Let $g \in \mathbb{Q}[X]$ be an irreducible and separable polynomial which has a real and a complex root in $\mathbb{C}$.

Show that in this situation $\text{Gal}(K|\mathbb{Q})$ is not abelian, where $K$ is the splitting field of $g$ over $\mathbb{Q}$. (*)

Find for each degree $n$ of $g$ an example to (*).

Now my first question is: How can I prove (*) and how can I find an example for each $n$? Is there a limit for $n$ so that $g$ has a complex and a real root?

  • $\begingroup$ Clearly, we need $n\ge 3$. - (Regarding the premises: Can you exhibit an irreducible inseparable polynomial in $\mathbb Q[X]$?) $\endgroup$ – Hagen von Eitzen May 11 '14 at 20:26

Let $r$ be a real root, and let $z$ be a non-real root of the polynomial. We need the fact the restriction of the complex conjugation to $K$, call it $\sigma$, is an element of $G=Gal(K/\Bbb{Q})$. Another fact that we need is that $G$ acts transitively on the roots. Thus there is another element $\tau\in G$ such that $\tau(r)=z$.

Leaving the rest to you with the suggestions:

  1. Calculate the effect of both $\sigma\tau$ and $\tau\sigma$ on the root $r$.
  2. What happens with $x^n-2$, $n>2$?
  • $\begingroup$ Is your polynomial $f_n:=X^n -2$, $n >3$ the example I needed? Because $f_n$ has real and nonreal roots. $\endgroup$ – user893458 May 11 '14 at 20:48
  • $\begingroup$ Yes it is, but do check everything, because when you present it as a solution, then YOU are making that claim. So you need to recall a tool proving that it is irreducible, and also justify the presence of real an non-real roots. Sounds like you are well on your way! $\endgroup$ – Jyrki Lahtonen May 11 '14 at 20:51
  • $\begingroup$ For irreducible I would say: Eisenstein's criterion for $p=2$. $\endgroup$ – user893458 May 11 '14 at 20:53
  • $\begingroup$ Correct. Well done :-) $\endgroup$ – Jyrki Lahtonen May 11 '14 at 20:54
  • $\begingroup$ The real root is $x=2^{(1/n)}$ and the complex roots are some $n$th roots of unity multiplied by $x$. $\endgroup$ – user893458 May 11 '14 at 20:55

As $g$ has nonreal roots, we see that complex conjugation $\psi\colon z\mapsto \bar z$ is a nontrivial automorphism of $K$. Let $\alpha$ be a real root and $\beta$ a complex root. By transitivity of the Galois group, there exists an automorphism $\phi$ with $\phi(\alpha)=\beta$. Then $\psi(\phi(\alpha))=\overline\beta$, but $\phi(\psi(\alpha)=\beta$.


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