TITLE: Let $I$ be a left ideal of a ring $R$. Prove that if $I$ is a direct summand then $I^2=I$.
QUESTION [9 upvotes]: Let $I$ be a left ideal of a ring $R$ with $1$. Prove that if $I$ is a direct
  summand then $I^2=I$.

Definition. An ideal $A$ is a direct summand of a ring $R$ iff there exists an ideal $B$ of $R$ such that $R=A\oplus B$
Help me.

REPLY [7 votes]: One good thing to learn from this is that for rings with identity, left (or right) ideals are summands iff they are generated by an idempotent.
Here's a sketch:
With $R=I\oplus J$, write $1=e+f$ in this decomposition. Multiplying on the left by $e$, we discover $e=e^2+ef\in I\oplus J$. But $e$ is uniquely expressed already as $e+0$, so $e=e^2$.
We claim that $I=Re$. If $i\in I$, then $i=ie+if$, but because $i$ is already uniquely expressed as $i+0$ in the sum, we have that $if=0$ and $ie=i$. So, $I\subseteq Re$. Conversely, $e\in I$ so $Re\subseteq I$.
This makes proving $I^2=I$ a breeze: obviously $I^2\subseteq I$ so the other containment is the only thing to prove. But $e=e^2\in I^2$, therefore $I=Re\subseteq I^2$.