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For the easiest case, assume that $L/E$ is Galois and $E/K$ is Galois. Under what conditions can we conclude that $L/K$ is Galois? I guess the general case can be a bit tricky, but are there some "sufficiently general" cases that are interesting and for which the question can be answered?

EDIT: Since Jyrki's reply seems to suggest that there is no general criterion on the groups. Can we say something if we put criterions on the fields? Assume say that $K=\mathbb{Q}$ or $K=\mathbb{Q}_p$?

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What kind of criterion are you looking for? For example, assuming that the extensions in question are finite, there cannot be such a criterion in terms of the two Galois groups. This follows from the fact that given two finite groups $H$ and $K$, it is easy to construct a chain of finite groups $N_2\le N_1\le G$ such that $N_1$ is normal in $G$, $N_2$ is normal in $N_2$, $G/N_1\simeq H$, $N_1/N_2\simeq K$, but $N_2$ is not normal in $G$. Realize $G$ as a group of automorphisms of a field $F$, and let $L,E,K$ be the respective fixed fields of $N_2,N_1,G$. Then $L/K$ is not Galois. – Jyrki Lahtonen Dec 4 '11 at 9:58
@Jyrki: I guess the answer is then that no such conditions exist. – pki Dec 4 '11 at 18:33

This isn't an answer, nor is it very general, just an illustration with an example. Let $K$ be a number field, $E/K$ finite Galois, and $L/E$ the Hilbert class field of $E$ (the maximal unramified abelian extension of $E$, which I'll take to be inside some algebraic closure $\overline{K}$ of $K$). Then $L/E$ is Galois by definition, and using its characterizing property, it can be proved that $L/K$ is Galois. Namely, let $\sigma:L\rightarrow\overline{K}$ be a $K$-monomorphism. Then $\sigma(E)\subseteq E$ because $E/K$ is Galois, so $\sigma(E)=E$. Thus $E\subseteq \sigma(L)$, and $\sigma$ sets up an isomorphism between the extensions $L/E$ and $\sigma(L)/E$. I don't mean that $L$ and $\sigma(L)$ are $E$-isomorphic, because, while $\sigma(E)=E$, it needn't be the case that $\sigma$ fixes $E$ pointwise. But, for example, $\sigma(L)/E$ must be Galois, and we have an isomorphism $\mathrm{Gal}(L/E)\cong\mathrm{Gal}(\sigma(E)/E)$ induced by $\sigma$. One can also verify that $\sigma(L)/E$ is unramified because $L/E$ is. So $\sigma(L)/E$ is abelian and unramified. This means $L\sigma(L)$ (compositum inside $\overline{K}$) is abelian and unramified over $E$. By maximality, $L\sigma(L)=L$, which implies that $\sigma(L)=L$. So we can conclude that $L/K$ is Galois.

I don't really know if there is a way to make this formal. Similar arguments can be used to prove that various class fields of $E$ are Galois over $K$. I guess the informal idea is that $L/E$ should be maximal (inside $\overline{K}$) with respect to some properties which are preserved by $K$-embeddings of $L$ into $\overline{K}$ (which necessarily send $E$ onto itself) and which are preserved under compositum. I admit this is vague, but I'm not sure how to make it more precise.

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If every $K$-automorphism of $E$ can be extended to a $K$-automorphism of $L$, then $L/K$ is Galois.

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Always. Galois = normal + separable. A tower of normal extensions is normal, and a tower of separable extensions is separable.

Edit: That's wrong. Separability is transitive, but not normality. See comments below.

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I agree with your statement about separable extensions, but the annoying thing about normality is that it doesn't have this property. Consider $\mathbf{Q} \subset \mathbf{Q}(\sqrt{2}) \subset \mathbf{Q}(\sqrt[4]{2})$, for example. – Dylan Moreland Dec 4 '11 at 1:45
Yes, normality is the villain. A more general question would just be when a tower of normal extensions gives a normal extension. At least for perfect fields this is equivalent to my original question. – pki Dec 4 '11 at 2:00

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