# Nonabelian Galois Group

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.

• 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.
– J.G
Jun 3 '16 at 1:53
• 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. Jun 3 '16 at 1:59
• 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.
– J.G
Jun 3 '16 at 2:25
• @user293121 Thank you, my mistake. Jun 3 '16 at 2:47
• Just a special case of the argument in this answer. Jun 3 '16 at 6:51

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 $\mathbb Q$.