I'm doing Problem 1.40 from Fulton's "Algebraic Curves":
1.40. Let $I = (X^2 - Y^3, Y^2 - Z^3) \subset k[X, Y, Z]$. Define $\alpha : k[X, Y, Z] \to k[T]$ by $\alpha(X) = T^9$, $\alpha(Y) = T^6$, $\alpha(Z) = T^4$.
(a) Show that every element of $k[X,Y,Z]/I$ is the residue of an element $A + XB + YC + XYD$, for some $A, B, C, D \in k[Z]$.
(b) If $F = A + XB + YC + XYD$, $A, B, C, D \in k[Z]$, and $\alpha(F) = 0$, compare like powers of $T$ to conclude that $F = 0$.
(c) Show that $\operatorname{Ker}(\alpha) = I$, so $I$ is prime, $V(I)$ is irreducible, and $I(V(I)) = I$.
The problem is fairly straightforward, except for the last few items, which is to show that $V(I)$ is irreducible and $I(V(I)) = I$. There aren't any assumptions on $k$, and even though this is in the Nullstellensatz section where Fulton writes "We assume throughout this section that $k$ is algebraically closed", he still explicitly calls out in the other problems from this section whether $k$ is algebraically closed or arbitrary.
If $k$ is algebraically closed, then $I$ being prime immediately implies the rest.
If $k$ is just assumed to be infinite, then we can (probably?) use the previous parts plus a theorem saying that if a polynomial $F$ evaluates to $0$ for every point, then $F = 0$.
However, if $k$ is completely arbitrary, I don't see how it's possible to show that $V(I)$ is irreducible or $I(V(I)) = I$, at least with the tools from Chapter 1. Am I missing something?
