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.

  • $\begingroup$ When you say “finite field extension”. Does that mean an extension of finite fields, or do you have an infinite field $K$ and a finite degree extension $L$? $\endgroup$ Mar 30, 2020 at 21:36
  • $\begingroup$ By that I meant that the degree of the extension is finite. I changed by question accordingly. $\endgroup$
    – l2poca
    Mar 31, 2020 at 7:12

1 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 $f\neq\mathrm{id}$.

  • $\begingroup$ If we assume that G is finite / countable, we can use the linked results to see that L cant be a union of L_f's. But I still don't know what we know about the order of G beforehand. The source of confusion is that in my textbook, this argument is used to show that G is finite, so in this proof, I can't use that fact yet. $\endgroup$
    – l2poca
    Mar 31, 2020 at 7:17
  • $\begingroup$ @l2poca What textbook? What section of the textbook? Actually, the results I linked to show that if a finite dimensional vector space over $K$ is not a union of fewer than $|K|$ proper subspaces, so the argument works at least for $|G|\lt |K|$. I need to see exactly what assumptions you have, because it is not true that a vector space is never a union of proper subspaces. $\endgroup$ Mar 31, 2020 at 15:56
  • $\begingroup$ It's from my German lecture notes. Also, this is at the beginning of the chapter about fields and all the previous theorems were concerned with the minimal polynomial, so there are no results yet that I can use that directly to get some information about the orders of $G$ and $K$. The only general statement about these finite extensions that we showed before this was that they are algebraic. But i don't see a way to apply this here. $\endgroup$
    – l2poca
    Mar 31, 2020 at 18:00
  • $\begingroup$ @l2poca: Well, you can bound the order of $G$ easily enough, but if this is being done in order to show that the order of $G$ is finite, then I don’t know what to tell you. It could simply be an error in the notes. But without actually looking at them, I cannot help you. $\endgroup$ Mar 31, 2020 at 18:20
  • 1
    $\begingroup$ Okay, I asked someone who is responsible for the lecture notes and I was told that this has indeed been a mistake and it was supposed to be shown that $|G|<\infty$ by other means before. $\endgroup$
    – l2poca
    Apr 1, 2020 at 17:11

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