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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 a1 = a, and inductively a_n+1 ∈(a_nRa_n) ∩ M. Then we get a sequence a1, a2,a3, . . . 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 a1 = 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.

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