zero set of an analytic functio of several complex variables

Question

In one variable complex theory, we have the result that zeroes of a non-zero analytic function are isolated. In several variable theory, this result does not hold. I read it somewhere that this fact can be proved using Hurwitz theorem. If anyone can help me with this.

Answer 1

I'll restrict to two variables because it's clear how to extend to more.

Let's say $f(x_0, y_0) = 0$. Define $f_k(x)=f(x,y_0+1/k)$. Since $f$ is continuous, $f_k(x)$ converges uniformly on compact sets to $g(x)=f(x,y_0)$. Now $g(x_0)=0$, so Hurwitz's theorem says that for $k$ large enough, there is an $x_k$ with $|x_0 - x_k| < \varepsilon/2$ and $f_k(x_k)=0$. Assuming $k>2/\varepsilon$, we see that $f(x_k,y_0+1/k)=0$, and $d((x_k,y_0+1/k),(x_0,y_0)) < \varepsilon/2 + \varepsilon/2 = \varepsilon$.

So there are zeroes of $f$ arbitrarily close to $(x_0,y_0)$, and since $(x_0,y_0)$ was an arbitrary zero, we see that no zero of $f$ is isolated.

As a side note, this assumes $f$ has zeroes: $f(z_1, z_2) = e^{z_1+z_2}$ is holomorphic and nonvanishing.