1. ## Cubic surface

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

Prove that $\displaystyle \mathcal{I}(V)=<y-x^2,z-x^3>$

thanks!

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

Prove that $\displaystyle \mathcal{I}(V)=<y-x^2,z-x^3>$

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

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

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

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

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

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