TITLE: A question on $SK_1$ of rings
QUESTION [5 upvotes]: Let $B$ be a commutative ring with unity and $B/nil(B):=B_{red}$, where $nil(B)$ is the nilradical of $B$. Is $SK_1(B)=SK_1(B_{red}) ?$ In particular, is it true when $B$ is an affine algebra over an algebraically closed field ?

REPLY [7 votes]: Yes.  An element in the kernel of $SK_1(B)\rightarrow SK_1(B_{red})$ is represented by a matrix $M\in GL_n(B)$ for some $n$.  Write $\overline{M}$ for the reduction of $M$ mod $nil(B)$.  Then $\overline{M}$ is a product of elementary matrices, all of which lift to elementary matrices over $B$.   Adjusting $M$ accordingly, we can assume that $\overline{M}$ is the identity.
It follows that the elements on the diagonal of $M$ are all $1$ mod $nil(B)$, hence all units in $B$.  This allows us to use elementary operations to convert $M$ to a diagonal matrix, which therefore (by Whitehead's lemma) represents the zero element of $SK_1(B)$.
(The same argument works if $B_{red}$ is replaced by $B/I$, where $I$ is any ideal contained in the Jacobson radical.)