Galois over Galois I am working on this exercise:
If $E$ is an intermediate ﬁeld of an extension $F/K$ of ﬁelds. Suppose $F/E$ and $E/K$ are Galois extensions, and every $\sigma\in Gal(E/K)$ is extendible to an automorphism of $F$, then show that $F/K$ is Galois.
I can see that any $\sigma$ extended over $F$ fixes elements in $K$ but not in $E-K$. But how to show it doesn't fix elements in $F-E$?
Hints only please, this is homework.
p.s. we use Kaplansky, he doesn't require Galois extensions to be finite dimensional.
 A: Here is a full solution of the problem.
Since the OP already solved it, I think there is no harm in writing a full solution of it here.
We denote by $Aut(K/k)$ the group of automorphisms of a field extension $K/k$.
Let $K/k$ be a not necessarily finite dimensional algebraic extension field.
If $K$ is normal and separable over $k$, we say $K/k$ is Galois.
If $K/k$ is Galois, we write $G(K/k)$ instead of $Aut(K/k)$.
We need the following characterization of a Galois extension field.
Lemma
An algebraic extension field $K/k$ is Galois if and only if the fixed subfield of $K$ by $Aut(K/k)$ is $k$.
Proof.
Suppose $K/k$ is Galois.
Let $\alpha \in K - k$.
Let $f(X)$ be the minimal polynomial of $\alpha$ over $k$.
Since $K/k$ is normal and separable, there exists a root $\beta$ of $f(X)$ such that $\alpha \ne \beta$ and $\beta \in K$.
Let $\sigma\colon k(\alpha) \rightarrow k(\beta)$ be the unique isomorphism such that $\sigma(\alpha) = \beta$. Since $K/k$ is normal, $\sigma$ can be extended to an automorphism $\sigma'$ of $K/k$.
Since $\sigma'(\alpha) = \beta$, we are done.
Conversely suppose the fixed subfield of $K$ by $G = Aut(K/k)$ is $k$.
Let $\alpha$ be an element of $K$.
Let $f(X)$ be the minimal polynomial of $\alpha$ over $k$.
Since $\sigma(\alpha)$ is a root of $f(X)$ for every $\sigma \in G$,
the set $S = \{\sigma(\alpha)\mid \sigma \in G\}$ is finite.
Let $\sigma_1, \cdots, \sigma_m$ be elements of $G$ such that $\sigma_1(\alpha), \cdots, \sigma_m(\alpha)$ are pairwise distinct and $S = \{\sigma_1(\alpha), \cdots, \sigma_m(\alpha) \}$.
Let $g(X) = (X - \sigma_1(\alpha))\cdots (X - \sigma_m(\alpha))$.
Since every coefficient of $g(X)$ is fixed by $G$, $g(X) \in k[X]$.
Since $g(\alpha) = 0$, $g(X)$ is divisible by $f(X)$.
Since each $\sigma_i(\alpha)$ is a root of $f(X)$, $g(X) = f(X)$.
Hence $\alpha$ is separable over $k$ and $K/k$ is normal.
This completes the proof of the lemma.
Now let $F/K, E/K$ be as in the problem.
By the lemma, it suffices to prove that the fixed subfield of $F$ by $Aut(F/K)$ is $K$.
Suppose $\alpha \in F$ is fixed by $Aut(F/K)$.
Since $G(F/E) \subset Aut(F/K)$, $\alpha$ is fixed by $G(F/E)$.
Hence $\alpha \in E$ by the lemma.
Since every element of $G(E/K)$ is extended to an element of $Aut(F/K)$, $\alpha$ is fixed $G(E/K)$. Hence $\alpha \in K$ by the lemma.
This completes the proof.
A: I think the above answer assumes that $F/K$ is algebraic and although most of the time when we consider galois extensions we assume that they are algebraic as well, this problem can be solved without assuming so and in a shorter way.
Let $u\in F-K$, we have to prove that there is a $K$-automorphism $\sigma:F\to F$ such that $\sigma(u)\neq u$. Lets consider two cases:


*

*$u\in E$. Since $E/K$ is galois there is a $K$-automorphism $\theta:E\to E$ such that $\theta(u)\neq u$, we extend this $\theta$ to a $K$-automorphism $\sigma:F\to F$ and we are done.

*$u\notin E$. Since $F/E$ is galois there is a $E$-automorphism $\sigma:F\to F$ such that $\sigma(u)\neq u$ and this is also a $K$-automorphism since $K\subset E$
