Order of a holomorphic function along an analytic hypersurface

Question

Let $Y\subset \mathbb{C}^n$ be an analytic subset, i.e. for every $x \in Y$ there exists a holomorphic function (germ) $g \in \mathcal{O}_x$ such that $Y$ is locally cut out by $x$. Let $f$ be a holomorphic function, and suppose that $Y$ is locally cut out by irreducible $g\in \mathcal{O}_x$ around $x$, then the order of $f$ at $x \in Y$ is defined to be the integer $a=ord_x f$ such that $f=g^a h$ in the factorization of $\mathcal{O}_x$ (since $\mathcal{O}_x$ is a UFD).

It is claimed, in many complex geometry books like Huybrechts and Griffiths, that $ord_x f$ is locally constant, yet I don't see why this is true. I suppose a proof would be along the lines of:

$Y$ at a point $y$ near $x$ is also locally cut out by $g\in \mathcal{O}_y$, and $h$ remains to be relatively prime to $g$ if $y$ is close enough, so we still have $f=g^{ord_x f} h$ in $\mathcal{O}_y$.

But my problem with the above proof is that $g$ may not be a locally defining function of $Y$ at $y$, in particular $g$ could be reducible in $\mathcal{O}_y$, since there are examples of irreducible holomorphic function germs becoming reducible in a neighborhood (see: Irreducibility of holomorphic functions in a neighborhood of a point). In this case we can only conclude $ord_y f \geq ord_x f$, i.e. the order is locally non-decreasing.

An explanation of how to reason that the order is locally constant would be great. Thank you.

Answer 1

It is not restrictive to assume $x=0$. I can prove that the order is locally constant if we assume $Y$ irreducible at every its point. Let $f$ be a holomorphic function, and suppose that $Y$ is locally cut out by irreducible $g \in \mathcal{O}_n$ around $0$. Let $$ f=g^a\cdot h, \qquad a=\operatorname{ord}_0 f, \qquad h\in \mathcal O_n. $$ In particular $g$ and $h$ are relatively prime in $\mathcal O_n$. Let $y$ close enough to $0$, so that $g$ and $h$ remain relatively prime in $\mathcal O_{n,y}$. Let $g_1\in \mathcal O_{n,y}$ be a irreducible defining function for $Y$, around $y$, and let $$ f=g_1^b\cdot h_1, \qquad b=\operatorname{ord}_y f, \qquad h_1\in \mathcal O_{n,y}. $$ Since $Y$ is irreducible, $g$ and $g_1$, describe the same locus around $y$, we see that $g$ factorize in $\mathcal O_{n,y}$ as $$ g=g_1^\ell\cdot g_2, $$ with $g_2 \in \mathcal O_{n,y}^*$. Since $g$ and $h$ are relatively prime in $y$, we see that $g_1 \not\mid h$, hence $$ b=a\cdot \ell. $$ It remains to show that $\ell=1$. It is not restrictive to assume that $g$ is a Weierstrass polynomial $g \in \mathscr{O}_{n-1}\left[z_n\right]$ i.e. $$ g=z_n^d+a_1z_n^{d-1}+\dots+a_d, \quad a_j\in\mathscr{O}_{n-1}, \ a_j(0)=0. $$ Since $g_1 \mid g$ and $g$ is regular in $z_n$, also $g_1$ must be regular in $z_n$. Then, we can apply Weierstrass preparation theorem see that (up to the product of a unit $u\in \mathscr{O}_{n,y}$), $g_1\in \mathscr{O}_{n-1,y}[z_n]$ and that it is monic, i.e. $$ g_1=z_n^s+b_1z_n^{s-1}+\dots+b_s, \quad b_j\in\mathcal{O}_{n-1,y}. $$ Under this assumption we have $g=g_1^\ell$, hence each of the coefficients $b_1,\dots,b_s$ can be written as sums of products of the powers of $a_1, \dots, a_d$. We proved that we can extend $g_1$ to a holomorphic function $\tilde g_1 \in \mathcal{O}_n$. By construction $g=\tilde g_1^\ell$ in $\mathcal{O}_n$. Since $g$ is irreducible we conclude that $\ell=1$.