How to show that the norm of a fractional ideal is well-defined? Sorry. This might probably be a really easy question, but I am only a beginner in algebraic number theory. So, please bear with me.
Let $K$ be an algebraic number field and $\mathcal{O}_K$ its ring of integers.
Preludium


*

*In my lecture we have defined the norm $N_{K/\mathbb{Q}}(x)$ for an element $x \in K$ in the usual way. In a following lecture we "extended" this to the norm of an ideal $\mathfrak{a}$ of $\mathcal{O}_K$ by setting
$$ \mathfrak{N}(\mathfrak{a}) := [\mathcal{O}_K : \mathfrak{a}] := \left| \mathcal{O}_K / \mathfrak{a} \right|$$
and showed that for the principal ideal $(a)$ with $a \in \mathcal{O}_K$ we always have $\mathfrak{N}((a)) = \left|N_{K/\mathbb{Q}}(a)\right|$.

*Later, we introduced the notion of a fractional ideal in $K$ being a non-zero, finitely generated $\mathcal{O}_K$-submodule of $K$. And we showed that a non-zero $\mathcal{O}_K$-submodule $\mathfrak{a}$ of $K$ is a fractional ideal iff there is a $c \in \mathcal{O}_K \setminus \{0\} $ such that $c\mathfrak{a} \subseteq \mathcal{O}_K$ (thus $c\mathfrak{a}$ is an ideal of $\mathcal{O}_K$).

*We now extended the norm $ \mathfrak{N}$ by defining for each fractional ideal $\mathfrak{a}$ of $K$ the norm
$$ \mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c)\right|}, $$
$c$ being an element of $\mathcal{O}_K \setminus \{0\}$ such that $c\mathfrak{a} \subseteq \mathcal{O}_K$ (exists according to 2.).
Question
How can I show that this latter definition is independent of the chosen $c$?
Edit
I changed the definition in (3) from $ \mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{N_{K/\mathbb{Q}}(c)} $ to $\mathfrak{N}(\mathfrak{a}) := \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c)\right|} $
and added the absolute value everywhere else, too. I think that was a typo in my lecture notes.
Thoughts (Is it better to add them as comments? I was not sure.)


*

*Let $c, c' \in \mathcal{O}_K \setminus \{0\}$ such that $c \mathfrak{a} \subseteq \mathcal{O}_K $ and $c' \mathfrak{a} \subseteq \mathcal{O}_K$. At the moment I am contemplating if and in which way $c$ and $c'$ are related. I was wondering whether one always has $c'=bc$ (or vice versa) for an $b \in \mathcal{O}_K$. Then we had
$$ \frac{[\mathcal{O}_K : c'\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(c')\right|} = \frac{[\mathcal{O}_K : bc\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(bc)\right|} = \frac{[\mathcal{O}_K : bc\mathfrak{a}]}{\left|N_{K/\mathbb{Q}}(b)\right| · \left|N_{K/\mathbb{Q}}(c)\right|}$$
and it remained to show that $[\mathcal{O}_K : bc\mathfrak{a}] = \left|N_{K/\mathbb{Q}}(b)\right| · [\mathcal{O}_K : c\mathfrak{a}]$ (maybe because $[\mathcal{O}_K : bc\mathfrak{a}] = [\mathcal{O}_K : (b)] · [\mathcal{O}_K : c\mathfrak{a}]$ ???).

 A: Thanks to the hints in the comments, I think I now can summarize an answer myself. But I still need help on how to prove the isomorphy in point (2) below. Please still feel free to point out mistakes or to post alternative (easier?) answers.
Let $c,c' \in \mathcal{O}_K \setminus \{ 0 \}$ such that $c\mathfrak{a} \subseteq \mathcal{O}_K$ and $c'\mathfrak{a} \subseteq \mathcal{O}_K$.


*

*We have the chain $ c'c\mathfrak{a} \subseteq c\mathfrak{a} \subseteq \mathcal{O}_K $ of $\mathcal{O}_K$-modules. Because of the isomorphy
$$ \frac{\mathcal{O}_K / c'c\mathfrak{a} }{ c\mathfrak{a} / c'c\mathfrak{a}} \cong \mathcal{O}_K / c\mathfrak{a} $$
of quotient modules it follows that $[\mathcal{O}_K : c'c\mathfrak{a}] = [\mathcal{O}_K : c\mathfrak{a}] · [c\mathfrak{a} : c'c\mathfrak{a}]$ (note: all indeces involved are indeed finite – something I do not prove here).

*We have the isomorphy $ c\mathfrak{a} / (c') · c\mathfrak{a} \cong c\mathfrak{a} · (c\mathfrak{a})^{-1} / (c') · c\mathfrak{a} · (c\mathfrak{a})^{-1}$ (why?) of $\mathcal{O}_K$-modules and therefore
$$ c\mathfrak{a} / c'c\mathfrak{a} = c\mathfrak{a} / (c') · c\mathfrak{a} \cong c\mathfrak{a} · (c\mathfrak{a})^{-1} / (c') · c\mathfrak{a} · (c\mathfrak{a})^{-1} = \mathcal{O}_K / (c') · \mathcal{O}_K = \mathcal{O}_K / (c'). $$
This implies $[c\mathfrak{a} : c'c\mathfrak{a}] = [\mathcal{O}_K:(c')]$.

*By (1) and (2) we now get
$$ \frac{[\mathcal{O}_K : c'c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c'c) \right|} = \frac{[\mathcal{O}_K : c\mathfrak{a}] · [c\mathfrak{a} : c'c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c') \right| · \left| N_{K/\mathbb{Q}}(c) \right|} = \frac{[\mathcal{O}_K : c\mathfrak{a}] · \overbrace{[\mathcal{O}_K:(c')]}^{=\left| N_{K/\mathbb{Q}}(c') \right|}}{\left| N_{K/\mathbb{Q}}(c') \right| · \left| N_{K/\mathbb{Q}}(c) \right|} = \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c) \right|}$$

*Since we analogously (just switch the roles of $c$ and $c'$ in (1-3)) can show $ \frac{[\mathcal{O}_K : c'c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c'c) \right|} =  \frac{[\mathcal{O}_K : c'\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c') \right|}$ we get
$$ \frac{[\mathcal{O}_K : c\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c) \right|} = \frac{[\mathcal{O}_K : c'\mathfrak{a}]}{\left| N_{K/\mathbb{Q}}(c') \right|} $$
and therefore the norm $\mathfrak{N}(\mathfrak{a}) $ of a fractional ideal $\mathfrak{a}$ in $K$ is indeed well-defined.
