In $\mathbb{C}[x, y, z]$ we have that $V=\{y-x^2, z-x^3)=\{(t, t^2, t^3) | t \in \mathbb{C}\}$. To show that $$I(V(y-x^2, z-x^3))=\langle y-x^2, z-x^3\rangle $$ can we use the Nullstellensatz??
EDIT:
To show that $\langle y-x^2, z-x^3\rangle$ is prime do we have to do the following??
We consider the homomorphism $\phi : \mathbb{C}[x, y, z] \rightarrow \mathbb{C}[x]$ with $\phi(x)=x, \phi(y)=x^2, \phi(z)=x^3$ and $\phi(a)=a, \forall a \in \mathbb{C}[x]$.
We consider a polynomial $p(x, y, z) \in \mathbb{C}[x, y, z]$.
We apply the euclidean division of $p(x, y, z)$ and $y-x^2$.
$$p(x, y, z)=g(x, y, z) (y-x^2)+h(x, y, z)$$
with $deg_y h(x, y, z) < deg_y (y-x^2) \Rightarrow deg_y h(x,y,z)=0 \Rightarrow h(x, y, z)=h(x, z)$
$$p(x. y, z)=g(x, y, z)(y-x^2)+h(x, z)$$
We apply the euclidean division of $h(x, z)$ and $z-x^3$.
$$h(x, z)=(z-x^3)a(x, z)+b(x, z)$$
with $deg_z (b(x, z)) <1 \Rightarrow deg_z (b(x, z))=0 \Rightarrow b(x, z)=b(x)$.
If $p(x, y, z) \in ker \phi$, $$\phi(p(x, y, z))=0 \Rightarrow \phi(g(x, y, z)(y-x^2)+(z-x^3)a(x, z)+b(x))=0 \Rightarrow \phi(g(x, y, z)) \phi((y-x^2))+\phi((z-x^3))\phi(a(x, z))+\phi(b(x))=0\Rightarrow \phi(b(x))=0 \Rightarrow b(x)=0$$
So if $p(x, y, z) \in ker \phi \Rightarrow p(x, y, z)=g(x, y, z)(y-x^2)+(z-x^3)a(x, z) \Rightarrow p(x, y, z) \in \langle y-x^2, z-x^3 \rangle$.
So $$ker \phi \subseteq \langle y-x^2, z-x^3 \rangle$$
Is the step :
If $p(x, y, z) \in ker \phi$, $$\phi(p(x, y, z))=0 \Rightarrow \phi(g(x, y, z)(y-x^2)+(z-x^3)a(x, z)+b(x))=0 \Rightarrow \phi(g(x, y, z)) \phi((y-x^2))+\phi((z-x^3))\phi(a(x, z))+\phi(b(x))=0\Rightarrow \phi(b(x))=0 \Rightarrow b(x)=0$$
correct??
@Mary Star: You can prove the result for any infinite field $k$. In fact, let $p(x,y,z)$ so that the polynomial $p(x,x^2,x^3)$ is the zero polynomial. Then $p(x,y,z)$ is in the ideal generated by $y-x^2$ and $z-x^3$. Indeed, consider the polynomial $p$ as an element in $k[x][y,z]$ and write $p$ as a sum of monomials in $(y-x^2)$ and $(z-x^3)$ with coefficients in $k[x]$. The free term will be $p(x,x^2,x^3)$.