[proofplan]
We first use the definition of the cyclic submodule $Rm$ to check that the map $\varphi: R \to Rm$ is well-defined and surjective. We then verify additivity and compatibility with scalar multiplication directly from the module axioms. The kernel is computed by unwinding the definitions of kernel and annihilator. Finally, since $\operatorname{Ann}_R(m)$ is an ideal, the quotient module is defined, and the First Isomorphism Theorem for modules identifies it with $Rm$.
[citetheorem:7850]
[/proofplan]
[step:Verify that the scalar action defines a map onto the cyclic submodule]
By definition,
\begin{align*}
Rm=\{am : a \in R\}.
\end{align*}
For each $r \in R$, the element $rm$ belongs to $Rm$, so the assignment
\begin{align*}
\varphi: R \to Rm, \quad r \mapsto rm
\end{align*}
is well-defined. If $y \in Rm$, then by definition of $Rm$ there exists $a \in R$ such that
\begin{align*}
y=am.
\end{align*}
Since $\varphi(a)=am=y$, every element of $Rm$ lies in the image of $\varphi$. Hence $\varphi$ is surjective.
[guided]
The codomain of $\varphi$ is not all of $M$ but the cyclic submodule $Rm$. Therefore the first point to check is that $\varphi(r)$ actually lands in that codomain. The cyclic submodule generated by $m$ is defined by
\begin{align*}
Rm=\{am : a \in R\}.
\end{align*}
Thus, for any $r \in R$, the element $rm$ is one of the listed elements of $Rm$, namely the element obtained by taking $a=r$. This proves that the map
\begin{align*}
\varphi: R \to Rm, \quad r \mapsto rm
\end{align*}
is well-defined.
The same definition also gives surjectivity. Let $y \in Rm$. Since $Rm$ consists exactly of the elements $am$ with $a \in R$, there is some $a \in R$ such that
\begin{align*}
y=am.
\end{align*}
Evaluating the map at this element $a$ gives
\begin{align*}
\varphi(a)=am=y.
\end{align*}
So every $y \in Rm$ is hit by $\varphi$, and $\varphi$ is surjective.
[/guided]
[/step]
[step:Check that the map preserves addition and scalar multiplication]
Let $r_1,r_2 \in R$. Using distributivity of the $R$-module action over addition in $R$, we have
\begin{align*}
\varphi(r_1+r_2)=(r_1+r_2)m=r_1m+r_2m=\varphi(r_1)+\varphi(r_2).
\end{align*}
Let $s,r \in R$. The scalar action on the $R$-module $R$ is multiplication in $R$, and the scalar action on $Rm$ is inherited from $M$. By associativity of scalar multiplication in the module $M$,
\begin{align*}
\varphi(sr)=(sr)m=s(rm)=s\varphi(r).
\end{align*}
Therefore $\varphi$ is an $R$-module homomorphism.
[/step]
[step:Identify the kernel with the annihilator of the element]
By definition of the kernel of $\varphi$,
\begin{align*}
\ker(\varphi)=\{r \in R : \varphi(r)=0_M\}.
\end{align*}
Since $\varphi(r)=rm$ for every $r \in R$, this becomes
\begin{align*}
\ker(\varphi)=\{r \in R : rm=0_M\}.
\end{align*}
By definition of the annihilator of the element $m$,
\begin{align*}
\operatorname{Ann}_R(m)=\{r \in R : rm=0_M\}.
\end{align*}
Hence
\begin{align*}
\ker(\varphi)=\operatorname{Ann}_R(m).
\end{align*}
[/step]
[step:Apply the module isomorphism theorem to obtain the quotient description]
Since $R$ is commutative, $\operatorname{Ann}_R(m)$ is an ideal of $R$.
[citetheorem:7850]
Hence the quotient $R/\operatorname{Ann}_R(m)$ is an $R$-module. The map $\varphi: R \to Rm$ is a surjective $R$-module homomorphism, and its kernel is $\operatorname{Ann}_R(m)$. By the First Isomorphism Theorem for modules, there is an $R$-module isomorphism
\begin{align*}
R/\ker(\varphi) \cong Rm.
\end{align*}
Substituting $\ker(\varphi)=\operatorname{Ann}_R(m)$ gives
\begin{align*}
R/\operatorname{Ann}_R(m) \cong Rm.
\end{align*}
This proves the claimed quotient description of the cyclic submodule generated by $m$.
[/step]
