Consider the ideal \(I = (x^2 - y^3,\; x - y^2) \subseteq \mathbb{C}[x, y]\). Find \(\operatorname{rad} I\), the radical of \(I\), expressing it as an intersection of prime ideals (do not try to find a generating set for \(\operatorname{rad} I\)).
Is \(I\) a radical ideal?
Solution
Let \[ I=(x^2-y^3,\ x-y^2)\subseteq\mathbb C[x,y]. \]
1. There is an isomorphism\[ \mathbb C[x,y]/I\cong \mathbb C[y]/(y^3(y-1)). \]
First quotient by the relation \(x-y^2\). The substitution homomorphism \[ \mathbb C[x,y]\longrightarrow\mathbb C[y], \qquad x\longmapsto y^2, \qquad y\longmapsto y \] has kernel \((x-y^2)\), so \[ \mathbb C[x,y]/(x-y^2)\cong\mathbb C[y]. \] Under this isomorphism the other generator becomes \[ x^2-y^3\longmapsto y^4-y^3=y^3(y-1). \] The claimed quotient follows.
2. The affine variety of \(I\) is\[ V(I)=\{(0,0),(1,1)\}. \]
Every point of \(V(I)\) satisfies \[ x=y^2. \] Substituting into \(x^2-y^3=0\) gives \[ y^4-y^3=y^3(y-1)=0. \] Hence \(y=0\) or \(y=1\), and then \(x=y^2\) gives respectively \(x=0\) or \(x=1\). Both points satisfy the two generators of \(I\).
3. The radical of \(I\) is\[ \sqrt I=(x,y)\cap(x-1,y-1). \]
By Hilbert’s Nullstellensatz over the algebraically closed field \(\mathbb C\), \[ \sqrt I=I(V(I)). \] By 2, \[ I(V(I)) =I(\{(0,0)\})\cap I(\{(1,1)\}) =(x,y)\cap(x-1,y-1). \] Each factor is maximal, hence prime, as requested.
4. The ideal \(I\) is not radical.
Set \[ h=y(y-1). \] Under the quotient description of 1, membership in \(I\) is equivalent to divisibility by \(y^3(y-1)\) in \(\mathbb C[y]\). The polynomial \[ y(y-1) \] is not divisible by \(y^3(y-1)\), so \(h\notin I\). However \[ h^3=y^3(y-1)^3 \] is divisible by \(y^3(y-1)\), hence \(h^3\in I\). Therefore \[ h\in\sqrt I\setminus I, \] so \(I\ne\sqrt I\) and \(I\) is not radical.