I'm trying to follow this proof from Noncommutative Algebra by Farb. The theorem is that the following are equivalent for a left $R$-module $M$:
$(1)$ $M$ is simple
$(2)$ $M$ is cyclic and generated by every nonzero element
$(3)$ $M\cong R/I$ for a maximal left ideal $I$
where we say $M$ is cyclic if $Rm=M$ for some $m\in M$.
The part I'm trying to follow is from $(2)$ implies $(3)$: for any nonzero $m$ define a map $R\to Rm$ by $r\mapsto rm$; this has kernel $I=\text{Ann}(m)$, so $M=Rm\cong R/I$. Then he says that to see $I$ is maximal, note if $I$ weren't maximal then there would be a nonzero element which does not generate $M$. I fail to see why this is true. Can anybody help?
First $(3)$ implies $(1)$. Suppose $M \cong R/I$, with $I$ a maximal ideal in $R$. Then as a ring $R/I$ is a field. Hence the only ideals of $R/I$ are the ring itself and $\{0\}$. But this means that the only subgroups of $R/I$ that are closed under multiplication by elements of $R$ are $R/I$ and $\{0\}$, and so $M$ is a simple module.
For the other direction ($(1)$ implies $(3)$), suppose $M$ is a simple module. Then we know that $M=Rm$, for every nonzero $m \in M$. Hence there is an isomorphism $R/I \cong M$, for some ideal $I \subseteq R$. Suppose $I$ is not maximal. Then there exists $r \in R/I$ such that $(r) \neq (1)$. Otherwise for all $r \in R/I$, the ideal $(r)$ is equal to $R/I$, implying that $R/I$ is a field. Hence there exists a nonzero proper submodule of $M$, contradicting the fact that $M$ is simple. To see this, suppose $r$ maps to $m \in M$. Then $Rm \neq M$, otherwise $(r)=R/I$. But this means $M$ has a nontrivial submodule, contradicting the fact that it is simple. Hence $I$ is a maximal ideal.