# Fixed field of an automorphism group is Galois

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.

• 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. Jan 2 at 19:18
• @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. Jan 2 at 19:55

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 $$K$$.