Let $M$ be an irreducible $R$-module (where $R$ is a commutative ring with a 1) and $N = \{x \in R:mx=0\}$ be the order ideal of $m \in M$. We also assume that there exists an $r \in R$ such that $mr \neq 0$. Show that $N$ is a maximal ideal in $R_{R}$.
The first thing I tried was to assume there exists an ideal $J$ such that $N \leq J \leq R_{R}$. Since we know there exists an element $r \in R$ such that $mr \neq0$, we know that there must exist a $y \in J$ such that $mx \neq 0$ (otherwise it would be in $N$). So we know that $N \neq J$. Since $R_{R}$ contains a 1, we must have that $m1=m =\neq 0 $ so $1 \in J$. This would directly imply that $J=R$. But is it the cause that $1$ is always in $J$? If its not, how would one proceed with the proof?
I figured it out.
let $\theta : R_{R} \to M$ be defined by $(a)\theta = ma$. We can show easily that $\theta$ is an $R$-module homomorphism. $Im(\theta)$ is a submodule of $M$ and since there exists an $r \in R$ such that $mr \neq 0$, $\text{Im}(\theta) \neq \{0\}$. $M$ is irriducible so $\text{Im}(\theta) = M$. The first isomorphism theorem tells us $R_{R}/\text{ker}(\theta) \cong M$. Notice that $\text{ker}(\theta) = \{ x \in R : mx = 0 \} = N$, so $R_{R}/N \cong M$. The corrospondance theorem therefore tells us, since $M$ is irriducible, $N$ is maximal.