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I want to find an irreducible polynomial $f(x)$ over $\mathbb Q$ and a finite nonnormal extension $K/\mathbb Q$ which contains at least two roots of $f(x)$ such that $\operatorname{Aut}(K/\mathbb Q)$ acts nontransitively on the roots of $f(x)$ in $K$. I could not find simple examples of this.

Update: Julian Rosen gave an answer to the original question above but now I want to impose the condition that $K$ is contained in the splitting field of $f(x)$ because I was looking for intermediate field extensions of the splitting field.

Pratyush Sarkar
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    Translating this into the language of permutation groups (assuming that we can solve the inverse Galois problem for an eventual example group $G$ :-). Let $X$ be a finite set, and $G\le Sym(X)$ be a transitive subgroup. Assume that a subgroup $H\le G$ has as its set of fixed points some subset $S\subset X$ with $|S|\ge2$. Does it necessarily follow that $N_G(H)/H$ acts transitively on $S$? – Jyrki Lahtonen Mar 13 '17 at 08:18
  • @user218931 I wonder why did you delete your answer...? – DonAntonio Mar 13 '17 at 09:14
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    @DonAntonio My answer was wrong, since my example did act transitively on the roots of $f(x)$ in $K$. – Claudius Mar 13 '17 at 09:28
  • See the linked questions for a way of locating counterexamples. – Jyrki Lahtonen Sep 06 '21 at 04:47

1 Answers1

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Take $f(x)=x^2-2$, $K=\mathbb{Q}(\sqrt[4]{2})$. Then $K$ contains two roots of $f$, but there is no automorphism exchanging them because only one of the roots is a square in $K$.

Julian Rosen
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    Thanks! That definitely works, but I was always looking for intermediate extensions of the splitting field of $f(x)$. Any ideas for examples on that? – Pratyush Sarkar Mar 13 '17 at 20:29