I'm investigating the splitting field of $x^{17}-1$. Clearly this is just $\mathbb{Q}(\zeta)$ where $\zeta$ is a complex 17th root of unity. I know that there is a unique subextension $F$ of order 2, as the Galois group of the extension is isomorphic to $\mathbb{Z}_{16}$. I'm curious as to what a basis element could be, ie $\alpha\in\mathbb{Q}$ such that $F=\mathbb{Q}(\sqrt{\alpha})$. One obviously exists, as the extension is of degree 2, but I'm having trouble finding it. I tried picking something like $\sum_{\sigma\in H} \sigma(\alpha)$ with $H\subset G(F/\mathbb{Q})$ the unique subgroup of order 8, but it isn't fixed under action of the Galois group of $F$. Does anyone have some hints on how I might proceed with this question?
Basis of Subextension
2
$\begingroup$
abstract-algebra
galois-theory
-
1maxymoo- that turned out to be quite helpful. Thank you. I guess I wasn't too far off- I just needed to push a little harder. – 2011-06-03
1 Answers
1
You said you wanted a hint, so here goes: read up on quadratic Gauss sums, for instance here. But if you only want a hint, don't read Section 5!