ideal membership and solution

arpit
163 ●5 ●12 ●18

I gave sage the following ring and the ideal

R.<x,y,z>=GF(2)[];
f=1 + z + y*z + y^2*z + z^2 + y*z^2; 
g=1 + x + y^2 + z^2;
I = R.ideal(f,  g)

I found that the function h below lies in the Ideal I using

h=1 + y + z + x*z + y*z + x*y*z + y^2*z + y*z^2;
h in I

I know that in general, finding polynomials $a(x)$ and $b(x)$ such that $h = a f+ b g$ might be hard, but can I find the solutions for $a$ and $b$ to a certain degree of these polynomials, if they exist? I was wondering if sage can check this more efficiently

add a comment

answered 6 years ago

rburing

10964 ●4 ●80 ●219 https://www.rburing.nl/

Yes, this can be done algorithmically. First you compute a Groebner basis $\langle g_1, \ldots, g_r \rangle$ of $I$ , and remember the expressions $g_i = a_i f + b_i g$ . Then you do multivariate polynomial division of $h$ by $\langle g_1, \ldots, g_r \rangle$ . This yields $h = c_1 g_1 + \ldots + c_r g_r = (c_1a_1 + \ldots c_ra_r)f + (c_1b_1 + \ldots c_rb_r)g.$

In your example one Groebner basis of $I$ is $\langle f,g,f_1,f_2,f_3 \rangle$ where $f_1 = S(f,g) = f-zg$ , $f_2 = S(f_1,f) - zg = (y-z)f - z(y+1)g$ , $f_3 = S(f_1,f_2) = (x - z(y-z))f + z(z(y+1) - x)g$ . Multivariate polynomial division of $h$ by $\langle f,g,f_1,f_2,f_3 \rangle$ yields $h = f + f_2 = (y-z+1)f - z(y+1)g.$

This can all easily be automated in Sage (but I don't have time to do it now).

Comments

@rburing thanks for the answer. This is useful. How do I do the last step of multivariate polynomial division in Sage?

arpit ( 6 years ago )

add a comment

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

ideal membership and solution

1 Answer

Comments

Your Answer

Question Tools

Stats

Related questions

ideal membership and solution savecancel

1 Answer

Comments

Your Answer

Question Tools

Stats

Related questions

ideal membership and solution