Question

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.

Answer 1

Let N = N1(R).Say Nn = 0. Then R/N has no nonzero nilpotent ideals. (For, if I/Nis 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.