Using a property to prove that Annihilator is maximal left ideal

Question

Let $M$ be a nonzero $R$-module.

Suppose the map $\phi_m: R/\text{Ann}(m)\to M$, $\phi_m(r+\text{Ann}(m))=rm$ is a well-defined $R$-module isomorphism for all $m\in M\setminus \{0\}$.

Using this property, prove that $\text{Ann}(m)$ is a maximal left ideal of $R$.

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My attempt: I am ok with showing $\text{Ann}(m)$ is a left ideal of $R$, so we are left with showing $\text{Ann}(m)$ is maximal left ideal.

Suppose $I$ is a left ideal of $R$ and $\text{Ann}(m)\subsetneq I$. We wish to show that $I=R$.

What I tried is let $x\in I\setminus\text{Ann}(m)$. Then $xm\neq 0_M$. I am not very sure how to continue...

Thanks for any help!

Answer 1

What you have is that for any nonzero $m,n$ in $M$, there exists an $r\in R$ such that $rm=n$. This is equivalent to $M$ being a simple $R$ module. (Prove this if you don't know it.)

By the module first isomorphism theorem, you have that all the kernels of your maps are maximal left ideals.