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vector space basis for a quotient module

asked 2019-01-25 00:08:46 -0500

arpit gravatar image

updated 2019-09-16 06:45:38 -0500

FrédéricC gravatar image

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.

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The submodule in the denominator is: $$ \begin{bmatrix} 0 & 0 & 0 & 1+x+y+xy & 1+y+z+yz & 1+x+z+xz\\ 1+z & 1+x & 0 & 0 & 0 & 0\\ 0 & 1+x& 1+y & 0 & 0 & 0 \end{bmatrix} \Bbb F_2[x,y,z]^6 \ ? $$

dan_fulea gravatar imagedan_fulea ( 2019-01-25 11:50:52 -0500 )edit

Thanks. Yes, I made the correction.

arpit gravatar imagearpit ( 2019-01-25 14:32:02 -0500 )edit

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answered 2019-01-25 17:04:38 -0500

dan_fulea gravatar image

I will start an answer, but the question is not clear to me at all.

First of all, what i understand.

Let $R$ be the ring $$ R = \Bbb F_2[x,y,z] $$ fixed once for all times here to have an easy typing. Now we consider a submodule of a free module over $R$. It is not clear for me, if we work with the submodule $$ L = [R\ R\ R] \begin{bmatrix} 0 & 0 & 0 & 1+x+y+xy & 1+y+z+yz & 1+x+z+xz\\ 1+z & 1+x & 0 & 0 & 0 & 0\\ 0 & 1+x& 1+y & 0 & 0 & 0 \end{bmatrix} $$ of the "free row module" $[R\ R\ R\ R\ R\ R]=R^6$, or rather with the submodule $$ M= \begin{bmatrix} 0 & 0 & 0 & 1+x+y+xy & 1+y+z+yz & 1+x+z+xz\\ 1+z & 1+x & 0 & 0 & 0 & 0\\ 0 & 1+x& 1+y & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} R\\R\\R\\R\\R\\R \end{bmatrix} $$ of the "free column module" $\begin{bmatrix} R\\R\\R \end{bmatrix} =R^3$, but in both cases we can do the same with the corresponding module $R^n$. (Here, $n$ is either $3$ or $6$.) The wiki page Gröbner basis tells us to look at the ring $R\oplus R^n$, with the one from $R\oplus0$ and so that multiplying two elements of $0\oplus R^n$ we get zero. In sage, we can use for this the ring Q from

sage: R.<x,y,z,E1,E2,E3> = PolynomialRing(GF(2))
sage: R
Multivariate Polynomial Ring in x, y, z, E1, E2, E3 over Finite Field of size 2
sage: Q = R.quotient( [e*ee for e in [E1,E2,E3] for ee in [E1,E2,E3]] ) 
sage: Q
Quotient of Multivariate Polynomial Ring in x, y, z, E1, E2, E3 
over Finite Field of size 2 
by the ideal (E1^2, E1*E2, E1*E3, E1*E2, E2^2, E2*E3, E1*E3, E2*E3, E3^2)

Above, i took $n=3$ and there were also some manual rearrangements of output....

We now "only have to implement the generators" of $R\otimes R^n$ correspondingly, and ask for the Gröbner basis. So here is the one possibility with $n=6$, if i correctly understood the submodule.

Then one can try:

sage: R.<x,y,z,E1,E2,E3,E4,E5,E6> = PolynomialRing(GF(2))
sage: E_List = [E1, E2, E3, E4, E5, E6]
sage: A = matrix(R, 3, 6, 
....:     [0,0,0,(1+x)*(1+y),(1+y)*(1+z),(1+z)*(1+x),
....:      1+z,1+x,0,0,0,0,
....:      0,1+x,1+y,0,0,0])
sage: gens = [e*ee for e in E_List for ee in E_List] + list(A*vector(R, E_List))
sage: Q = R.quotient(gens)
sage: J = R*gens
sage: J.groebner_basis()
Polynomial Sequence with 24 Polynomials in 9 Variables

(I have to submit... loosing connection soon...)

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Asked: 2019-01-25 00:08:46 -0500

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Last updated: Aug 07