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Let $K$ be the splitting field of a separable irreducible polynomial $f(x) \in F[x]$ of degree $n$ and let $G = Gal(K/F)$. If for each $g \in G$, there is a root $\alpha$ of $f$ such that $g(\alpha) = \alpha$, prove that $K=F$.

We know that $G$ permutes the roots of $f$. Since $f$ is irreducible separable over $F$, $G$ is isomorphic to a transitive subgroup of $S_n$, i.e., $G$ has some $g$ that is an $n$-cycle. But at the same time, $g$ also fixes some root of $f$ and so $n=1$.

Is that all? I've thought long and hard about this only to come up with a literal one-liner.

Saurav Maheshkar
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oleout
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  • A subgroup of $S_n$ need not have an $n$-cycle to be transitive. Consider ${1,(12)(34),(13)(24),(14)(23)}\le S_4$. Of course, that subgroup does not have the prescribed property either. – Jyrki Lahtonen Apr 01 '22 at 05:21
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    Instead, try and use the fact that the average number of fixed points of a transitive subgroup of $S_n$ is equal to $1$. Prove that first if you haven't seen it! The identity permutation has more than the average number of fixed points, so some other permutation must have ____?? – Jyrki Lahtonen Apr 01 '22 at 05:24
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    @JyrkiLahtonen I will think about the result you mentioned. To finish off what you said, the other automorphisms (elements of $G$) cannot all have at least one fixed root. Hence the only way this can work is if we only have the identity. – oleout Apr 01 '22 at 05:26
  • @JyrkiLahtonen I think I got confused with the case where the degree of $f$ is prime (say, $p$). In this case, $p$ divides $|G|$ would imply that $G$ has a $p$-cycle. Thank you so much for your clarification. – oleout Apr 01 '22 at 05:29
  • Correct. That other result is often encountered when studying group actions more generally. It depends on your background how easy it will be to prove that lemma. A starting point (essentially the orbit-stabilizer formula) is that any root will be fixed by exactly $|G|/n$ permutations. Then rewrite that in terms of pairs $(\alpha,g)$ such that $g(\alpha)=\alpha$. – Jyrki Lahtonen Apr 01 '22 at 05:34
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    @JyrkiLahtonen I found it, there's a name for it: Burnside's lemma. The number of orbits defined by the action of automorphism on the roots is $$\frac{1}{|G|}\sum_{g \in G}{\alpha:g(\alpha)=\alpha}.$$ Since $G$ is transitive, this number is $1$ but the size of ${\alpha: \mathrm{id}(\alpha) = \alpha}$ is already $n$. – oleout Apr 01 '22 at 05:42
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    Correct! Discussed in many threads on this site also. Like here. – Jyrki Lahtonen Apr 01 '22 at 05:44

1 Answers1

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Suppose $K\neq F$.

Let $\{\alpha_1,\ldots,\alpha_n\}$ be the distinct roots of mimimal pol. of $\alpha$ over $F$ ($\alpha=\alpha_1$), and $G=\{\sigma_1,\ldots,\sigma_n\}$.

Take any $1\neq \sigma\in\mbox{Gal}(K/F)$. Since $\sigma$ fixes some $\alpha_i$, so $\mbox{Stab}(\alpha_i)$ is non-trivial subgroup.

Since $\alpha_i\notin F$, so $\alpha_i$ is not fixed by $\mbox{Gal}(K/F)$, so $\mbox{Stab}(\alpha_i)$ is proper subgroup.

For any other $\alpha_j$,there is $\tau\in\mbox{Gal}(K/F)$ such that $\tau(\alpha_i)=\alpha_j$. Then $$ \tau\mbox{Stab}(\alpha_i)\tau^{-1}=\mbox{Stab}(\tau(\alpha_i))=\mbox{Stab}(\alpha_j). $$ Thus, the given condition says: every $\sigma\in\mbox{Gal}(K/F)$ is in some conjugate of $\mbox{Stab}(\alpha_i)$, i.e. $$ \mbox{Gal}(K/F)=\cup_{\tau} \tau \mbox{Stab}(\alpha_i)\tau^{-1}. $$ But, a finite group can never be union of conjugates of a proper subgroup. Q.E.D.

Maths Rahul
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