[proofplan]
We first verify directly that the annihilator $I$ is a two-sided ideal, so the quotient ring $R/I$ is defined. We then define the proposed action of a coset $r+I$ on an element $m \in M$ and prove that it is independent of the chosen representative $r$. The module axioms follow from the original $R$-module axioms, and faithfulness follows because any coset acting as zero on all of $M$ must have a representative lying in $I$.
[/proofplan]
[step:Show that the annihilator is a two-sided ideal]
Recall that
\begin{align*}
I=\{a \in R : am=0_M \text{ for every } m \in M\}.
\end{align*}
Since $0_Rm=0_M$ for every $m \in M$, we have $0_R \in I$.
Let $a,b \in I$. For every $m \in M$, additivity of the scalar action gives
\begin{align*}
(a-b)m=am-bm=0_M-0_M=0_M.
\end{align*}
Hence $a-b \in I$, so $I$ is an additive subgroup of $R$.
Let $a \in I$ and $r \in R$. For every $m \in M$,
\begin{align*}
(ra)m=r(am)=r0_M=0_M,
\end{align*}
so $ra \in I$. Also,
\begin{align*}
(ar)m=a(rm)=0_M,
\end{align*}
because $rm \in M$ and $a$ annihilates every element of $M$. Thus $ar \in I$. Therefore $I$ is a two-sided ideal of $R$, and the quotient ring $R/I$ is defined.
[guided]
We need $R/I$ to be a ring before we can speak about an $R/I$-module. For that, $I$ must be a two-sided ideal of $R$.
By definition,
\begin{align*}
I=\{a \in R : am=0_M \text{ for every } m \in M\}.
\end{align*}
The zero element $0_R$ belongs to $I$ because the scalar action of $0_R$ satisfies $0_Rm=0_M$ for every $m \in M$.
Next take $a,b \in I$. This means that $am=0_M$ and $bm=0_M$ for every $m \in M$. Using additivity of the scalar action in the scalar variable, for each $m \in M$ we get
\begin{align*}
(a-b)m=am-bm=0_M-0_M=0_M.
\end{align*}
Thus $a-b$ annihilates every element of $M$, so $a-b \in I$. Hence $I$ is an additive subgroup of $R$.
Now we verify closure under multiplication from both sides. Let $a \in I$ and $r \in R$. For every $m \in M$, associativity of the module action gives
\begin{align*}
(ra)m=r(am).
\end{align*}
Since $a \in I$, we have $am=0_M$, so
\begin{align*}
(ra)m=r0_M=0_M.
\end{align*}
Therefore $ra \in I$. For right multiplication, again using associativity of the module action,
\begin{align*}
(ar)m=a(rm).
\end{align*}
The element $rm$ lies in $M$, and $a$ annihilates every element of $M$, so $a(rm)=0_M$. Hence $ar \in I$. Therefore $I$ is a two-sided ideal, and the quotient ring $R/I$ is well-defined.
[/guided]
[/step]
[step:Define the quotient action and prove it is well-defined]
Define a map
\begin{align*}
\alpha:(R/I)\times M \to M,\quad (r+I,m)\mapsto rm.
\end{align*}
We must prove that $\alpha$ is independent of the representative $r$ of the coset $r+I$.
Suppose $r,s \in R$ satisfy $r+I=s+I$. Then $r-s \in I$. Hence, for every $m \in M$,
\begin{align*}
(r-s)m=0_M.
\end{align*}
Using additivity of the scalar action,
\begin{align*}
rm-sm=(r-s)m=0_M.
\end{align*}
Thus $rm=sm$ for every $m \in M$. Therefore $\alpha(r+I,m)$ is well-defined.
[/step]
[step:Verify the left module axioms for the induced action]
Let $r,s \in R$ and $m,n \in M$. The addition law in $R/I$ is $(r+I)+(s+I)=(r+s)+I$, and multiplication is $(r+I)(s+I)=rs+I$.
For additivity in the module variable,
\begin{align*}
(r+I)(m+n)=r(m+n)=rm+rn=(r+I)m+(r+I)n.
\end{align*}
For additivity in the scalar variable,
\begin{align*}
((r+I)+(s+I))m=((r+s)+I)m=(r+s)m=rm+sm=(r+I)m+(s+I)m.
\end{align*}
For compatibility with multiplication in $R/I$,
\begin{align*}
((r+I)(s+I))m=(rs+I)m=(rs)m=r(sm)=(r+I)((s+I)m).
\end{align*}
Finally, since $M$ is a unital left $R$-module,
\begin{align*}
(1_R+I)m=1_Rm=m.
\end{align*}
Thus $M$ is a unital left $R/I$-module under the induced action.
[guided]
The action has already been shown to be well-defined, so we may compute using any representatives of the cosets. Let $r,s \in R$ and $m,n \in M$.
First, the action must be additive in the module variable. Using the original $R$-module structure on $M$, we have
\begin{align*}
(r+I)(m+n)=r(m+n)=rm+rn=(r+I)m+(r+I)n.
\end{align*}
Second, it must be additive in the scalar variable. The sum of cosets in $R/I$ is represented by the sum of representatives:
\begin{align*}
(r+I)+(s+I)=(r+s)+I.
\end{align*}
Therefore
\begin{align*}
((r+I)+(s+I))m=((r+s)+I)m=(r+s)m=rm+sm=(r+I)m+(s+I)m.
\end{align*}
Third, the scalar multiplication must be compatible with multiplication in the quotient ring. Since the product of cosets is represented by the product of representatives,
\begin{align*}
(r+I)(s+I)=rs+I.
\end{align*}
Hence, using associativity of the original $R$-module action,
\begin{align*}
((r+I)(s+I))m=(rs+I)m=(rs)m=r(sm)=(r+I)((s+I)m).
\end{align*}
Finally, the identity element of $R/I$ is $1_R+I$. Since $M$ is a unital left $R$-module,
\begin{align*}
(1_R+I)m=1_Rm=m.
\end{align*}
These four verifications prove that the induced action makes $M$ into a unital left $R/I$-module.
[/guided]
[/step]
[step:Prove that the quotient action is faithful]
Let $r \in R$ and suppose the coset $r+I \in R/I$ annihilates every element of $M$ under the induced action. Then, for every $m \in M$,
\begin{align*}
0_M=(r+I)m=rm.
\end{align*}
Therefore $r \in \operatorname{Ann}_R(M)=I$. Hence $r+I=I=0_{R/I}$.
Thus the only element of $R/I$ that acts as zero on all of $M$ is $0_{R/I}$. Equivalently,
\begin{align*}
\operatorname{Ann}_{R/I}(M)=\{0_{R/I}\}.
\end{align*}
So $M$ is faithful as a left $R/I$-module.
[/step]
