I have a problem in understanding the inertia group of an infinite extension. I am studying it in this context.

Let $K$ be a field, $v$ a discrete valuation on $K$, and $\mathcal{O}_v$ the discrete valuation ring of $v$. Let $k$ be the residue field of $v$, we assume $k$ to be perfect. Let $K_s$ be a separable closure of $K$ and $\bar{v}$ an extension of $v$ to $K_s$. The discrete valuation ring of $\bar{v}$ will be $\mathcal{O}_\bar{v}$ and the residue field $\bar{k}$.

Then $K_s/K$ is a galois extension with galois group $G=Gal(K_s/K)$

Now I see two ways to define the inertia group.

  1. The subgroup $I \subset G$ given by the elements $\sigma\in G$ such that $$\bar{v}(\sigma(a)-a)\geq0$$ for every $a\in \mathcal{O}_\bar{v}$.
  2. The inverse limit of the inertia groups $$I=\varprojlim I_{F/K}(v_F)$$ where the inverse limit is taken over the finite galois extensions $F/K$, and $v_F$ is the restriction of $\bar{v}$ to $F$.

I am convinced that these two definitions are equivalent.

Now comes my actual question: is it true that the subfield of $(K_s)^I\subset K_s$ given by the elements fixed by the action of $I$ is the maximal unramified extension of $K$ contained in $K_s$? In case, how is it proved?


1 Answer 1


So it's been a while since I've thought about valuations, but this is too long for a comment anyway.

Firstly, I don't think $\overline{\nu}$ is a discrete valuation. For example, if $\pi$ is the uniformizer of $K$, then $\sqrt[n]{\pi}$ should have valuation $1/n$, so there isn't any "minimum positive valuation", which is what you'd want your uniformizer to have.

However, you can still define the valuation ring as the set of elements of $K_s$ for which $\overline{\nu}(x) \ge 0$, in which case you should get a non-noetherian integral domain of dimension 1, and where you can think of the maximal ideal as being determined by a compatible inverse system of uniformizers for $F/K$ as $F$ ranges over all finite separable extensions of $K$.

Also, I think in your first definition of the inertia group, you want $\overline{\nu}(\sigma(a) - a) > 0$. (ie, in the case of finite extensions you want the elements of the inertia group to fix every element mod the uniformizer), so the inverse limit should correspond to "$> 0$", since you want the valuation of $\sigma(a)-a$ to be 0 mod every uniformizer in the "compatible inverse system" describing your maximal ideal of $\mathcal{O}_{\overline{\nu}}$.

Anyway, the answer to your question I think is yes, and should come from the fact that the galois connection makes the fixed field of your inverse limit just the direct limit of fixed fields of the individual inertia groups.

This follows from the fundamental theorem of (infinite) galois theory, which says that intermediate extensions of say $K_s/K$ correspond to closed subgroups of $G$, and that this gives a contravariant equivalence of categories from the category of subextensions of $K_s/K$ and closed subgroups of $G$, with all morphisms given by inclusions. Let $E$ denote the inverse of this equivalence (which associates to any closed subgroup the corresponding fixed field), then you have $$E(I) = E(\lim_{\leftarrow} I_{F/K}(\nu_F)) = \lim_{\rightarrow} E(I_{F/K}(\nu_F))$$ where the rightmost thing is a direct limit (ie, union/composite) of all the maximal unramified subextensions of $F/K$ as $F$ ranges over finite galois extensions of $K$.

  • $\begingroup$ I totally agree on the mistakes I did. Thank you for the corrections and the answer. $\endgroup$
    – Pgatti
    Jun 7, 2014 at 9:37

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