Content deleted Content added
Line 88:
:If the coefficients of a polynomial belong to a field {{mvar|F}}, all the factors of the square-free factorization have their coefficients in {{mvar|F}}. This results from the fact that the main tool of the algorithms is the polynomial greatest divisor. In your example, the input polynomial is the product of distinct irreducible quadratic factors. This implies that the results of the square-free factorization and the distinct-degree factorization are both the input polynomial. [[User:D.Lazard|D.Lazard]] ([[User talk:D.Lazard|talk]]) 19:15, 14 July 2020 (UTC)
::As I posted in the above section, for f(x) = x^32 + x^22 + x^2 + x + 1 with coefficients in GF(2^16), square free factorization, f '(x) = 1, and the first step of distinct degree factorization, <math>g=\gcd(f, x^{q}-x)</math> = 1. So those two methods are not working in GF(q), but instead only in GF(p), and the size of the coefficients being used to hold 0 or 1 doesn't matter. [[User:Rcgldr|Rcgldr]] ([[User talk:Rcgldr|talk]]) 01:07, 15 July 2020 (UTC)
:::Please, read the definition of the distinct-degree factorization. It is not the complete factorization. In your example the result of the distinct-degree factorization of f(x) is f(x) itself, independently whether the coefficients are considered to belong to GF(q) or GF(2). This does not means that the algorithm does not work This means that you confuse the specifications of the square-free factorization and of the complete factorization. [[User:D.Lazard|D.Lazard]] ([[User talk:D.Lazard|talk]]) 08:45, 15 July 2020 (UTC)
| |||