1. ## Cubic surface

We define the cubic $V$ in $\mathbb{R}^3$ by the equations: $y-x^2=0, z-x^3=0$.

Prove that $\mathcal{I}(V)=$

thanks!

2. Originally Posted by roporte
We define the cubic $V$ in $\mathbb{R}^3$ by the equations: $y-x^2=0, z-x^3=0$.

Prove that $\mathcal{I}(V)=$

thanks!
let $J=.$ clearly $J \subseteq \mathcal{I}(V).$ now suppose $f(x,y,z) \in \mathcal{I}(V).$ then $f(x,x^2,x^3)=0.$

since $f \in \mathbb{R}[x,y,z],$ we have: $f(x,y,z)=\sum {c_{ijk}}x^i y^j z^k,$ for some $c_{ijk} \in \mathbb{R}.$ thus:

$f(x,y,z)=\sum c_{ijk}x^i(y-x^2 + x^2)^j (z-x^3 + x^3)^k
$

$=p(x,y,z)(y-x^2) + q(x,y,z)(z-x^3) + f(x,x^2,x^3)$

$=p(x,y,z)(y-x^2) + q(x,y,z)(z-x^3) \in J.$

thus $\mathcal{I}(V) \subseteq J$ and we're done.