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Let $R$ be a unital ring with Jacobson radical $J(R)$. If for any right ideal $I$ of $R$ we have $I+J(R)=I^2+J(R)$ could we deduce that $I$ is the sum of an idempotent right ideal and a nil right ideal? (By the hypothesis, of course $I\subseteq I^2+J(R)$.)

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  • $\begingroup$ This is not clear. Are you assuming that for all right ideals $I$ we have $I+J(R)=I^2+J(R)$, or are you assuming that this holds for some special ideal $I$? $\endgroup$ – Crostul Aug 30 '16 at 15:35
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The answer is no, a counterexample follows.

Let $R$ be any DVR with unique maximal ideal $J$ (which is also the Jacobson radical of $R$). In particular the equality $$I+J=J=I^2+J$$ is trivially satisfied for all ideals $I$. Moreover, recall that all ideals have the form $J^n$ for some $n \ge 0$. The unque idempotent ideal is thus $J^0=R$, and the unique nil ideal is $0$ (there are no nonzero nilpotent elements). Hence the sum of an idempotent ideal and a nil ideal is necessarily $R$, which turns out to be distinct from $I$ in general.

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  • $\begingroup$ Thanks for the answer! If we add the condition that $J(R)$ is nil, is the answer to my question "yes"? $\endgroup$ – karparvar Aug 30 '16 at 17:23
  • $\begingroup$ Well, $\{0\}$ is an idempotent ideal too, so I wouldn't say $R$ is the unique idempotent ideal... but the same conclusion follows! $\endgroup$ – rschwieb Aug 30 '16 at 18:12

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