TITLE: Are these quotient modules isomorphic?
QUESTION [6 upvotes]: Let $K$ be an algebraic number field and $\mathcal{O}_K$ its ring of integers. For a non-zero ideal $\mathfrak{a}$ of $\mathcal{O}_K$ and an element $c \in \mathcal{O}_K \setminus \{0\}$ I wonder whether we always have an isomorphism
   $$ \mathfrak{a} / c \mathfrak{a} \cong \mathcal{O}_K / (c) $$
  as $\mathcal{O}_K$-modules. 

Using the inverse (fractional) ideal $\mathfrak{a}^{-1}$, one could naïvely "calculate"
$$ \mathfrak{a} / (c) \mathfrak{a} \cong \mathfrak{a}\mathfrak{a}^{-1} / (c)\mathfrak{a}\mathfrak{a}^{-1} = \mathcal{O}_K / (c)\mathcal{O}_K = \mathcal{O}_K / (c). $$
But (again) I do not know how to justify the isomorphism.
Additonal question
Is it easier to somehow only show $ [\mathfrak{a} : c \mathfrak{a}] = [\mathcal{O}_K : (c)]$ ? This would help me, too :-).

REPLY [2 votes]: Here is another proof for the more general $I/IJ \cong R/J$ proposed in the answer above:
The statement certainly holds if $I$ is principal. Locally, $R$ is a PID, hence the statement is true after localization at any maximal ideal.
$R/J$ is artinian, hence semi-local. Over a semi-local ring, we can test two finitely-presented modules to be isomorphic by testing them to be isomorphic locally (Without any global homomorphism given!). This is Exercise 4.13 in Eisenbud's "Commutative Algebra with a View...".
Thus $R/J \cong I/IJ$ as $R/J$-modules and subsequently also as $R$-modules.