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Topic review  Polynomial division over GF(p^n) 
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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.
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.




Posted: Fri Jul 07, 2017 3:59 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 presentation over 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.unikl.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]=x1 [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
> (x1)*(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 different if 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 C
With this approach, the elements are not represented as a power of the primitive element 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; x1 > factorize (f); [1]: _[1]=1 _[2]=x+(a3) _[3]=x+(a+3) [2]: 1,1,1 > division(f,g); [1]: _[1,1]=x1 [2]: _[1]=2 [3]: _[1,1]=1 > f == (x1)*g +2; 1
Without going into detail, this is due to the fact that the polynomials have another presentation over Galoisfields and different algorithms / implementations are involved. Also factorize is not at your disposal.
Note that in general [i]f/g[/i] only gives the quotient without remainder. Most likely that is not what you want, but you are not lost here. There is the command [b]division[/b] [url]http://www.singular.unikl.de/Manual/latest/sing_226.htm#SEC266[/url] which performs the task. It is important to use a global odering [b]dp[/b] 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]=x1 [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
> (x1)*(x+1) + 2; x2+1
> a16; a16 > number(2); a16 [/code] (The third value [i]L[3][/i] is a unit matrix in global ordering, but the result is different if you would work in [b]ring rads = (49a,),x,ds;[/b]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 C [/code] With this approach, the elements are not represented as a power of the primitive element [i]a[/i], 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; x1 > factorize (f); [1]: _[1]=1 _[2]=x+(a3) _[3]=x+(a+3) [2]: 1,1,1 > division(f,g); [1]: _[1,1]=x1 [2]: _[1]=2 [3]: _[1,1]=1 > f == (x1)*g +2; 1 [/code]




Posted: Fri Jul 07, 2017 3:40 pm 





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.
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;` [/code]
Any help would be appreciated.




Posted: Thu Jul 06, 2017 11:43 am 





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