Let $K_0\subseteq F$ be an arbitrary field extension and $K$ be the fixed field of ${\rm Aut}_{K_0}(F)$, where $\mbox{Aut}_{K_0}(F):=(K_0)':=\{\sigma\in \mbox{Aut}(F) \,: \sigma|_{K_0}=\mbox{id}_{K_0}\}$ and $K:=(K_0)'':= \{v\in F:f(v)=v,\forall f\in \mbox{Aut}_{K_0}(F)\}$.

Similarly, we define $K':=\{\sigma\in \mbox{Aut}(F) \,: \sigma|_{K}=\mbox{id}_{K}\}$ and $K'':=\{v\in F:f(v)=v,\forall f\in \mbox{Aut}_{K}(F)\}$.

We say $F$ is Galois over $K$ if $K''=K$, and $F$ is Galois over $K_0$ if $(K_0)''=K_0$.

My book says that,

$F$ is Galois over $K$, $K_0\subseteq K,$ and Aut$_K(F)=\mbox{Aut}_{K_0}F$.

I have difficulty in showing that $F$ is Galois over $K$ and $K_0$.

$\color{darkred}{\mathbf{Edit:}\mbox{ I found the words in the book actually doesn’t mean that }F \mbox{ is Galois over } K_0. “K_0\subseteq K”\mbox{ just simply means } “K_0\subseteq K” …}$

Here is my attempt:

$K\subseteq K''$ is trivial because every $f\in \mbox{Aut}_K(F)$ fixes $K$. Similarly, we have $K_0\subseteq (K_0)''=K$.

Let $v\in K''$. Then $\forall g\in \mbox{Aut}_{K}(F), g(v)=v.$ In order to show $K''\subseteq K$, we need to show that $\forall f\in \mbox{Aut}_{K_0}(F)$, $f(v)=v.$ Equivalently, we need to show that $\mbox{Aut}_{K_0}(F)\subseteq \mbox{Aut}_{K}(F)$, which I cannot prove.

So, my question is how to show that $F$ is Galois over $K$ and $K_0$. Thanks for help.

  • 1
    $\begingroup$ You are probably assuming that $F/K_0$ is a finite extension? $Aut(F/K_0)$ needs to be a finite group, or at least every element $a\in F$ must have a finite orbit. Unclear which definition of Galois you are using. It is quite easy to show that $K/F$ is normal and separable. It remains to check that it is a finite extension. $\endgroup$
    – reuns
    Jan 2 at 19:18
  • $\begingroup$ @reuns Hmm, the book doesn't require the extension to be finite. I have written the definition of Galois of the book in the description of the OP. Nevermind, I found out a way to solve this. It is just a game of words. I shall keep this OP here in case some other future learners face the same problem. $\endgroup$
    – Sam Wong
    Jan 2 at 19:55

1 Answer 1


I suddenly understood the logic behind it. It is just a game of words.

Let us prove $\mbox{Aut}_{K_0}(F)\subseteq \mbox{Aut}_K(F)$.

Let $\varphi \in \mbox{Aut}_{K_0}(F)$. If we can show that $\varphi$ fixes every point in $K$, then $\varphi \in \mbox{Aut}_K(F)$.

Let $k\in K$. Recall that $K:=(K_0)'':= \{v\in F:f(v)=v,\forall f\in \mbox{Aut}_{K_0}(F)\}$.

Therefore, $\varphi(k)=k$, which implies that $\varphi$ fixes every point in $K$.

By the reasoning I wrote in the OP, we can conclude that $F$ is Galois over $K$.


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