Elimination Example

Compute a lexicographic Gröbner basis of the ideal

(1)
\begin{align} I = \langle f_1:=xy^2-xz+y,f_2:=xy-z^2,f_3:=x-yz^4\rangle \in \mathbb{Q}[x,y,z] \end{align}

a) Does the system of polynomial equations $\{f_1=0,f_2=0,f_3=0\}$ have finitely many solutions? If so, find all the solutions over the complex numbers.
b) Compute a basis of $\mathbb{Q}[x,y,z]/I$.
c) Find all weight vectors $\omega\in\mathbb{R}^n$ such that $\operatorname{in}_\omega(I)=\operatorname{in}_\succ(I)$ where $\succ$ is the lexicographic order used in part a). What kind of geometric structure does this set have?


Solution:

a)/b) So here is a code in Macaulay2 which computes a Gröbner basis of $I$ with respect to the lexicographic order $x\succ y\succ z$ and a basis of the quotient ring $\mathbb{Q}[x,y,z]/I$:

R = QQ[x,y,z, MonomialOrder => Lex];
I = ideal (x*y^2-x*z+ y, x*y-z^2, x- y*z^4)
gens gb I
M = R/I
basis M

(Side: For some reason, the command "basis R/I" did not work on my computer. Does anyone know the reason for this?)
The resulting Gröbner basis is
| z7-z4-z2 yz5-yz2-y y2+z5-z3-z2 x-yz4 |

Since the first polynomial is univariate and the system is triangularized, it has only finitely many solutions over the complex numbers. Therefore, the quotient ring is finite dimensional and a basis as a $\mathbb{Q}$-vector space is given by the following monomials:
| 1 y yz yz2 yz3 yz4 z z2 z3 z4 z5 z6 |

c) A weight order given by $\omega\in\mathbb{R}^n$ gives the same initial ideal as the lexicographic order if and only if it gives the same initial term for every element of the reduced Gröbner basis computed above.
Therefore, we get a number of strict inequalities with integer coefficients for $\omega$. So the set of all weight vectors giving the same initial ideal is an open polyhedral cone.
In the above example, the defining inequalities are:

(2)
\begin{align} 7\omega_3>4\omega_3, 7\omega_3>2\omega_3 \\ \omega_2+5\omega_3>\omega_2+2\omega_3, \omega_2+5\omega_3>\omega_2\\ 2\omega_2>5\omega_3, 2\omega_2>3\omega_3, 2\omega_2>2\omega_3\\ \omega_1>\omega_2+4\omega_3 \end{align}

This system of inequalities can be reduced to

(3)
\begin{align} \omega_3>0, \omega_2>\frac52 \omega_3, \omega_1>\omega_2+4\omega_3 \end{align}
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License