endomorphism of the right R-module $eR$ Let $e$ be a non-zero idempotent of a ring R, I want to find the unique endomorphism of the right R-module $eR$ such that $f(e)e=ere$ for a fix $r\in R$. I tried to solve this problem and I have found to possible endomorphism:
first one: $f(es)=ers$ for every $es\in eR$
second one: $f(es)=eres$ for every $es\in eR$
So I want to understand which is the wrong answer.
In my opinion the wrong answer is the first one because in my opinion it isn't an endomorphism of $eR$ but I can't found the mistake in the following proof:
$f(et\centerdot s)=f(ets)=erts=ert\centerdot s= f(et)\centerdot s $
$f(es+et)=f(e(s+t))=er(s+t)=ers+ert=f(es)+f(et)$
Can anyone help me?

Note the second map works $f(e1)=f(e) = ere$ so $f(e)e=ere²=ere$. The first map is a priori not well defined: you have to show that if $es=es'$ then $ers=ers'$, and this might not be true. It is if $e$ is a central idempotent, but in such case both maps agree. 
