# Left ideal $I$ is a direct summand of $R$ iff $I = Rr$, $r^2 = r$

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!

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

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 $$r$$.
• I'm having trouble seeing how this works when $(r)$, $(1 - r)$ are left ideals. I assume $(r) = Rr$ etc. – Robert Lewis May 5 at 18:04