Let $K/F$ be a finite Galois extension of fields with Galois group $G$. Let $H$ be a subgroup of $G$. Then there is $\alpha \in K$ such that $H=\{\sigma \in G : \sigma \alpha = \alpha\}$.

My proof is this:

Consider the fixed field $E$ of $H$. Let $\alpha \in E$ be such that if $\alpha \in E' \subseteq K$ then $E\subseteq E'$. Then $H = \{\sigma \in G : \sigma \alpha = \alpha \}$ since if $\sigma \in H$, then sigma fixes $\alpha$. Conversely, if $\sigma \in G$ fixes alpha, then by choice of $\alpha$, $\sigma$ is contained in some subgroup $H'$ with fixed field $E'$ containing $\alpha$ and $E' \supseteq E$. In particular, since $E \subseteq E' \iff H' \subseteq H$, we have $\sigma \in H$. The result follows.

Does this work? I feel like I might be missing something since I (apparently) did not use the finiteness assumption on the field extension.

People have posted (better) solutions but I'm still interested in knowing whether what I came up with is correct.

  • $\begingroup$ Correct me if I'm wrong, but did you not use the finiteness assumption to apply Galois Correspondence? $\endgroup$
    – Andrew
    Jul 19 '15 at 0:40
  • $\begingroup$ How do you an $\alpha$ with this property exists? $\endgroup$
    – Bernard
    Jul 19 '15 at 0:44
  • $\begingroup$ @ Bernard. It's clear it exists. If $F$ is any subgroup of $K$ with $E$ not contained in $F$ and $F$ not contained in $E$ then $E-F$ is nonempty. Also, there has to be an $\alpha$ in $E$ not contained in any proper subfield of $E$ since $E$ is strictly larger than any proper subfield. $\endgroup$
    – TuoTuo
    Jul 19 '15 at 0:47
  • $\begingroup$ @ Andrew If it is the case that the Fundamental Theorem of Galois Theory is only true for finite extensions, then yes, I suppose I did. $\endgroup$
    – TuoTuo
    Jul 19 '15 at 0:48
  • $\begingroup$ $H$, being a subgroup of $G$, represents an intermediate field $F'$ with $F\subseteq F' \subseteq K$. Since these are all Galois extensions, then $F'$ is generated by a single element over $F$ (by the primitive element theorem). This is your $\alpha$. $\endgroup$
    – Arthur
    Jul 19 '15 at 0:53

You used the assumption of finiteness of the extension when you assumed that exist $ \alpha $ as you choose. The existence of such $ \alpha $ is ensured by the Primitive element theorem and not necessary true for infinite extensions. From then on, your solution is absolutely correct.


$E=K^H$, being a subextension of the Galois extension $K/F$ is a finite separable extension of $K$, hence by the Primitive element theorem, there exists $\alpha\in E$ such that $E=K(\alpha)$. Now, it isresults from the Galois correspondence that: $$H=\{\sigma\in G\mid \sigma(\alpha)=\alpha\}.$$ 


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.