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.
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Expert's answer
2013-02-22T06:39:45-0500
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.
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