Let $I$ be a left ideal of a ring $R$. Prove that if $I$ is a direct summand then $I^2=I$.

If the ideal $I$ is a direct summand of $R$, then there is an ideal $J$ with $R=I\oplus J$, with $I \cap J = 0$. Put $1= e + f$, with $e \in I$ and $f \in J$. Then, $e=e\cdot 1=e(e+f)=e^2+ef$. Since $ef \in I \cap J$, $ef=0$, and hence $e$ is an idempotent, that is $e^2=e$ (and likewise $f^2=f$). Of course $I^2 \subseteq I$. If $r \in I$, then $r=r\cdot1=r\cdot(e+f)=re \in I^2$.

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 $I=Re\subseteq I^2$.