# Answer to Question #17655 in Abstract Algebra for Tsit Lam

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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