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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
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Jan 18, 2023 at 11:08 history asked Marcos CC BY-SA 4.0