Find generators of the ideal of an algebraic set The following exercise comes from Fulton's Algebraic Curves:

Let $k$ be an infinite field and $V = \{(t,t^2,t^3) \, ; \, t \in k \} \subset \mathbb{A}^3(k)$. Show that $V$ is algebraic and compute $I(V)$ by giving a finite set of generators.

Now, showing $V$ is algebraic is easy by noting that $V = V(X^2-Y, X^3-Z)$, but I can't find a way to describe $I(V)$ explicitly. Obviously $(X^2-Y)(X^3-Z) \in V$, but how can I find the actual generators of this ideal? 
 A: The ideal is generated by $y-x^2$ and $z-x^3$ for assume that $f(x,y,z) \in I$ that is $$f(t,t^2,t^3)=0$$ identically. We need to show that $$f \in(y-x^2,z-x^3)$$ now since our generators are linear in $y$ and $z$ we can subtract multiples of these to reduce to case where $f$ does not contain the variables $y$ or $z$. But the the assumption becomes that $f(t)=0$ identically. So we are done.
A: Rene has given a perfectly fine answer to your question. 
Here is another approach that works when $k$ is algebraically closed (I realize that the assumption is a little weaker: you only assume that $k$ is infinite). Let $X=\{(t, t^2, t^3): t\in k\}$ to denote your algebraic set.  You already know that $X=V(y-x^2, z-x^3)$. Then, by Nullstellensatz, $I(X) =I(V(y-x^2, z-x^3)) = \sqrt{(y-x^2, z-x^3)}$ where I am using the notation $\sqrt{J}$ to denote the radical of the ideal $J$. But note that $(y-x^2, z-x^3)$ is a prime ideal. Indeed, the quotient ring
$$
k[x, y, z]/(y-x^2, z-x^3] \cong k[x, y]/(y-x^2) \cong k[x]
$$
is an integral domain. Here we used (twice!) the fact $R[t]/(t-a)\cong R$. Since $(y-x^2, z-x^3)$ is prime, it is in particular a radical ideal, i.e.  $\sqrt{(y-x^2, z-x^3)} = (y-x^2, z-x^3)$. Thus, $I(X)=(y-x^2, z-x^3)$.
