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 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.)

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.)

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 represents the zero element of $SK_1(B)$.

(The same argument works if $B_{red}$ is replaced by $B/I$, where $I$ is any radical ideal.)

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 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.)