# Why can the zero set of a collection of holomorphic functions be written as the zero set of finitely many?

I've been reading the first few sections of Griffiths and Harris, and they state without proof (about halfway down page 14 in my 1994 Wiley Classics Library edition) that

the zero locus of an arbitrary collection of holomorphic functions in a polydisc is in fact given by a finite number of holomorphic functions in a smaller polydisc.

I'd like to know how to prove this.

If $\mathcal{O}_n$ denotes the ring of germs of holomorphic functions about $0$ in $\mathbb{C}^n$ then I can show (by induction, using the Hilbert basis theorem and Weierstrass preparation theorem) that $\mathcal{O}_n$ is Noetherian, but this isn't quite enough. Indeed, if $\{f_\alpha\}$ are holomorphic functions on a polydisc $U$ about zero then there exist $g_1, \ldots, g_k \in \mathcal{O}_n$ with $(f_\alpha)=(g_i)$ and, shrinking $U$ if necessary, I get that the zero set of the $f_\alpha$ is contained in the zero set of the $g_i$, but I can't see how to prove the reverse inclusion.

• For every $i$, you have a representation $$g_i = \sum_{\kappa = 1}^k h_\kappa \cdot f_{a_\kappa}$$ of the germ with some $h_\kappa \in \mathcal{O}_n$. Thus, in a small enough polydisk $P$, you have $g_i = \sum h_\kappa \cdot f_{a_\kappa}$ in $\mathcal{O}(P)$. – Daniel Fischer Sep 26 '15 at 15:02
• @Daniel Fischer Right, this shows that the zero set of the $f_\alpha$ is contained in the zero set of the $g_i$. I want the other direction. It may be that Noetherian-ness is simply not enough, and that a new idea is needed to prove the result I want. – Jez Sep 26 '15 at 15:07
• Ah, right, sorry. And a priori, it would be possible that the domains of the coefficients in the representations $f_a = \sum h_{a,i} g_i$ shrink to $\{0\}$. – Daniel Fischer Sep 26 '15 at 15:15
• @DanielFischer Yes, that's exactly my problem. Maybe I should have been more explicit. – Jez Sep 26 '15 at 15:17
• Nah, it's just me being slow. – Daniel Fischer Sep 26 '15 at 15:17

You can find a proof in Chapter IV, Section D, Theorem 2 of Gunning and Rossi, Analytic functions of several complex variables. They prove, more strongly, that if $f_{\alpha}$ is any set of holomorphic functions on $U$, $p$ is any point of $U$ and $m \geq 0$ is any integer, then there is an open neighborhood $U'$ of $p$ on which the ideal $(f_{\alpha})$ has a finitely generated free resolution resolution of length $m$. You just need $m=1$.
Roughly, the proof proceeds as follows. Let $V$ be the zero locus of the $f_{\alpha}$. By earlier results, they can shrink $U$ and then write $V$ as $\bigcup V_i$ where each $V_i$ is irreducible of dimension $d_i$. They reduce to proving the result for each $V_i$ separately. They then shrink further and perform a Noether normalization like transformation for each $V_i$, to get what they call an admissible representation. At this point, they have only shrunk finitely many times, and they are able to use Weierstrass factorization without shrinking further.