Let $E$ be finite dim ext of a field $F$and let $G$ be a gp of $F$-automorphisms of $E$ s.t $[E:F]=|G|$ Show that $F$ is the fixed field of $G$.

Question

Question: Let $E$ be finite dimensional extension of a field $F$and let $G$ be a group of $F$-automorphisms of $E$ such that $[E:F]=|G|$. Show that $F$ is the fixed field of $G$. $\DeclareMathOperator{\Aut}{Aut}$

My Thoughts: Let $K=\mathcal F(\Aut(E/F))$, thus we have $\Aut(E/K)\subseteq\Aut(E/F)$, but since $F\subseteq K$, we have $\Aut(E/F)\subseteq\Aut(E/K)$, so we get equality. So, $$ |\Aut(E/F)|=[E:F]\geq[E:K]=|\Aut(E/K)|=|\Aut(E/F)|. $$ Thus $[E:F]=[E:K]$ so $K=F$ and so $F=\mathcal F(\Aut(E/F))$, and since $E$ is finite dimensional and $[E:F]=|G|$, we have $F=\mathcal F(G)$.

But I am questioning my first sentence and my last sentence. I am trying to do this without resorting to too much Galois Theory, but any help is greatly appreciated! Thank you.

Answer 1

Ah, I have found a reference for the only needed fact: Proving the number of K-automorphisms is bounded by the degree of the extension by considering field embeddings.

The linked question asks for a proof for the bounding $|\mathrm{Aut}(L/K)|$ from above by $[L:K]$. It is a rather elementary argument as it turns out. Anyway, I will freely use this fact now.

Back to your question. Consider $E/F$ finite and $G\subseteq\mathrm{Aut}(E/F)$ with $|G|=[E:F]$. We then have $$ |G|\le|\mathrm{Aut}(E/\mathcal F(G))|\le[E:\mathcal F(G)] $$ by definition of $\mathcal F(G)$ and using the fact from above. But, $[E:\mathcal F(G)]\le[E:F]$ and $|G|=[E:F]$ implying that $[E:\mathcal F(G)]=[E:F]$ which is only possible for $\mathcal F(G)=F$.

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By the way: In this case we have also have $|\mathrm{Aut}(E/F)|=[E:F]$, by the given inclusions and bounds, implying at once that $E/F$ is actually a Galois extension (and $G$ its Galois group).

Answer 2

We can use the standard result that if $G$ is a finite subgroup of group of automorphisms of any field $E$ then the the fixed field $E_G$ of $G$ has the property that $[E:E_G] =|G|$. The proof of this result involves showing that $[E:E_G]$ is finite and does not exceed $|G|$. Once we know that the extension $E/E_G$ is finite we can apply a little bit of Galois theory to get $[E:E_G] =|G|$.

We are given that $F\subseteq E_G$ and thus we get $F\subseteq E_G\subseteq E$. Now you can use the tower law to get $$[E_G:F] =\frac{[E:F]} {[E:E_G]} =\frac{|G|} {|G|} =1$$ and thus $E_G=F$.