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 ?



A question on $SK_1$ of rings
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.)


