Let $I$ be a minimal ideal of a ring $R$. Then, $I$ is a direct summand of $R$ if and only if $I^2\neq \{0\}$. Prove this statement:

Let $I$ be a minimal ideal of a ring $R$.  Then, $I$ is direct summand of $R$ if and only if $I^2\neq \{0\}$. 

(where direct summand means that we consider $I$ as a submodule of the $R$-module $R$.)
 A: If $I$ is a minimal right ideal, then $I^2=0$ or else $I=eR$ for some idempotent $R$. In any ring with unity, $(1-e)R\oplus eR$ is possible, making it a summand.
Actually there is a sub-bit of information that's good to take away: a right ideal (of a ring with 1) is a summand iff it is of the form $eR$ for an idempotent $e$.

The standard proof for seeing $I^2=0$ or $I^2=I$ goes as follows. If $I^2\neq 0$, there is $a\in I$, $aI\neq 0$ (consequently $aI=I$). Choose $e\in I\setminus \{0\}$ such that $ae=a$. Notice that then $a(e^2-e)=0$, and that $e^2-e\in I$. Now we examine $(e^2-e)R$. If this were nonzero, it would be all of $I$, but then $aI=a(e^2-e)R=0$, a contradiction. Therefore $(e^2-e)R=0$, but that implies $e^2-e=0$. Thus $e^2=e\neq 0$ is a nonzero idempotent which generates $I$.
Verifying that $(1-e)R\oplus eR=R$ is very straightforward as well. 
Firstly, $r=r-er+er=(1-e)r+er\in (1-e)R+eR$. 
Secondly, if $x\in (1-e)R\cap eR$, then $x=er=(1-e)s$ for some $r,s\in R$. But since this implies $eer=e(1-e)s=0$, we have $0=eer=er=x$, proving that $x=0$ and that $(1-e)R\cap eR=\{0\}$. 
