Show that the degree of any finite extension of $F$ is a power of $p$

Question

Let $F$ be a field of characteristic zero such that for some prime $p$, every extension $E$ of $F$ with $E\neq F$ has $[E:F]$ divisible by $p$. Show that the degree of any finite extension of $F$ is a power of $p$

My Attempt:
We have $[E:F]=pk$ for some $k$.
Thus $order(Gal(E,F))=pk$.
So from Sylow's theorem, we have a subgroup $H$ such that $|H|=p$ and
index between $Gal(E,F)$and $H$ = $m$
index between $\{e\}$ and $H$ = $p$

So from the Galois correspondance we get a subfield $F\subset K\subset E$ so that
$[E:K]=p$
$[K,F]=k$

Now by considering the given condition for $K$, we get $k=[K,F]=pm$ for some $m$
As $[E:F]=[E:K][K:F]$,
We get $pk=p^2m$,
So can we do it unitil it reaches $p^k\times1?$ If so why?

Also I know that to apply Galois correspondance, the field extension should be both normal and separable. and Characteristic zero implies it is separable.
But Why should that be normal?

Appreciate your help

Answer 1

You are mostly on the right track.

Hint: You cannot apply the fundamental theorem of Galois theory directly to an arbitrary extension, $E$.

Suppose by way of contradiction, then that there is some proper extension of $F$, $E$, for which $[E:F]$ is not a power of $p$. Then write $[E:F]=p^kn$ where $n$ is chosen to be $\_\_\_\_\_\_$ to $p$.

Let $N$ be the $\_\_\_\_\_$ of $E$. Since $[N:F]=[N:E][E:F]$ is divisible by $\_\_\_\_\_\_$ by Sylow's theorem, there exists a subgroup of $\mathrm{Gal}(N:F)$ of order $\_\_\_\_\_\_$. By the fundamental theorem of Galois theory, this means that there exists a field intermediate $F$ and $N$, $K$, s.t. $[N:K]=|\mathrm{Gal}(N:K)|=\_\_\_\_\_\_$, so $[K:F]=\_\_\_\_\_\_$, a contradiction.

Full Explanation:

The number $n$ is chosen to be coprime to $p$. Since $[E:F]$ is not a power of $p$, $n>1$. $N$ is chosen to be the normal closure of $E$ over $F$. As we are working over a field of characteristic $0$, $N$ is then automatically a Galois extension of $F$. We choose $K$ so that $[N:K]=|\mathrm{Gal}(N:K)|=p^k$. Then $[K:F]=n$ which is not divisible by $p$ even though $n>1$ means that $K$ is a proper extension of $F$, a contradiction.