[proofplan] We use the scalar multiplication map from the left regular module $R$ to $M$. Its image is exactly the cyclic submodule $Rm$, and its kernel is exactly the annihilator $\operatorname{Ann}_R(m)$. We then construct the induced map from the quotient $R/\operatorname{Ann}_R(m)$ to $Rm$ explicitly and verify that it is a well-defined bijective $R$-[module homomorphism](/page/Module%20Homomorphism). [/proofplan]

[step:Verify that the annihilator is a left ideal] Let \begin{align*} A := \operatorname{Ann}_R(m) = \{r \in R : rm = 0_M\}. \end{align*} We first show that $A$ is a left ideal of $R$. Since $0_R m = 0_M$, we have $0_R \in A$. If $r,s \in A$, then $rm = 0_M$ and $sm = 0_M$, so additivity of the module action gives \begin{align*} (r - s)m = rm - sm = 0_M - 0_M = 0_M. \end{align*} Thus $r - s \in A$, so $A$ is an additive subgroup of $R$. If $a \in R$ and $r \in A$, then associativity of the module action gives \begin{align*} (ar)m = a(rm) = a0_M = 0_M. \end{align*} Hence $ar \in A$. Therefore $A$ is a left ideal of $R$, so the quotient left $R$-module $R/A$ is well-defined. [/step]

[step:Construct the quotient map induced by scalar multiplication]Define the scalar multiplication map \begin{align*} \varphi: R \to M, \qquad r \mapsto rm. \end{align*} Here $R$ is regarded as the left regular $R$-module. For $r,s,a \in R$, the module axioms give \begin{align*} \varphi(r+s) = (r+s)m = rm + sm = \varphi(r) + \varphi(s) \end{align*} and \begin{align*} \varphi(ar) = (ar)m = a(rm) = a\varphi(r). \end{align*} Thus $\varphi$ is an $R$-module homomorphism. Its image is \begin{align*} \operatorname{im}\varphi = \{rm : r \in R\} = Rm, \end{align*} which is the cyclic submodule generated by $m$. Its kernel is \begin{align*} \ker \varphi = \{r \in R : \varphi(r) = 0_M\} = \{r \in R : rm = 0_M\} = A. \end{align*}[/step]

[guided]The natural map to consider is scalar multiplication by the fixed element $m$. Define \begin{align*} \varphi: R \to M, \qquad r \mapsto rm. \end{align*} The domain is the left regular module $R$, meaning that $R$ acts on itself by left multiplication. We verify that $\varphi$ respects both addition and the left $R$-action. For $r,s,a \in R$, the distributive and associative module axioms give \begin{align*} \varphi(r+s) = (r+s)m = rm + sm = \varphi(r) + \varphi(s) \end{align*} and \begin{align*} \varphi(ar) = (ar)m = a(rm) = a\varphi(r). \end{align*} So $\varphi$ is an $R$-module homomorphism. Now we identify the two pieces of data attached to this homomorphism. Its image consists exactly of all scalar multiples of $m$: \begin{align*} \operatorname{im}\varphi = \{\varphi(r) : r \in R\} = \{rm : r \in R\} = Rm. \end{align*} This is precisely the cyclic submodule generated by $m$. Its kernel consists exactly of the scalars that kill $m$: \begin{align*} \ker \varphi = \{r \in R : \varphi(r) = 0_M\} = \{r \in R : rm = 0_M\} = \operatorname{Ann}_R(m) = A. \end{align*} Thus the quotient by the annihilator is the quotient by the kernel of the scalar multiplication map.[/guided]

[step:Show that the induced map $R/\operatorname{Ann}_R(m) \to Rm$ is an isomorphism] Define \begin{align*} \Phi: R/A \to Rm, \qquad r + A \mapsto rm. \end{align*} This map is well-defined: if $r + A = s + A$, then $r - s \in A$, so $(r-s)m = 0_M$, and hence $rm = sm$. For $r,s,a \in R$, the [quotient module](/page/Quotient%20Module) operations and the module axioms give \begin{align*} \Phi((r+A)+(s+A)) = \Phi((r+s)+A) = (r+s)m = rm + sm = \Phi(r+A)+\Phi(s+A) \end{align*} and \begin{align*} \Phi(a(r+A)) = \Phi(ar+A) = (ar)m = a(rm) = a\Phi(r+A). \end{align*} Thus $\Phi$ is an $R$-module homomorphism. The map $\Phi$ is surjective because every element of $Rm$ has the form $rm$ for some $r \in R$, and then $rm = \Phi(r+A)$. It is injective because \begin{align*} \Phi(r+A)=0_M \iff rm=0_M \iff r \in A \iff r+A=A. \end{align*} Therefore $\Phi$ is a bijective $R$-module homomorphism. Hence \begin{align*} R/\operatorname{Ann}_R(m) \cong Rm \end{align*} as left $R$-modules. [/step]