Find radical of ideal using algebraic set

Question

I want to find the radical of the ideal $I=(y-x^2,y^2)$ in $\mathbb{C}[x,y]$. The vanishing points of the first polynomial are $V(y-x^2)$ the parabola $y=x^2$ in $\mathbb{C}$ and $V(y^2)=0\in \mathbb{C}$. But I'm not sure how to connect these two together. Then I can just apply Hilbert's theorem since $\mathbb{C}$ is algebraically closed.

Answer 1

$V(y^2)$ is the line $y = 0$. Thus $V(y - x^2, y^2)$ is the intersection of the two curves $y = x^2$ and $y = 0$, which is simply the point $(0,0)$. Thus $\sqrt{I} = \mathcal I(V(I)) = \mathcal I(\{(0,0)\}) = (x, y)$.

You can also do this purely algebraically. Since $y^2 \in I$, therefore $y \in \sqrt{I}$. Since $x^2 = y - (y - x^2) \in \sqrt{I}$, therefore $x \in \sqrt{I}$. Thus $\sqrt{I}$ is either $(x,y)$ or $\mathbb{C}[x,y]$ and since $I$ contains no units (every element of $I$ has total degree $\ge 1$), we can conclude that $\sqrt{I} = (x,y)$.