---
tag: 5EZRN3812E
kind: statement
role: proposition
name: Zeros of holomorphic functions are not isolated
msc: [32A60, 32A10]
uses: [507C34W0ZT, FSGES670AZ]
about: [507C34W0ZT]
sources:
  - url: https://en.wikipedia.org/wiki/Function_of_several_complex_variables
---

## Statement

Let $n \ge 2$, let $D \subseteq \C^n$ be an open set, and let $f$ be a [[holomorphic function|507C34W0ZT]] on $D$.
If $p \in D$ and $f(p) = 0$, then every neighbourhood of $p$ contains a point $q \ne p$ with $f(q) = 0$.

## Remarks

Suppose instead that $p$ is the only zero of $f$ in a small open ball $B$ around $p$.
Then $1/f$ is holomorphic on $B \setminus \set{p}$, which is connected.
By [[Hartogs's extension theorem|FSGES670AZ]] with $K = \set{p}$, it extends to a holomorphic function $g$ on $B$.
Then $g f = 1$ on $B \setminus \set{p}$, hence at $p$ by continuity, which contradicts $f(p) = 0$.

Example: $f(z_1, z_2) = z_1^2 + z_2^2$ vanishes at the origin, and its zeros near the origin form the two complex lines $z_1 = i z_2$ and $z_1 = -i z_2$.

The hypothesis $n \ge 2$ is necessary: $f(z) = z$ on $\C$ has the isolated zero $0$.

## Explanations

### L1 human

In one complex variable, a nonzero holomorphic function has isolated zeros, like the roots of a polynomial.
In two or more variables this is never the case: a zero always lies on a whole curve, surface or larger set of zeros.
So one cannot hope to describe zero sets by listing points.

### L2 human

The zero set of a holomorphic function that is not identically zero is an analytic hypersurface, of complex dimension $n - 1$ at each of its points.
This proposition is the first step toward that fact.
Divisors in several variables are made of hypersurfaces instead.