Cancellation in quotient of fractional ideals When reading about fractional ideals of rings of integers, I came upon the following footnote:

For fractional ideals $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ with $\mathfrak{a} \supset \mathfrak{b}$, $$\displaystyle ^{\mathfrak{a}\mathfrak{c}}/_{\mathfrak{b}\mathfrak{c}} \simeq \ ^{\mathfrak{a}}/_{\mathfrak{b}}$$ as $\mathcal{O}_K$-modules.

This was not obvious to me, so I tried to prove it, however did not succeed. I think it must be connected to the unique product decomposition in Dedekind domains. I also found this question, where someone was also not sure how to prove this isomorphism, but did not succeed either.
Any help is greatly appreciated! Thanks in advance! 

Here is the main statement:

Proposition: Let $R$ be a Dedekind domain (for instance the ring of integers $R=\mathcal O_K$ of some finite extension $K$ of $\Bbb Q$).
  Let $\mathfrak{a}$, $\mathfrak{b}$ and $\mathfrak{c}$ be non-zero fractional ideals of $R$ with with $\mathfrak{a} \supset \mathfrak{b}$.
  Then there is an isomorphism of $R$-modules $$\dfrac{\mathfrak{a}\mathfrak{c}}{\mathfrak{b}\mathfrak{c}} \cong \dfrac{\mathfrak{a}}{\mathfrak{b}}.$$

We first need a little result:

Lemma: let $I,J$ be non-zero integral ideals of a Dedekind domain $R$. Then there is $x \in \mathrm{Frac}(R)$  such that $xI+J=R$.

This lemma is proved here ; we just take $x:=b/a \in \mathrm{Frac}(R)$ from the proof there.

Proof of proposition.
— Since $\frak a, b, c$ are fractional ideals, there are elements $\alpha, \beta, \gamma \in \mathrm{Frac}(R)$ such that 
$\alpha \frak a, \beta \frak b, \gamma \frak c$ are all integral ideals of $R$.
We clearly have isomorphisms of $R$-modules
$$
\dfrac{\frak a}{\frak b} \cong \dfrac{\alpha \beta \frak a}{\alpha \beta \frak b},
\qquad
\dfrac{\frak ac}{\frak bc} \cong \dfrac{\alpha \beta \gamma \frak ac}{\alpha \beta \gamma \frak bc}
$$
We may therefore assume that all of $\frak a, b, c$ are integral ideals (i.e. contained in $R$) : replace them by $\alpha \beta \frak a, \alpha \beta \frak b, \gamma \frak c$ respectively.
— Then, by the lemma, there is $x \in \mathrm{Frac}(R)$  such that  $x \mathfrak{c}^{-1} + \mathfrak{ba}^{-1} = R$, i.e.
$$x \frak{a + bc = ac}.$$
Consider the morphism
$f :  \frak{a} \to \frak{ac/bc}$ be defined by
$f(r) = [xr]$. It is clearly surjective. Moreover, the kernel of $f$ is
$$\mathfrak{a} \cap x^{-1}\mathfrak{bc} = (\mathfrak{c}^{-1}    \cap    x^{-1}\mathfrak{ba}^{-1}) \mathfrak{ac}   =    x^{-1}(x\mathfrak{c}^{-1} \mathfrak{ba}^{-1})\mathfrak{ac} = \frak b$$
The first equality comes from (13) in this answer and the second comes from (14), using $x \mathfrak{c}^{-1} + \mathfrak{ba}^{-1} = R$ (notice that the answer holds in our setting with fractional ideals of the Dedekind domain $R$, not only for integral ideals in $R$).
As a conclusion, $f$ induces an isomorphism of $R$-modules $\frak{a/b \cong ac/bc}$ as claimed.
$\hspace{1.5cm}\blacksquare$
