Why are there no discrete zero sets of a polynomial in two complex variables? Why is the zero set in $\mathbb{C}\times\mathbb{C}$ of a polynomial $f(x,y)$ in two complex variables always non-discrete (no zero of $f$ is isolated)?
 A: Write $f(x,y)$ as a polynomial in $y$ with coefficients in $\mathbb C[x]$, so that $f(x,y)=\sum_{i=0}^nf_i(x)y^i$. If $x\in\mathbb C$ is such that $f_n(x)\neq0$, then there exist $n$ values of $y$ such that $f(x,y)=0$. It follows that if $X\subseteq \mathbb C\times\mathbb C$ is the zero locus of $f$, then the first projection $\pi_x:X\to\mathbb C$ has an image with finite complement. This implies, in particular, that $X$ is uncountable, and no uncountable subset of $\mathbb C\times\mathbb C$ is discrete.
A: This actually holds for any analytic function of two complex variables on an open set. Suppose $f(x_0,y_0) = 0$. Then viewed as a function of $y$, $f(x_0,y)$ must have an isolated zero of some order $k$ at $y = y_0$. By complex analysis (the residue theorem will work for example), for some small $r$ one has 
$${1 \over 2\pi i} \int_{|z - y_0| = r} {f'(x_0, z) \over f(x_0,z)}\,dz = k$$
But the integrand above is a continuous function of $x$ near $x = x_0$ and takes integer values. Hence for $x$ close enough to $x_0$ one also has
$${1 \over 2\pi i} \int_{|z - y_0| = r} {f'(x, z) \over f(x,z)}\,dz = k$$
This in turn implies that whenever $x$ is close enough to $x_0$, $f(x,y)$ (viewed as a function of $y$) has $k$ zeroes inside $|z - y_0| = r$. Since this holds for arbitrarily small $r$, we conclude the zero of $f(x,y)$ at $(x_0,y_0)$ is not isolated.
This is actually a watered-down version of the proof of the famous Weierstrass Preparation Theorem (which incidentally also implies this fact in pretty short order).
