Lemma 5.5.5116B90FD9752

Proof. Choose coordinates as in lemma 5.5.1, so that $f$ does not vanish identically on any coordinate axis through $z_0=0$. Fix $j$, and write points as $(z',w)$ with $w=z_ j$. By theorem 5.5.2 in the variable $w$, there is a polydisc $P_ j=P'_ j\times \{ |w|<\rho \} $ in the domain of $f$ on which $f=gW$, where $g$ is a holomorphic unit and $W$ is a monic polynomial in $w$ of some degree $m\geq 1$ whose coefficients are holomorphic on $P'_ j$ and whose roots lie in $\{ |w|<\rho _1\} $ for some $\rho _1<\rho $. We shall find a holomorphic function $h$ on $P'_ j$, not identically zero, such that every zero of $f$ in $P_ j$ lying over $\{ h\neq 0\} $ belongs to $V_{\mathrm{reg}}$. Then $S\cap P_ j\subset \pi _ j^{-1}(N_ j)$ with $N_ j=\{ h=0\} $, which is null by lemma 5.5.1 if $n\geq 2$, and empty if $n=1$, when $P'_ j$ is a point. Any polydisc $P$ centered at $0$ and contained in $P_1\cap \cdots \cap P_ n$ then has the required property.

For $z'\in P'_ j$, let $\lambda _1(z'),\ldots ,\lambda _ m(z')$ be the roots of $W(z',\cdot )$, repeated according to multiplicity, and put $p_ k(z')=\sum _ l\lambda _ l(z')^ k$ for $k\geq 0$. By the residue theorem,

so $p_ k$ is holomorphic on $P'_ j$, by differentiation under the integral sign. Let $s$ be the largest number of distinct roots of $W(z',\cdot )$ for $z'\in P'_ j$, and let $h$ be the determinant of the $s\times s$ matrix $(p_{a+b})_{0\leq a,b\leq s-1}$; it is holomorphic on $P'_ j$. For fixed $z'$, this matrix is $A^{\mathsf T}A$, where $A$ is the $m\times s$ matrix with entries $A_{la}=\lambda _ l^ a$.

To compute $h(z')$, note that the $b$th column of $A^{\mathsf T}A$ is $\sum _ l\lambda _ l^ bv_ l$, where $v_ l=(\lambda _ l^0,\ldots ,\lambda _ l^{s-1})$ is the $l$th row of $A$. Expanding the determinant by multilinearity in the columns, discarding the terms in which some $v_ l$ occurs twice, and grouping the remaining terms according to the set $L$ of indices that occur, we find that each group contributes $(\det A_ L)^2$. Indeed, if $L=\{ l_1<\cdots <l_ s\} $, the terms of this group are indexed by the permutations $\pi $ of $\{ 1,\ldots ,s\} $, the term of $\pi $ taking $\lambda _{l_{\pi (b+1)}}^ bv_{l_{\pi (b+1)}}$ from the $b$th column. Reordering the columns multiplies the determinant by $\operatorname {sign}(\pi )$, so the group contributes $\det A_ L\sum _\pi \operatorname {sign}(\pi )\prod _{b=0}^{s-1}\lambda _{l_{\pi (b+1)}}^ b$, and this sum is the Leibniz expansion of $\det A_ L$. Thus we obtain the Cauchy–Binet formula

where $L$ runs over the $s$-element subsets of $\{ 1,\ldots ,m\} $, $A_ L$ is the square matrix formed by the rows of $A$ with indices in $L$, and the last equality is the Vandermonde formula. If $W(z',\cdot )$ has fewer than $s$ distinct roots, every set $L$ contains two indices with equal roots, and $h(z')=0$. If $W(z',\cdot )$ has exactly $s$ distinct roots $\mu _1,\ldots ,\mu _ s$, of multiplicities $k_1,\ldots ,k_ s$, then the nonzero terms are those of the $k_1\cdots k_ s$ sets $L$ containing exactly one index of each root, and each of them equals $\prod _{c<c'}(\mu _ c-\mu _{c'})^2\neq 0$; hence $h(z')\neq 0$. By the choice of $s$, the function $h$ is not identically zero.

Now let $h(z'_0)\neq 0$, and let $(z'_0,w_0)\in P_ j$ be a zero of $f$. Then $W(z'_0,w_0)=0$, since $g$ is a unit, and $W(z'_0,\cdot )$ has exactly $s$ distinct roots $\mu _1,\ldots ,\mu _ s$, of multiplicities $k_1,\ldots ,k_ s$; say $w_0=\mu _1$. Choose disjoint closed discs $E_ c$ centered at $\mu _ c$ in $\{ |w|<\rho \} $, and a connected neighborhood $Y$ of $z'_0$ in $P'_ j$ such that $W(z',\cdot )$ has no zeros on the boundary circles $\partial E_ c$ for $z'\in Y$; this is possible because $W(z'_0,\cdot )$ has no zeros on these compact circles and $W$ is continuous. By the argument principle, the number of roots of $W(z',\cdot )$ in $E_ c$, counted with multiplicity, is $(2\pi \mathrm i)^{-1}\int _{\partial E_ c}\partial _ wW/W\, \mathrm dw$, a continuous integer-valued function on $Y$, hence equal to $k_ c$. Thus each $E_ c$ contains at least one root of $W(z',\cdot )$. Since $W(z',\cdot )$ has at most $s$ distinct roots, each $E_ c$ contains exactly one, $\mu _ c(z')$, of multiplicity $k_ c$, and since $k_1+\cdots +k_ s=m$, there are no other roots. By the residue theorem,

which is holomorphic on $Y$, and $W(z',w)=\prod _ c(w-\mu _ c(z'))^{k_ c}$ for $z'\in Y$, both sides being monic polynomials in $w$ with the same roots and multiplicities. Near $(z'_0,w_0)$ the function $U=\prod _{c\geq 2}(w-\mu _ c(z'))^{k_ c}$ does not vanish, so $f=(gU)(w-\mu _1(z'))^{k_1}$, where $gU$ is a holomorphic unit and $\mathrm d(w-\mu _1(z'))\neq 0$. That is, $(z'_0,w_0)\in V_{\mathrm{reg}}$. $\square$