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It's not generaly true that roots of the minimal polynomial $f_\alpha$ over $\mathbb{Q}$ of some algebraic number $\alpha$ can be a $\mathbb{Q}$-basis for $\mathbb{Q}(\alpha)$. As a counterexample, one can consider $x^2-2=0$

But it's said to be true for the roots of n-th cyclotomic polynomial $\Phi_n$, The task is to prove

$\{e^{\frac{2\pi i k}{n}}:\gcd(k,n)=1\}$ form the basis for $\mathbb{Q}$-vector space $\mathbb{Q}(e^{\frac{2\pi i}{n}})$

Oolong Milktea
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1 Answers1

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It is true that $\zeta_p$ ($p$-prime) generates a normal basis for $\Bbb{Q}(\zeta_m,\zeta_p)/\Bbb{Q}(\zeta_m)$ when $p\nmid m$,

the proof is that a non-trivial $\Bbb{Q}(\zeta_m)$-linear relation between the $\zeta_p^1,\ldots,\zeta_p^{p-1}$ would mean $\zeta_p$ is a root of a polynomial $\in \Bbb{Q}(\zeta_m)[x]$ of degree $\le p-2$.

Thus $\prod_{p| n}\zeta_p $ and hence $\zeta_n$ generate a normal basis for $\Bbb{Q}(\zeta_n)/\Bbb{Q}$ whenever $n$ is square-free.

If $n$ is not square-free then it fails because $\sum_{a\le n,\gcd(a,n)=1} \zeta_n^a = \mu(n)=0$

reuns
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  • Would you mind making the last sentence clearer. I come across a question which ask me to find all the subextension of $\mathbb{Q}(e^\frac{2\pi i}{72})$ and the corresponding subgroup of $\mathbb{Q}$ and I think a good choice of basis may be useful. – Oolong Milktea Dec 26 '19 at 12:22
  • If $k\ge 2$ then $\sum_{a\le p^k, p\ \nmid\ a} \zeta_{p^k}^a = 0$ – reuns Dec 26 '19 at 12:31
  • I'm clear now. Thanks a lot. – Oolong Milktea Dec 27 '19 at 03:49