2019-08-20 01:53:52 -0500 | received badge | ● Popular Question (source) |

2019-08-09 22:49:27 -0500 | commented question | sub-module membership test over the polynomial ring $\mathbb{Z}_2[x,y,z]$ |

2019-08-07 17:21:32 -0500 | 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? |

2019-04-09 23:42:55 -0500 | 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 z + xyz, 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?I can find the resultant as y^2 |

2019-04-08 15:53:45 -0500 | received badge | ● Nice Question (source) |

2019-04-07 15:40:40 -0500 | asked a question | parametric solution for a system of polynomial equations I have the following system of equations, 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? |

2019-03-29 11:47:01 -0500 | 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 00:29:57 -0500 | asked a question | ideal membership and solution I gave sage the following ring and the ideal I found that the function h below lies in the Ideal I using 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 |

2019-03-10 06:28:01 -0500 | received badge | ● Popular Question (source) |

2019-01-25 14:32:02 -0500 | commented question | vector space basis for a quotient module Thanks. Yes, I made the correction. |

2019-01-25 00:16:59 -0500 | asked a question | codimension of an ideal or free submodule Is there an option to calculate the codimension of an ideal Sage? For example, I have the following ideal $I=(1+xy, x+y)$ in $\mathbb{Z}_{2}\left[x,y\right]$ which is a polynomial ring over the field $\mathbb{Z}_2$. How do I calculate the codimension for this simple example? I would like to generalize to free submodules if possible |

2019-01-25 00:09:16 -0500 | received badge | ● Associate Editor (source) |

2019-01-25 00:08:46 -0500 | asked a 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 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. |

2019-01-09 17:46:11 -0500 | asked a question | generators of annihilator of an ideal of a polynomial ring I have an ideal in the polynomial ring $\mathbb{F}_2[x,y,z]$ given by $I=(1+ x + y + xy; 1+y + z + yz; 1 + x + z + xz )$. The annihilator of this ideal is generated by $f_{xy}=\sum_{n,m\in \mathbb{Z}} x^n y^m$, $f_{xz}=\sum_{n,m\in \mathbb{Z}} x^n z^m$, $f_{zy}=\sum_{n,m\in \mathbb{Z}} z^n y^m$ i.e. $f_{xy} I=((1+1+1+1)f_{xy},(1+1+z+z)f_{xy},(1+1+z+z)f_{xy})=(0,0,0)$ and so on. Can I find these generators in sage? More generally, given an ideal like the above in $\mathbb{F_2}[x,y,z]$ can I find the generators of its annihilator? |

2018-11-01 21:01:06 -0500 | commented answer | solution of a system of equations in algebraic closure of GF2 Thank you. Also, you used the equations $x^n+x$ as tautological equations for arbitrary $\mathbb{GF}(n)$. I know that for $n=4$, one has $x^4+x$ but for higher powers (i.e. $\alpha$) of 2 in $n=2^\alpha$, the form of the polynomial should be different. https://en.wikipedia.org/wiki/Finite_... Is there a general form of this polynomial for an arbitrary $n$? |

2018-10-30 19:53:41 -0500 | marked best answer | solution of a system of equations in algebraic closure of GF2 How do I look for solutions of a system of equations in a particular field? For example, the following set of equations in variables ${a,b,c,d,e,f}$ $1 + a + c + e = 0, b + d + f = 0, 1 + a e + c e + a c = 0, b e + a f + c f + e d + b c + a d = 0, b f + f d + b d = 0$ have the solution $a=1,b=1,c=1,d=\omega,e=1,f=\omega^2$ where $\omega$ is the $3^{rd}$ root of unity. This lies in an extension of $\mathbb{GF}_2$, $\mathbb{GF}_4= \frac{\mathbb{GF}_2[u]}{u^2+u+1}$ where $\mathbb{GF}_2[u]$ is a polynomial ring with variable $u$ and one representation of $\mathbb{GF}_4$ is ${0,1,\omega,\omega^2}$. I have a set of equations and I want to know whether there exist solutions of these equations in an extension of Galois field $\mathbb{GF}_2$ and what are they? Is there a way to check this in Sage? |

2018-10-30 19:53:38 -0500 | commented answer | solution of a system of equations in algebraic closure of GF2 This is a very nice answer. Thanks very much. Can I check only the existence of a solution in the algebraic closure for example, without adding extra equations? I assume the answer to it is No as otherwise sage might have just solved it as this doesn't seem to be a problem whose solution's existence is much easier compared to the solution itself. |

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