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! 
 A: 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.
$\hspace{1.5cm}\blacksquare$
