# An automorphism in Galois group must map imaginary root to itself or conjugate pair.

Suppose we have a Galois group of a splitting field of a rational polynomial over $$\mathbb{Q}.$$ Is it true that any automorphism of the Galois group must map an imaginary root to itself or its conjugate pair? I think this is true because we can extend any automorphism $$\sigma$$ to an automorphism of $$\mathbb{C}.$$ And if $$\alpha$$ is an imaginary root, then $$(x - \alpha)(x - \bar{\alpha})$$ is the real minimal polynomial of $$\alpha$$ over $$\mathbb{R}.$$ Thus, since automorphisms must map roots of a minimal polynomial to roots of the minimal polynomial, $$\alpha$$ must map to either $$\alpha$$ or $$\bar{\alpha}.$$ Is this correct? Is there a simpler explanation?

While $$m(x)=(x-\alpha)(x-\overline{\alpha})$$ is the minimal polynomial of $$\alpha$$ over $$\Bbb{R}$$, there is no need for an automorphism $$\sigma$$ to fix the coefficients of $$m(x)$$. If $$K/\Bbb{Q}$$ is the Galois extention in question, then the coefficients of $$m(x)$$ will belong to the intermediate field $$M=K\cap \Bbb{R}$$, but that field is often a lot bigger than $$\Bbb{Q}$$, and hence the coefficients of $$m(x)$$ need not be fied points of $$\sigma$$. This means that $$\sigma(\alpha)$$ need not be a zero of $$m(x)$$. The automorphism $$\sigma$$ only needs to respect those polynomial relations that have all their coefficients in $$\Bbb{Q}$$.
The first counterexample that comes to mind is that of the fifth cyclotomic polynomial $$\Phi_5(x)=x^4+x^3+x^2+x+1.$$ Its zeros are $$\zeta_5=e^{2\pi i/5}$$, $$\zeta_5^2$$, $$\zeta_5^3$$ and $$\zeta_5^4$$. Applying Eisenstein's criterion to $$\Phi_5(x+1)$$ shows that $$\Phi_5(x)$$ is irreducible over $$\Bbb{Q}$$. The standard argument (ask, if you have not seen it) then tells that $$K=\Bbb{Q}(\zeta_5)$$ has an automorphism $$\sigma$$ such that $$\sigma(\zeta_5)=\zeta_5^2$$. But $$\overline{\zeta_5}=\zeta_5^4$$ (plot those roots of unity on the complex plane to see this, if it is not clear).
It is even possible that $$\sigma$$ does not map complex conjugate pairs to other such pairs. This will happen when the Galois group is not abelian. An example is the splitting field of $$x^3-2$$ over $$\Bbb{Q}$$, i.e. $$K=\Bbb{Q}(\root3\of2,e^{2\pi i/3})$$. By the argument mentioned above $$K$$ has an automorphism $$\sigma$$ such that $$\sigma(e^{2\pi i/3}\root3\of 2)=\root3\of2$$ is real. The automorphism $$\sigma$$ thus must map the conjugate root $$e^{-2\pi i/3}\root3\of2$$ to one of the other two roots (in this case both are actually possible).