Re: PolynomialMod

• To: mathgroup at smc.vnet.net
• Subject: [mg121356] Re: PolynomialMod
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Mon, 12 Sep 2011 04:20:50 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com

```On 11 Sep 2011, at 20:15, Artur wrote:

> Mayby 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 x
> 2
>  + 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
>

I think what you want to do is to use PolynomialReduce and not =

In[11]:= Last[
PolynomialReduce[1 + 3*x + 5*x^2 + 5*x^3 + 5*x^4 + 3*x^5 + x^6,
x^3 - b*x - c, x]]

Out[11]= x^2*(b^2 + 5*b + 3*c + 5) +
x*(3*b^2 + 2*b*c + 5*b + 5*c + 3) +
3*b*c + c^2 + 5*c + 1

Andrzej Kozlowski

```

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