I have the following problem. Let $R$ be a ring and let $e$ be an idempotent in $R$. We consider $R$ as an $R$-module. Let $R \cdot e$ be the submodule generated by $e$ (since it happens in the $R$-module $R$, it is the same as the left ideal generated by $e$). Show that $R = R\cdot e \oplus Ann R(e)$ (i.e. $R = R·e+AnnR(e)$ and $R·e∩AnnR(e) =\{0\}$), where $AnnR(m) := \{r ∈ R \; | \; r= 0\}$ is the annihilator of $m$ in $R$.
This is my attempt at a proof: $R$ is simple when the submodules of $R$ are $\{0\}$ and $\{R\}$. As $e$ is idempotent I figured $e=0$ or $e=1$. Therefore $R·e$ must be $\{0\}$ and $\{R\}$ therefore simple.
However I found in my notes that "If $R$ is simple, then $R$ is generated by any nonzero element: If $r∈R\setminus \{0\}$ then the submodule generated by $e$ is nonzero, and then must be $R$.
Therefore $R·e$ is nonzero so $\{r∈R \; | \; re \text{ nonzero}\}$ and $AnnR(e)=\{r∈R \; | \; re = 0\}$. Therefore the intersection must be $\{0\}$.
I'm not sure if I have contradicted myself in this question, or if my method was along the right lines.
Why should you have $e \in \{0,1\}$? And why should $R$ be simple?
It's as easy as that:
We have $r=re+(r-re)$, which shows $R = Re+\operatorname{Ann}_R(e)$.
$Re \cap \operatorname{Ann}_R(e)=0$ might be still easier, because it is straight forward.