I'm reading Serre's local field, chapter 1 section 5, Norm and Inclusion Homomorphisms.
In particular, he defines the operation $\chi_A$ from the category of finite length $A$-modules to the collection of ideals of $A$ where $A$ is Dedekind domain.
Here goes: since $M$, a finite length $A$-module, admits composition serries $$0=M_0\subseteq...\subseteq M_m=M$$ with $M_i/M_{i-1}\cong A/p_i$ where $p_i$ is non-zero prime ideal of $A$. We define $$\chi_A(M)=\prod_{i=1}^mp_i$$ We know by Jordan-Holder Theorem this operation is well-defined.
Then he mentions an example as follows:
When $M=b/a$ where $a,b$ are fractional ideals with $a\subset b$, one has $\chi_A(M)=ab^{-1}$. In particular, $\chi_A(A/a)=a$ if $a\subset A$.
I want to have a better grasp on this concept so decided to word by word show the claim, starting with the particular case. Here is my attempt:
Let us first exam the claim $\chi_A(A/a)=a$. Note since $a\subseteq A$, it admits prime factorization, say $a=Q_1Q_2...Q_n$ where $Q_i$'s are all prime ideals with possible repetition, then we have the composition series $$0=a/a\subseteq Q_1Q_2...Q_{n-1}/a\subseteq Q_1...Q_{n-2}/a\subseteq...\subseteq Q_1/a\subseteq A/a$$ In particular, we claim that $(Q_1...Q_{k-1}/a)/(Q_1...Q_k/a)\cong A/Q_k$. Indeed, note $(Q_1...Q_{k-1}/a)/(Q_1...Q_k/a)\cong Q_1...Q_{k-1}/Q_1...Q_k$ and so it suffices to show that equal $A/Q_k$.
To show this, we claim that if $I$ and $J$ are distinct comaximal ideals of $A$, then $I/IJ\cong A/J$. Indeed, note the map $\phi:I\to A/J$ is surjective where $\phi(a)=a+J$. Indeed, take $x+J\in A/J$ be arbitrary, we have $x=i+j$ where $i\in I$ and $j\in J$, hence $x-i\in J$, i.e. $x+J=i+J$ and so $\phi(i)=x+J$ as desired. Now, it is evident that the kernel of this map is $IJ$, i.e. we indeed have $I/IJ\cong A/J$. Thus, if $Q_{k-1}$ and $Q_k$ are distinct, we have $Q_1....Q_{k-1}/Q_{1}...Q_{k-1}Q_k$ satisfy the claim in this paragraph, i.e. $Q_1...Q_{k-1}/Q_1...Q_k\cong A/Q_k$.
Next, we deal with $I$ and $J$ are the same. This is the same as showing $P/P^2\cong A/P$ where $P$ is prime. First, consider the localization at $P$, we have $PA_P$ becomes principal ideals, i.e. $PA_P=\pi A_P$ and so $P^2A_P=\pi^2 A_P$. On the other hand, consider the map $\phi:PA_P\to A_P/PA_P$ given by $x\mapsto \frac{x}{\pi}$ with the kernel equal $P^2A_P$.
Here I got stuck as not sure how to continue, I could show that $PA_P/P^2A_P$ can be embedded into the module $A_P/PA_P$ and I think $A_P/PA_P$ is isomorphic to $A/P$ based on arguments here. In addition, note since $A/P$ is a simple module while $PA_P/P^2A_P$ is a non-zero submodule, we must have $PA_P/P^2A_P$ equal $A/P$. However, even this, I cannot conclude $PA_P/P^2A_P$ is just the same as $P/P^2$, can we?
Thus, here goes my questions,
Q1: How could we show the two examples holds (the one I just showed partially and the more general case)?
Q2: Can someone tell me what this operation $\chi_A$ is? Also, if you could point out some text explain this and relative concepts I would be much appreciated.