$\DeclareMathOperator{\ann}{Ann}$
I can show that if $M$ is a simple left $R$-module for a unital ring $R$, then for any non-zero $m\in M$, (a) $Rm=M$; and (b) $\ann(m)$ is a maximal left ideal. My proof for (b) is that if $I$ is an ideal that strictly contains $\ann(m)$, we have that $Im=$ is a submodule of $M$ not equal to 0, so $Im=M$. Then for any $r\in R$, there exists some $i\in I$ such that $rm=im$. This means $r-i\in \ann(m)\subseteq I$, so $r\in I$. Hence $I=R$.
Does something similar hold for the annihilator of $M$ overall (which will be a two-sided ideal)? I noticed that I did not use the fact that $I$ is an ideal in the above, except to argue that $Im=M$.
(Ultimately I'm trying to show the equivalence of various definitions of the Jacobson radical, specifically that the intersection of annihilators of right simple modules equals the intersection of annihilators of left simple modules).
Best Answer
Not always, no.
Take an infinite dimensional vector space $V$, and let $E$ be its ring of linear transformations acting on it in the natural way. Then $V$ is a simple $E$ module that has annihilator $\{0\}$, which is not maximal in the ring, since there is also a proper ideal of transformations with finite dimensional images.