3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $x^n-2$, $n>2$?
$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $\phi(\psi(\alpha)=\beta$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.