Automorphism group of finite field extension has a trivial stabilizer

Question

We have this theorem.

Let $L|K$ a field extension with $[L:K]<\infty$ and $G=\text{Aut}(L|K)$. We let $G$ act on $L$. Then there is a trivial stabilizer.

The proof is the following, I would like to get help with the vector space argument (the $\neq$ part), I don't see how you can argue like that without knowing beforehand that the automorphism group is finite (in fact, this result is used to show that it's finite in my lecture notes later).

If $G=\{\text{id}\}$ we are done, so assume $G\neq\{\text{id}\}$ and $L\neq K$. For $f\in G$ let $L_f$ be the set of all $y\in L$ that are fixed under $f$. Obviously, for all $f$ we have that $L_f \leq L$ is a field. Moreover, if $f\neq \text{id}$, $L_f$ will be a proper subfield of $L$. We can consider the $L_f$ to be subspaces of the $K$-vector space $L$. Thus, $$L\neq \bigcup_{f\in G\setminus\{\text{id}\}}L_f,$$ so that there exists a $y\in L$ that lies in no nontrivial $L_f$ and thus, the stabilizer of $y$ is trivial.

Answer 1

“Finite field extension” is ambiguous: it could mean that it is an extension in which $K$ and $L$ are finite fields, or it could mean that it is an extension of finite degree, $[L:K]\lt\infty$, but where $K$ is an infinite field (that is, $|K|$ is not finite).

I answer assuming the latter, because it seems more compatible with the argument.

Because $f\neq \mathrm{id}$, it follows that there must exist some $x\in L$ such that $f(x)\neq x$. Therefore, $$L_f = \{x\in L\mid f(x)=x\}\neq L.$$ That means that $L_f$ is a proper subset (and hence a proper subfield, and hence a proper subspace) of $L$. But when we work over an infinite field, a vector space cannot be the union of finitely many proper subspaces (or here, or here, with a mention of Pete Clark’s note in the Monthly about this). Thus, $$L\neq \bigcup_{f\in G\setminus\{\mathrm{id}\}} L_f$$ because $L$ cannot be the union of finitely many proper subspaces.

So if we pick $y\in L$ that is not in the union, it cannot lie in any stabilizer.

Now, if $K$ is finite, then the extension $L/K$ is cyclic because all finite extensions of finite fields are cyclic. The argument above doesn’t work, because a finite dimensional vector space could be a union of proper subspaces. In this case, though, there is an element $y$ such that $L=K(y)$, and then it is immediate that $y$ cannot lie in any stabilizer, because if $y\in L_f$, then $f(y)=y$ and hence $f(x)=x$ for all $x\in L$, contradicting $f\neq\mathrm{id}$.