Let $G$ be a finite group of automorphisms of $E$ and set $F=Fix(G)$, Then why is $E:F$ always separable ?
I have a feeling that it has something to do with the idea that if $E:F$ is separable hence all elements adjoined in $E$ are separable and so their min. poly has simple roots, and so if $E:F$ were not separable it would mean that $\alpha\in E$ appears twice in its min poly and so the co-efficents of the min poly would contain some power of $\alpha$ and then would not be fixed by G.
I'm not sure if I'm thinking in the right vein at all though , I'm finding it difficult to formulate the reason concisely in my head .
Could anyone clear this up for me please ?
Let $\alpha\in E$ and let $S$ be the orbit of $\alpha$ under the action of $G$. Then the polynomial $$f(x)=\prod_{s\in S}(x-s)$$ has coefficients in $F$: each coefficient is a symmetric function in the elements of $S$, and any element of $G$ permutes the elements of $S$ and hence fixes that symmetric function. But by definition, the roots of $f$ are all distinct. So $f$ is a polynomial with coefficients in $F$ with distinct roots and $\alpha$ as a root. Since $\alpha\in E$ is arbitrary, this means $E$ is separable over $F$.