Answer to Question #17655 in Abstract Algebra for Tsit Lam
Show that for any ideal I ⊆ rad R, the natural map GLn(R) → GLn(R/I) is surjective.
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 Mn(R)and its ideal Mn(I) ⊆ Mn(rad R) = rad Mn(R),we see that U(Mn(R)) → U(Mn(R)/Mn(I))= U(Mn(R/I)) is onto; that is, GLn(R) → GLn(R/I)is onto.
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