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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 :-).

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    $\begingroup$ for sure they are when $K$ has class number $1$, since if $\mathfrak a$ is principal then $\mathcal O_K\simeq \mathfrak a$ as $\mathcal O_K$-modules. $\endgroup$ – Ferra Aug 21 '15 at 13:09
  • $\begingroup$ It may help to note that if $\mathfrak a$ and $(c )$ are coprime, then this follows from the Chinese Remainder Theorem. $\endgroup$ – Mathmo123 Aug 21 '15 at 13:47
  • $\begingroup$ May I also invite you to have a look at point (2) of my answer (to my own question) over here? Although this being of own interest to me, that is the original context where the question arose. Maybe it is even easier to answer it in that particular setting? $\endgroup$ – puck29 Aug 21 '15 at 16:00
  • $\begingroup$ @Mathmo123: If we had $c \notin \mathfrak{a}$, could we somehow deduce $\mathfrak{a}$ and $(c)$ being coprime? $\endgroup$ – puck29 Aug 21 '15 at 16:59
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    $\begingroup$ @puck29 not necessarily. For example, in $\mathbb Z$, $6\notin (9)$. However, if $\mathfrak a$ is prime, then this will be true. $\endgroup$ – Mathmo123 Aug 21 '15 at 17:24
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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 $$R/(c) \cong \mathfrak{a} / c\mathfrak{a} $$

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  • $\begingroup$ Thanks @Crostul :-). I still have to understand some things. It's clear to me why $f$ is surjective. As for the kernel, at the moment I only get $(c) \subseteq \ker(f)$. But why does $\alpha x \in c\mathfrak{a}$ already imply $x \in (c)$? $\endgroup$ – puck29 Aug 21 '15 at 15:37
  • $\begingroup$ By the way: When deducing $ \alpha \notin c\mathfrak{a} $ we assume w.l.o.g. that $c$ is not a unit of $\mathcal{O}_K$, right? And that's okay, because otherwise my question would be trivial ;-). $\endgroup$ – puck29 Aug 21 '15 at 16:05
  • $\begingroup$ I also would be interested to read a proof of the stated lemma later. Maybe someone can point me to a good source? $\endgroup$ – puck29 Aug 21 '15 at 17:14
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    $\begingroup$ @puck29 : see e.g. lemma 1.8 in Introduction to Algebraic K-theory (Milnor)... $\endgroup$ – Watson Feb 16 '18 at 10:21
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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 $$R/J=(xI+J)/J\simeq xI/xI\cap J=xI/xIJ\simeq I/IJ.$$

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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.

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