Assume now ideals in R satisfy DCC. Then Nn=Nn+1 for some n. We finish by showing that M:=Nn is zero. To see this, assume instead M<>0. Then there exist ideals A⊆M with MAM<>0 (for instance A=M). Among such ideals A, choose a B that is minimal. Then MBM<>0, so MbM<>0 for some b∈B. Since MbM⊆B and M(MbM)M=MbM<>0, we must have MbM=B. In particular, there exists an equation
b=i=1∑rmibmi′,
where mi,mi∈M.
Now consider the ideal J⊆N generated by {mi,mi′}:1≤i≤r. Since N is a sum of nilpotent ideals, J lies in a finite sum of nilpotent ideals, so Jk=0 for some k. Since b∈JbJ, we now have by repeated substitution b∈JkbJk=0, a contradiction.
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