Timeline for is every positive real cyclotomic number the norm of a cyclotomic?
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 15, 2021 at 13:15 | comment | added | Dima Pasechnik | @VictorOstrik - $\sqrt{2}=\zeta_8+1/\zeta_8$. In general, see math.stackexchange.com/a/282779/39797 | |
| Dec 14, 2021 at 19:17 | vote | accept | Dima Pasechnik | ||
| Apr 27, 2021 at 6:10 | comment | added | François Brunault | If $X_0$ is a solution then all other solutions are given by $X=X_0 \cdot (t-i)/(t+i)$ with $t \in \mathbb{Q}^{\mathrm{ab}} \cap \mathbb{R}$. | |
| Apr 27, 2021 at 1:13 | history | became hot network question | |||
| Apr 26, 2021 at 22:00 | comment | added | GH from MO | You don't need Artin-Schreier theory. If you conjugate $a$ by $\sigma$, all the $b_k$'s get conjugated by $\sigma$, because complex conjugation commutes with $\sigma$. Then use that $z\bar z$ is positive for every nonzero $z\in\mathbb{C}$. | |
| Apr 26, 2021 at 21:51 | answer | added | Will Sawin | timeline score: 6 | |
| Apr 26, 2021 at 21:19 | history | edited | Dima Pasechnik | CC BY-SA 4.0 |
English fix
|
| Apr 26, 2021 at 20:30 | history | edited | Dima Pasechnik | CC BY-SA 4.0 |
added 84 characters in body
|
| Apr 26, 2021 at 20:28 | comment | added | Dima Pasechnik | yes, indeed, $a$ is totally positive, by a bit of Artin-Schreier theory. | |
| Apr 26, 2021 at 20:26 | comment | added | Dima Pasechnik | a square with $i$ in it is not real. | |
| Apr 26, 2021 at 20:18 | comment | added | GH from MO | Your $a$ is totally positive, because it is the sum of totally positive elements. | |
| Apr 26, 2021 at 19:33 | comment | added | pavl0 | For the sum of squares, if $i$ is in the field, then $a=\left(\dfrac{a+1}{2}\right)^2+\left(i\dfrac{a-1}{2}\right)^2$. I'd check "A Historical View of the Pythagoras Numbers of Fields" by D. Leep | |
| Apr 26, 2021 at 19:02 | comment | added | Dima Pasechnik | my $a=\sum_{k=1}^m b_k\overline{b}_k$, $b_k$ cyclotomic. Is $a$ totally positive? (References on the topic would be much appreciated, too) | |
| Apr 26, 2021 at 17:34 | comment | added | Victor Ostrik | The number $X\bar X$ is totally positive (any of its Galois conjugates is positive). Thus any $a$ which is positive but not totally positive would be a counterexample, e.g. $a=\sqrt{2}$. | |
| Apr 26, 2021 at 17:12 | history | asked | Dima Pasechnik | CC BY-SA 4.0 |