# Are half of the $2^n$-th roots of the unit rationally independent?

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?

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.
• 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. Commented Feb 22, 2021 at 8:23
• @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 $m$? Commented Feb 22, 2021 at 12:22
• 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. Commented Feb 22, 2021 at 13:07
• 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 $\mathbb{Q}$. Commented Feb 22, 2021 at 13:14
• @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. Commented Feb 22, 2021 at 13:29