# Basis of Subextension

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?

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did you look at this question? math.stackexchange.com/questions/33535/… –  maxymoo Jun 3 '11 at 5:51
What you tried should work, e.g., if you take $\zeta$ for your $\alpha$. Your get a sum $\eta$ of 8 17th roots of unity; if you let $\eta'$ be the sum of the other 8 (non-real) 17th roots of unity, and look at $\eta+\eta'$ and at $\eta\eta'$.... –  Gerry Myerson Jun 3 '11 at 6:10
maxymoo- 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. –  Jadmrial Jun 3 '11 at 7:53