Finding generators of an ideal / showing an ideal is prime 
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.
 A: The trick is to replace  $y$ by $x^2$ and $z$ by $x^3$ in the quotient ring $k[x,y,z]/\langle y-x^2, z-x^3\rangle$ so that  the quotient becomes: 
$k[x,y,z]/\langle y-x^2, z-x^3\rangle =k[x,x^2,x^3]=k[x]$.
Hence, since the quotient is a domain, your wish has come true : the ideal $\langle y-x^2, z-x^3\rangle$ is prime.  
A generalization
If you have a quotient of a polynomial ring $k[x_0,x_1,..., x_n]$ of the special form 
$$A=k[x_0,x_1,\dots, x_n]/\langle x_0-g(x_1,\dots,x_n),f_1(x_0,\dots, x_n),\dots,f_i(x_0,\dots, x_n),\dots\rangle$$ you can apply the same trick of replacing every occurrence of $x_0$ by $g(x_1,\dots,x_n)$ thus obtaining 
$$A=k[x_1,\dots, x_n]/\langle f_1(g(x_1,\dots,x_n),\dots, x_n),\dots,f_i(g(x_1,\dots,x_n),\dots, x_n),\dots\rangle.$$ In the example you submitted I implicitly used that  twice (but only wrote the end result), exploiting  the rule for successive quotients
 $$k[x,y,z]/\langle f,g\rangle=(k[x,y,z]/\langle f\rangle)/\langle \bar g\rangle.$$  Edit: a geometric interpretation
  We have an embedding $\mathbb A^1 \to \mathbb A^3: t\mapsto (t,t^2,t^3)$ with image an irreducible afine variety $ V$ known as the affine twisted cubic curve.
The calculation above says that  this variety is the ideal-theoretic (or scheme-theoretic) intersection  of the affine quadrics $y=x^2$ and $z=x^3$.
It is interesting  to notice that the closure $\bar V$ of $V$ in $\mathbb P^3$ is no longer a scheme-theoretic complete intersection even though it is  set-theoretically the intersection of a quadric and a cubic surface.
A: 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 $(\star)$?
