Let $D \subset \mathbb C^n$ be a domain and let $f \in \mathscr O(D)$, $f \not\equiv 0$ be a holomorphic function. Define $$ V_f = \bigl\{ z \in D : f(z) = 0 \bigr\}. $$ Let $p \in V_f$. Suppose that $f$ is irreducible in the ring of germs $\mathscr O_p$. Is it true that there exists a neighborhood $U$ of point $p$ such that $f$ is irreducible in $\mathscr O_q$ for all $q \in V_f \cap U$? Maybe I should use somehow the property that if two functions in $\mathscr O_p$ are relatively prime then they will be relatively prime in $\mathscr O_q$ for $q$ close to $p$?

Update. If $f$ is reducible in $\mathscr O_q$ then $f = f_1 f_2$ in a neighborhood of $q$ with $f_1(q)=f_2(q)=0$. This implies that $f(q)=0$ and $\frac{\partial f(q)}{\partial z_k}=0$, $k=1$, $\dots$, $n$. If at point $p$ some $\frac{\partial f(p)}{\partial z_k} \neq 0$ then the statement is true.

Update 2. Suppose that $f$ divides all $\frac{\partial f}{\partial z_k}$ in $\mathscr O_p$ so that we have $$ \frac{\partial f}{\partial z_k} = fh_k, \quad k =1,\ldots,n, $$ in a neighborhood of $p$ with holomorhic $h_k$, $h_k(0) = 0$. Differentiating these equalities we obtain $$ \frac{\partial^2 f}{\partial z_k \partial z_l} = \frac{\partial f}{\partial z_l} h_k + f \frac{\partial h_k}{\partial z_l}, $$ this implies $\frac{\partial^2 f(p)}{\partial z_k z_l} = 0$ for all $k$, $l$. We can continue this process showing that all derivatives of $f$ at $p$ are equal to zero so that $f \equiv 0$.

The only remaining case when the statement may be false is when all $\frac{\partial f}{\partial z_k}(p)=0$ and $f$ doesn't divide some $\frac{\partial f}{\partial z_k}$ in $\mathscr O_p$.


1 Answer 1


Irreducible germs do not propagate: take $D=\mathbb{C}^3$ and $f=y^2-xz^2$: in the origin $f$ is irreducible, but it is reducible for every $p=(x_0,0,0)$ with $x_0\neq 0$: $$f=f_1f_2=(y-\sqrt{x}z)(y+\sqrt{x}z)$$ observe that $f_j(p)=0$ so $f_j\not\in\mathcal{O}_p$, in particular they are not invertibles in $\mathcal{O}_p$. In order to prove irreducibility first note that $f$ is a Weierstrass polynomial in $\mathcal{O}_{(x,z),0}[y]$, since $\mathcal{O}_{(x,z),0}$ is a UFD and $x\in\mathcal{O}_{(x,z),0}$ is a prime which divides each coefficient of $f$, but $x^2$ doesn't divide $xz^2$, we have by the Eisenstein criterion that $f=y^2-xz^2$ is irreducible

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    $\begingroup$ Very interesting. Does it mean that proposition 1.1.35 of Huybrechts "Complex Geometry" is not correct? It states "Let $f \in \mathscr O_{\mathbb C^n,0}$ be irreducible. Then for sufficiently small $\varepsilon$ and $z \in B_\varepsilon(0)$ the induced element $f \in \mathscr O_{\mathbb C^n,z}$ is irreducible" $\endgroup$
    – Appliqué
    Oct 16, 2014 at 20:54
  • $\begingroup$ The proof of the proposition you are quoting shows that $f$ is irreducible in $\mathcal{O}_z$ for $z$ sufficiently small and $z\not\in Z(f,\frac{\partial f}{\partial z_1})$; this is a sufficient condition, the example I gave shows that it is not necessary because $0\in Z(y^2-xz^2,2y)$ $\endgroup$
    – Gabriele
    Oct 16, 2014 at 21:27
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    $\begingroup$ I found a similar doubt on mathoverflow mathoverflow.net/questions/75581/… $\endgroup$
    – Gabriele
    Oct 16, 2014 at 21:31

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