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 Author: redtrumpet [ Thu Jul 06, 2017 11:43 am ] Post subject: Polynomial division over GF(p^n) Why is it not possible to divide polynomials over GF(p^n), if n >= 2? I get the following error:Code:> ring r = (7^2,a),x,dp;> poly f = x2+1;> poly g = x+1;> f/g;   ? not implemented   ? error occurred in or before STDIN line 14: `f/g;`Any help would be appreciated.

 Author: gorzel [ Fri Jul 07, 2017 3:40 pm ] Post subject: Re: Polynomial division over GF(p^n) Without going into detail, this is due to the fact that the polynomials have another presentationover Galoisfields and different algorithms / implementations are involved. Also factorize is not at your disposal. Note that in general f/g only gives the quotient without remainder. Most likely that is not what you want, but you are not lost here. There is the command division http://www.singular.uni-kl.de/Manual/latest/sing_226.htm#SEC266which performs the task. It is important to use a global odering dp as you do.Code: factorize(f);   ? not implemented   ? error occurred in or before STDIN line 6: `factorize(f);`> list L = division(f,g);> L;[1]:   _[1,1]=x-1[2]:   _[1]=a16[3]:   _[1,1]=1> typeof(L[1]);matrix> typeof(L[1][1,1]);poly> typeof(L[2]);ideal> f==g*L[1][1,1] + L[2][1];1> (x-1)*(x+1) + 2;x2+1> a16;a16> number(2);a16(The third value L[3] is a unit matrix in global ordering, but the result is differentif you would work in ring rads = (49a,),x,ds;Try it!)You may also define this finite field as an quadratic extension of Z_7 by the minimal polynomial displayed from the ring itself.Code:> setring r;> basering;// coefficients: ZZ/49[a]//   minpoly        : 1*a^2+6*a^1+3*a^0// number of vars : 1//        block   1 : ordering dp//                  : names    x//        block   2 : ordering CWith this approach, the elements are not represented as a power of the primitiveelement a, but now f/g and factorize work.Code:> ring ra49 = (7,a),x,dp; minpoly = a2+6a+3;// compare with above> a16;2> number(2);2> poly f = x2+1;> poly g = x+1;> f/g;x-1> factorize (f);[1]:   _[1]=1   _[2]=x+(-a-3)   _[3]=x+(a+3)[2]:   1,1,1> division(f,g);[1]:   _[1,1]=x-1[2]:   _[1]=2[3]:   _[1,1]=1> f == (x-1)*g +2;1

 Author: redtrumpet [ Fri Jul 07, 2017 3:59 pm ] Post subject: Re: Polynomial division over GF(p^n) Thank you for your answer, this definitely solves my problem But is this kind of pitfall anywhere documented? I tried to find something, but using google to find problems with Singular is kind of hard.

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