# Normal field extension separable over its fixed field

Let us have a field $K\supseteq E$ and $G$ be its group of automorphisms over $E$. Let the fixed field of $G$ be $K^G$. I would like to show that $K$ is separable over $K^G$.

I know that for algebraic extensions normal and separable iff Galois and $K$ is Galois over $K^G$. However, I want to show it a more direct way. I was wondering if there is someway to show that if some irreducible polynomial $f$ was not separable, than some something in its splitting field (which must be contained $K$) that is not in $K^G$ must be also fixed by $G$ leading to a contradiction. However, I cannot get this to work out. This would make sense if $f$ had only a single root repeated, but that is not true in general. (If need be we can also include the assumption that $K$ is normal of $E$, but I do not know if this is necessary)

Any help or direction is greatly appreciated.

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What wxactly do you mean by "$G$ be its group of automorphisms over $E$" ? Do you mean $G=Aut(K/E)$, or maybe it is a subgroup of $Aut(K/E)$ ? –  Belgi Feb 7 '13 at 22:18
The former, that $G=\operatorname{Aut}(K/E)$. –  user45150 Feb 7 '13 at 22:26

Pick $\alpha \in K$, need to show that $\alpha$ is separable over $K^G$. In particular, we just need to show that $\alpha$ is the root of a polynomial $f$ with no repeated roots, because it's minimal polynomial must divide $f$.

Let $\{\sigma_1, \ldots, \sigma_n\} \subseteq G$ be the largest set of elements in $G$ with the property that $\sigma_1(\alpha), \ldots, \sigma_n(\alpha)$ are all distinct. Now prove the following:

1. If $\tau \in G$ show that $\tau$ must permute $\{\sigma_1(\alpha), \ldots, \sigma_n(\alpha)\}$.
2. Define $f = \prod_i(x - \sigma_i(\alpha))$. Show that $f(\alpha) = 0$.
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I seem to remember the following argument from Kaplansy's Fields and Rings. I assume all degrees are finite.

Let $\alpha \in K$, and let $f \in K[x]$ be its minimal polynomial over $K^G$, so that $f$ is irreducible in $K^G[x]$.

Let $\alpha_1 = \alpha, \alpha_2, \dots, \alpha_k$ be the distinct roots of $f$ in $K$, and let $$g = (x - \alpha_1) (x - \alpha_2) \dots (x - \alpha_k).$$ Since $G$ permutes the $\alpha_i$, we have that $g \in K^G[x]$. Since $g$ divides $f$, and $f$ is irreducible in $K^G[x]$, we have $f = g$, so the minimal polynomial of $\alpha$ over $K^G$ has distinct roots, and $K / K^G$ is separable.

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