If I have an analytic function in plane $F(x,y)$ that is zero on a curve $y=f(x)$, is it true that $F=(yf(x))^n h$, where $h$ is nonzero on the curve? More general, can be somethink said about factorisation of analytic functions? How much is it determined by its zero set? Thx

$\begingroup$ You are using $f$ for two different things... $\endgroup$ – Mariano SuárezÁlvarez Oct 14 '10 at 14:58

$\begingroup$ Thank you very much. Does somethink like that hold also over the reals? Peter $\endgroup$ – Peter Franek Oct 14 '10 at 20:38

$\begingroup$ Yes it does.... $\endgroup$ – Thierry Zell Oct 14 '10 at 20:45
You need a combination of Weierstrass preparation and Puiseux series expansion to factor the analytic function, but it is indeed possible. Keep in mind that this is a local factorization near a point of your choice, that the factors may be complex valued and singular (=Holder continuous) at the point, but they are analytic outside the point. Better than writing here a lengthy explanation let me point you at a paper where I wrote all the details since I could not find them in the literature, although this stuff must be well known. See Section 2 of this paper.