PomeLemma 9.3.12.D245F9712FB1 Let $M$ be a complex manifold, let $(v,w)$, $w=(w_1,\ldots ,w_ k)$ with $k\geq 0$, be holomorphic coordinates on an open set $U\subset M$, and let $N=\{ w_1=\cdots =w_ k=0\} \subset U$. Then for every $y\in U$ the ideal $I_{N,y}$ of germs vanishing on $(N,y)$ is generated by the germs of $w_1,\ldots ,w_ k$; it is the unit ideal if $y\notin N$. Moreover, for $y\in N$ the germ $(N,y)$ is irreducible, and its dimension is $\dim _ yM-k$, the dimension of $N$ at $y$.
Lemma 9.3.12D245F9712FB1
Proof. If $y\in U\setminus N$, then some $w_ l$ does not vanish at $y$, so it is a unit, and $(N,y)$ is empty because $N$ is closed in $U$; both ideals are the unit ideal. Let $y\in N$, and let $g$ be holomorphic on a polydisc centered at $y$ in the coordinates $(v,w)$, vanishing on $N$. Then $g(v,0)=0$, and
\[ g(v,w)=\int _0^1\frac{\mathrm d}{\mathrm dt}g(v,tw)\, \mathrm dt=\sum _ lw_ l\int _0^1\partial _{w_ l}g(v,tw)\, \mathrm dt, \]
where the integrals are holomorphic by differentiation under the integral sign. Hence the germ of $g$ lies in the ideal generated by $w_1,\ldots ,w_ k$; conversely, the $w_ l$ vanish on $N$. Consequently, the map $g\mapsto g(v,0)$ identifies the quotient of $\mathcal O_{M,y}$ by this ideal with the ring of germs of holomorphic functions of $v$ at $v(y)$, which is an integral domain (section 9.2); so $I_{N,y}$ is prime. Since $y\in N$, the germ $(N,y)$ is nonempty, and it is irreducible by proposition 9.3.1. All points of $N$ near $y$ are regular points of $N$ of dimension $\dim _ yM-k$, so this is the dimension of $(N,y)$, by proposition 9.3.9. $\square$
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