If $R$ is a simple ring and $n \in \Bbb Z$, show that either $nR = 0$ or $nr=0$ implies that $r=0$ If  $R$ is a simple ring , $n \in \Bbb Z$, show that either $nR = 0$, or $nr=0$ implies that $r=0$ for $r \in R$.
Since $nR$ is an ideal of $R$, then either $nR=0$, or $nR = R$. In this second case, I am not sure why it is the case that if $nr = 0$, that necessarily $r = 0$. 
Side note, I know a theorem that states that a commutative ring $R$ is simple iff it is a field, but in this case I don't have commutativity. I also know that if $\text{char}(R)= k$, then $kR=0$, but I don't know that the characteristic is non-zero, and not sure about the case where $n$ is not a multiple of $k$.

Consider the set of elements $r \in R$ with $nr = 0$. This is a two-sided ideal, so it must be either $0$ or all of $R$.
