Proposition 9.5.21497A0492453

Proof. The germ of $B\cup X_{\mathrm{sing}}$ is again different from $(X,x)$, since $(X,x)$ is irreducible and $X_{\mathrm{sing}}$ has smaller dimension by corollary 9.3.13(1). Since every analytic germ $(A,x)$ satisfies $V(I_{A,x})=(A,x)$, the ideals of these two germs differ, and we can choose a holomorphic germ $f$ that vanishes on $B\cup X_{\mathrm{sing}}$ but not identically on $X$. Take a local parametrization of $(X,x)$ over a polydisc $\Delta '\subset \mathbb C^ d$, $d\geq 1$, with covering locus $X_\delta $, and for $z'\in \Delta '\setminus \{ \delta =0\} $ let $N(z')$ be the product of the values of $f$ at the $q$ points of $X$ over $z'$. This function is holomorphic and bounded, so it extends holomorphically to $\Delta '$ by lemma 9.2.9. It is not identically zero. Otherwise, choose a small ball in $\Delta '\setminus \{ \delta =0\} $ over which the $q$ points are given by holomorphic sections $s_1,\ldots ,s_ q$; the product of the functions $f\circ s_ i$ would vanish identically there, so one of these functions would vanish on an open set. Then $f$ would vanish on a nonempty open subset of the connected manifold $X_\delta $, hence on $X_\delta $ by the identity theorem, and on $X$ by density, which is not the case.

The function $g=N\delta $ is holomorphic and not identically zero on $\Delta '$. Choose a vector $v\in \mathbb C^ d$ at which the first nonzero homogeneous term of $g$ does not vanish, and $\rho >0$ such that $g(tv)\neq 0$ for $0<|t|<\rho $ and the ball $U=\{ |z'|<\rho |v|\} $ lies in $\Delta '$. Then $Y=X\cap \pi ^{-1}(\{ tv:|t|<\rho \} )$ is an analytic subset of $X\cap \pi ^{-1}(U)$. Over the punctured disc, $Y$ consists of points of $X_\delta $ at which $f\neq 0$, and it is a covering of the punctured disc; over $0$ it consists of the single point $x$. Thus $Y\setminus \{ x\} $ is a complex manifold of dimension one, and by properness of $\pi $ its points accumulate at $x$, so $x$ is not an isolated point of $Y$. Let $(C,x)$ be a branch of $(Y,x)$ of positive dimension. Its regular points near $x$ lie in the one-dimensional manifold $Y\setminus \{ x\} $, so $(C,x)$ has dimension one. Let $\gamma \colon \Delta \to C$ be the parametrization of proposition 9.5.1. For $t\neq 0$, the point $\gamma (t)$ lies in $C\setminus \{ x\} \subset Y\setminus \{ x\} $, hence in $X_\delta \subset X_{\mathrm{reg}}$ and outside $\{ f=0\} \supset B$. $\square$