The following question was on my midterm, I didn't get a chance to solve it because I ran out of time. Now that I'm reviewing for my final, I decided to take a crack at it but I can't seem to figure it out.
Let $q\in\mathbb{Q}[x]$ be an irreducible polynomial of degree $\geq3$. Suppose that q has precisely two nonreal roots in a splitting field K. Show that $Gal(K/\mathbb{Q})$ is nonabelian.
What I got so far is that:
Let $\alpha\in Gal(K/\mathbb{Q})$ and let $a$ be one of the nonreal roots. Then $\bar{a}$ must be the other nonreal root. By homomorphism, $\alpha$ can only map $a$ to $a$ or $\bar{a}$ and $\alpha$ must fix all other roots. Therefore, it seems that there are only two automorphisms in $Gal(K/\mathbb{Q})$. But any group with two elements is always abelian. Now, I'm thinking, I don't understand what an automorphism is. Can someone help me out?
Thanks in advance!
Best Answer
Let the two nonreal roots be $r,s$, let a real root be $t$. Write $\sigma$ for complex conjugation: then $\sigma(r)=s$, $\sigma(s)=r$, $\sigma(t)=t$. Now Galois groups are transitive, so there's an automorphism $\tau$ with $\tau(r)=t$. Then $\sigma(\tau(r))=\sigma(t)=t$, while $\tau(\sigma(r))=\tau(s)\ne t$.