Timeline for roots of reciprocal polynomials
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 9, 2015 at 2:29 | comment | added | Dave Witte Morris | @JoeSilverman is right. My comment only applies to the original statement of the question, which did not require the polynomial to be irreducible. For the corrected statement of the question, we can say that S is the set of algebraic units that are Galois conjugate to their reciprocal. It's not immediately obvious to me whether this set is dense. | |
| May 8, 2015 at 23:25 | comment | added | Fedor Petrov | @JoeSilverman $x^3+2x^2+1$ divides $(x^3+2x^2+1)(x^3+2x+1)$ which is reciprocal. But of course, it is reducible. | |
| May 8, 2015 at 23:21 | comment | added | Joe Silverman | @DaveWitteMorris I don't think that $S$ is the set of algebraic units. Not every unit is the root of a reciprocal polynomial. For example, the roots of $x^3+2x^2+1$ are units. Being reciprocal means that the reciprocals of each root is a Galois conjugate of one of the other roots. | |
| May 8, 2015 at 23:18 | comment | added | Fedor Petrov | Well, are not polynomials of the form $x^2((x+1/x-a)^2-2b^2)=(x^2-ax+1)^2-2b^2x^2$ irreducible often enough? | |
| May 8, 2015 at 22:59 | comment | added | Igor Rivin | Sorry, I forgot the "irreducible" assumption in the hypotheses, otherwise it's trivial... | |
| May 8, 2015 at 22:40 | comment | added | Dave Witte Morris | Another way of looking at this is that $S$ is the set of algebraic units (that is, the set of algebraic integers whose reciprocal is also an algebraic integer). It is easy to see that this is dense in $\mathbb{R}$. (For example, take $\{\alpha^m \beta^n\}$ where $\alpha$ and $\beta$ are multiplicatively independent.) | |
| May 8, 2015 at 22:34 | history | answered | Fedor Petrov | CC BY-SA 3.0 |