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Let $R$ be a non-commutative ring. Suppose $N(R)$ and $J(R)$ respectively denote the set of nilpotent elements and the Jacobson radical of $R$.

Consider the following statements:

(1) $N(R) \subseteq J(R)$,

(2) $a^2=0 \implies a \in J(R)$.

Clearly, $(1) \implies (2)$.

Does $(2) \implies (1)$ ?

The answer to this question should be negative. Please help us to find a suitable counter example.

P.S. This is an open question. If somebody can solve it then he/she will be duly credited.

ABh
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  • Do you mean to quantify "For all $a\in R$, $a^2=0\implies a\in J(R)$? – rschwieb Sep 12 '17 at 18:21
  • Nil radical is always contained in Jacobson radical. Given $a\in R$, $a^2=0$ implies $a$ is in the nil radical so it is in the Jacobson radical. What is your question –  Sep 12 '17 at 18:21
  • @arnab Are you assuming a commutative ring? – rschwieb Sep 12 '17 at 18:22
  • @rschwieb We can at least wait for users response. I do not see why it is an emergency to mark it as a duplicate. –  Sep 12 '17 at 18:27
  • I am assuming $R$ to be noncommutative. @rschwieb – ABh Sep 12 '17 at 18:34
  • I can't say I've seen anyone control where elements of a particular nilpotency index land. Seems hard... unless someone finds a clever example. It also implies that every nilpotent element has a power in $J(R)$... – rschwieb Sep 12 '17 at 19:16

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