# Algebraic study of a curve

Consider the parametric curve $C=\{(u^3,u^4,u^5)\,|\,u\in K\}$, where $K$ is an algebraic closed field with characteristic $0$. I'd like to prove that its ideal $$I(C)=\{f\in K[x,y,z]\,\mbox{such that}\,f(C)=\{0\}\}$$ is prime and has height $2$.

I managed to show that $$C=\{(x,y,z)\in K^3 \,|\,p_i(x,y,z)=0\ \forall i=1,2,3\}$$ with $$p_1(x,y,z)=xy^3-z^3,\ p_2(x,y,z)=x^2y-z^2,\ p_3(x,y,z)=x^5z-y^5\,.$$ Now I think that $I(C)=(p_1,p_2,p_3)$. But I don't know how to proceed.

1) Since $f:\mathbb A^1\to \mathbb A^3:u\mapsto (u^3,u^4,u^5)$ is continuous, its image $C$ is irreducible because $\mathbb A^1$ is.
So the closure $\bar C$ is irreducible too, by general topology.
Hence the ideal $I(C)=I(\bar C)$ is prime, being the vanishing ideal of an irreducible closed set (Fulton, Algebraic curves, Chap.I, §5, Prop.1, page 15).
2) Actually $C$ is closed (see here or there).
Hence the corestriction to $C$ of the map $f$ above is a homeomorphism $\mathbb A^1\to C$, so that $C$ has same dimension as $\mathbb A^1$, namely $1$.
This implies (if you know the rudiments of dimension theory over a field) that the height of $I(C)$, which is the codimension of $C$ in $\mathbb A^3$, is $2$ .