Timeline for Is the set of certain polynomials finite or infinite?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 29, 2014 at 5:54 | vote | accept | user64494 | ||
| Oct 28, 2014 at 9:31 | comment | added | Vesselin Dimitrov | @GerryMyerson: Yes, and also to the usual transfinite diameter of $[-2,2]$ being $1$. Fekete's theorem states more generally that a compact set $K \subset \mathbb{C}$ stable under complex conjugation contains only finitely many Galois orbits of algebraic integers if its transfinite diameter is smaller than $1$, while conversely, every open neighborhood of $K$ contains infinitely many such orbits if the transfinite diameter exceeds $1$. | |
| Oct 28, 2014 at 8:54 | vote | accept | user64494 | ||
| Oct 28, 2014 at 8:54 | |||||
| Oct 28, 2014 at 5:35 | answer | added | Vesselin Dimitrov | timeline score: 13 | |
| Oct 28, 2014 at 5:34 | comment | added | Gerry Myerson | I think this is related to a concept called the monic integer transfinite diameter. | |
| Oct 28, 2014 at 5:21 | comment | added | user64494 | @ Vesselin Dimitrov: Can you kindly elaborate your comment as an answer, giving references? Note that the interval under consideration is a proper subset of $[-2,2]$. | |
| Oct 28, 2014 at 5:00 | comment | added | Vesselin Dimitrov | As to the proof of the statement, it is an application of pigeonholing. Write the roots as $2\cos(2\pi \alpha)$ and consider for $n \in \mathbb{N}$ the monic polynomials with simple roots at $2\cos(2\pi n \alpha)$. Show that it is an integer polynomial with coefficients bounded uniformly in $n$, hence some two such polynomials must coincide, and the conclusion follows. | |
| Oct 28, 2014 at 4:57 | comment | added | Vesselin Dimitrov | It is finite. This is a well known problem and an analog of Kronecker's theorem: the integer monic polynomials having all their roots in $[-2,2]$ are exactly the products of the minimal polynomials of algebraic integers of the form $2\cos(2\pi \, m/d)$. The finiteness then follows once you know also that all these numbers having $(m,d) = 1$ for a given $d$ are algebraic conjugates. | |
| Oct 28, 2014 at 4:35 | history | asked | user64494 | CC BY-SA 3.0 |