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I am stuck on proving a left ideal $I$ of a ring $R$ is a direct summand of $R$ if and only if $I = Rr$ with $r^2 = r$. Could you help me with that? Any help will be very appreciated. Thanks!

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    $\begingroup$ Indication : consider $J = (1-r)$ $\endgroup$ – DLeMeur May 1 at 20:05
  • $\begingroup$ I added the "ring-theory" tag to your post. Cheers! $\endgroup$ – Robert Lewis May 5 at 19:56
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Hints:

  1. $R(1-r)$ is your candidate for a complement

  2. In the other direction, if $1=a+b$ where $a\in I$ and $b\in J$ and $I\oplus J=R$, $a$ is your candidate for $r$.

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  • $\begingroup$ I'm having trouble seeing how this works when $(r)$, $(1 - r)$ are left ideals. I assume $(r) = Rr$ etc. $\endgroup$ – Robert Lewis May 5 at 18:04
  • $\begingroup$ @RobertLewis yes that’s how I interpreted it too. What is giving you trouble? $\endgroup$ – rschwieb May 5 at 23:02

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