Prove that if $I$ is a right ideal of $R$ and $I$ is a direct summand of $R$ then $I=eR$ for some idempotent $e \in R$

Question

Prove that if $I$ is a right ideal of $R$ and $I$ is a direct summand of $R$ then $I=eR$ for some idempotent $e \in R$ Also, $R$ is a unital ring.

So $I$ is a direct summand, this means there is another ideal $J$ of $R$ such that $R = I \oplus J$ and $J+I = R$ and $J \cap I = \emptyset$... I'm a little bit lost here, would appreciate some help!

Answer 1

Let $$1=e+(1-e)$$ where $e \in I$ and $1-e \in J$.

Note that $$e(1-e)=(1-e)e= e-e^2$$ belongs to both $I$ and $J$, because they are right ideals. Thus, $$e-e^2=0$$ which shows that $e$ is idempotent.

Now, clearly $eR \subseteq I$. To show the converse, for all $x \in I$ $$x-ex = (1-e)x \in I \cap J$$ Thus $x-ex=0$, i.e. $x=ex \in eR$.

Answer 2

Let $1$ be the unit of $R$, and write $1=e+f$ for some $e\in I$, $f\in J$. If $x\in I$, we have $$x=1x=ex+fx=ex,$$ so $I=eR$. Furthermore, letting $x=e$ in the above calculation shows that $e$ is an idempotent.

Additionally, note that this is not true if we do not assume that $R$ is unital. For then we can consider $R=C_0(\{0,1\}\times \mathbb R)$, $I=C_0(\{0\}\times\mathbb R)$ and $J=C_0(\{1\}\times\mathbb R)$. Then we have $R=I\oplus J$, but there are no idempotents in $R$.