/I) \to K_0(R, I) \to K_0(R) \to K_0(R/I),$$ where $K_1$ is K1 of a ring. With Milnor's relative $K_2$ or Quillen's higher groups, the sequence continues to the left.

Example: take $I = R$. Then $R/I = 0$, so $GL(I) = GL(R)$ and $E(R, I) = E(R)$, which gives $K_1(R, R) = K_1(R)$. Also $D(R, R) = R \times R$, and $K_0(R \times R) = K_0(R) \oplus K_0(R)$, so $K_0(R, R)$ is the second copy of $K_0(R)$.

Non-example: the subgroup generated by the matrices $e_{ij}(x)$ with $x \in I$ alone, without taking the normal closure in $E(R)$, can be strictly smaller than $E(R, I)$. So it cannot replace $E(R, I)$ in the definition of $K_1(R, I)$.

The group $K_0(R, I)$ depends only on $I$ viewed as a ring without identity; this is excision for $K_0$. For $K_1$ the analogous statement fails in general.