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 ?
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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.)
answered Sep 13, 2020 at 14:48
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$\begingroup$ Thank you sir for the response. It is clear to me that the map $SK_1(B) \rightarrow SK_1(B_{red})$ is injective. But why that map is {\bf surjective } ? I understand the preimage of a determinant 1 matrix is invertible. But why the preimage has determinant 1 is not what is clear to me. $\endgroup$ Commented Sep 14, 2020 at 7:04
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3$\begingroup$ You can multiply with a diagonal matrix that is congruent to the identity and get the determinant equal to one. $\endgroup$ Commented Sep 14, 2020 at 8:30
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$\begingroup$ Thank you sir very much for the clarification. $\endgroup$ Commented Sep 15, 2020 at 4:43