Consider the equations \[\begin{align} x^2+2y^2&=2\\ x^2-xy+y^2&=1. \end{align}\]
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 \(y>x\).
Find a Gröbner basis for \(\mathbb{C}[x]\cap I\).
Find all solutions to these equations that lie in \(\mathbb{C}^2\).
Find a vector space basis for \(\mathbb{C}[x,y]/I\).
Solution
Put \[ f_1=x^2+2y^2-2, \qquad f_2=x^2-xy+y^2-1. \] We use lexicographic order with \(y>x\).
1. A Gröbner basis for \(I=(f_1,f_2)\) is\[ \boxed{ G=\left\{ y^2+\frac12x^2-1, \ xy-\frac12x^2, \ x^3-\frac43x \right\}.} \]
Set \[ g_1=y^2+\frac12x^2-1=\frac12f_1. \] Also \[ f_2-g_1=\frac12x^2-xy, \] so \[ g_2=xy-\frac12x^2\in I. \] From these relations one obtains \[ g_3=x^3-\frac43x\in I; \] for example, \[ 4xg_1-4yg_2+3xg_2=3x^3-4x=3g_3. \] Conversely \(f_1=2g_1\) and \[ f_2=g_1-g_2, \] so \((g_1,g_2,g_3)=I\).
Their leading monomials for lex order \(y>x\) are \[ \operatorname{LM}(g_1)=y^2, \qquad \operatorname{LM}(g_2)=xy, \qquad \operatorname{LM}(g_3)=x^3. \] We verify Buchberger’s criterion. First, \[ S(g_1,g_2)=xg_1-yg_2 =\frac12x^3-x+\frac12x^2y. \] Using \(g_2\) gives \(x^2y\equiv\frac12x^3\), so \[ S(g_1,g_2)\equiv\frac34x^3-x\equiv0 \] modulo \(g_3\). Next, since \(y^2\) and \(x^3\) are relatively prime, \(S(g_1,g_3)\) reduces to \(0\) by Buchberger’s product criterion. Finally, \[ S(g_2,g_3)=x^2g_2-yg_3 =-\frac12x^4+\frac43xy. \] Reducing first by \(g_2\) and then by \(g_3\) gives \[ -\frac12x^4+\frac23x^2 \equiv -\frac12\left(\frac43x^2\right)+\frac23x^2=0. \] Hence \(G\) is a Gröbner basis.
2. A Gröbner basis for the elimination ideal \(I\cap\mathbb C[x]\) is\[ \boxed{\left\{x^3-\frac43x\right\}.} \]
For lex order \(y>x\), the elimination theorem says that the elements of a Gröbner basis involving only \(x\) form a Gröbner basis for \(I\cap\mathbb C[x]\). In the basis from 1, the only such polynomial is \(g_3\).
3. The solution set in \(\mathbb C^2\) is\[ \boxed{ (0,1),\ (0,-1),\ \left(\frac2{\sqrt3},\frac1{\sqrt3}\right),\ \left(-\frac2{\sqrt3},-\frac1{\sqrt3}\right).} \]
Every solution satisfies \[ x^3-\frac43x=x\left(x^2-\frac43\right)=0. \] If \(x=0\), then \(g_1=0\) gives \(y^2=1\), hence \(y=\pm1\).
If \(x\ne0\), then \(g_2=0\) gives \[ y=\frac x2. \] The equation \(g_3=0\) gives \(x^2=4/3\), so \[ x=\pm\frac2{\sqrt3}, \qquad y=\pm\frac1{\sqrt3} \] with matching signs. Each listed point satisfies \(g_1,g_2,g_3\), hence the original equations.
4. A vector-space basis of \(\mathbb C[x,y]/I\) is\[ \boxed{\{1,x,x^2,y\}.} \]
For a Gröbner basis, the residue classes of the standard monomials, namely those not divisible by any leading monomial, form a basis of the quotient. Here the forbidden leading monomials are \[ y^2,\qquad xy,\qquad x^3. \] The monomials not divisible by any of these are exactly \[ 1,\ x,\ x^2,\ y. \] Therefore their residue classes form a basis.