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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?

cyc
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  • Suppose $x$ is in the fixed field of $\mathrm{Aut}KF$. Since $F/E$ is Galois, $x\in E$. Let $\sigma\in\mathrm{Aut}{K}E$. By extending $\sigma$ to $\mu\in\mathrm{Aut}_{K}F$, we find that $\mu(x)=x$. But $x\in E$, so $\sigma(x)=\mu(x)=x$. Since $E/K$ is Galois, we have $x\in K$. – 19021605 Jun 29 '25 at 07:27

1 Answers1

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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 $K$.

tghyde
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    Why is it true that every $\sigma\in \mathrm{Aut}_KF$ restricts to an element $\tilde{\sigma}$ of $\mathrm{Aut}_KE$? I don't see a reason for $\sigma(E)\subseteq E$. – Daniel Teixeira Feb 13 '23 at 19:57
  • @DanielTeixeira In the case of finite extension $F/K$, $E/K$ is Galois iff $E$ is the splitting field of a separable polynomial in $K[x]$. An argument similar to the one here: https://math.stackexchange.com/questions/3265327/restriction-of-element-in-galois-group can be used to show $\sigma(E)=E$. – 19021605 Jun 29 '25 at 07:22