Proving $y-x^2, z-xy$ generate the ideal of the twisted cubic This question comes from Hartshorne's exercise 1.2.
He defines $Y:=\{(t,t^2,t^3)\in \mathbb{A}^3\mid t\in k\}$ and asks us, among other things, to find generators for the ideal $I(Y):=\{f\in k[x,y,z]\mid f(p)=0\,\,\,\forall p\in Y\}$.
To me it's obvious that the natural candidates are $y-x^2, z-xy$. 
I've already verified that $Y=Z(J)$, where $J:=(y-x^2, z-xy)$. By the Nullstellensatz, I could only conclude that $I(Y)=\sqrt{J}$. I guess it's true that $J$ is a radical ideal (and this would solve the problem), but I don't know how to prove it.
Using the mere definition of radical didn't work for me. Is there some other way?
Obs.: here $k$ is an algebraically closed field.
 A: Here's another way to use your explicit generators. Suppose $f(x,y,z)$ vanishes on the twisted cubic. Since the $z$ coefficient of $z-xy$ is a unit, we may write 
$$
f(x,y,z)=(z-xy)g(x,y,z)+h(x,y).
$$
Similarly, the $y$ coefficient of $y-x^2$ is a unit, so we may write 
$$
f(x,y,z)=(z-xy)g(x,y,z)+(y-x^2)q(x,y)+r(x).
$$
Now $f(t,t^2,t^3)=0$ for any $t$, hence $r(t)=0$ for any $t$. If the base field is infinite this implies $r(x)$ is identically zero, so that $f$ is in the ideal as desired.
A: I don't see how these candidates are natural or obvious. I would go for $x^2-y$ and $x^3-z$. 
And in fact $y-x^3\notin I(Y)$ because plugging in $t=2$ yields $(2,4,8)$ which is not a zero of $y-x^3$. Whatever you have done to verify that $Y=Z(y-x^3, z-xy)$ is must be wrong because 
$$(2,8,16)\in Z(y-x^3,z-xy)\qquad\text{ but }\qquad(2,8,16)\notin Y.$$

In stead consider $f\in I(Y)$, so that $f(t,t^2,t^3)=0$ for all $t\in k$. If $f=\sum_{i,j,k}c_{ijk}x^iy^jz^k$ then in the quotient $k[x,y,z]/(x^2-y,x^3-z)$ we have
$$\overline{f}\equiv\sum_{i,j,k}c_{ijk}x^{i+2j+3k}\pmod{(x^2-y,x^3-z)},$$
and also $\overline{f}(t,t^2,t^3)=0$ for all $t\in k$. But then
$$\sum_{i,j,k}c_{ijk}t^{i+2j+3k}=0,$$
and if $k$ is infinite this implies $\overline{f}=0$ and so $f\in(x^2-y,x^3-z)$. This proves the inclusion 
$$I(Y)\subset(x^2-y,x^3-z),$$
and the converse inclusion is clear.

Edit: With the generators corrected from $(y-x^3,z-xy)$ to $(y-x^2,z-xy)$ the argument above works just as well; note that
$$x^3-z=-x(y-x^2)-(z-xy)
\qquad\text{ and }\qquad
z-xy=x(x^2-y)-(x^3-z).$$
