$Rx/R\alpha x $ is isomorphic to $R/\alpha R$ Where $R$ is a principal ideal and $x$ is a basiselement in the $R$ module $M=R^d$ where $d\in\mathbb{N}$ and $\alpha$ is a non-unit and not zero and a element in $R$. With $Rx$ defining the subideal in $M$ for which we have $Rx=\{a\in M ; a=rx , r\in R\}$ and similarily  $R\alpha x=\{a\in M ; a=r\alpha x , r\in R\}$ and $\alpha R=\{b\in R:b=r\alpha\}$ an ideal in $R$. I am looking for an explicit isomorphism.
Note:
A basis of a $R$ module $M$ is a family of elements  $(a_i)$ that constructs the ring, i.e $\sum Ra_i=M$ such that every such sum has a unique representation. One can prove that if a module has a finite basis then every other basis must have the same cardinality.
For context I refer to this question:
The torisionmodule of a $R$ module (where $R$ is PID) is isomorphic to a direct sum of ideals $\bigoplus_{j=1}^sR/\alpha_j R$
 A: No.
For example let $R = \mathbb Z^2$  and $x=(0,0)$ and $\alpha = (0,1)$. 
Then $Rx = R \alpha x  = Rx/ R \alpha x = \{0\}$. 
but $\alpha R = \{0\} \times \mathbb Z$ and so $R/\alpha R = \mathbb Z \times \{0\}$.
Edit: If you want $x$ to be a basis element of $R$ then take   $R = \mathbb Z^3$  and $x=(1,0,0)$ and $\alpha = (0,0,1)$. Then we have $\alpha x=(0,0,0)$ and $$Rx = \mathbb Z \times \{0\} \times \{0\} \qquad R\alpha =  \{0\} \times \{0\}\times \mathbb Z$$
$$R\alpha x =  \{0\} \times \{0\}\times \{0\} $$and
$$\frac{Rx}{R \alpha x} =Rx = \mathbb Z \times \{0\} \times \{0\}  \qquad \frac{R }{R  x}  = \{0\}\times \mathbb Z \times \mathbb Z$$
A: Since $x$ is a member of a basis of a free module, you know that the map $\mu_x\colon R\to Rx$, $\mu_x(r)=rx$ is an isomorphism (only that $x$ has no torsion, that is, zero annihilator, would be sufficient).
Consider the canonical map $\pi\colon Rx\to Rx/R\alpha x$; then the kernel of $\pi\circ\mu_x$ is
$$
\ker(\pi\circ\mu_x)=\{r\in R:rx\in R\alpha x\}
$$
Clearly $R\alpha$ is contained in $\ker(\pi\circ\mu_x)$. Suppose $rx\in R\alpha x$, that is, $rx=s\alpha x$, for some $s\in R$.
Since $x$ has no torsion, this implies $r=s\alpha$, so $r\in R\alpha$.
