# Answer to Question #24894 in Abstract Algebra for Hym@n B@ss

Question #24894

Show that Nil*R is precisely the set of all strongly nilpotent elements of R.

Expert's answer

First assume

*a*is not in Nil**R*.Then there exists an*m*-system*M*containing*a*and notcontaining 0. Take*a*1 =*a*, and inductively*a_n*+1*∈*(*a_nRa_n*)*∩ M.*Then we get a sequence*a*1*, a*2*,a*3*, . . .*of the desired type which is never 0. Therefore,*a*isnot strongly nilpotent. Conversely, if*a*is not strongly nilpotent,there exists a set*M*=*{ai*:*i ≥*1*}*of nonzeroelements such that*a*1 =*a*and*a_n*+1*∈**a_nRa_n*(*∀**n*). Thus,*M*is an*m*-system. Since 0 is not in*M*,then*a*is not inNil**R*.Need a fast expert's response?

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