Problem: Let $R$ be a ring with $1$ and $e\in R$ such that $e^2=e$. If $R$ is finite and does not contain any nonzero nilpotent element then show that $Re=eRe$.
I think I solved the problem yet I do not use the finiteness of $R$. Thus, I doubt that my solution might be wrong.
Solution:
Claim 1: $R=ker(f)\oplus im(f)$ where $f$ is a homomorphism (with respect to $+$) from $R$ to $eR$ by $f(r)=er$.
For any $r\in R$, we can write $r=(r-er)+er$ where $r-er\in ker(f)$ and $er\in im(f)$. Thus, clearly we have $R=ker(f)+ im(f)$. Now let $k\in ker(f)\cap im(f)$ then $k=ex$ for some $x\in R$ and $ek=0$.
$$0=ek=eex=ex=k.$$
Thus the sum is direct as claimed.
Claim 2: Set $K=\{r-er \mid r\in R\}$. Then $K=ker(f)$.
Note that $K\leq ker(f)$ as $r-er\in ker(f)$ for each $r\in R$. Thus, the sum $K+im(f)$ is also direct. On the other hand, $r=(r-er)+er$ for all $r\in R$, and hence $R=K\oplus im(f)$.
Now pick $x\in ker(f)$. Then $x=k+i$ where $k\in K$ and $i\in im(f)$. Then $i=x-k\in ker(f)$, and hence $i=0$. We obtain $x=k$ and $ker(f)=K$ as desired.
Claim 3: $eR=eRe$.
We have $R=ker(f)\oplus eR$, and hence $Re=ker(f)e\oplus eRe$.
Pick $t\in ker(f)e$ then $t=re-ere$ for some $r\in R$.
$$t^2=((re)^2-re^2re-erere+ere^2re)=0$$
and hence $t=0$ by hypothesis. Then $ker(f)e=0$, and $Re=eRe$ as desired. Similarly, we would have $eR=eRe$.