Proposition 9.5.60E64BF4DC7C7

Proof. If $d=0$, then $f$ has no zeros. Let $d\geq 1$ and let $x$ be a zero of $f$. We first show that $\dim (\{ f=0\} ,x)=d-1$. For each branch $X_0$ of $(X,x)$, the germ $\{ f=0\} \cap X_0$ is a subgerm of $X_0$ different from it, so $\dim (\{ f=0\} ,x)\leq d-1$ by proposition 9.3.9. Fix a branch $X_0$, and choose linear forms $\ell _1,\ldots ,\ell _{d-1}$ on the ambient space of a local model, vanishing at $x$, inductively such that $\ell _ j$ does not vanish identically on any positive-dimensional branch of the germ at $x$ of $\{ f=0\} \cap X_0\cap \{ \ell _1=\cdots =\ell _{j-1}=0\} $. At each step this excludes finitely many proper linear subspaces, and, by proposition 9.3.9, it lowers the dimension of this germ by at least one as long as it is positive. Since the initial dimension is at most $d-1$, the point $x$ is isolated in the common zero set of $\Phi =(f,\ell _1,\ldots ,\ell _{d-1})$ on $X_0$.

Suppose that $\dim (\{ f=0\} ,x)<d-1$; then $d\geq 2$, and by upper semicontinuity $\dim (\{ f=0\} ,y)\leq d-2$ for all zeros $y$ of $f$ near $x$. By lemma 9.5.5, applied to the germ $(X_0,x)$, which is pure-dimensional, there is a neighborhood $X'_0$ of $x$ in a small representative of $X_0$, on which this bound holds, such that $\Phi $ maps $X'_0$ onto a ball $B\subset \mathbb C^ d$ centered at $0$. Hence $\Phi $ maps $\{ f=0\} \cap X'_0$ onto the set $B\cap \{ w_1=0\} $, which has positive $(2d-2)$-dimensional Lebesgue measure; here $w_1$ is the first coordinate of $\mathbb C^ d$. On the other hand, by corollary 9.3.13(2), $\{ f=0\} \cap X'_0$ is a finite union of complex submanifolds of dimensions at most $d-2$ of open subsets of the ambient space, on each of which $\Phi $ has rank at most $d-2$; by lemma 9.3.15(1), its image under $\Phi $ is $\mathcal H^{2d-2}$-null, hence Lebesgue null in $\{ w_1=0\} \cong \mathbb R^{2d-2}$. This contradiction shows that $\dim (\{ f=0\} ,x)=d-1$.

Now let $V_1$ be a branch of $(\{ f=0\} ,x)$. By corollary 9.3.6, there are regular points $y$ of $V_1$ arbitrarily close to $x$ that do not lie on the other branches. Near such a point, $\{ f=0\} $ coincides with $V_1$, which is a manifold of dimension $\dim V_1$ by purity. The hypotheses hold at $y$, so the first part gives $\dim (\{ f=0\} ,y)=d-1$, and therefore $\dim V_1=d-1$. $\square$