Does irreducibility of germ at a point implies local irreducibility of the function?

Question

Let $M$ be a complex manifold, $U$ be an open subset of $M$, $x\in U$ and $f$ be a $\mathbb{C}$ valued holomorphic function on U. Assume that the germ of $f$ at $x$, denoted as $f_{x}$ is irreducible in $\mathcal{O}_{M,x}$. Does this imply that $f$ is irreducible in some sub-neighbourhood of $x$ in $U ?$

Answer 1

If you mean by irreducible in an open $V$ that it defines an irreducibles germ at each point of $V$, then the answer is yes. See for example $\textit{Huybrechts's Complex Geometry}$:

$\textbf{Proposition 1.1.35}$ Let $f\in \mathcal{O}_{\mathbb{C}^n,0}$ be irreducible. Then for sufficiently small $\varepsilon$ and $z\in B_{\varepsilon}(0)$ the induced element $f\in \mathcal{O}_{\mathbb{C}^n,z}$ is irreducible.

$\textbf{Edit}$ In the notation of the Proposition set $V=B_{\varepsilon}(0)$. I claim that $f$ is irreducible on $V$, i.e. if $f=gh,$ then $g$ is nowhere vanishing on $V$ or $h$ is nowhere vanishing on $V$. Indeed, since $f_0=g_0h_0$ and $f_0$ is irreducible, $g_0$ or $h_0$ must be a unit in $\mathcal{O}_{\mathbb{C}^n,0}$, say $g_0$ is a unit, i.e. $g(0)\neq 0$. But the set $A:=\{z\in V : g(z)\neq 0\}$ is open and closed in $V$: It is clearly open and given any $z\in V-A$, we have $g(z)=0$ and we must therefore have that $h(z)\neq 0$ (because $f_z$ is irreducible in $\mathcal{O}_{\mathbb{C}^n,z}$). Now pick an open neighbourhood $W\subset V$ of $z$ on which $h$ is non-zero. Then clearly $g(w)=0$ for all $w\in W$, i.e. $W\subset V-A$. This shows that $V-A$ is open. Since $V$ is connected and $0\in A$ (so that it is non-empty), we deduce that $A=V$, so that $g$ is nowhere vanishing on $V$.