Let $F \supset E \supset K$, $E/K$ and $F/E$ Galois and every $\sigma \in \mathrm{Aut}_K(E)$ extendible to $F$. Then $F/K$ is Galois.

Question

Let $F$ be an extension field of a field $K$. Let $E$ be an intermediate field such that $E$ is Galois over $K$, $F$ is Galois over $E$, and every $\sigma\in\mathrm{Aut}_{K}E$ is extendible to $F$. Show that $F$ is Galois over $K$.

The definition of a Galois extension $F/K$ in the book is that the fixed field of $\mathrm{Aut}_{K}F$ is $K$. So I just need to show that $\mathrm{Aut}_{K}F$ does not fix elements of $F\setminus K$. If $x\in E\setminus K$, then since $E/K$ is Galois, there exists $\sigma\in\mathrm{Aut}_{K}E$ such that $\sigma(x)\neq x$. By the hypothesis, $\sigma$ extends to $F$. If $x\in F\setminus E$, then since $F/E$ is Galois, there exists $\tau\in\mathrm{Aut}_{E}F$ such that $\tau(x)\neq x$ and also $\tau\in\mathrm{Aut}_{K}F$. It seems that this completes the argument, am I right?

Answer 1

Suppose $x$ is in the fixed field of $\mathrm{Aut}_KF$. Since $F/E$ is Galois, $x\in E$. Every $\sigma\in \mathrm{Aut}_KF$ restricts to an element $\tilde{\sigma}$ of $\mathrm{Aut}_KE$. Furthermore, our assumption about $K$-linear automorphisms of $E$ lifting to $F$ implies that the restriction map $\rho:\mathrm{Aut}_KF \rightarrow \mathrm{Aut}_KE$ is surjective. Thus, $x\in E$ is fixed by all elements of $\mathrm{Aut}_KE$, hence is an element of $K$.