Lemma 5.5.1AD4CF8180187

Proof. We prove the first two assertions together by induction on $n$. For $n=1$, every zero of $f$ is isolated: at a zero $a$, the Taylor expansion has a first nonzero term $c(z-a)^ m$, since otherwise $f$ would vanish near $a$ and hence on $\Omega $ by the identity theorem; so $f(z)=(z-a)^ mg(z)$ with $g(a)\neq 0$. Hence the zero set meets every compact subset of $\Omega $ in a finite set, and since $\Omega $ is a countable union of compact sets, it is countable, hence null. Moreover, $f$ does not vanish identically on a disc $D\subset \Omega $, by the identity theorem, which proves the second assertion for $n=1$.

Let $n\geq 2$, and let $P=P'\times D$ be as in the statement; after permuting the coordinates, which preserves Lebesgue measure, we may assume that $j=n$, and we write $w=z_ n$ and $w_0$ for the center of $D$. The functions $c_ k(z')=\partial _ w^ kf(z',w_0)/k!$ are holomorphic on $P'$, and, by the one-variable Taylor expansion of $w\mapsto f(z',w)$ on the disc $D$ centered at $w_0$, this function vanishes identically on $D$ if and only if all $c_ k(z')$ vanish. Since $f$ does not vanish identically on the nonempty open set $P$, by the identity theorem, some $c_ k$ is not identically zero on the polydisc $P'$, and its zero set $N\subset P'$ is null by the first assertion in dimension $n-1$. This proves the second assertion in dimension $n$, since the set in question is contained in $N$. For $z'\notin N$, the zero set of $w\mapsto f(z',w)$ in $D$ is null by the case $n=1$. By Tonelli's theorem, applied to the indicator function of the relatively closed zero set of $f$ in $P$, this zero set is null; and countably many such polydiscs, for instance those with rational centers and radii contained in $\Omega $, cover $\Omega $.

For the last assertion, translate $a$ to the origin and let $P_ m$ be the first nonzero term of the homogeneous expansion of $f$ at $0$; it exists because $f$ does not vanish identically near $0$, and $m\geq 1$ because $f(0)=0$. The set $\{ P_ m\neq 0\} $ is open and dense in $\mathbb C^ n$, because a polynomial vanishing on a nonempty open set is zero: its derivatives at a point $b$ of that set vanish, and the polynomial is the sum of its finite Taylor expansion at $b$. Choose $v_1\in \{ P_ m\neq 0\} $, and inductively $v_ j\in \{ P_ m\neq 0\} $ outside the linear span of $v_1,\ldots ,v_{j-1}$; this is possible because the complement of a proper linear subspace is open and dense, and the intersection of two dense open sets is dense. Then $v_1,\ldots ,v_ n$ is a basis, and $f(tv_ j)=t^ mP_ m(v_ j)+O(t^{m+1})$ by (5.2.2), so $t\mapsto f(tv_ j)$ does not vanish identically near $t=0$. Taking $v_1,\ldots ,v_ n$ as coordinate axes proves the assertion. $\square$