Timeline for An algebraic number is not a root of unity?
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 2, 2017 at 3:37 | answer | added | Noam D. Elkies | timeline score: 11 | |
| Jul 10, 2014 at 20:04 | comment | added | dvitek | @AlexDegtyarev I don't care if you cite my contribution or not. Dunno how Gerry feels... | |
| Mar 2, 2014 at 9:48 | vote | accept | Alex Degtyarev | ||
| Mar 2, 2014 at 1:35 | answer | added | Victor Ostrik | timeline score: 15 | |
| Feb 28, 2014 at 15:06 | comment | added | Emil Jeřábek | This meta question was asked before: meta.mathoverflow.net/questions/402/… | |
| Feb 28, 2014 at 14:56 | answer | added | ACL | timeline score: 14 | |
| Feb 28, 2014 at 12:17 | comment | added | Alex Degtyarev | @drvitek Accidentally, I have a meta question. Should I use this in a paper, how do I refer to something that I learned at MO? | |
| Feb 28, 2014 at 12:15 | comment | added | Alex Degtyarev | @drvitek Thanks, I got it, too. Yes, it looks like the claim holds. | |
| Feb 28, 2014 at 11:54 | comment | added | dvitek | @@GerryMyerson The paper that we're looking for is Newman's "Some Results on Roots of Unity, with an Application to a Diophantine Problem" in which he characterizes all rational $x, y$ with $\sin(\pi x)\sin(\pi y) = 1/4$, which is equivalent via prosthaphaeresis (cosine-addition, but I just wanted an excuse to use that word) to the equation I gave (which is in fact missing some 2's). In particular, none of the solutions have denominators greater than 12, so it looks like you're good. I have access to the PDF; please let me know if you can't get a hold of it. | |
| Feb 28, 2014 at 11:47 | vote | accept | Alex Degtyarev | ||
| Mar 2, 2014 at 9:48 | |||||
| Feb 28, 2014 at 11:46 | comment | added | Alex Degtyarev | @GerryMyerson Thanks, I'll take a look. | |
| Feb 28, 2014 at 11:31 | comment | added | Emil Jeřábek | @AlexDegtyarev: If $\lambda$ is a root of unity, then $\mathbb Q(\lambda)$ contains all algebraic conjugates of $\lambda$, one of which is the other root of $\chi$. | |
| Feb 28, 2014 at 11:29 | answer | added | Filippo Alberto Edoardo | timeline score: 12 | |
| Feb 28, 2014 at 11:19 | comment | added | Gerry Myerson | I'm away from my references. The two cosines paper was, if I remember right, by Morris Newman. My paper was called (something like) Rational products of sines of rational angles. | |
| Feb 28, 2014 at 11:15 | comment | added | Alex Degtyarev | @GerryMyerson Could you give me a ref? | |
| Feb 28, 2014 at 11:11 | comment | added | Gerry Myerson | The equation of @drvitek can be rewritten as a rational product of cosines of two rational angles. But these were all found and listed many years ago. Indeed, 20 years ago I published all rational products of three and of four cosines of rational angles. | |
| Feb 28, 2014 at 11:09 | comment | added | Alex Degtyarev | @SebastianSchoennenbeck Actually, I'm confused. Is it really obvious that $\mathbb{Q}[\xi]\subset\mathbb{Q}[\lambda]$? If I do not assume the coefficients of $\chi$ in the field, how do I know that both roots are there? | |
| Feb 28, 2014 at 11:06 | comment | added | Alex Degtyarev | @FilippoAlbertoEdoardo I was badly interested in $n=7$ and $9$. Now, I am curious about all $n>6$, but I think the case of prime $n$ would imply the rest, at least concerning the original geometric problem. | |
| Feb 28, 2014 at 10:57 | comment | added | Alex Degtyarev | @SebastianSchoennenbeck Oops! You are right: I seem to have missed the fact that the inclusion $\mathbb{Q}[\xi]\subset\mathbb{Q}[\lambda]$ is obvious. Thanks, I'll try to think along these lines! | |
| Feb 28, 2014 at 10:53 | comment | added | Alex Degtyarev | @drvitek Of course, but this looks even worse (= transcendental) | |
| Feb 28, 2014 at 10:52 | comment | added | Filippo Alberto Edoardo | In your examples, you take $n=7$ or $n=3^2$: since it makes things easier, I am wondering if it is enough for you that $n$ be a prime-power or whether you might want to take things like $n=60. | |
| Feb 28, 2014 at 10:52 | comment | added | Sebastian Schoennenbeck | If $\lambda '$ is a root of $\chi$ then $\mathbb{Q}(\xi,\lambda ')$ has degree at most 2 over $\mathbb{Q}(\xi)$ so $\lambda '$ would have to be either a power of $\xi$ or a primitive $4n$-th root of unity. But the minimal polynomial of the latter is given by $\lambda^2-\xi$ hence this is not the case. So what remains to show is that no power of $\xi$ is a root of this polynomial. | |
| Feb 28, 2014 at 10:47 | comment | added | dvitek | This isn't a way of testing algebraic numbers for being roots of unity, but a little bit of algebra reveals that this problem is exactly equivalent to, writing $\xi = exp(2\pi i\ m/n),$ showing that there do not exist $p, q$ with $$\cos(\pi\ m/n) + \cos(\pi\ p/q) = 1/2.$$ | |
| Feb 28, 2014 at 10:19 | history | asked | Alex Degtyarev | CC BY-SA 3.0 |