Lemma 6.12.223099F8BA856

Proof. Let $g=\sum _\alpha c_\alpha w^\alpha $ be the power series at $p$ of a holomorphic function, as in (5.2.1). If $g=w_ i^ ag'$ near $p$, then multiplying the series of $g'$ by $w_ i^ a$ shows that $c_\alpha =0$ whenever $\alpha _ i<a$. Conversely, if $c_\alpha =0$ whenever $\alpha _ i<a_ i$ for some $i\leq \ell $, then $g=\prod _ iw_ i^{a_ i}\cdot \sum _\alpha c_\alpha w^{\alpha -\sum _ ia_ ie_ i}$, where $e_ i$ is the $i$th unit vector, and the last series converges on the same polydiscs as that of $g$. This proves the last assertion, and it shows that $k_ i<\infty $, since otherwise all $c_\alpha $ of $h$ would vanish. Hence $h=u\prod _{i\leq \ell }w_ i^{k_ i}$ with $u$ holomorphic near $p$, and $u$ is not divisible by any $w_ i$, $i\leq \ell $, by the maximality of $k_ i$.

It remains to show that $u(p)\neq 0$. Suppose, to the contrary, that $u(p)=0$. The zero set of $u$ near $p$ lies in that of $h$, hence in $\{ w_1\cdots w_\ell =0\} $; so $\ell \geq 1$. We work in the coordinates $w$, with $p=0$. Let $P$ be the first nonzero homogeneous term of $u$; it has positive degree $m$. Choose $v\in \mathbb C^ n$ with $P(v)\neq 0$ and $v_ i\neq 0$ for all $i$, which is possible since the polynomial $P(v)v_1\cdots v_ n$ is not identically zero. Then $t\mapsto u(tv)$ has a zero of order $m$ at $t=0$; choose $\varepsilon >0$ such that it has no other zeros in $|t|\leq \varepsilon $, and such that all points $y+tv$ considered below lie in the neighborhood of $0$ where the hypothesis on the zero set holds. Since $(y,t)\mapsto u(y+tv)$ is continuous and does not vanish on the compact set $\{ 0\} \times \{ |t|=\varepsilon \} $, there is a connected neighborhood $N$ of $0$ such that $u(y+tv)\neq 0$ for $y\in N$ and $|t|=\varepsilon $. By the argument principle, the number of zeros of $t\mapsto u(y+tv)$ in $|t|<\varepsilon $, counted with multiplicity, is

a continuous integer-valued function of $y\in N$; it equals $m\geq 1$ at $y=0$, hence everywhere on $N$. So for every $y\in N$ there is $t$ with $u(y+tv)=0$, and then $y_ i+tv_ i=0$ for some $i\leq \ell $, that is, $u(\sigma _ i(y))=0$ for the linear projection $\sigma _ i(y)=y-(y_ i/v_ i)v$ onto the hyperplane $H_ i=\{ y_ i=0\} $. Hence the product of the holomorphic functions $u\circ \sigma _ i$, $i\leq \ell $, vanishes on $N$. If none of them vanished identically on $N$, each would be nonzero on a dense open subset of $N$, by the identity theorem, and so would be their product. Thus some $u\circ \sigma _ i$ vanishes on $N$; since $\sigma _ i(y)=y$ for $y\in H_ i$, the function $u$ vanishes on $H_ i\cap N$, which is a neighborhood of $0$ in $H_ i$. By the first paragraph, applied with $a=1$, $u$ is then divisible by $w_ i$ near $0$: all coefficients $c_\alpha $ of $u$ with $\alpha _ i=0$ vanish, since they are the coefficients of the restriction of $u$ to $H_ i$. This contradiction shows that $u(p)\neq 0$. $\square$