The automorphism group of $E(\alpha)/F(\alpha)$ Suppose $E/F$ is a Galois extension. Let $\alpha$ be an (for simplicity, suppose $\alpha$ is algebraic over $F$) element of some extension $K$ of $E$. I have two questions:

*

*Is $E(\alpha)/F(\alpha)$ Galois?


*If the answer to the first question is affirmative, then what can we say about the Galois group of $E(\alpha)$ over $F(\alpha)$?
Here's the example I worked on to get a sense of this problem:
Let $F = \mathbb{Q}$ and $E = \mathbb{Q}(\sqrt[3]{2}, \omega)$, where $\omega$ is a third root of unity. $E/F$ is a Galois extension since $E$ is the splitting field of $x^3-2$ over $F$. I took $\alpha = i$ and managed to show that $E(\alpha)/F(\alpha)$ is Galois, but I had to use a lot of properties that are specific to this example and did some tedious calculations, so I couldn't generalize the result. I also proved that the Galois group $\operatorname{Gal}(E(\alpha)/F(\alpha)) \cong S_3$ just like $\operatorname{Gal}(E/F)$. Is this a coincidence, or does it hold in general? (Hence my second question.)
 A: Apologies if you already know what follows.
Let us assume for simplicity that all the extensions have finite degree.
It holds in general that, if $L/F$ is Galois and $K/F$ is arbitrary, $LK/K$ is Galois. You cannot say much of the Galois group of $LK/K$ in general.
However, one may prove that if $L/F$ is Galois, then $Gal(L/F)\simeq Gal(LK/K) \iff L\cap K=F$.
The main ingredient to prove that  the notion of linearly disjoint extensions. Two extensions $E_1/F$ and $E_2/F$ sitting in a big extension $\Omega/F$ are linearly disjoint if any $F$-basis of $E_1$ is an $E_2$-basis of $E_1E_2$ (I slightly modified the official definition because we work with finite extensions. The more general setting is a bit more subtle). One may show that the definition is symmetric. In this case, we have $E_1\cap E_2=F$.
In general, this necessary condition is not sufficient, but one may show that if at least one of the extensions is Galois, then it is also sufficient.
The result above follows more or less directly from the fact that you have linear disjunction.
This is the case in your example (it takes a bit of work to check that the intersection is $\mathbb{Q}$ though).
You should find all definitions and proofs in any good book on Galois theory.
