Let $R$ be a ring and $M$ a $R$-module. For $r\in R$ define $f:M\to M$ by $f(s)=sr$.
Show that $f$ is injective if and only if $r$ is not a right zero divisor.
I have done a similar problem to this in the past, involving surjectivity instead of injectivity and came up with:
Suppose $f$ is surjective and $gf=0$. To prove $f$ is not a right divisor of zero, we need to show that $g=0$, i.e. that $g(m)=0$ for all $m$. So let $m$ be in $M$. Since $f$ is surjective, $m=f(n)$ for some n in $M$. Thus $g(m)=g(f(n))=(gf)(n)=0(n)=0$.
However I am stuck with how to show this problem using injectivity, and was wondering could anybody provide guidance on this?
$f$ is injective if and only if
$f(s)=0 \Rightarrow s=0$ if and only if
$sr=0 \Rightarrow s=0$ if and only if
$r$ is not a right zero divisor.