$Z_{xy}(I) \times \mathbb{C} = Z(I) = \{(a, b, c) \in \mathbb{C}^2: f(a, b, c) = 0$ for all $f \in I\}$ iff $\operatorname{rad}I = JA$ For an ideal $I$ in $A = \mathbb{C}[x, y, z]$ set $$Z_{xy}(I) = \{(a, b) \in \mathbb{C}^2: f(a, b, z) = 0\text{ for all }f \in I\}.$$
Let
$$J = \{f(x, y): f(a, b) = 0\text{ for all }(a, b) \in Z_{xy}(I)\}.$$
Prove $$Z_{xy}(I) \times \mathbb{C} = Z(I) = \{(a, b, c) \in \mathbb{C}^2: f(a, b, c) = 0\text{ for all }f \in I\} \iff \operatorname{rad}I = JA.$$
I'm a bit confused about the definition of $J$. I'm not sure where the $f$ is supposed to come from since in the other sets in this problem the function $f$ took on 3 arguments rather than 2.
 A: I'll give a geometrical interpretation first. An ideal $I$ of $A$ is always finitely generated (Hilbert's Basis theorem says $A$ is noetherian), so, $I$ can be seen as generated by a finite family of polynomials $\lbrace p_i(x,y,z) \rbrace_{i=1}^r$. Those polynomials give a system of algebraic equations:
$$\left\lbrace\begin{matrix}p_1(x,y,z)= 0\\\vdots\\p_r(x,y,z) = 0\\\end{matrix}\right.$$
Its solutions form the set $Z(I)$, the algebraic variety defined by $I$. It's a set of points in $3$D complex space (not much harm is done if you picture it as $\mathbb{R}^3$).
Now, the set $Z_{xy}(I)$, as you've defined it, it's the set of $(a,b)$ such that the line $\lbrace x= a, y = b \rbrace$ (it's a complex line, i. e., the same as $\mathbb{C}$, but again, not much truth is lost when imagining it as a straight line) is contained in the set $Z(I)$.
The set $Z_{xy}(I) \times \mathbb{C}$ is just the union of lines $\lbrace x= a, y = b \rbrace$ contained in $Z(I)$, so it's always true that $Z_{xy}(I) \times \mathbb{C} \subset Z(I)$. The opposite inclusion means that, with every point $(a,b,c)\in Z(I)$, the whole line $\lbrace x= a, y = b \rbrace$ is contained in $Z(I)$. It's a geometric condition on the set $Z(I)$, which can be expressed on a condition involving the ideal $I$.
Now, the ideal $I$ can't always be recovered from the set $Z(I)$, because the set of polynomials which are zero on the set $Z(I)$ may not always be equal to $I$. It is equal to $\text{rad}(I)$.
We can prove the first implication. If $Z(I) = Z_{xy}(I) \times \mathbb{C}$, then, the set of polynomials which are zero in $Z(I)$, which is equal to $\text{rad}(I)$, must be equal to the set $X$ of polynomials which are zero in $Z_{xy}(I) \times \mathbb{C}$.
By definition, the set $X$ consists on those polynomials $p(x,y,z)$ such that $p(a,b,z) = 0$ for all $(a,b)\in Z_{xy}(I)$. But this is just the ideal generated by $J$ in $A$. (Polynomials in two variables can be thought as polynomials in three variables as well.)
Now, the second implication. Suppose that $\text{rad}(I) = JA$. Then, the sets of zeros of those ideals are the same, but we also know that $Z(I) = Z(\text{rad}(I))$ (this is always true, for every ideal in $A$). We deduce that
$$Z(I) = Z(JA)$$
So we need to prove that $Z(JA) = Z_{xy}(I)\times \mathbb{C}$. But polynomials in $JA$ are just polynomials which are zero on the lines $\lbrace x=a,y= b\rbrace$ contained in $Z(I)$, that is, polynomials null over $Z_{xy}(I)\times \mathbb{C}$.
The moral of this problem is that, if the variety $Z(I)$ is made of lines parallel to the $z$ axis, then it's the solution of a system of equations independent of $z$ (that is, the ideal $\text{rad}(I)$ has a system of generators independent of $z$, as in $JA$).
A: I post this answer because the point of view should be slightly different,  although the main argument is the same. I'll denote with $(S)$ the ideal  generated by any subset $S$ in a ring; in particular $\operatorname {rad} (S)$ is the radical of $(S)$, while if $\mathfrak a$ is already an ideal,  its radical is written just $\operatorname{rad}\mathfrak a$.
Consider the maps $\pi:\Bbb C^3\to \Bbb C^2:(a,b,c)\mapsto (a,b)$ and $\iota:\Bbb C^2\to \Bbb C^3:(a,b)\mapsto  (a,b,0)$. Obviously for any set of the  form $X\times \Bbb C$ ($X\subseteq  \Bbb C^2$) holds $\pi^{-1}(\iota^{-1}(X\times \Bbb C))=X\times \Bbb C$; plus $\pi$ is induced by the inclusion $p:\Bbb C[x,y]\to \Bbb C[x,y,z]$, while $\iota$ is induced by the projection $i:\Bbb C[x,y,z]\to \Bbb C[x,y]$. It is known that, if $f:R\to R'$ is a morphism of $\Bbb C$-algebras, the induced map $f^*$ between spectra  is continuous; in particular, if $\mathfrak  a\subseteq R$  is an ideal, ${f^*}^{-1}(Z(\mathfrak a))=Z(f(\mathfrak a))$.
Under the hypothesis that  $Z(I)=Z_{xy}(I)\times \Bbb C$, you see that $Z_{xy}(I)$ is the restriction of $Z(I)$ to $\Bbb C^2\times \{0\}$, that is to say $\iota^{-1}(Z(I))=Z(i(I))$; then, by Hilbert's nullstellensatz, $J=\operatorname{rad}(i(I))$. Said in the previous paragraph that $Z(I)=\pi^{-1}(\iota^{-1}(Z(I)))=Z(p(i(I)))$, again by the nullstellensatz, $\operatorname {rad}I=\operatorname {rad}(p(i(I)))$; thus $\operatorname {rad}I=\operatorname {rad}(p(\operatorname{rad}(i(I))))=\operatorname {rad}(p(J))=\operatorname {rad}AJ$ (not sure if in this setting $J$ radical forces $AJ$ to be radical). Conversely, for  any extended ideal $I$ (i.e.  $I=I'A$ for an ideal $I'\subseteq \Bbb C[x,y]$), it is quite immediate that $Z(I)=Z(I')\times \Bbb C$.
