TITLE: Zero set of holomorphic function locally isomorphic to zero set of finitely many polynomial?
QUESTION [7 upvotes]: Let $f\colon \Omega \to \mathbb{C}$ be a holomorphic function on an open subset $\Omega$ of $\mathbb{C}^n$. Denote by $Z(f)$ its zero set. It is clear that there exists $f$ such that $Z(f)$ is not the zero set of finitely many polynomials.
My question is: Does the above remain true if one allows to locally change the embedding of $Z(f)$ by a biholomorphic map. More precisely, does there exist an $f$ such that there is a point $p\in Z(f)$ with no open neighbourhood $U\subseteq \Omega$ that admits a biholomorphic map $\phi\colon U \to V$ such that $Z(f\circ \phi^{-1})=Z(P_1,...,P_k)$ for some polynomials $P_i$ on $V$.
It is clear that if $Z(f)$ is a complex submanifold, then for every point such a $\phi$ always exists. So, the zero set needs to be singular and $n>1$.
Essentially, my question is: Is every analytic hypersurface $X\subseteq \mathbb{\Omega}$ locally equivalent to an algebraic variety (up to biholomorphic changes of the embedding).
If not are there explicit examples? What happens in the case of $Z(f_1,...,f_l)$ for multiple analytic functions.

REPLY [2 votes]: In "Local Properties of Analytic Varieties" H. Whitney gives an example (Example 14.1) of classes of analytic varieties that have points that are not locally equivalent to an algebraic variety.
The examples are the varieties in $\mathbb{C}^3$ given by
$f(x,y,z):=x\cdot y \cdot (y-x)\cdot (y- (3+z)x)\cdot(y-\gamma(z)\cdot x)$,
where $\gamma\colon \mathbb{C}\to\mathbb{C}$ is any transcendental function such that $\gamma(0)=4$. The argument why these are not locally algebraic relies on some further results about tangent cones in the paper.