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2020-04-13 00:57:46 +0200 asked a question Finding polynomial solutions that belong to an ideal

I have a system of equations in which the variables belong to a certain ideal of a polynomial ring over a field. We can call this ideal $I$ and its generators $c_1$, $c_2$ and $c_3$. Let's take the ring to be $\mathbb{F}_2[x_1,x_2,x_3]$ and the ideal to be $I=<x_1x_2x_3-1,x_2-x_1,x_1-1>$ . Let's say the equations are $g_1+x_1w_2+x_2w_3 = x_1-1$, $g_2+x_2w_1+x_3w_3 = x_2-1$ and $g_3+x_1w_1+w_2 = x_3-1$ and one needs to find solution to the above set of equations with variables $g_i$ and $w_i$ inside the ideal $I$. There are of course more variables than equations here. One obvious solution is $g_1=x_1-1,g_2=x_2-1, g_3=x_3-1$. How does one find the full set of solutions? I thought of implementing this as a syzygy problem where I take $x_1-1$ and so on on the left but that seems to be not ideal since it is not clear whether I find all solutions or not. To simplify the problem, we can choose a cut-off for degree of polynomial solutions, for example, 1 or 2.

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2020-04-04 19:23:39 +0200 asked a question a basis for quotient module/vector space

I have a ring (field) $R$, a polynomial ring $R[x_1,x_2,...,x_n]$ and a quotient module (vector space) $R[x_1,x_2,...,x_n]/I$ where $I$ is an ideal of $R[x_1,x_2,...,x_n]$ . For the case, when $R$ is a field, the basis of the quotient vector space can be found to consist of the cosets of monomials which are not divisible by leading monomials of grobner basis of $I$. A similar theorem exists for the case when $R$ is a ring with effective coset representatives. For example, the ring of integers. Can I find these coset representatives that form the basis for the quotient module/vector space in sage?

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2020-02-18 19:30:48 +0200 asked a question intersection of two free submodules

I have two modules found as follows

 F=GF(2);R.<x,y,z> = PolynomialRing(F)
 f1 = 1+z;g1=1+y;h1=0;
 I1 = Ideal([f1,g1,h1])
 M1 = I1.syzygy_module(); M1

[ 0 0 1]

[y + 1 z + 1 0]

 F=GF(2);R.<x,y,z> = PolynomialRing(F)
 f2 = 0;g2=1+y;h2=1+x;
 I2 = Ideal([f2,g2,h2])
 M2 = I2.syzygy_module(); M2

[ 1 0 0]

[ 0 x + 1 y + 1]

Is it possible to find the intersection of two such submodules $M_1$ and $M_2$ in sage? Another possibility would be to find the syzygy of the module generated by vectors (f1,g1,h1) and (f2,g2,h2).

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2019-08-10 05:49:27 +0200 commented question sub-module membership test

over the polynomial ring $\mathbb{Z}_2[x,y,z]$

2019-08-08 00:21:32 +0200 asked a question sub-module membership test

I have a submodule of the module $\mathbb{Z}_2[x,y,z]^3$, which can be specified by its 6 generators that are the columns below \begin{array}{cccccc} 0 & 0 & 0 & 1+x+y+xy & 1+y+z+yz & 1+x+z+xz\newline 1+z & 1+x & 0 & 0 & y+z & 0\newline 0 & 1+x& 1+y & x+y & 0 & z^2 \end{array}

How can I implement the sub-module membership test in sage? For example, I want to check whether \begin{array}{c} x+z \newline x+y\newline y+z \end{array} belongs to the above submodule or not?

2019-08-07 23:53:57 +0200 edited question vector space basis for a quotient module

For my question, let's say I have the following quotient module as an example,

\begin{align} & \frac{\left(\mathbb{Z}_{2}\left[x,y,z\right]\right)^{3}}{\left(\begin{array}{cccccc} 0 & 0 & 0 & 1+x+y+xy & 1+y+z+yz & 1+x+z+xz\newline 1+z & 1+x & 0 & 0 & 0 & 0\newline 0 & 1+x& 1+y & 0 & 0 & 0 \end{array}\right)} \end{align}

where $\mathbb{Z}_{2}\left[x,y,z\right]^{3}$ is a polynomial ring in variables $x,y,z$ over field $\mathbb{Z}_2$. I am interested in calculating the Groebner basis of the submodule in the denominator using Sage and I can do the rest. I am finally interested in finding the vector space basis of the quotient module or its dimension. If that is also possible directly using Sage, it will be great.

2019-04-10 06:42:55 +0200 commented answer parametric solution for a system of polynomial equations

@slelievre, thanks for the answer. But I want to also know how to apply it more generally i.e. can I also find the variety in the cases when I don't know the form of the parametrization? for example, if the polynomials are
P2=x + y + z + yz
Q2=1 + y + x
y + z + xz + xyz,
I can find the resultant as
z^2 + y^2 + z^2 + y + z + 1 but how do I find the one dimensional variety in this case when I don't know a suitable parametrization to begin with?

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2019-04-07 22:40:40 +0200 asked a question parametric solution for a system of polynomial equations

I have the following system of equations,

1+x+y+z==0, 1+xy+yz+xz==0

which I want to solve in the extension field of GF(2) for example. There is a parametric solution of these equations in terms of the parameter s as x=1+s, y=1+$\omega$ s, z=1+$\omega^2$s where $s$ is the parameter and $\omega^2+\omega+1=0$. How can I modify the following for Sage to be able to output parametric solutions like this one?

R.<x,y,z> = PolynomialRing(GF(4))
I = R.ideal([1 + x + y + z, 1 + x*y + y*z + x*z])
2019-03-29 17:47:01 +0200 commented answer ideal membership and solution

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

2019-03-28 06:29:57 +0200 asked a question ideal membership and solution

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