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ideal membership and solution

asked 2019-03-28 06:29:57 +0200

arpit gravatar image

I gave sage the following ring and the ideal

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

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answered 2019-03-28 13:56:30 +0200

rburing gravatar image

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).

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

arpit gravatar imagearpit ( 2019-03-29 17:47:01 +0200 )edit

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Asked: 2019-03-28 06:29:57 +0200

Seen: 542 times

Last updated: Mar 28 '19