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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!

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Everything after "by homomorphism" is completely wrong. I'm not sure how you've come to believe it: perhaps you think that $\alpha$ has to fix $\Bbb{R}$? –  Chris Eagle Dec 6 '12 at 11:09
    
When I learned field theory, it was said that the only homomorphism of $\mathbb{R}$ is the identity. –  user44322 Dec 6 '12 at 11:10
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@user44322 continuous homomorphism that is. –  user38268 Dec 6 '12 at 11:17
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@BenjaLim: Continuous isn't needed. –  Chris Eagle Dec 6 '12 at 11:31
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@user44322: That's true, but so what? You aren't working with $\Bbb{R}$. You're working with $\Bbb{Q}$ and with $K$. –  Chris Eagle Dec 6 '12 at 11:32

1 Answer 1

up vote 5 down vote accepted

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 $\tau(\sigma(r))=\tau(s)\ne t$.

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