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I have to referee a paper not really in my field and need some answers concerning the prime radical of a ring and nilpotent ideals.

The definition of a strong nilpotent element already have appeared in this question:

strong nilpotent elements

Recall that an element x in a noncommutative ring R is strongly nilpotent if any chain $x_1=x, x_2, ... $, with $x_{n+1}\in x_n R x_n$ terminates at zero.

  1. My question is the following: is this definition equivalent to $RxR$ to be a nilpotent ideal of R?

  2. Could you also please give a precise reference for the statement that the prime radical coincides to the set of all strong nilpotent elements?

From my understanding, in non commutative rings, an ideal is called locally nilpotent if any finitely generated sub-ideal of it is nilpotent.

  1. Is the prime radical $P(R)$ the largest locally nilpotent ideal of $R$? If yes can someone provide a reference for this to me.

Thank you in advance for your help!

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  • $\begingroup$ Reminder (en.wikipedia.org/wiki/Nilpotent_ideal): if $I$ is a (2-sided) ideal, $I$ nilpotent means that there exists $k$ such that any product of $k$ elements of $I$ is 0. $\endgroup$ – YCor Jan 24 '16 at 16:43
  • $\begingroup$ @ YCOr: Thanks! I've seen that but not being able to get definition of a locally nilpotent ideal. I found definition of a locally nilpotent ring,en.wikipedia.org/wiki/Locally_nilpotent I assume that an ideal can be considered as a ring without unit. $\endgroup$ – nusernew Jan 24 '16 at 16:53
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  1. I believe that the counterexample given in the accepted answer to the question you link (strong nilpotent elements) shows that there are strongly nilpotent elements for which $RxR$ is not nilpotent.

  2. One precise reference for the statement that the prime radical coincides with the set of all strong nilpotent elements is Proposition 1 of Section 3.2 of Lectures on Rings and Modules by J.Lambek.

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  • $\begingroup$ @ Vladimir. Thanks for the reference! I will check the counterexample! $\endgroup$ – nusernew Jan 24 '16 at 16:59
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    $\begingroup$ Yes, you are right. Clearly if $RxR$ is nilpotent then $x$ is naively strong nilpotent as defined in the previous question. But the element $x$ in the example is strong nilpotent and not naively strong nilpotent. Thus $RxR$ cannot be nilpotent. $\endgroup$ – nusernew Jan 24 '16 at 17:07

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