Consider the following problem: This question is from Thomas Hungerford Section 3 Problem 15.
If F is algebraic Galois over K, then F is algebraic Galois over E, where E is an intermediate field.
Is " algebraic Galois " here algebraic and Galois.
I tried it by assuming that it is algebraic and Galois. So, if F is algebraic over K then F is algebraic over E as $K\subset E$. But If F is Galois over K then $Aut_K F = K $. Now by the definition of $Aut_K E$ and as $E \subset F$ if an element of F is not fixed then clearly any element of E will not be fixed.
Is the proof correct?
(b) If F is Galois over E, E is Galois over K and F is a splitting field over E of a family of polynomials in K[x] then F is Galois over K.
From the given condition of splitting I can deduce that F is galois over E but that is already given. SO, I am not able to deduce anything in this part and help is needed.
abstrac
The proof is essentially based on the fundamental theorem of Galois theory, which states that there is a bijection between intermediate field between $F$ and $K$ and the subgroups of $Gal(F/K)$, given that the extension $F/K$ is algebraic Galois.
a) Let $F/E/K$ be the intermediate field and let $\Phi: E\mapsto Gal(F/E)$ be the bijection.
b) Let $H$ be a subgroup of $Gal(F/K)$ and let $\Psi: H \mapsto F^H$ be the inverse mapping of $\Phi$.
Then suppose $F/E/K$ is intermediate field, then
c) $\Psi\circ\Phi(E) = E$
since $\Phi$ and $\Psi$ are inverses. Then combine a) b) c) we know that $E=\Psi\circ\Phi(E) = \Psi(Gal(F/E))=F^{Gal(F/E)}$. This shows that $F/E$ is Galois.