Prove if $F \subset E$ is a finite dimensional field extension then $\operatorname{Gal}(E/F)$ is finite

Question

My idea is that since $F \subset E$ is finite dimensional then the set of roots (call it $R$) of $E$ is finite so $\operatorname{Sym}(R)$ is finite and since the Galois group can be embedded in the symmetric group, $\operatorname{Gal}(E/F)$ must be finite, but I'm struggling with the details.

Answer 1

A Galois automorphism is determined by its restriction on the basis since it is linear. If $(e_1,...,e_n)$ is a basis, the image of $e_I$ is an element of the set $S_i$ of root of its minimal polynomial. Let So the number of Galois automorphisms is finite since it has the cardinality of a subset of $S_1\times...\times S_n$.