composing Galois extensions

Question

If $k\to E$ is a Galois extension with Galois group $G$ and $E\to F$ is the separable closure of $E$ then is $F$ the separable closure of $k$? If $\mathrm{Gal}(F/E)$ is the absolute Galois group of $E$ and $G$ is the Galois group of $k\to E$ then what's the Galois group of $k\to F$? Is it $\mathrm{Gal}(F/E)\times G$?

Answer 1

The answer to your first question is yes. We need to prove that $F$ is a separably closed field that is separable over $k$. But we know it's separably closed because it's assumed to be a separable closure of $E$, and we know it's separable over $k$ because a separable extension of a separable extension is separable. (Edited: changed the word "algebraic" to "separable".)

For your second question, we do get that $\mathrm{Gal}(F/k)/\mathrm{Gal}(F/E) \cong \mathrm Gal(E/k)$ (with the quotient map given by restricting automorphisms of $F$ to automorphisms of $E$), but it doesn't need to be a direct product, or even a split extension. Here's an example, with caveat below:

Let $k = \mathbb Q$, $E$ a cubic Galois extension, and $F = \overline{\mathbb Q}$. Then $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$ is an incredibly complicated object, and it is too optimistic to hope that it will decompose as a direct (or semidirect) product of $G = \mathrm{Gal}(E/\mathbb Q) \cong \mathbb Z/3\mathbb Z$ and $\mathrm{Gal}(\overline{\mathbb Q}/E)$. If it were, then the fixed field of $G$ would be a degree-3 subextension of $\overline{\mathbb Q}$. But the Artin-Schreier theorem tells us that the only finite-degree subextensions of algebraically closed fields have degree 2, a contradiction.

Caveat: when studying infinite Galois extensions, it's more natural to study the Galois group not just as a group, but as a topological group, endowed with its profinite (Krull) topology. This is because in the infinite case, the only subgroups of Galois groups that correspond to intermediate fields are the closed subgroups. So when I talked about the fixed field of $G$ above, I really meant the fixed field of the topological closure of $G$ in $\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$. Fortunately, profinite groups are Hausdorff spaces, so finite subgroups are necessarily closed, and there's nothing to worry about here.

Answer 2

Yes $F$ is the separable closure of $k$. Separable closures are themselves separably closed, so $F$ is separably closed. It's also separable over $k$ since it's Galois over $E$ and $E$ is Galois over $k$, and hence is the separable closure of $k$.

The Galois group of $F/k$ need not be a direct product. In general you have an exact sequence:

$$1\rightarrow Gal(F/E)\rightarrow Gal(F/k)\rightarrow Gal(E/k)\rightarrow 1$$

which need not be split. For example, you can let $k = \overline{\mathbb{Q}}(x)$, then its absolute Galois group is the free profinite group of countable rank. Such a group cannot be a nontrivial direct product by Proposition 8.7.7 of Ribes/Zalesskii's book Profinite Groups.