I'm trying to solve the following problem:
Let $R$ be a semisimple ring. Prove that a left ideal $L$ of $R$ is minimal iff it can be written as $L=Re$ where $e$ is a primitive idempotent of $R$.
Here,
An idempotent $e\neq0$ is primitive if it cannot be written as $e=e_1+e_2$, where $e_1,e_2$ are nonzero idempotents such that $e_1e_2=0$.
I know since $R$ is semisimple, $L$ is a left ideal iff $L=Re$ for some idempotent $e$. But I couldn't use that to solve the question. I'd like someone to give me a hint, please. Thanks in advance!
Suppose $L=Re$ is a minimal left ideal and $e$ is not primitive. If $e=e_1+e_2$ where $e_1,e_2 \neq 0$ and $e_1e_2=0$ Then $e_1 e=e_1(e_1+e_2)=e_1$. What does that say about $Re_1$?
For the other direction, suppose that $e$ is a primitve idempotent. We need to show that $Re$ is a minimal left ideal. If we have another ideal $Rf \subset Re$, where $f$ is an idempotent, we can write $f=re$ for $r \in R$. Then $fe=re^2=re=f$ and $(ef)^2=efef=ef^2=ef$, so $ef$ is also idempotent.
Try to show, using the same kind of method, that $e-ef$ is also an idempotent. Then we can write $e=(e-ef)+ef$, a sum of two idempotents and $(e-ef)ef=0$, so we can use that $e$ is primitve and work from there.