Let $R$ be a ring with identity element such that every ideal of which is idempotent or nilpotent. Is it true that the Jacobson radical $J(R)$ of $R$ is nilpotent?
If $R$ is Noetherian and $J(R)$ is idempotent the Nakayama lemma yields $J(R)=0$. So, for Noetherian rings whose Jacobson radicals are nonzero we have an affirmative answer to the raised question.
Thanks for any help or suggestion!
Best Answer
Let $a$ be an element of $R$. If $\left<a\right>=\left<a\right>^2$, then it is clear that the exists $e^2=e\in \left<a\right>$ such that $\left<a\right>=\left<e\right>$. Now, if $0\not= a\in J (R)$, then $\left<a\right>$ is nilpotent or idempotent. If $\left<a\right>$ is idempotent, by above argument there exists $e^2=e\in\left<a\right>$ such that $\left<a\right>=\left<e\right>$. Since $e\in J (R)$ , $1-e$ is a unit idempotent and so $1-e=1$. Thus, $\left<a\right>=0$, a contradiction. Therefore, every element of $J (R)$ in nilpotent and so $J (R)$ is nil. If $J (R)$ is finitely generated $J (R)$ is nilpotent.