# Can a holomorphic function of several variables have just one zero?

Of course “several” in the title means $$n$$ strictly greater than 1 and the function is defined on some open subset of $$\mathbb{C}^n$$.

I tried to use the Weierstrass preparation theorem because it’s the only result on analysis of several variables I know but I couldn’t find a contradiction (I believe the answer is “no”).

The answer is indeed no, which might be surprising since the real analogue of this question has a different outcome ($$f(x) = ||x||^2$$ is smooth and only has one zero).

The crucial result in this proof is Hartog's Theorem:

Let $$n \geq 2$$ and $$\Omega \subset \mathbb C^n$$ be open. Suppose $$K \subset \Omega$$ is compact and $$\Omega \setminus K$$ is connected. Then any holomorphic function on $$\Omega \setminus K$$ can be extended to all of $$\Omega$$.

Remark: This is not true in one variable: The function $$z \mapsto e^{1/z}$$ cannot be extended to all of $$\mathbb C$$ because it has an essential singularity at zero.

We are now going to prove that no holomorphic function of several variables has just one zero.
Toward a contradiction, let $$\Omega \subset \mathbb C^n$$ be a (connected) domain and let $$f: \mathbb{\Omega} \rightarrow \mathbb C$$ be holomorphic with $$f(z) = 0$$ if and only if $$z = z_0$$. Then the function $$1/f$$ is holomorphic in $$\Omega \setminus \{z_0\}$$. By Hartog's Theorem, $$1/f$$ admits a holomorphic extension to $$\Omega$$. But then, $$f(z_0) \cdot 1/f(z_0) = 1$$ by continuity of the product of continuous functions, which is a contradiction to $$f(z_0) = 0$$.

The answer is no for a function defined on an open subset of the plane. It actually comes down to the following algebra fact :

Let $$A$$ be a normal ring, then $$A=\cap_{ht(\mathfrak{p})=1} A_{\mathfrak{p}}.$$

where if $$\mathfrak{p}$$ is a prime ideal of $$A$$, the height of $$\mathfrak{p}$$ denoted by $$ht(\mathfrak{p})$$ is the maximal number of prime ideals contained in $$\mathfrak{p}$$ that form a strictly increasing chain $$\mathfrak{p_1} \subsetneq\cdots\subsetneq \mathfrak{p_s}\subsetneq \mathfrak{p}.$$

And $$A_{\mathfrak{p}}$$ is the localization of $$A$$ with respect to $$\mathfrak{p}$$.

A proof can be found in Matsumura's "Commutative Algebra".

Now, you may suppose that your open subset of the plane is connected, hence the ring of holomorphic functions on $$U$$ is normal (this is an exercise). Denote it by $$A$$ and denote by $$K$$ its fraction field i.e., the field of meromorphic functions on $$U$$.

My claim is that if $$f$$ is holomorphic and has only one point as zero locus, then $$1/f$$ is holomorphic, which yealds to a contradiction.

But then I only have to prove that $$1/f$$ is in all the $$A_{\mathfrak{p}}$$ for all primes of height one.

To do this, I say that the elements of a certain prime of height one of $$A$$ are exactly the holomorphic functions vanishing on a certain codimension $$1$$ subset $$Z$$ of $$U$$ (this is the analytic version of the Nullstellensatz, again a good exercise).

Therefore $$1/f$$ is not in any $$A_{\mathfrak{p}}$$.

• As is, your argument does not work since the result you quote from Matsumura's book (12.4. (i)) requires some noetherianity assumption. – Gaussian Nov 21 '19 at 13:12
• Indeed, there is a step missing in the argument. When localized at any maximal ideal $m$, $A_m$ is noetherian. Then apply the argument to $A_m$ and use the fact that $A=\cap A_m$. – Carot Nov 21 '19 at 23:57
• Indeed, it fills the gap. – Gaussian Nov 22 '19 at 7:38

The vanishing set of an holomorphic function $$\Bbb{C}^{n}\to\Bbb{C}$$ is an union of analytic hypersurfaces (locally biholomorphic to $$\Bbb{C}^{n-1}$$)

Use the n-polydisk Cauchy integral formula to show holomorphic implies analytic. Let $$f(w,z)=F_w(z)=\sum_n g_n(w) z^n, \qquad w\in \Bbb{C}^{n-1},z\in \Bbb{C}$$ be analytic non-constant and $$f(0,0)=0$$. The $$g_n$$ are analytic with a common non-zero radius of convergence.

If $$F_0'(0) \ne 0$$ it is easy : $$F_w'(0)\ne 0$$ for $$w$$ small thus with $$r$$ small enough from the residue thoerem $$h(w) = \frac1{2i\pi} \int_{|z|=r} \frac{z F_w'(z)}{F_w(z)}dz, \qquad f(w,h(w))=0$$

Otherwise $$F_0'(0) =0$$, assume $$F_0^{(k)}(0)\ne 0$$ then from the argument principle the number of zeros of $$F_w$$ near $$z=0$$ is the number of times the curve $$F_w(r e^{it}),t\in [0,2\pi]$$ encloses $$0$$, for $$w$$ small it is $$\le k$$ and it is continuous in $$w$$.

There may be several hypersurfaces passing through $$w=0$$ but for $$w\ne 0$$ small the hypersurfaces separate and replacing $$\int_{|z|=r}$$ by $$\int_\gamma$$ we get the equations of our hypersurfaces.

The remaining case is $$F_0$$ identically zero : thus $$f(z,w) = z^m f_2(z,w)$$ and if $$f_2(0,0)=0$$ then $$f_2$$ satisfies the previous hypothesis.

Suppose for simplicity that $$f$$ is holomorphic in an open set $$U\subset \mathbb C^2$$ with $$(0,0)\in U.$$ Assume that $$f$$ has an isolated zero at $$(0,0).$$ Then we can choose $$r>0$$ such that $$f$$ is holomorphic in $$D(0,2r)^2$$ and $$f\ne 0$$ in $$D(0,2r)^2\setminus \{(0,0)\}.$$

For $$w\in D(0,r)$$ define

$$g(w)=\frac{1}{2\pi i}\int_{|z|=r} \frac{D_zf(z,w)}{f(z,w)}\,dz.$$

The argument principle shows that $$g(0)$$ is a nonzero integer $$m,$$ while $$g(w)= 0$$ for $$0<|w| Note also that $$g$$ is a continuous function of $$w.$$ Thus $$g(0) = m = \lim_{z\to 0} g(w)=0,$$ which is a contradiction.