Structure of functions with multiplicity 2 Let $\Omega$ be a domain in $\mathbb{C}^{n}$ and $f:\Omega\mapsto\mathbb{C}$ be a holomorphic function. Define $ord_{a}f$ to be the total order of the function $f$ at the point $a$ (also called the multiplicity of the zero of $f$ at $a$). Now, suppose $ord_{z}f=2$ for all $z\in Z(f)$, where $Z(f)$ stands for the zero set of $f$ in $\Omega$. Does this force the following$?$:
For each $z\in Z(f)$, $\exists$ a neighborhood $U_{z}$ of $z$ in $\Omega$ s.t. $f=\psi^{2}g$ on $U_{z}$, where $\psi,g:U_{z}\mapsto\mathbb{C}$ are holomorphic functions, $g$ is nonvanishing on $U_{z}$, $ord_{z}\psi=1$ for all $z\in Z(\psi)=Z(f)\cap U_{z}$.
 A: Yes.  Suppose near $z$ the subvariety $Z(f)$ is defined by $\psi$ (as it is a hypersurface there is always a single defining function).  That is for every point $p \in U_z$, the germ of $\psi$ at $p$ generates the ideal of germs at $p$ of holomorphic functions vanishing on $Z(f)$.  If the order of $f$ is uniformly 2 on $Z(f)$, it must be that all partials of $f$ must vanish uniformly on $Z(f)$.  Note that $f$ itself is in the ideal, so it means that
$$
f = g \psi
$$
This means that any partial derivative is locally in the ideal.  In other words:
$$
\frac{\partial f}{\partial z_j}
=
\frac{\partial g}{\partial z_j}
\psi
+
g
\frac{\partial \psi}{\partial z_j}
$$
But for some $j$ we know that $\frac{\partial \psi}{\partial z_j}$ does not vanish identically on $Z(f)$, because $\psi$ generates the ideal
even at regular points of $Z(f)$, so $\psi$ must have order 1 there, so at least one derivative of $\psi$ is nonzero on an open dense set of the regular points of $Z(f)$.  Hence it must be that $g$ vanishes identically on the regular points of $Z(f)$ and hence on all of $Z(f)$.  Thus $g$ is in the ideal and
$$
g = h \psi
$$
And you are done.
