Question #17655

Show that for any ideal I ⊆ rad R, the natural map GLn(R) → GLn(R/I) is surjective.

Expert's answer

First consider the case *n *=1. For *x **∈** R*, we have that *x **∈** *U(*R*) iff *x **∈** *U(*R/I*). Therefore, U(*R*)*→ *U(*R/I*) is surjective. Applying this to the matrix ring M*n*(*R*)and its ideal M*n*(*I*) *⊆** *M*n*(rad *R*) = rad M*n*(*R*)*,*we see that U(M*n*(*R*)) *→ *U(M*n*(*R*)*/*M*n*(*I*))= U(M*n*(*R/I*)) is onto; that is, GL*n*(*R*) *→ *GL*n*(*R/I*)is onto.

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