Answer to Question #17267 in Abstract Algebra for Tsit Lam
Choose d such that d ◦ ra =0; then choose c such that c ◦ (r − dr)b = 0. The above equation shows that c ◦ d is a left quasi-inverse of r(a+ b), as desired. To show that rad R is also a right ideal,we must check that a ∈rad R and s ∈ R imply as ∈rad R. For any r ∈ R, sra is left quasiregular, so r(as) is also left quasi-regular. This shows that as ∈rad R.
Next, we show that rad R is a quasi-regular ideal.It suffices to show that every element a ∈rad R is left quasi-regular. Since a2 ∈ Ra is left quasi-regular, there exists b ∈ R such that b ◦ a2 = 0. But then
(b ◦ (−a)) ◦ a =b ◦ ((−a) ◦ a) = b ◦ a2 = 0
shows that a is left quasi-regular. By definition, rad R contains every quasi-regular left ideal. Let I be any quasi-regular right ideal. For a ∈ I, aR is right quasi-regular, therefore quasi-regular. We also see that Ra is (left) quasi-regular, so I ⊆rad R, as desired.
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