# 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.

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?

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 $K$.