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I've been dealing with the following situation:

Let $R\subseteq S$ be an extension of Dedekind rings, where $Quot(R)=:L \subseteq E:=Quot(S)$ is a $G$-Galois extension. Let $\mathfrak{p}$ be a prime of $R$, and $\mathfrak{q}$ a primes of $S$ above $\mathfrak{p}$. Let $D_{\mathfrak{q}}$ denote the decomposition group, and $I_{\mathfrak{q}}$ the inertia group, of $\mathfrak{q}$ over $\mathfrak{p}$.

However, unlike in the classic case, I allow the residue fields of $\mathfrak{p}$ to be infinite, with positive characteristic. So the extension of residue fields may be inseparable.

It seems that the paper I'm reading implicitly assumes:

$|I_{\mathfrak{q}}|=e[\kappa(\mathfrak{q}):\kappa(\mathfrak{p})]_ i $ (the ramification index times the inseparability degree of the residue extension)

$|D_{\mathfrak{q}}|=e[\kappa(\mathfrak{q}):\kappa(\mathfrak{p})]$

$|G|=re[\kappa(\mathfrak{q}):\kappa(\mathfrak{p})]$ (where $r$ is the number of primes above $\mathfrak{p}$)

Is that right? I keep hitting walls when I try to prove it.

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1 Answer 1

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I believe that's right, at least when $S$ is finitely generated over $R$. See Serre's Local Fields page 21-22 (in the English translation); he states his assumptions on page 13.

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  • $\begingroup$ Serre proves that the module-finiteness condition for integral closure holds when the extension of fraction fields is separable, so in the context of the question $S$ is automatically module-finite over $R$. These matters can also be considered from the viewpoint of valuation theory, allowing for the possibility of non-archimedean valuations which are not discretely-valued. See 3.6/5,6 in "Non-archimedean analysis" by Bosch, Guntzer, Remmert. $\endgroup$
    – BCnrd
    Mar 31, 2010 at 0:18

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