Consider the following monomial ideal: \[I=\langle x^3yz,\ xy^2z,\ xy^3z^2,\ x^3y^2z^2,\ x^4,\ y^4,\ z^3,\ x^2yz^4 \rangle. \] Find a minimal basis for \(I\).
We observe that \(x^3y^2z^2\) and \(xy^3z^2\) are divisible by \(xy^2z\), so they can be removed. We are left with \[\{ x^3yz,\ xy^2z,\ x^4,\ y^4,\ z^3 \}\] where no monomial divides any other. This is the desired minimal basis.
Determine if the set \(G = \{ x^2y-y+1, xy^2-3x+2 \}\) in \(\mathbb{Q}[x,y]\) is a Gröbner basis for the lexicographic ordering.
We compute the S-polynomial \[\begin{align*} &S(x^2y-y+1,xy^2-3x+2)\\ =&y(x^2y-y+1)-x(xy^2-3x+2)\\ =&3x^2-2x-y^2+y. \end{align*}\] Applying the division algorithm, we see that \(3x^2-2x-y^2+y\) is also the remainder. Thus, Buchberger’s criterion fails and we do not have a Gröbner basis.
Find a Gröbner basis for the ideal \(I = \langle x^2-y,\ xy-x \rangle \subseteq \mathbb{Q}[x,y]\) for the lexicographic ordering.
Set \(g_1=x^2-y\), \(g_2=xy-x\), and \(G_0=(g_1,g_2)\). We find \(S(g_1,g_2)=yg_1-xg_2=x^2-y^2\). Set \(g_3=\overline{S(g_1,g_2)}^G_0=-y^2+y\). Let \(G_1=(g_1,g_2,g_3)\).
Now we have:
Thus \((g_1,g_2,g_3)\) is a Gröbner basis.
The set \[G = \{ x^2y + 4xy^6 - 3yz,\ 4xy^2+z^8,\ 5y^2z+z^3,\ 6z^2 \} \] in \(\mathbb{Q}[x,y]\) is a Gröbner basis using the lexicographic ordering (you do not need to check this). Find a Gröbner basis for \(\langle G \rangle\).
By computing the remainder of the second and third generators modulo the other polynomials, we find \[G' = \{ x^2y + 4xy^6 - 3yz,\ 4xy^2,\ 5y^2z,\ 6z^2 \} \] which is also a Gröbner basis. Now we divide the first generator by the other three to obtain \[G'' = \{ x^2y - 3yz,\ 4xy^2,\ 5y^2z,\ 6z^2 \}. \] Finally, we rescale so the leading coefficients are all \(1\): \[G''' = \{ x^2y - 3yz,\ xy^2,\ y^2z,\ z^2 \} \] This is now a reduced Gröbner basis.
The set \[G = \{ x^2 - z^4,\ xz - z^3,\ y^3 - z^3,\ yz - z^4,\ z^7 - z \} \] in \(\mathbb{Q}[x,y]\) is a Gröbner basis using the lexicographic ordering (you do not need to check this). Determine if \(f = x^4+xz\) is an element of the ideal \(I=\langle G \rangle\).
The division algorithm produces: \[x^4+xz= (x^2 + z^4)(x^2 - z^4)+(1)(xz - z^3)+(z)(z^7 - z) + (z^3 + z^2).\] Since the remainder is \(z^3+z^2\) and \(G\) is a Gröbner basis, we conclude that \(f \notin I\).
Find all real solutions \((x,y,z) \in \mathbb{R}^3\) of the following system: \[\begin{align*} xyz+2y-4x &= 4 \\ y^2z^2-4y^2+yz &= 2 \\ z^4-3z^2 &= 4 \ . \end{align*}\] (Hint: you do not need to compute a Gröbner basis!)
The quadratic formula gives \(z^2=4\) or \(z^2=-1\). Since we only want real solutions, we are interested only in \(z=\pm 2\). If \(z=2\), the second equation becomes \(2y=2\). Thus \(y=1\). The first equation is now \(-2x+2=4\) so \(x=1\). If \(z=-2\), the second equation becomes \(-2y=2\) so \(y=-1\). The first equation is now \(-2x-2=4\), so \(x=-3\). Thus the only two solutions are \((-3,-1,-2)\) and \((1,1,2)\).
Consider the ideal \(I = \langle G \rangle \subseteq \mathbb{Q}[x,y,z]\) where \[G = \{ x^2+x+2z-1,\ xz-x+2z-2,\ y-z,\ 2z^2-z-1\}\] is a Gröbner basis using the lexicographic ordering (you do not need to check this). Find \(I \cap \mathbb{Q}[y,z]\) and \(I \cap \mathbb{Q}[z]\).
By the Elimination Theorem, we have \[I \cap \mathbb{Q}[y,z] = \langle y-z,\ 2z^2-z-1 \rangle\] and \[I \cap \mathbb{Q}[z] = \langle 2z^2-z-1 \rangle.\]
Consider the ideal \(I = \langle G \rangle \subseteq \mathbb{C}[x,y,z]\) where \[G = \{ xy+x-z,\ xz-x+y+z-1,\ y^2+yz+z^2-1\}\] is a Gröbner basis using the lexicographic ordering (you do not need to check this). Describe the set of all partial solutions \((y,z)=(a_2,a_3)\) that extend to full solutions \((a_1,a_2,a_3) \in \mathbb{V}(I)\).
Writing the elements of \(G\) as polynomials in \(x\) with coefficients in \(\mathbb{C}[y,z]\) we have \((y+1)x-z\), \((z-1)x+y+z-1\), and a polynomial with no \(x\) terms. Thus, a solution \((a_2,a_3)\) of \(y^2+yz+z^2-1\) extends if \((a_2,a_3) \notin \mathbb{V}(y+1,z-1)\). Note that \((-1,1)\) is a partial solution since it satisfies \(y^2+yz+z^2-1\). Thus, it remains to check if \((-1,1)\) extends. However, \(xy+x-z=x(-1)+x-1=-1 \ne 0\) in this case. Thus the whole curve \(\mathbb{V}(I)\) extends except the point \((-1,1)\).