# Let $G$ be a finite group of automorphisms of $E$ and set $F=Fix(G)$, Then why is $E:F$ always separable?

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 ?

## 1 Answer

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 $$F$$.