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?


No, this is not true. There are several reasons why this need not happen. One of the more pressing ones is that then all the automorphisms would be of order two (or one).

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).

  • $\begingroup$ A somewhat related thread with slightly similar confusion. If any one thinks these are duplicate, I won't stand in the way. I'm not sure new content is created here :-/ $\endgroup$ – Jyrki Lahtonen May 1 at 18:07

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