Timeline for Are half of the $2^n$-th roots of the unit rationally independent?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 20, 2022 at 12:08 | vote | accept | Taras Banakh | ||
| Feb 22, 2021 at 13:29 | comment | added | GH from MO | @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. | |
| Feb 22, 2021 at 13:14 | comment | added | David E Speyer | 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 $\mathbb{Q}$. | |
| Feb 22, 2021 at 13:07 | comment | added | Vlad Matei | 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. | |
| Feb 22, 2021 at 12:22 | comment | added | Taras Banakh | @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 $m$? | |
| Feb 22, 2021 at 8:23 | comment | added | GH from MO | 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. | |
| Feb 22, 2021 at 8:15 | history | answered | Vlad Matei | CC BY-SA 4.0 |