TITLE: Finding generators of an ideal / showing an ideal is prime
QUESTION [7 upvotes]: Suppose I have an ideal like $\mathrm{rad}(\langle y - x^2, z - x^3 \rangle)$. How do I go about finding generators for the ideal? 

If I could show that $\langle y - x^2, z - x^3 \rangle$ is a prime ideal then obviously $y-x^2$ and $z-x^3$ would generate $\mathrm{rad}(\langle y - x^2, z - x^3 \rangle)$, but I don't see how to do this.
Thanks.

REPLY [3 votes]: You could start by showing the ideal $I=(y-x^2,z-x^3)$ is in fact radical so that finding generators for its radical becomes easy!
So let $f\in k[x,y,z]$ be such that there is an $n\geq0$ such that $f^n\in I$.
It is easy to see that there exist $g$, $h\in k[x,y,z]$ and $r\in k[x]$ such that 
$$(\clubsuit)\qquad\qquad f(x,y,z)=g(x,y,z)\cdot(y-x^2)+h(x,y,z)\cdot(z-x^3)+r(x),$$
and using Newton's formula for powers of trinomials— we see immediately from this that $$f(x,t,z)^n\equiv r(x)^n\mod I,$$ so that in fact $r(x)^n\in I$. Suppose for a second that we manage to prove that 

$(\star)\qquad\qquad$ we have $k[x]\cap I=0.$

Then, since $r(x)^n$ is in that intersection, we have $r(x)^n=0$ and therefore $r(x)=0$. Looking back at $(\clubsuit)$ we see now that $f\in I$, and we are happy.
Can you prove $(\star)$?