Re: PolynomialMod
- To: mathgroup at smc.vnet.net
- Subject: [mg121349] Re: PolynomialMod
- From: Artur <grafix at csl.pl>
- Date: Mon, 12 Sep 2011 04:19:34 -0400 (EDT)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <201109111128.HAA12187@smc.vnet.net> <52BDE91D-1ACE-471D-A992-DF8902ADBE7D@mimuw.edu.pl>
- Reply-to: grafix at csl.pl
Maybe I was used wrong example Example 1. f = 1 + 3 x + 5 x2 + 5 x3 + 5 x4 + 3 x5 + x6; If we have smaller order equation for example x^5-x-1=0 that x^5=x+1 multiply both sides by x x^6=x^2+x if we substitute in f x^6->x^2+x and x^5->x+1 we have f = 1 + 3 x + 5 x2 + 5 x3 + 5 x4 + 3 (x+1) +(x^2+x); now after reduction f=4 + 7 x + 6 x^2 + 5 x^3 + 5 x^4 That same result we will obtain uses PolynomialMod[1 + 3 x + 5 x^2 + 5 x^3 + 5 x^4 + 3 x^5 + x^6, x^5 - x - 1] of course f=0 is divisible by my p from previous example but we can do similar cycle of substitutions as in Example 1 but if f<>0 isn't divisible in my example let symbolically x^3-b x-c==0 x^3=b x+c x^4=b x^2+c x x^5=b x^3+c x^2=b (b x+c)+c x^2=b^2 x+b c+c x^2 x^6=b^2 x^2+b c x+c x^3=b^2 x^2+b c x+c (b x+c)=b^2 x^2+2 b c x+c^2 and apply these equations to starting function f f=1 + 3 x + 5 x2 + 5 (b x+c) + 5 (b x^2+c x) + 3 (b^2 x+b c+c x^2) + (b^2 x^2+2 b c x+c^2) after reduction we have finally result (Collect by x) f=(1 + 5 c + 3 b c + c^2) + (3 + 5 b + 3 b^2 + 5 c + 2 b c) x + (5 + 5 b + b^2 + 3 c) x^2 generally g=0 and f is some function (not necessary zero) Artur Jasinski W dniu 2011-09-11 16:48, Andrzej Kozlowski pisze: > f = 1 + 3 x + 5 x2 + 5 x3 + 5 x4 + 3 x5 + x6; > > p = -1 + (-(3/2) - (I Sqrt[23])/2) x + (3/2 - (I Sqrt[23])/2) x2 + x3;
- References:
- PolynomialMod
- From: Artur <grafix@csl.pl>
- PolynomialMod