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Let $f(x)$ be an irreducible polynomial in $\mathbb{Q}[x]$ with both real and nonreal roots. Show that its Galois group is nonabelian. Can the condition that $f$ is irreducible be dropped? If not, provide an example to show this.

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  • $\begingroup$ Not entirely sure how to do this, but remember the Galois group is a transitive subgroup of $S_n$, $n \geq 3$ which contains a transposition, so that might get you started. $\endgroup$
    – J.G
    Jun 3 '16 at 1:53
  • $\begingroup$ I'm not quite sure this is true. Take an irreducible polynomial over Q of degree 3, which has a real root and two nonreal roots. Let K be its splitting field. Then [K : Q] = 3 and so the order of the Galois group is 3, so it's abelian. Apologies if I'm missing something. $\endgroup$
    – florence
    Jun 3 '16 at 1:59
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    $\begingroup$ I believe such a polynomial would have Galois group $S_3$. If you adjoin the real root, you clearly don't have the complex ones in your extension, so the splitting field is not degree 3. $\endgroup$
    – J.G
    Jun 3 '16 at 2:25
  • $\begingroup$ @user293121 Thank you, my mistake. $\endgroup$
    – florence
    Jun 3 '16 at 2:47
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    $\begingroup$ Just a special case of the argument in this answer. $\endgroup$ Jun 3 '16 at 6:51
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For the second part, consider $f(x)=(x^2-2)(x^2+1)$, whose splitting field $\mathbb Q(\sqrt 2,i)$ has degree $4$ over $\mathbb Q$.

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