7
$\begingroup$

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 $\mathscr O_p$.

$\endgroup$
5
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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 '14 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 '14 at 21:27
  • 1
    $\begingroup$ I found a similar doubt on mathoverflow mathoverflow.net/questions/75581/… $\endgroup$ – Gabriele Oct 16 '14 at 21:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.