$Rx/Rax \simeq R/Ra$ whenever $\mathrm{Ann}(x)=\{0\}$ 
Let $R$ be a PID and $M$ an $R$-module. Let $x\in M$ be such that $\mathrm{Ann}(x)=\{0\}$. Prove that for any $a\in R$ we have $Rx/Rax \simeq R/Ra$. Indicate in your argument where you use the fact that $\mathrm{Ann}(x)=\{0\}$.

My Proof Attempt:
Proof. Let $\varphi:Rx\to R/Ra$ such that $$rx\longmapsto r+Ra.$$
We require that $Ann(x)$ consists of only zero. Otherwise, $\varphi$ may not be surjective since if $\exists r\in Ann(x)$, $r\neq 0$, then $rx=0\longmapsto Ra$. Hence, an element $r+Ra$ may never be "reached."
Now, $\varphi$ is also a homomorphism since with each $r\in R$ "surviving" multiplication by $x$ means that the map behaves exactly as the canonical epimorphism.
$$\pi:R\to R/Ra.$$ 
Hence, $\varphi$ is an epimorphism and Now we just need to check that $ker(\varphi)=Rax$ and the proof follows by the 1st Isomorphism Theorem. Clearly $ker(\varphi)\supset Rax$ since $$\varphi(rax)=ra+Ra=Ra.$$ 
So conversely let $y\in ker(\varphi)$. $\implies y=rx$ for some $r\in R$ is mapped to $Ra \implies r\in Ra$. Hence, $y\in Rax$.
This completes the proof.

If anything, I'd be very appreciative if somebody may comment on whether I'm using the hypothesis, $Ann(x)=\{0\}$, correctly here. I had another idea that if there were a nonzero element in $Ann(x),$ then $0$ may be mapped to a non-identity element i.e.
$rx=0\longmapsto r+Ra$. But I'm not sure which argument to use.
Thank you again for any help.
 A: Your idea is not pursued correctly. The map $\varphi\colon rx\mapsto r+Ra$ need not be well defined, because it can happen that $rx=sx$ and in this case you need to prove that $r+Ra=s+Ra$.
However, you're in a very special case! You know that $\operatorname{Ann}(x)=\{0\}$, and $rx=sx$ implies $(r-s)x=0$, so $r-s=0$. Thus the map $\varphi$ is indeed well defined. 
Here's where the assumption is used and in a very decisive role. Without it, you can't even define $\varphi$.
The map $\varphi$ is clearly a homomorphism. Now it's just sufficient to compute its kernel. If $\varphi(rx)=0+Ra$, then $r\in Ra$; therefore $r=sa$ for some $s\in R$. Hence $rx=sax\in Rax$.
Conversely, $Rax\subseteq\ker\varphi$.
The homomorphism theorem then says that $Rx/Rax=Rx/\ker\varphi\cong R/Ra$, because $\varphi$ is surjective.
On the other hand, going backwards is preferable. Consider $\psi\colon R\to Rx/Rax$ defined by $\psi(r)=rx+Rax$. This is obviously a surjective homomorphism. If $r\in\ker\psi$, then $rx\in Rax$, which means $rx=sax$, for some $s\in R$. Since $\operatorname{Ann}(x)=\{0\}$, we conclude that $r=sa\in Ra$. Conversely $Ra\subseteq\ker\psi$.
A: $\require{AMScd}\newcommand\Ker{\operatorname{Ker}}\newcommand\Ann{\operatorname{Ann}}$Let consider the canonical $R$-module homomorphism
\begin{align}
&\xi:R\to Rx&&r\mapsto rx
\end{align}
Then clearly, $\xi$ is surjective, and since $\Ker\xi=\Ann_R(x)=\{0\}$, $\xi$ is an isomorphism.
Let $\pi:R\to R/Ra$ be the canonical projection.
Then $\Ker(\pi\circ\xi^{-1})=\xi[Ra]=Rax$, hence $\pi\circ\xi^{-1}:Rx\twoheadrightarrow R/Ra$ factors through the projection $Rx\to Rx/Rax$ giving rise to an isomorphism $Rx/Rax\to R/Ra$ making the following diagram commutative:
\begin{CD}
R@>\xi >\sim >Rx\\
@V\pi VV @VVV\\
R/Ra@<<\sim < Rx/Rax
\end{CD}
A: Consider $f:R\rightarrow Rx/Rax$ defined by $f(u)=p(ux)$ where $p:Rx\rightarrow Rx/Rax$ is the quotient map, $f$ is  surjective. 
$f(a)=0$, $Kerf$ is an ideal, since $R$ is a $PID$, $Ker f=(b)$, we deduce that $a=cb$, and $(a)\subset (b)$. We also have $bx=dax$ since $f(b)=0$. This implies that $(b-da)x=0$ since $Ann(x)=0$, we deduce that $b=da$, $(b)\subset (a)$ and $(b)=(a)$.
