Assume that $R$ is a unital ring with nil Jacobson radical $J(R)$. If each right ideal of $R/J(R)$ is idempotent, is it necessarily true that each right ideal of $R$ is idempotent or nil?

If $I$ is a right ideal of $R$ contained in $J(R)$ then the result follows. If $I$ is a right ideal of $R$ containing $J(R)$ and $x\in I$ then $x+ J(R)=∑x_t y_t +J(R)$ (a finite sum over $t$ with $x_t,y_t\in I$). Hence $x-∑x_t y_t$ is a nilpotent element as it belongs to $J(R)$...

That is all my try! Thanks for any pushing suggestion!


Let $F$ be a field, $S=F\times F$, and $M=F\times\{0\}$ and use the trivial extension $R=S\times M$ where $(s,m)+(s',m')=(s+s',m+m')$ and $(s,m)(s',m')=(ss', sm'+ms')$.

Clearly $J(R)=\{0\}\times M$ is nil (in fact it squares to zero.) Also $R/J(R)\cong F\times F$ has all right ideals idempotent.

Now look at $I:=(\{0\}\times F, M)\lhd R$. It's not nil (since it's not contained in $J(R)$) and it's not idempotent (since $I^2=(\{0\}\times F,\{0\})$.)

  • $\begingroup$ I guess any right ideal of your $R$ is the sum of a nil and an idempotent. Right? $\endgroup$ – karparvar Aug 4 '16 at 22:08
  • $\begingroup$ @karparvar I'm not totally sure, offhand. Maybe! Are you going to use this example in research? $\endgroup$ – rschwieb Aug 5 '16 at 0:34
  • $\begingroup$ With an attribution of course! I would appreciate you (Ryan Schwiebert) in advance! $\endgroup$ – karparvar Aug 5 '16 at 7:59
  • $\begingroup$ Sure, that sounds fine. I hadn't ever heard of such an example, and I was particularly pleased to think of this solution. Hopefully there is not a much simpler example that I'm overlooking... $\endgroup$ – rschwieb Aug 5 '16 at 10:50
  • $\begingroup$ I'm very curious to see the end result of your efforts. $\endgroup$ – rschwieb Aug 5 '16 at 11:03

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