field extension $K\subset L$ and Uniqueness of $G$ such that $K=L^{G}$

Question

Let $K\subseteq L$ be a algebraic field extension. If there exists subgroups $G_{1},G_{2}\subseteq$ Aut$(L)$ such that $K=L^{G_{1}}=L^{G_{2}}$, does $G_{1}=G_{2}$ holds?

($L^{G}=\{a\in L:\sigma a =a$ for all $\sigma \in G\})$

Answer 1

No: Take $K=\Bbb F_p$ and $L$ its algebraic closure $\overline {\Bbb F_p}$. Then $\text{Aut}(L)=\hat {\Bbb Z}$. Take $\varphi$ to be the Frobenius automorphism $x\mapsto x^p$, then the subgroup of $\text{Aut}(L)$ generated by $\varphi$ is isomorphic to $\Bbb Z$ and hence not the whole group $\text {Aut}(L)$, but they have the same fixed field $\Bbb F_p$.

However, by the Fundamental Theorem of Galois theory the result is true for finite separable extensions. More generally, if $L$ is an infinite algebraic separable extension, then $G=\text{Aut}(L/K)$ becomes a topological group, where a fundamental system of neighborhoods of $1$ is given by the subgroups $\text{Aut}(L/M)$, where $M$ ranges over all the finite subextension of $K$ in $L$. If we now restrict ourselves to closed subgroups $G_1, G_2\le G$, then again the result is true, as in this case the subextensions correspond one-one to the closed subgroups of $G$.

Answer 2

There is a general result, I believe due to Artin, which can be used as a foundation for Galois theory. It says that if $L$ is a field, and $G$ is a finite group of automorphisms of $L$, then $L / L^{G}$ is Galois, with Galois group $G$.

So if your subgroups are finite, the answer is yes, as $$ G_{1} = \operatorname{Gal}(L/K) = G_{2}. $$

In the infinite-dimensional case, however, the answer is negative, see for instance this MSE post (in particular point (iii) there). This has to do with the fact that in the infinite case, to get the one-to-one Galois correspondence you need to restrict yourself to subgroups of the Galois group that are closed in the Krull topology; all subgroups that share the same closure will also share the same set of fixed points.

PS The example just posted by @walcher is exactly the same of the post I quoted, indeed possibly the most natural one.