Are these quotient modules isomorphic?

Question

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 :-).

Answer 1

The answer is yes, but this is not trivial, and the isomorphism is not canonical (it depends on some choice). Actually this is true for every Dedekind domain $R$. Firstly, I state you a lemma:

Let $R$ be a Dedekind domain, $\mathfrak{a}, \mathfrak{b} \subset R$ ideals of $R$. Then there exists $\alpha \in \mathfrak{a}$ such that $$\alpha \mathfrak{a}^{-1} + \mathfrak{b} = R$$

Apply this lemma to $\mathfrak{a}, (c)$. So there exists $\alpha \in \mathfrak{a}$ such that $$\alpha \mathfrak{a}^{-1} + (c) = R$$ multiplying by $\mathfrak{a}$ we get the relation $$(\alpha ) + (c)\mathfrak{a} = \mathfrak{a}$$ from which we deduce that $\alpha \notin c\mathfrak{a}$.

Now, define the map $$\begin{matrix} f : &R& \to & \mathfrak{a} / c\mathfrak{a} \\ &x& \mapsto & \alpha x+ c\mathfrak{a} \end{matrix}$$ should be surjective, and the kernel should be $(c)$. Hence $f$ induces an isomorphism $$R/(c) \cong \mathfrak{a} / c\mathfrak{a} $$

Answer 2

Let $R$ be a Dedekind domain and $I,J$ two nonzero ideals of $R$. Then $$I/JI\simeq R/J.$$

In this answer I've shown that there is an ideal $I'$ such that $I\simeq I'$ and $I'+J=R$. But $I\simeq I'$ is equivalent to $\exists x\in Q(R)$, $x\ne 0$ such that $I'=xI$. (Here $Q(R)$ denotes the field of fractions of $R$.) From $xI+J=R$ we get $xI\cap J=xIJ$. Then $$R/J=(xI+J)/J\simeq xI/xI\cap J=xI/xIJ\simeq I/IJ.$$

Answer 3

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.