maximal ideal implies empty algebraic set For an ideal $I$ in $A = \mathbb{C}[x, y, z]$ set $Z_{xy}(I) = \{(a, b) \in \mathbb{C}^2: f(a, b, z) = 0$ for all $f \in I\}$. Prove that $I$ maximal implies $Z_{xy}$ is empty.
I was thinking of doing this by contradiction assuming that $(x_0, y_0) \in Z_{xy}$ and then showing that $I$ cannot be maximal as a result.
 A: That idea works! If $(x_0, y_0) \in Z_{xy}(I)$, then $f(x_0, y_0, z)=0$ for all $f \in I$. This implies that $I \subseteq (x-x_0, y-y_0)$: for instance, if $f \in I$, write
$$ f(x, y, z)=\sum_{i,j,k} a_{ijk}(x-x_0)^i(y-y_0)^jz^k, $$
and use that $(x_0, y_0) \in Z_{x,y}(I)$ to conclude that $\sum_k a_{00k}z^k=0$. Therefore
$$ f(x, y, z)=\sum_{\substack{i,j,k}{(i,j) \neq (0,0)}} a_{ijk}(x-x_0)^i(y-y_0)^jz^k \in (x-x_0, y-y_0). $$
It follows that $I \subseteq (x-x_0, y-y_0)$, and so $I$ cannot be maximal (as the right hand side is not maximal).
A: If $I$ is maximal, then you have $I = (x-a, y-b, z-c)$ for some complex numbers $a,b, c$.
This is true because $A/I$ is a field, in particular, it's a field extension of $\mathbb{C}$ (finite by Hilbert's Nullstellensatz), which must be equal to $\mathbb{C}$ because $\mathbb{C}$ is algebraically closed. Take $a = [x], b = [y], c= [z]$, with $[x]$ being the residue class of $x$ modulo $I$ (that is, an element of $A/I \cong \mathbb{C}$) and the same thing for $y, z$. Then, the polynomials $x-a, y-b, z-c$ are in the kernel of the natural map $A\rightarrow A/I$, so the ideal $(x-a, y-b, z-c)$ which they span is contained in $I$. Since $(x-a,y-b,z-c)$ is also a maximal ideal, they must be equal: $I=(x-a,y-b,z-c)$.
Knowing that, the theorem is trivial, because for the polynomial $f(x,y,z) =z-c$, there are no $x_0, y_0$ such that $f(x_0,y_0,z) = z-c$ is equal to $0$ for all $z$.
Geometrical meaning: A point $(x_0,y_0)$ is in $Z_{xy}(I)$ if and only if the line $\lbrace x= x_0, y = y_0\rbrace$ is contained in the zeros of $I$ (which can never happen because a point can't contain a line).
