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

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?

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.

• Hi, I believe you have some typos and mistakes in the proof of the second part. (1) I think wherever you talk about the field $L$, you actually mean $L^H$, in the last two paragraphs. For instance, the Galois correspondence gives you $|\mathrm{Aut}(L^H/K)| = 1$, not $\mathrm{Aut}(L/K) = 1$. Similarly, $|\mathrm{Aut}(L_f/K)| = \frac{|\mathrm{Aut}(L^H/K)|}{|\mathrm{Aut}(L^H/L_f)|}$, as $\mathrm{Aut}(L^H/L_f)$ is normal in $\mathrm{Aut}(L/K)$. – Brahadeesh Jun 4 at 6:53
• (2) There is a difference between $\beta$ being "not separable" and $\beta$ being "purely inseparable". What you have shown at the end is that $\min(\beta,K)$ has a unique distinct root, namely $\beta$ itself, and hence $\beta$ is purely inseparable; but you concluded something weaker, namely that $\beta$ is merely not separable. The reason I'm emphasising this is that one cannot say that $K \subset L^H$ is purely inseparable if one has only shown that each $\beta \in L^H$ is not separable over $K$. – Brahadeesh Jun 4 at 6:57
• (3) Your proof can be streamlined further: once you have that $|\mathrm{Aut}(L^H/K)| = 1$, you need not go through the Galois theory in the last paragraph. Since $K \subset L^H$ is a normal extension, every embedding of $L^H$ over $K$ into an algebraic closure of $L$ is actually an automorphism of $L^H$, hence $|\mathrm{Aut}(L^H/K)| = 1$ implies that there is only one embedding of $L^H$ over $K$ into an algebraic closure of $L$. Hence, the separable degree of $L^H/K$ is $1$, and so $L^H/K$ is purely inseparable. – Brahadeesh Jun 4 at 7:01
• (4) Lastly, with the above simplification, the same proof works for infinite extensions as well, because there is no more counting to be done. The part where we concluded that $|\mathrm{Aut}(L^H/K)| = 1$ is true for infinite extensions as well, because what is more generally true under the given conditions is that $\mathrm{Aut}(L^H/K)$ is isomorphic to $\mathrm{Aut}(L/K)/\mathrm{Aut}(L/L^H)$. – Brahadeesh Jun 4 at 7:03