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    Gröbner basis computation and variety of an ideal generated by two polynomials

    P-APASP09G

    Area
    Applied Algebra
    Topics
    Commutative Algebra, Gröbner Bases
    Status
    draft

    Let \(f_1 = x^2 y^2 - xy\) and \(f_2 = x^3 y - 1\).

    1. Calculate a Gröbner basis for \(\langle f_1, f_2 \rangle \cap k[x]\) and for \(\langle f_1, f_2 \rangle\), where \(\langle f_1, f_2 \rangle\) is the ideal in \(k[x, y]\) generated by \(f_1\) and \(f_2\).

    2. What is the variety \(V(f_1, f_2) = \{(a, b) : f(a, b) = 0,\; f \in \langle f_1, f_2 \rangle\}\) for \(k = \mathbb{C}\)?

    Solution

    Use lexicographic order with \[ y>x, \] so that the elimination theorem applies to the subring \(k[x]\). Let \[ f_1=x^2y^2-xy, \qquad f_2=x^3y-1. \]

    1. The ideal generated by \(f_1,f_2\) is\[ I=(y-x,\ x^2-1). \]

    First, \[ yf_2-xf_1 =(x^3y^2-y)-(x^3y^2-x^2y) =y(x^2-1), \] so \[ y(x^2-1)\in I. \] Then \[ x\,y(x^2-1)-f_2 =x^3y-xy-(x^3y-1) =1-xy, \] so \[ xy-1\in I. \] Consequently \[ f_2-x^2(xy-1)=x^2-1\in I. \] Finally, \[ y(x^2-1)-x(xy-1)=x-y, \] so \(y-x\in I\). Thus \[ (y-x,x^2-1)\subseteq I. \] Conversely, \[ f_1=xy(xy-1) \] and \[ xy-1=x(y-x)+(x^2-1), \] so \(f_1\in(y-x,x^2-1)\). Also \[ f_2=x^2(xy-1)+(x^2-1), \] so \(f_2\) belongs to the same ideal. Hence equality holds.

    2. The set\[ \mathcal G=\{y-x,\ x^2-1\} \]is a Gröbner basis for \(I\) with respect to lex order \(y>x\).

    The leading monomials are \[ \operatorname{LM}(y-x)=y, \qquad \operatorname{LM}(x^2-1)=x^2. \] Their least common multiple is \(x^2y\), and the only \(S\)-polynomial is \[ S=x^2(y-x)-y(x^2-1)=y-x^3. \] Reducing first by \(y-x\) gives \[ x-x^3=-x(x^2-1), \] which reduces to zero by \(x^2-1\). Buchberger’s criterion therefore shows that \(\mathcal G\) is a Gröbner basis.

    3. A Gröbner basis for the elimination ideal\[ I\cap k[x] \]is\[ \{x^2-1\}. \]

    By 2, \(\mathcal G\) is a Gröbner basis for lex order \(y>x\). The elimination theorem says that the elements of \(\mathcal G\) involving only \(x\) form a Gröbner basis for \(I\cap k[x]\). The only such element is \(x^2-1\).

    4. For \(k=\mathbb C\),\[ V(f_1,f_2)=\{(1,1),(-1,-1)\}. \]

    By 1, a point \((a,b)\) lies in the variety exactly when \[ b-a=0, \qquad a^2-1=0. \] Thus \(a=\pm1\) and \(b=a\), giving precisely the two displayed points.