Show that rad (R/rad R) = 0, and that, if I is an ideal in R, then, viewing I as a ring, rad I = I ∩ rad R.
Let m ⊇rad R be a typical modular maximal left ideal. Clearly m/radR is a modular maximal left ideal of R/rad R. Since ∩m= rad R, we see that rad (R/rad R) = 0. For any ideal I⊆ R, consider a ∈ I ∩ rad R. For any b ∈ R, ba has a left quasi-inverse r ∈ R. But then r = rba − ba ∈ I, so ba is left quasi-regular as anelement of the ring I. In particular, this shows that a ∈rad I. Conversely, consider any a ∈rad I. Since (Ra)2⊆ Ia, (Ra)2 is quasi-regular, so(Ra)2⊆rad R. The image of Ra,being nilpotent in R/rad R, must be zero, since rad (R/radR) = 0. Therefore, Ra ⊆rad R. In particular, Ra isquasi-regular, so a ∈ I ∩ rad R.