$I(Z(y-x^{2},z-x^{3}))=(y-x^{2},z-x^{3})$

Question

Why is this:

$I(Z(y-x^{2},z-x^{3}))=(y-x^{2},z-x^{3})$

"obvious"?

I can easily see $Z(y-x^{2},z-x^{3}) = \lbrace (k,k^{2},k^{3}) \vert k \in\mathbb{A}^{1} \rbrace$

but I have some problems when I need to take the ideal of this.

Answer 1

$(y-x^2,z-x^3)$ is a radical ideal, in fact prime ideal, because $k[x,y,z]/(y-x^2,z-x^3)=k[x]$. Now use Hilbert's Nullstellensatz. (Alternatively, a direct computation also works.)