Question #17273

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.

Expert's answer

Let m *⊇** *rad *R *be a typical modular maximal left ideal. Clearly m*/*rad*R *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/*rad*R*) = 0. Therefore, *Ra **⊆** *rad *R*. In particular, *Ra *isquasi-regular, so *a **∈** I ∩ *rad *R*.

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