# Let $E \supset F \supset G$ and $G = E^{\rm{Aut}(E/G)}$. Then is it true that $F = E^{\rm{Aut}(E/F)}$?

Let $$E/F$$ be a field extension. Let us say $$E/F$$ is semi-Galois if $$F$$ is the fixed field of $$\rm{Aut}(E/F)$$.

Now consider a field tower $$E/F/G$$. If $$E/G$$ is semi-Galois, then is $$E/F$$ necessarily semi-Galois? What would be a counterexample if this is false?

Let $F$ be an algebraic closure of $\mathbb{Z}/2\mathbb{Z}$. Then,$F(X)/F$ is semi-Galois while $F(X)/F(X^2)$ is not a semi-Galois.
Characteristic zero counterexample: $$\mathbb{C}(z) \supset \mathbb{C}(z^3-z) \supset \mathbb{C}$$.
Verification that $$\mathbb{C}(z) / \mathbb{C}$$ is semi-Galois: The only periodic rational functions are constants, so $$f(z)$$ is fixed by $$z \mapsto z+1$$ if and only $$f$$ is constant.
Verification that $$\mathbb{C}(z) / \mathbb{C}(z^3-z)$$ is not semi-Galois Write $$t = z^3-z$$. This field extension is finite, so Galois and semi-Galois are equivalent. This is a cubic extension with minimial polynomial $$x^3-x-t=0$$. The discriminant of this polynomial, $$4 - 27 t^2$$, is not a square, so this cubic extension is not Galois.