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$A$ is a Dedekind domain.

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How to compute $\chi_A (M)$ in the above example? What's the series using?

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    There is something that bugs me about this, $\chi$ is a product of prime ideals and thus an ideal of $A$, but $\mathfrak{a}\mathfrak{b}^{-1}$ in the case where $\mathfrak{a}$ and $\mathfrak{b}$ are ideals will not be an ideal of $A$, only a fractional ideal. – 2017-01-22
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    I didn't notice it and it also bugs be now, It is excerpted from p16 of Serre's Locak Fields. – 2017-01-22
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    You know a lot of the stuff from that era is just not that clearly written. Anyway I think the dot there is important I think he means $\chi=[\mathfrak{b}:\mathfrak{a}]=\{a|a\mathfrak{b}\subseteq \mathfrak{a}\}$. More.... – 2017-01-22
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    Assume that $\mathfrak{b}/\mathfrak{a}$ is simple the it is $A/\mathfrak{p}$ so what is $\mathfrak{p}$, it is (I think) $[\mathfrak{b}:\mathfrak{a}]$. If you could prove that $[\mathfrak{c}:\mathfrak{b}][\mathfrak{b}:\mathfrak{a}]=[\mathfrak{c}:\mathfrak{a}]$ you would be done. Although I dont see the proof at the moment. – 2017-01-22
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    OK so in fact $[\mathfrak{b}:\mathfrak{a}]=\mathfrak{a}\mathfrak{b}^{-1}$. You can find this in Zariski and Samuel Commutative algebra I page 274. (note that the notation there changes the order from mine). – 2017-01-22

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Since $A$ is a Dedekind domain the fractional ideals are a group and remembering that $A=(1)$ we have that

$$\frac{\mathfrak{b}}{\mathfrak{a}}=\frac{A}{\mathfrak{a}\mathfrak{b}^{-1}}$$

This we take a maximal chain $$\mathfrak{a}=\mathfrak{a}_0\subseteq \mathfrak{a}_1 \subseteq \cdots \mathfrak{a}_n=\mathfrak{b}$$ and by definition

$$\chi(\mathfrak{b}/\mathfrak{a})=\prod \mathfrak{a}_i\mathfrak{a}_{i+1}^{-1}=\mathfrak{a}\mathfrak{b}^{-1}$$

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    Well I may be a little responsible, if $\mathfrak{a}$ and $\mathfrak{b}$ are ideals of $A$ then yeah its a problem however that case does not arise since $\mathfrak{b}/\mathfrak{a}$ can never be simple ( there are always higher powers of the primes in $\mathfrak{b}$ as submodules, or better multiplication by a new prime). So for the simple case to arise you need fractional ideals and so then you do get an ideal in $A$. – 2017-01-22