Timeline for Vanishing of a product of cyclotomic polynomials in characteristic 2
Current License: CC BY-SA 4.0
13 events
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Jan 18, 2023 at 22:24 | history | became hot network question | |||
Jan 18, 2023 at 15:15 | comment | added | LSpice | @PeterTaylor, re, sure, and you do so in your answer, to good effect. But, since the question specifies characteristic $2$ from the outset, it can hardly be said to be wrong if written as it is. (That is, to turn on a very fine point, I would argue that it could be $\frac{x^i - 1}{x - 1}$, but not necessarily that it should be, since as written we already get your desiderata (a) and (b) in the setting of the question.) | |
Jan 18, 2023 at 15:13 | comment | added | Peter Taylor | @LSpice, in my opinion it's cleaner to give the definition without specifying the characteristic of $\mathbb{F}$ and then specialise later. The way the question's phrased seems to me to fit this model, except that $p_{n,j}$ isn't necessarily a polynomial except in characteristic 2. | |
Jan 18, 2023 at 14:55 | comment | added | LSpice | @PeterTaylor, re, we're in characteristic $2$. | |
Jan 18, 2023 at 14:46 | answer | added | Peter Taylor | timeline score: 4 | |
Jan 18, 2023 at 14:03 | vote | accept | Marcos | ||
Jan 18, 2023 at 13:04 | answer | added | Ofir Gorodetsky | timeline score: 5 | |
Jan 18, 2023 at 12:48 | comment | added | Peter Taylor | Should $\frac{x^i+1}{x+1}$ be $\frac{x^i-1}{x-1}$ everywhere, to (a) get the cyclotomics mentioned in the title; (b) get polynomials rather than rational polynomials? | |
Jan 18, 2023 at 12:03 | history | edited | Marcos | CC BY-SA 4.0 |
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Jan 18, 2023 at 12:02 | comment | added | Marcos | @OfirGorodetsky ups, right... it is everything at $x=1$ | |
Jan 18, 2023 at 11:55 | comment | added | Ofir Gorodetsky | Isn't $p_{n.j}(0)$ always $1$, because $(x^i+1)/(x+1)$ is $1$ at $0$? | |
Jan 18, 2023 at 11:25 | history | edited | Marcos | CC BY-SA 4.0 |
added 6 characters in body
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Jan 18, 2023 at 11:08 | history | asked | Marcos | CC BY-SA 4.0 |