Question #24884

Let N1(R) be the sum of all nilpotent ideals in a ring R. If N1(R) is nilpotent, show that N1(R) = Nil*R.

Expert's answer

Let *N *= *N*1(*R*).Say *Nn *= 0. Then *R/N *has no nonzero nilpotent ideals. (For, if *I/N*is a nilpotent ideal of *R/N*, then *Im **⊆** N *for some *m*, and hence *Imn **⊆** Nn *= 0. This implies that *I **⊆** N*.) Therefore, *N *is semiprime, so *N *=Nil**R*.

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