Let $R$ be a ring with $1_R$, $M$ a simple left $R$-module and $m\in M$ is a non-zero element.
We would like to prove that the annihilator $\mathrm{Ann}_R(m)\subseteq R$ is a maximal left ideal of $R$.
It may be something obvious but unfortunately I can't see it.
To prove that it is a left ideal is easy. Now I stuck at the maximality. The first thought is to assume that $$\mathrm{Ann}_R(m) \subsetneqq J \subseteq R.$$ So, it suffices to show that in this case $J=R$.
We know that since $$\mathrm{Ann}_R(m) \subsetneqq J\implies\exists j\in J \setminus \mathrm{Ann}_R(m) \implies jm\neq 0_M \implies jm=n,$$ for some $n\neq 0_M$.
So if we take an element $x\in R$ we should show that $x\in J$. But how?
Thank you.
$M$ is simple, so it's cyclic, hence $M=Rm$. But $n=jm\neq 0$ is also in $M$, so $Rn=M$. As $m\in M$, there is an $r\in R$ such that $m=rn=rjm$, meaning $rj-1\in\text{Ann}_{R}(m)$. Yet since $\text{Ann}_{R}(m)\subset J$, we have $rj-1\in J$, but $rj\in J$ so we also have $1\in J$.