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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?

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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.

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    $\begingroup$ 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. $\endgroup$
    – GH from MO
    Commented Feb 22, 2021 at 8:23
  • $\begingroup$ @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 $m$? $\endgroup$ Commented Feb 22, 2021 at 12:22
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    $\begingroup$ 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. $\endgroup$
    – Vlad Matei
    Commented Feb 22, 2021 at 13:07
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    $\begingroup$ 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 $\mathbb{Q}$. $\endgroup$ Commented Feb 22, 2021 at 13:14
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    $\begingroup$ @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. $\endgroup$
    – GH from MO
    Commented Feb 22, 2021 at 13:29

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