Galois theory for arbitrary field extension

Question

I know that for an arbitrary field extension $E/F$ there may be distinct subgroups $G$ and $H$ of $Aut(E/F)$ with the same fixed field: $E^{G} = E^{H}$. When can we conclude that for subextensions $E/E_{1}$ and $E/E_{2}$, $Aut(E/E_{1}) = Aut(E/E_{2})$ implies $E_{1} = E_{2}$?

Answer 1

If $E/F$ is a finite Galois extension, this is a standard result from Galois theory. If $E/F$ is a profinite Galois extension, and $E_1$ and $E_2$ are closed subgroups, then this is a standard result from profinite Galois theory.

Is there another case you are interested in? Infinite extensions, or in finite but non-Galois extensions?

Answer 2

In general, this is not true, not even when the base field is trivial.

Consider $E= \mathbf Q[\sqrt[3] 2]$, $F=\mathbf Q$. Since $\sqrt[3]{2}$ is the unique cubic root of two in $E$, it is fixed by any automorphism, so $\operatorname{Aut}(E/F)=\operatorname{Aut}(E)$ is trivial, even though $E\neq F$.

Similarly, if $E=\mathbf R$, then $\operatorname{Aut}(E)=\{\operatorname{id}_{\mathbf R}\}=\operatorname{Aut}(E/K)$, where $K$ is any of the $2^{\mathfrak c}$ distinct subfields of $\mathbf R$.