I am reading Principles of Algebraic Geometry by Griffith and Harris. Here the authors define an analytic variety in a domain as follows :
A subset $V$ of an open set $U \subset \mathbb{C}^n$ is an analytic variety in $U$ if, for any $p \in U$, there exists a neighbourhood $U’$ of $p$ in $U$ such that $V \cap U’$ is the common zero locus of a finite collection of holomorphic functions $f_1, \dotsc, f_k$ on $U’$ i.e. $$V \cap U’=\{p’ \in U’ \ : \ f_1(p’)=\dotsc=f_k(p’)=0\}. $$
In particular, $V$ is an analytic hypersurface if $V$ is locally the zero locus of a single non-zero holomorphic function $f$. The irreducibility is defined in the usual way : An analytic variety $V \subset U \subset \mathbb{C}^n$ is said to be irreducible if $V$ cannot be written as the union of two analytic varieties $V_1, V_2 \subset U$ with $V_1, V_2 \ne V$.
It follows from the definition of analytic variety that if $f \in \mathbb{C}[z_1, \dotsc, z_n]$ and $U$ is a domain (open + connected set) in $\mathbb{C}^n$, then $V=Z(f) \cap U$ is an analytic variety (here $Z(f)$ is the set of zeros of $f$).
Here is my question : If $f$ is irreducible in $\mathbb{C}[z_1, \dotsc, z_n]$, then does it imply that $V=Z(f) \cap U$ is an irreducible variety?
I have just started learning algebraic geometry. I know that if $f$ is reducible, then $Z(f) \cap U$ is not irreducible. Indeed, if $f=f_1 \dotsc f_k$, then $Z(f)=Z(f_1) \cup \dotsc \cup Z(f_k)$ and so, $V=Z(f) \cap U=(Z(f_1) \cap U) \cup \dotsc \cup (Z(f_k) \cup U)$. Since each $V_i=Z(f_i) \cap U$ is a variety and $V_i \subset U$ with $V_i \ne V$, therefore, $V=V_1 \cup \dotsc \cup V_k$ is not irreducible variety.
Let if possible, $V=V_1 \cup V_2$ for two analytic varieties $V_1, V_2 \subset U$ with $V_1, V_2 \ne V$. Take any $p \in U$. By definition, there exists a neighbourhood $W$ of $p$ in $U$ and holomorphic functions $f_1, f_2$ on $W$ such that $f_1$ vanishes identically on $V_1 \cap W$ and $f_2$ vanishes identically on $V_2 \cap W$. If this so happens that, $f$ divides $f_1$ and $f_2$ then irreducibility of $f$ implies that $f$ divides $f_1$ or $f$ divides $f_2$. How to proceed from here.