Find a reduced Groebner basis

Question

Problem: Let ideal $I = \langle f_1,f_2,f_3 \rangle \subset \mathbb{Q}[x,y,z]$ in which $f_1 = x-3y-4z, f_2 = -x+y+6z-2, f_3 = x-z+2$.

1. Find a reduced Groebner basis of $I$ with lexicographic ordering.
2. Show that $g = -2y+14z-10-x^3+3x^2z-6x^2-3xz^2+12xz+z^3-6z^2 \in \mathbb{Q}[f_1,f_2,f_3]$.

My attempt: I have computed \begin{cases} &x-3y-4z\\ &-x+y+6z-2\\ &x-z+2 \end{cases}

$\Leftrightarrow$\begin{cases} x&-3y-4z\\ &-2y+2z-2\\ & 3y+3z+2 \end{cases}

$\Leftrightarrow$\begin{cases} x-3y&-4z\\ -2y&+2z-2\\ &6z-1 \end{cases}

$\Leftrightarrow$\begin{cases} x+\frac{11}{6}\\ y+\frac{5}{6}\\ z-\frac{1}{6} \end{cases}

Let $g_1=x+\frac{11}{6},g_2=y+\frac{5}{6},g_3=z-\frac{1}{6}$, then $G = \{g_1,g_2,g_3\}$ is a reduced Groebner basis. But I have used Maple to computed the reduced Groebner basis of $I$ and the result is $1$. Is my attempt true? Thank all!

Answer 1

This is a system of linear equations whose unique solution is $x = -11/6$, $y = -5/6$ and $z = 1/6$, so it is easy to check that your answer is correct. (It is.)

But I have used Maple to computed the reduced Groebner basis of I and the result is 1.

I do not know why this would happen. Maybe your input into Maple contained a typo.

Answer 2

Well, I've tried Singular:

ring r = 0, (x,y,z), lp;

ideal i = x-3y-4z, -x+y+6z-2, x-z+2;

std(i);

The result is

6z-1,

y-5z,

x-z+2.

For number 2, divide $g$ into the Groebner basis. Groebner bases have the property that $g$ lies in the ideal generated by the Groebner basis if and only if the remainder of the division is zero.