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I'm trying to understand how this works in terms of Galois theory and local class field theory. Assume we have an extension of local fields $E/L/K$ s.t. $E/L$ and $L/K$ are abelian. I'm interested in recognizing when $E/K$ is Galois. Clearly, $E/K$ is Galois if and only if $E$ is always fixed by an extension of an $L/K$ automorphism to $E$, but this is tricky to compute.

I overheard a brief conversation that this can be done through Galois groups and some group actions that occur in the tower, but I haven't found anything explicit through google. We should be able to see from how $Gal(L/K)$ acts on something whether or not the extension is Galois. I'm having trouble seeing what the action should be. I hope someone who knows what I'm talking about could write it down explicitly. Since the Galois groups should correspond through local class field theory to very concrete objects which are quotients of $E^\times$, $L^\times$ and $K^\times$. I was wondering how this action on the Galois side is expressed on the local field side?

I'm interested in this since it clearly would provide a tool for constructing some solvable extensions of e.g. $\mathbb{Q}_p$. I apologize for being fuzzy, but I don't know how to be more explicit.

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If $E/K$ is Galois, then $\mathrm{Gal}(L/K)$ acts on $\mathrm{Gal}(E/L)$ by lifting to $\mathrm{Gal}(E/K)$ and conjugating (and this doesn't use that $L/K$ is abelian, just that $E/L$ is). –  Keenan Kidwell Jan 14 '12 at 0:34
    
Is there a way to translate this through the class field theory correspondence? Pick an $\alpha\in K^\times/N(L^\times)$ and assume you know what both $N(L^\times)$ and $N(E^\times)$ are? If $E/K$ is Galois, then $\alpha$ should act on the subgroup $L^\times/N(E^\times)\cong Gal(E/L)$ of $Gal(E/K)$. Is there a way to use this to recognize when an extension is Galois? –  asdf Jan 14 '12 at 1:20
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3 Answers

When you have a finite abelian extension $E|L$ of a finite galoisian extension $L|K$ of local fields, the extension $E|K$ is going to be galoisian if and only if the subgroup $N_{E|L}(E^\times)\subset L^\times$ is $Gal(L|K)$-stable.

This is somewhat similar to the following purely algebraic fact : Let $K$ be any field, $L$ a finite galoisian extension of $K$ of group $G=\mathrm{Gal}(L|K)$, and suppose that $L^\times$ has an element of order $n$ for some $n>1$. Then abelian extensions $E$ of $L$ of exponent dividing $n$ correspond bijectively to subgroups $H$ of $L^\times/L^{\times n}$ under the map $E\mapsto\mathrm{Ker}(L^\times/L^{\times n}\to E^\times/E^{\times n})$, the reciprocal being $H\mapsto L(\root n\of H)$ (Kummer theory). In such a situation, the abelian extension $E=L(\root n\of H)$ of $L$ is galoisian over $K$ if and only if the subgroup $H\subset L^\times/L^{\times n}$ is $G$-stable.

One has a similar (purely algebraic) statement in Artin-Schreier theory of abelian extension of exponent $p$ in characteristic $p$, and indeed in Witt's theory of abelian extensions of exponent dividing $p^n$.

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Would $Gal(L/K)$-stability be checked in practice, since we know $K^\times/N(L^\times)$ is isomorphic? Will functoriality translate the $Gal(L/K)$-action to an action by the elements of the group $K^\times/N(L^\times)$ on $L^\times$, since we want to check if $N_{E/L}(E^\times)$ is stable. I just want to see this at a level where it can be done in practice not just in principle. Thanks very much. –  asdf Jan 14 '12 at 15:01
    
More concretely. What does the action of $K^\times/N(L^\times)$ on $L^\times$ look like? How is it compute in practice? I'm pretty sure the functoriality would translate this into something that one can actually computed with. –  asdf Jan 14 '12 at 15:05
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Not an answer, but perhaps something useful from an apprentice in CFT:

Considerations very closely related to your question led Andre Weil to discover the "Weil group"; see his paper on Class Field theory (1955).

Let $G$ be the Galois group of $L/K$ and $H$ the Galois group of $E/L$.

The multiplicative group $L^*$is naturally a $G$-module and so one can consider the cohomology group $H^2(G, {L^*})$. Now $E^*$is not a $G$-module in any natural way. But it does give rise to a $G$-module which should be useful in answering your question.

Namely, the subgroup $NE^*$ of norms from $E$ to $L$ is a $G$-module. So one can consider the quotient $G$-module $M = {L^*}/{NE^*}$ and the cohomology group $H^2(G, M)$.

If $E/K$ is Galois, then its Galois group $\Gamma$ would be sit in an exact sequence $$0 \to H \to \Gamma \to G \to 0.$$ The conjugation action of $\Gamma$ on its normal subgroup $H$ (identified with its image in $\Gamma$) factorises via the quotient ${\Gamma}/{H} = G$. So this gives $H$ a $G$-module structure.

The group $\Gamma$ gives rise to an element $\gamma$ of $H^2(G, H)$ (via a standard construction). Local class field theory provides us with

  • a fundamental class $u_{L/K}$ in $H^2(G, {L^*})$, and
  • an isomorphism $H \cong M$.

The latter yields a map $L^* \to M \cong H$ of $G$-modules which provides a map $$t: H^2(G, L^*) \to H^2(G, H).$$

If $E/K$ were Galois, then the Galois group $\Gamma$ would have a constraint, namely, $$\gamma = t(u_{L/K}).$$

This discussion presupposes that $E.K$ is Galois and gives a constraint; it does not answer your question on how to check if $E/K$ is Galois!!

The fundamental class is what gives rise to the local Weil group.

By the way, every finite Galois extension of $Q_p$ (or a $p$-adic local field) is solvable.

All of this can be found in Cassels-Frohlich or Serre's Local Fields or Milne's notes on Class Field Theory.

Let us wait for a master in LCFT for an answer!!

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Thanks I should review group extensions, since they come in handy here as you point out. The easiest case would be looking at the smallest non-abelian group which is $D_3\cong S_3$. In this case we would have $L/K/\mathbb{Q}_p$, where the first extension is quadratic and the second cubic. $p=2,3$ arethe tricky cases since we can have wild ramification. However, with $p=2$ we only have $3$ extensions of $\mathbb{Q}_2$ that are non-Galois, since they are tamely ramified, so there's a unique $D_3$ extension. With $p=3$ one could fix a quadratic extension and then enumerate the cubics. –  asdf Jan 14 '12 at 2:03
    
Apparently, I can't edit what I wrote, since there are some typos. With $p=2$ there are $3$ degree $3$ extensions of $\mathbb{Q}_2$ that aren't Galois, so they must be conjugate and generate a unique $D_3$ extension. When $p=3$ we can fix a quadratic extension $K/\mathbb{Q}_3$ and then enumerate the cyclic cubic extensions of $K$. To construct a $D_3$ extension (or even generate all) one would have to recognize which of these towers won't give a Galois extension. –  asdf Jan 14 '12 at 2:11
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If $L/K$ is abelian and $K/k$ cyclic, then $L/k$ is going to be normal if the Galois group at the gottom acts on the class group attached to $L/K$ via class field theory, and it will be abelian if action fixes the classes. This is due to Hasse and follows easily from the standard results in class field theory. This is Prop. 1.2.8 in my survey on class field towers (I apologize for the outdated content. All of this needs to be rewritten).

There are similar constructions in Kummer theory going back to Kummer; this can be used in several proofs of the Kronecker-Weber theorem and should be e.g. in Washington's book; see also Prop. 2.2.1 in the survey

If the base extension is abelian and not cyclic, stuff happens. There are many articles investigating this problem, but nothing as simple as in the cyclic case.

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