Suppose we define an arbitrary field extension $K/F$ to be <b>Galois</b> if, for all subextensions $L$ of $K/F$, we have $K^{\operatorname{Aut}(K/L)} = L$. In words: for any element $x$ of $K \setminus L$, there exists an automorphism $s$ of $K$ such that $s(l) = l$ for all $l$ in $L$, but $s(x) \neq x$. (Note that in case $K/F$ is algebraic, this is indeed a characteristic property of Galois extensions.) What are the transcendental Galois extensions? 

In my rough notes [Transcendental Galois Theory][1], I show that if $F$ has characteristic $0$ and $K$ is algebraically closed, then $K/F$ is Galois in the above sense.  [Actually, these notes are somewhat incomplete.  Having been unable to complete the proof of the conjecture below, I left out some of the more straightforward details.  If anyone wants to see more detail on anything in these notes, please let me know.]

I also conjectured: if $K/F$ is Galois, then either $K/F$ is algebraic, normal and separable, or $F$ has characteristic $0$ and $K$ is algebraically closed. Is this true?

Comment: It is easy to see that if $K/F$ is not algebraic, then $K$ must have characteristic $0$.  It is possible to modify the question a bit so that the positive characteristic case is not ruled out, but I would like to understand what's going on in characteristic $0$ first!

In my notes, I show that an affirmative answer follows from a certain (arguably) less weird conjecture about Galois closures of subfields of rational function fields.  If there is any interest, I will reproduce this conjecture here explicitly.  

[1]: http://math.uga.edu/~pete/galois.pdf