relative K1 depends on the Ring

Let

be rings with ideals such that

is a Bijection

Then in general we can't conclude that

Let denote your favorite field,

and the ring morphism

Then both maps have the same kernel

and are split epis with one sided inverse

resp.

by relative general linear group depends only on the ideal

Consider

Then define

which represents an element in .

It follows that hence .

we want to show, that .

Note that

is commutative, hence the determinant map of k1 of a ring

is welldefined.

But then it follows that (omitted)