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 $U ?$

  • $\begingroup$ This is quite the same question as your previous one $\endgroup$ – reuns Dec 23 '17 at 19:34
  • $\begingroup$ Yes that question can be boiled down to this one (Specifically through the way i proposed). But there are other ways one can address that question whereas this one is interesting on its own. I somehow feel that this answer should follow from very standard arguments of the existing literature. That's why i posted it. $\endgroup$ – Sam Dec 23 '17 at 19:50
  • $\begingroup$ Did you try in the case of two complex variables ? I think it reduces to a problem of radius of convergence of the involved analytic functions and how to relate it to the location of singularities. $\endgroup$ – reuns Dec 23 '17 at 20:22
  • $\begingroup$ Can you give a sketch of what you are trying to say? $\endgroup$ – Sam Dec 23 '17 at 20:29
  • $\begingroup$ $V \subset \mathbb{C}^2$ bounded open, $f $ analytic non-constant $V_2 \supset \overline{V} \to \mathbb{C}$. A zero of $f$ on $V$ would be a locally analytic curve $C$ defined by finitely many analytic charts $\phi_j : U_j \subset \mathbb{C} \to \mathbb{C}^2$. The $(U_j,\phi_j)$ define $C \cap V$ in the sense that following $C$ by analytic continuation, or we stay in the charts, or we arrive outside $V$. The main point is "finitely many charts". Finally, decompose $\{ (z_1,z_2\in V,f(z_1,z_2)= 0\}$ into finitely many such zeros. $\endgroup$ – reuns Dec 23 '17 at 20:58

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 $V$.

  • $\begingroup$ Well, here $f_{0}\in\mathcal{O}_{\mathbb{C}^{n},0}$ being irreducible gives a neighbourhood of $0$, namely $B_{\epsilon}(0)$ s.t. for each $p\in B_{\epsilon}(0)$, $f_{p}$ is irreducible in $\mathcal{O}_{\mathbb{C}^{n},p}$. But how this ensures a neighbourhood of $0$, say $V$ s.t. the function $f$ itself is irreducible globally on $V$.(I am calling here a function $f$ irreducible on $V$ if $f=g.h$, where $g,h$ are non-zero, non-unit elements of $\mathcal{O}(V)$ i.e. not identically $0$ and not non-vanishing on $V$). $\endgroup$ – Sam Dec 24 '17 at 6:12
  • $\begingroup$ @Sam This is implied by the Proposition, see the edit. $\endgroup$ – user363120 Dec 24 '17 at 15:38
  • $\begingroup$ Ah right! Thanks a lot. $\endgroup$ – Sam Dec 24 '17 at 21:13

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