Is any direct summand of a PI-ring (polynomial identity ring) necessarily idempotent as a right ideal?

The answer is yes for a special case of PI-rings, namely any direct summand of a commutative ring would be an idempotent ideal.

Thanks for any suggestion!


A right ideal of a ring with identity which is a direct summand is always idempotent: if it is $eR$, then $e=e^2\in (eR)^2$, so $eR\subseteq (eR)^2$, and $(eR)^2\subseteq eR$ trivially.

(Are you really asking about PI rings without identities?)

  • $\begingroup$ Thanks for your answer, but if $R$ does not possess identity what is the strategy? $\endgroup$ – karparvar Apr 27 '16 at 12:35
  • $\begingroup$ @karparvar I don't know. It rarely comes up in this context for me $\endgroup$ – rschwieb Apr 27 '16 at 21:20

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