Consider the equations \[\begin{align} x^2+y&=-2\\ 2xy&=y^2-2y. \end{align}\]
Let \(I\) be the ideal of \(\mathbb{C}[x,y]\) generated by these equations. Find the Gröbner basis for \(I\) relative to lexicographic order where \(x>y\).
Find a Gröbner basis for \(\mathbb{C}[y]\cap I\).
Find all solutions to these equations that lie in \(\mathbb{C}^2\).
Find a vector space basis for \(\mathbb{C}[x,y]/I\).
Solution
Let \[ f_1=x^2+y+2, \qquad f_2=2xy-y^2+2y. \] We use lexicographic order with \(x>y\).
1. The ideal \(I=(f_1,f_2)\) has Gröbner basis\[ G=\left\{ g_1=x^2+y+2, \quad g_2=xy-\frac12y^2+y, \quad g_3=y^3+12y \right\}. \]
First \(g_1=f_1\) and \(g_2=f_2/2\), so \(g_1,g_2\in I\). The \(S\)-polynomial of \(g_1\) and \(g_2\) is \[ S(g_1,g_2)=yg_1-xg_2. \] Reducing first by \(g_2\) using \[ xy\equiv \frac12y^2-y, \qquad xy^2\equiv \frac12y^3-y^2, \] one obtains \[ S(g_1,g_2)\equiv \frac14y^3+3y =\frac14(y^3+12y). \] Hence \(g_3\in I\).
The leading monomials are \[ \operatorname{LM}(g_1)=x^2,\qquad \operatorname{LM}(g_2)=xy,\qquad \operatorname{LM}(g_3)=y^3. \] It remains, by Buchberger’s criterion, to check the three \(S\)-pairs reduce to zero. The first does by the calculation above. For the other two, \[ S(g_1,g_3)=y^3g_1-x^2g_3 =y^4+2y^3-12x^2y. \] Using \(x^2\equiv-y-2\) modulo \(g_1\) gives \[ S(g_1,g_3)\equiv y^4+2y^3+12y^2+24y, \] and then \(y^3\equiv-12y\) modulo \(g_3\) makes this zero. Likewise, \[ S(g_2,g_3)=y^2g_2-xg_3 =-\frac12y^4+y^3-12xy. \] Reducing \(y^4\) and \(y^3\) by \(g_3\) and \(xy\) by \(g_2\) gives \[ 6y^2-12y-12\left(\frac12y^2-y\right)=0. \] Thus every \(S\)-pair reduces to zero, so \(G\) is a Gröbner basis. This proves part (a).
2. A Gröbner basis for the elimination ideal \(I\cap\mathbb C[y]\) is\[ \boxed{\{y^3+12y\}}. \]
For lex order \(x>y\), the elimination theorem says that if \(G\) is a Gröbner basis of \(I\), then \[ G\cap\mathbb C[y] \] is a Gröbner basis of \(I\cap\mathbb C[y]\). From 1, \[ G\cap\mathbb C[y]=\{g_3\}. \] This proves part (b).
3. The possible \(y\)-coordinates of solutions are\[ y=0,\qquad y=2\sqrt3\,i,\qquad y=-2\sqrt3\,i. \]
Every common zero of \(I\) must vanish on the elimination polynomial \[ g_3=y^3+12y=y(y^2+12). \] Its roots over \(\mathbb C\) are exactly the three displayed values.
4. The complete solution set in \(\mathbb C^2\) is\[ \boxed{ (i\sqrt2,0),\ (-i\sqrt2,0),\ (-1+i\sqrt3,2i\sqrt3),\ (-1-i\sqrt3,-2i\sqrt3). } \]
If \(y=0\), then \(g_1=0\) gives \[ x^2+2=0, \] so \(x=\pm i\sqrt2\).
If \(y\ne0\), then \(g_2=0\) implies \[ x=\frac y2-1. \] For \(y=\pm2i\sqrt3\), this gives \[ x=-1\pm i\sqrt3. \] Conversely, substituting these four pairs into \(g_1,g_2,g_3\) gives zero, so they are all the common zeros. This proves part (c).
5. A vector-space basis of \(\mathbb C[x,y]/I\) is\[ \boxed{\{1,x,y,y^2\}}. \]
For a Gröbner basis, the residue classes of monomials not divisible by any leading monomial form a basis of the quotient. By 1 the initial ideal is \[ \operatorname{in}(I)=(x^2,xy,y^3). \] The monomials not divisible by \(x^2\), \(xy\), or \(y^3\) are exactly \[ 1,\ x,\ y,\ y^2. \] Therefore their residue classes form a \(\mathbb C\)-basis of \(\mathbb C[x,y]/I\). This proves part (d).