If $E/K$ is a field extension we use the notation $\def\Aut{\operatorname{Aut}}\Aut(E/K)$ for the set of field automorphisms of $E$ that are the identity over $K$. It's immediate that the set $\Aut(E/K)$ is a group with the composition operation. If $H$ is a subgroup of $\Aut(E/K)$, we define $E^H=\{x \in E : \sigma (x) = x\,$ for all $\sigma \in H \}$. It can be easily verified that $E^H$ is an intermediate field of $E/K$ and we call it the field fixed by $H$.
It's known that if $E/K$ is a Galois extension with galois group $G$ then $E^G=K$. I'm interested in the converse of this claim. That is:
Let $E/K$ be an algebraic field extension and $G=\Aut(E/K)$. Is it true that $E^G=K$ implies $E/K$ Galois? What if $E/K$ is separable? What if $K$ is $\mathbb{Q}$? What if $E/K$ is finite?
I have been thinking for a while but I didn't find a counterexample nor a proof. I've tried examples with $K=\mathbb{Q}$ and $K=\mathbb{F}_2$. The problem is I've found $K \subsetneq E^G$ every time that $E/K$ was not Galois.
Thanks!
I am pretty sure Galois includes separability as a criteria. I will assume you mean by Galois "normal and separable". This is an equivalent condition to your question:
If $E/K$ is not galois, then there is some element $a \in E, a \notin K$ such that one of its conjugate roots over $K$ is not in $E$. Then let $x_i = g_i(a)$ where $g_i()$ are the automorphisms in $G$. The polynomial with coefficients the elementary polynomials in $x_i$ has roots in $E$ and coefficients in the fixed field of $G$. If this fixed field were K, then all the conjugate roots of a would be contained in E. This is a contradiction.
I am new to writing proofs in latex, so if you don't understand or want to make some edits, please feel free to ask me/edit it yourself.