Consider the following proposition with its relative proof:
Let $k$ be an algebraically closed field of characteristic $0$.
a) If $L$ is a subfield of $k$, then every elements of $\operatorname {Aut} (L)$ extends to an element of $\operatorname{Aut} (k)$
b) $\operatorname{Fix}_k\left(\operatorname{Gal}(k/L)\right)=L$
Proof: $a)$ Let $S$ be a transcendence basis for $k/L$, then every element $\sigma\in\operatorname{Aut}(L)$ extends naturally to an element $\widetilde\sigma\in\operatorname{Aut}(L(S))$; since $k$ is an algebraic closure of $L(S)$, we use the isomorphism extension theorem to conclude that $\widetilde\sigma$ extends to an element of $\operatorname{Aut}(k)$.
$b)$ Obviously it is enough to prove that for every $x\in k\setminus L$ there is an element $\sigma\in\operatorname{Gal}(k/L)$ such that $\sigma(x)\neq x$. We distinguish two cases:
- $x$ is transcendental over $L$. Consider the field $L(x)$, then the assignment $x\mapsto-x$ induces a unique element $\sigma\in\operatorname{Gal}(L(x)/L)$ that moves $x$. By the point $a)$ this $\sigma$ extends to an element of $\operatorname{Gal}(k/L)$.
- $x$ is algebraic over $L$. Since in characteristic
$0$ every irreducible polynomial is separable, if
$f=\textrm{min}\left(x,L\right)$ then there exists in $k$
(remember that $f$ splits over $k$) a root $y$ of $f$ such
that $x\neq y$. Let $M$ be the splitting field of $f$ over $L$
and look at the inclusions
$$L\subseteq L(x)\subseteq M\subseteq k$$ $M$ is normal over $L$ and the canonical $L$-isomorphism $\sigma:L(x)\longrightarrow L(y)$ can be viewed as an immersion $\sigma: L(x)\longrightarrow k$. By the characterization of normal extensions there is an element $\tau\in\operatorname{Gal}(M/L)$ such that $\tau_{|L(x)}=\sigma$ (in particular $\tau(x)=y$), therefore by the point $a)$ $\tau$ extends to an element of $\operatorname{Gal}(k/L)$ which moves $x$.
Now my question: Is the point $b)$ of the proposition true also when the characteristic of $k$ is $p\neq 0$? The above proof uses the separability of all irreducible polynomials!
Edit: Moreover I'd like to know if there are shorter or more elegant proofs of the proposition.
Thanks in advance.