# Answer to Question #17273 in Algebra for sanches

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*.Need a fast expert's response?

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