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# 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 &supe; rad R be a typical modular maximal left ideal. Clearly m/radR is a modular maximal left ideal of R/rad R. Since &cap;m= rad R, we see that rad (R/rad R) = 0. For any ideal I&sube; R, consider a &isin; I &cap; rad R. For any b &isin; R, ba has a left quasi-inverse r &isin; R. But then r = rba &minus; ba &isin; I, so ba is left quasi-regular as anelement of the ring I. In particular, this shows that a &isin; rad I. Conversely, consider any a &isin; rad I. Since (Ra)2 &sube; Ia, (Ra)2 is quasi-regular, so(Ra)2 &sube; rad R. The image of Ra,being nilpotent in R/rad R, must be zero, since rad (R/radR) = 0. Therefore, Ra &sube; rad R. In particular, Ra isquasi-regular, so a &isin; I &cap; rad R.

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