Lemma 9.2.9AA4164301998

Proof. Let $f$ be holomorphic on $M\setminus A$ and locally bounded near $A$. If $\dim M=0$, then $A=\varnothing $ and there is nothing to prove. Fix $a\in A$. By lemma 9.2.8 there is a holomorphic function $g\not\equiv 0$ on a connected neighborhood of $a$ vanishing on $A$; choose linear coordinates $(z',w)$ centered at $a$ in which $g$ is distinguished in $w$. By the Weierstrass preparation theorem and (9.2.1), there is a polydisc $\Delta '\times \Delta $ centered at $a$, with $\Delta =\{ |w|<r\} $, on which $g$ is a unit times a Weierstrass polynomial $W$ and $f$ is bounded off $A$, such that all roots of $W(z',\cdot )$, $z'\in \Delta '$, have modulus at most some $\rho <r$. For fixed $z'\in \Delta '$, the function $f(z',\cdot )$ is holomorphic on $\Delta $ outside the finitely many points of $A$ in the fiber, which are roots of $W(z',\cdot )$, and it is bounded near them; by the removable singularity theorem in one variable it extends holomorphically to $\Delta $. For $\rho <s<r$ the function

is holomorphic, by differentiation under the integral sign, since $f$ is holomorphic near $\Delta '\times \{ |\zeta |=s\} $, a set disjoint from $A$. By Cauchy's formula for the extended functions $f(z',\cdot )$, it agrees with $f$ outside $A$. Letting $s\uparrow r$, we obtain a holomorphic extension of $f$ to $\Delta '\times \Delta $. Since $A$ is nowhere dense, two holomorphic extensions on overlapping open sets agree, being continuous functions that coincide on a dense subset; so the local extensions glue to a holomorphic function on $M$, and this extension is unique.

If $M\setminus A$ were disconnected, the function equal to $0$ on one nonempty union of its connected components and to $1$ on the others would be bounded and holomorphic on $M\setminus A$. Its holomorphic extension is continuous and takes values in $\{ 0,1\} $ on the dense set $M\setminus A$, hence everywhere; since $M$ is connected, it is constant, a contradiction.

Now let $b,c\in \mathcal O_ n$, $c\neq 0$, with $|b|\leq C|c|$ on a connected neighborhood $U$ of the origin on which both are defined. The set $\{ c=0\} $ is a proper analytic subset of $U$, and the holomorphic function $b/c$ on $U\setminus \{ c=0\} $ is bounded by $C$. By the first part it extends to a holomorphic function $h$ on $U$, and $ch=b$ on the dense set $U\setminus \{ c=0\} $, hence on $U$; so $b\in c\, \mathcal O_ n$. Finally, let $b/c\in \operatorname {Frac}(\mathcal O_ n)$, with $c\neq 0$, satisfy a monic equation $(b/c)^ m+a_1(b/c)^{m-1}+\cdots +a_ m=0$ with $a_ j\in \mathcal O_ n$. Multiplying by $c^ m$, we obtain the identity $b^ m+a_1b^{m-1}c+\cdots +a_ mc^ m=0$ between germs, which holds on a connected neighborhood of $0$ on which the $a_ j$ are bounded. At the points of this neighborhood where $c\neq 0$, the number $b/c$ is a root of a monic polynomial with bounded coefficients, so $|b/c|$ is bounded there by (9.2.1); at the points where $c=0$ the identity gives $b^ m=0$. Hence $|b|\leq C|c|$ on this neighborhood for some constant $C$, and by what we have just proved, $b/c\in \mathcal O_ n$, which proves that $\mathcal O_ n$ is integrally closed. $\square$