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The following question was motivated by this MO-post.

I hope that the answer should be known to experts (because of very simple formulation)...

Problem. Let $n\ge 2$. Is the set of complex numbers $\{e^{i\pi k/2^n}:0\le k<2^n\}$ linearly independent over the field of rationals?

Taras Banakh
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Denote by $\omega$ the order $2^{n+1}$-th primitve root of unity $\omega=e^{i\pi/2^n}$. The linear dependence of the above set would imply that there is a polynomial of degree at most $2^n-1$ with $\mathbb{Q}$ coefficients which vanishes on $\omega$. But its minimal polynomial is the cyclotomic polynomial $\Phi_{2^{n+1}}(X)=X^{2^n}+1$ and we are done.

Vlad Matei
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    Perhaps it is more direct to say that the OP's set generates $\mathbb{Q}(\omega)$ as a vector space over $\mathbb{Q}$. This vector space is of dimension $2^n$ by the irreducibility of $\Phi_{2^{n+1}}(X)=X^{2^n}+1$ over $\mathbb{Q}$, so the OP's set is in fact a basis. – GH from MO Feb 22 '21 at 08:23
  • @GHfromMO Does the same holds for any $m$ instead of $2^n$? I means that the set ${e^{i\pi k/m}:0\le n<m}$ is linearly idependent? Or there are some requirements on $m$? – Taras Banakh Feb 22 '21 at 12:22
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    If $n$ has an odd prime $p$ factor note that the above set contains ${1,\omega,\ldots, \omega^{p-1}}$ where $\omega$ is a primitive $2p$th root of unity. We have $\omega^p=-1$ and we can obviously factor $\omega^p+1$ to get $\omega^{p-1}-\omega^{p-2}+\ldots-\omega+1=0$. Thus you only have powers of $2$ for independence. – Vlad Matei Feb 22 '21 at 13:07
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    The degree of the minimal polynomial of $e^{2 \pi i/m}$ is $\phi(m)$. So the corresponding statement is that ${ e^{2 \pi i j/m} : 0 \leq j < \phi(m) }$ is linearly independent over $\mathbb{Q}$. – David E Speyer Feb 22 '21 at 13:14
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    @TarasBanakh: See David E Speyer's comment. The minimal polynomial of $e^{2\pi i/m}$ is the $m$-th cyclotomic polynomial $\Phi_m(X)$ whose roots are the primitive $m$-th roots of unity. – GH from MO Feb 22 '21 at 13:29