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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?

Jadmrial
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    did you look at this question? http://math.stackexchange.com/questions/33535/finding-a-unique-subfield-of-mathbbq-zeta-of-degree-2 – maxymoo Jun 03 '11 at 05:51
  • 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 03 '11 at 06:10
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    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 03 '11 at 07:53

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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!

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Pete L. Clark
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