Let $E/K$ be a field extension, let $p(x)$ be an irreducible polynomial in $K[x]$ which splits in $E$ with roots $\alpha_1$, $\alpha_2$, etc., and let $\sigma$ be an automorphism of $E$ which fixes $K$. Then $\sigma$ fixes $p(x)$ and so permutes the roots. The proposition I am trying to prove is that if $\sigma$ fixes one root $\alpha_1$ of $p(x)$ then it must also fix the other roots $\alpha_i$.

At first I thought the statement is false for $x^3-2$ in $Q[x]$, but it's not. Are there perhaps additional conditions needed to make this statement true?

  • 2
    $\begingroup$ the statement IS false for $x^3-2$ in $Q[x]$. $\endgroup$
    – mercio
    Jul 15, 2013 at 10:46
  • 1
    $\begingroup$ In general, it will fail for any polynomial with a mixture of real and non-real roots, as complex conjugation will only fix the real roots. $\endgroup$ Jul 15, 2013 at 10:48

1 Answer 1


This fails exactly when the the order of the Galois group $G$ exceeds the degree of $p(x)$, call it $n$. Remember that the action of $G$ is transitive, so the order of a point stabilizer is (for any root $\alpha$) $$ \left\vert\operatorname{Stab}_G(\alpha)\right\vert=\frac{|G|}n>1, $$ iff $|G|>n$. So under those circumstances there exist a non-trivial automorphism $\sigma$ fixing $\alpha$. Non-triviality implies that $\sigma(\alpha')\neq\alpha'$ for some other root $\alpha'$ of $p(x)$.

OTOH, if $|G|=n$, then any point stabilizer is trivial, and the claim is true.

Yet another way of saying the same things is that the claim is true iff $\alpha$ is a primitive element of $E/K$ (keeping the assumption that $E$ is the splitting field of the minimal polynomial of $\alpha$). As then $E=K(\alpha)$ this is almost a tautology.


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