If $L/K$ normal and $H = \operatorname{Aut}(L/K)$, then $L/L^H$ is separable and $L^H/K$ is purely inseparable.

Question

I need to prove the following:

Let $L/K$ be a normal field extension. Denote by $H=\operatorname{Aut}(L/K)$ the Galois group of the extension, and by $L^H$ the fixed field of $H$ in $L$. Prove that $L/L^H$ is separable, and that $L^H/K$ is purely inseparable.

For now, we tried to use the fact that an extension $M/F$ is separable (purely inseparable) if the number of homomorphisms $\phi:M\to \overline{M}$ that preserve $k$ is $[M:F]$ ($1$). We did not manage to conclude any results, and moreover this works only for finite extensions. How should we proceed?

Answer 1

For the first part take any $\alpha \in L$. Then let $\{\alpha_1,\dots \alpha_n\}$ be the set of distinct elements obtained by $\text{Aut}(L/K)$ acting on $\alpha$. Note that this set is finite, as the extension is algebraic. Now consider:

$$h(x) = \Pi_{i=1}^{n} (x-\alpha_i)$$

Now it's not hard to see that $h(x)$ is fixed by $\text{Aut}(L/K)$, as it only permutes the factors on the right and so we have that $h(x) \in L^H[x]$. Moreover it's irreducible, as if $g = \min(\alpha,L^H)$ then by the transitivity of the Galois group on the set of distinct elements we have that $(x-\alpha_i)$ is a factor of $g$ for any $i$. Hence we conclude that $h = \min(\alpha,L^H)$ and as it's separable $\alpha$ is separable over $L^H$ and so we conclude that $L^H \subseteq L$ is a separable extension.

However I'm not able to prove the second part for infinite extensions. Anyway here's proof for finite extensions.

We first prove that the extension $\text{Aut}(L/K) = \text{Aut}(L/L^H)$. As $K \subseteq L^H \subseteq L$ we have that every automorphism of $L$ fixing $L^H$, also fixes $K$ and so $\text{Aut}(L/L^H) \subseteq \text{Aut}(L/K)$. However from the condition we have that any automorphism on $L$ fixing $K$, also fixes $L^H$ and so we must have $\text{Aut}(L/K) \subseteq \text{Aut}(L/L^H)$. From here we conclude that $\text{Aut}(L/K) = \text{Aut}(L/L^H)$

This will give us that $K \subseteq L^H$ is also a normal extension and by Galois correspondence we have that $|\text{Aut}(L/K)| = 1$. (Here's the part where I need finiteness).

Now let $\beta \in L^H$ and consider $f = \min(\beta,K)$. Let $L_f$ be the splitting field of $f$ over $K$. As $K \subseteq L^H$ is normal we must have $L_f \subseteq L^H$. But then $|\text{Aut}(L_f/K)| = \frac{|\text{Aut}(L/K)|}{|\text{Aut}(L/L_f)|} = 1$, as $\text{Aut}(L/L_f)$ is normal in $\text{Aut}(L/K)$. But now $\text{Aut}(L_f/K)$ acts transitively on the roots of $f$ and so we must have that the only root of $f$ is $\beta$. So if $\beta \not \in K$, then $\deg f \ge 2$ and as it's only root is $\beta$ we have that $f$ isn't separable and hence $\beta$ isn't separable. From here we conclude that $K \subset L^H$ is purely inseparable extension.

Answer 2

This is in Lang's Algebra, 2002, p. 251, Proposition 6.11. It is not assumed that $L/K$ is Galois. The first part of the proof is identical to Stefen4024's proof.

For the second part, let $\beta\in L^H$ be separable over $K$. Then the identity map on $K$ may be extended to an embedding $\tau$ of $K(\beta)$ in $L$. Since $L/K$ is normal, this embedding is an automorphism of $K$, but $\tau\beta=\beta$ by definition and hence $\tau$ is the identity on $K(\beta)$. Therefore, $L^H/K$ is purely inseparable.