$G$ be a finite subgroup of $Aut(K)$ , $F:=\{x\in K : f(x)=x , \forall f \in G\}$ . Then is $[K:F]$ finite ?

Question

Let $K$ be a field , $Aut(K)$ be the group of all field homomorphisms (sending unity to unity) from $K$ to $K$ , let $G$ be a finite subgroup of $Aut(K)$ , let $F$ be the fixed field of $G$ i.e. $F:=\{x\in K : f(x)=x , \forall f \in G\}$ . Then is it true that $K/F$ is a finite extension of fields ?

Answer 1

Yes.

Every $\alpha\in K$ is the root of a polynomial $\in F[X]$ of degree $\le |G|$. This is because the coefficients of $\prod_{f\in G}(X-f(\alpha))$ (aka. the elementary symmetric polynomials in the $n$ conjugates $f(\alpha)$, $f\in G$), are $G$-invariant and hence $\in F$.