Timeline for Factoring $x^p H(x) + x^q B(x) + T(x)$ over a finite field
Current License: CC BY-SA 4.0
6 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
May 4, 2019 at 13:53 | history | edited | Seva | CC BY-SA 4.0 |
added 218 characters in body; edited title
|
May 4, 2019 at 6:53 | comment | added | alpoge | Ah! I tacitly assumed the discriminant was nonzero when I saw split completely (I’m not sure why). Alas! | |
May 4, 2019 at 6:30 | comment | added | Seva | @alpoge: Replacing $x^p$ with $x$, you change the polynomial. The new polynomial will have the same roots as the original one, but the multiplicities can be affected, and full reducibility is not necessarily preserved (consider $x^{p-1}(x-1)$). | |
May 3, 2019 at 21:32 | comment | added | alpoge | Sorry, I am probably confused and if so forgive me for what is likely a dumb remark, but for x\in \F_p you may replace the x^p term by simply x, and then you are asking that x^q B(x) + (x H(x) + T(x)) have q + \deg B roots mod p, i.e. you are asking for conditions on the leading \deg B (< 0.1 (p-1) ) coefficients of a polynomial over \F_p (of degree < p) of the form const * \prod_{\alpha\in S} (x - \alpha), where S\subseteq \F_p. [This recovers the 2^p vs. p^{0.1 p} comparison, of course.] It would be particularly nice if \deg B = p-1-q, in which case up to translation you’d have x^{p-1} - 1. | |
May 3, 2019 at 13:50 | answer | added | Fedor Petrov | timeline score: 3 | |
May 3, 2019 at 13:22 | history | asked | Seva | CC BY-SA 4.0 |