Maximal separable sub-extensions and Galois groups. I'm reading Neukirch's ANT book and at page 56 (bottom) it seems that the author is assuming the following fact:

Let $L|K$ be a finite and normal extension of fields and consider the maximal separable sub-extension $K(\theta)|K$. Then $\text{Gal}(L|K)=\text{Gal}(K(\theta)|K)$.

Why is this result true?
Many thanks in advance
 A: I will only deal with the case $\operatorname{char} K = p > 0$, the other case being trivial.
Let $\phi \in \operatorname{Aut}_K(L)$; then note that $\phi(K(\theta)) \subseteq K(\theta)$. Indeed, if $x$ is separable, then so is $\phi(x)$, because they have the same minimal polynomial. Thus, we get a natural map
\begin{align*}
f \colon \operatorname{Aut}_K(L) &\to \operatorname{Aut}_K(K(\theta))\\
\phi &\mapsto \phi\big|_{K(\theta)}.
\end{align*}
A standard theorem of Galois theory shows that this map is surjective since $L/K(\theta)$ is normal (see e.g. Rotman's Galois Theory, Thm. 51(a)); we only need to show injectivity.
Since $K(\theta) \to L$ is purely inseparable, for any $x \in L$ there exists $n \in \mathbb N$ such that $x^{p^n} \in K(\theta)$. Note that
$$\phi(x^{p^n}) = \phi(x)^{p^n}.$$
Since $p$-th roots are unique, we conclude that knowing $\phi$ on $K(\theta)$ determines $\phi$ on $L$. That is, $f$ is injective. $\square$
Remark. In this case, you can actually use the above to construct a very explicit inverse map to $f$: Let $\phi \in \operatorname{Aut}_K(K(\theta))$, and let $x \in L$. For $n$ such that $x^{p^n} \in L$, there is at most one $y \in L$ such that
$$y^{p^n} = \phi(x^{p^n}).$$
Note that $x$ and $y$ have the same minimal polynomial, since $x^{p^n}$ and $y^{p^n}$ do (maybe choose $n$ minimal). Thus, this $y$ exists in $L$ since $L$ is normal. We then extend $\phi$ to $L$ by setting $\phi(x) = y$ for this $x$.
(Omitted: check that this does not depend on $n \gg 0$, that this gives a $K$-linear automorphism of $L$, and that the construction is actually a double-sided inverse of $f$.)
