gepo

541 Reputation

4 Badges

16 years, 18 days

MaplePrimes Activity


These are answers submitted by gepo

ok, I got it.

Thank you.

That  is interesting.

but how the groebner basis works for the undertermined root.

for example:

 

alias(alpha=RootOf(Z^3+Z+1));
F := {alpha*x*y+alpha^2+1, alpha^2*x^2-1};
G := Groebner[Basis](F, plex(x,y), characteristic=2);

 

 

Yes, the problem is how to compare a real number and the root?

 

Thanks, Alex.

Thank you so much. I am trying to understand your procedure.

 

Thanks again.

from the following, it said

GF(4):=GF(2)/x^2+x+1;

So we have a field with 22 = 4 elements, namely one representation of \mathbb{F}_4. The fields elements are 0, 1, x, and x+1 (choosing their cannonical representatives). In the ring of integers, the elements 0,1,2 and 3 are chosen to represent the field. To map between both sets the substitution map, substituting x with 2 is chosen, so we have

0 <-> 0
1 <-> 1
2 <-> x
3 <-> x+1.

A polynomial Ring Z[y[1..3]]; can be used to represent \mathbb{F}_4[y_1, y_2, y_3]. Therefore, we map Z[y[1..3]]; to Z/(4)[1..3]]; and interpret each integer via the above described map. So the polynomial 3y[1] + 1y[2] + 5y[3] - 1 y[1]y[2]; is first mapped to 3y[1] + 1y[2] + 1y[3] +3 y[1]y[2]; which means the polynomial (x + 1)y1 + y2 + y3 + (x1)y1y2.

------------------------------------------------------------------------------

Is that right? Make all the elements in GF(4) correspond with a integer number from 0 through 3.

 

Thank you.

GB([x^2-2*x*z+5, x*y^2+y*z^3, 3*y^2-8*z^3],4);

       3      8    9     3      7    8      5   2      3
  [-7 z  - 4 z  + z , y z  + 8 z  - z  + 3 z , y  - 7 z ,

           3      7      8      5   2
        x z  + 3 z  - 3 z  + 6 z , x  - 2 x z + 5]

GB({x^2+2*y, x^3+5},4);

                                3         2
                         [-3 + y , x - 6 y ]

Alec

---------------------------------------------------------------------------

I cannot understand why the coefficients are not in GF(16);

Thank you.

and for the following example, how to compute the groebner basis?

F := [x^2-2*x*z+5, x*y^2+y*z^3, 3*y^2-8*z^3] over GF(2^4)

which is different from your example: the coefficients are all integers, not root of some irreducible polynomials.

alias(a=RootOf(_Z^4+_Z+1)):
Groebner:-Basis({x^2+a*y,x^3+(a^2+1)},
tdeg(y,x),characteristic=2);

Thank you.

I am not quite clear that.
for example, for a polynomial "5x^2+3y^2" over a field GF(2^4), all the coefficients should be transformed into the number in Field GF(16), i.e., coefficients (5 and 3) should be changed to the number in GF(16)? if so, how to change?

1.Well, I understand how to compute Groebner Basis over R, or GF(p), p is a prime number and actually I have implemented the B's algorithm and made the result of B's algorithm unique. The problem now is I am not quite clear how to compute Groebner Basis over GF(2^m).

2,Do you know the differences between ideal over R(real number) and ideal over R[x]?

Thank you

I benefit a lot from your guys answers. Thanks a lot.

BY the way, do you know how to computer the Groebner Basis over GF(2^m) manually?  or could you give me a link about that?

Yes, I also tried that. I met the same problems.

Maybe there are other ways to issue it.

 

Thank you, Alex. I really appreciate your help.

Well, the problem is I have already completed part of the programming. So now it is difficult for me to transfer my codes to sage or other platform.

If maple cannot offer such operations, I will try to implement it in maple.  But now I have some problems.

I am not quite understanding the concept of "a polynomial over Galois Field"? whether"over some field" means the coefficients of the polynomial should be in some field? For example, make 4x^2+5x+6 over GF(4). Whether this means the coefficients 4,5,6 should be changed in GF(4). Here, we should notice that GF(4) is not Z_4.

Thank you so much. But What i want to do is to do some programming using maple.

Any other ideas?

o,yes, i made a mistake.

 

thanks a lot.

to  Robert Israel:

yes, when you want to newline, shift+enter instead of enter can get:

> F:=proc(L)

>local LL;

>  LL:=subsop(1=NULL,LL);
> end proc:

> L:=[x,y,z,d];
                               L := [x, y, z, d]

>L:= F(L);
                                   [y, z, d]
 >L:= F(L);
                                    [z, d]
 >L:= F(L);
                                      [d]
 >F(L);
                                      []
 

actually, it is the same as:

F := proc(L) local LL; LL:=subsop(1=NULL,LL);(wihtout "enter")end proc;

 

1 2 3 4 Page 3 of 4