Let $\Omega$ be an open subset of $\mathbb{C}^n$ and let $f$ be analytic in $\Omega$. Assume $P\in\mathbb{C}[z\_1,\ldots,z\_n]$ is a polynomial whose irreducible factors are all of multiplicity one.


If $f$ vanishes on the zero set $Z(P)$ of $P$, how to justify that $f/P$ is analytic in $\Omega$ ?


(asked on [MSE](https://math.stackexchange.com/questions/3164508), but didn't get a convincing answer)



Factorization of an analytic function in $\mathbb{C}^n$
*This is a copy of [my answer](https://math.stackexchange.com/a/4654372/312406) from math.SE*


I think you could argue stalk wise. Take a point $x \in \Omega$. By general theory, the ring $\mathcal O\_{\mathbb C^n, x}$ of convergent power series at $x$ is a UFD. Since $P$ is square-free, and $f$ vanishes at $\{P =0\}$, $P$ is a factor of $f$ in $\mathcal O\_{\mathbb C^n,x}$ by the analytic Nullstellensatz. So there is a quotient $\overline{g\_x} = f/P$ in $\mathcal O\_{\mathbb C^n,x}$. This is represented by an actual holomorphic function $g\_x \in \mathcal O(U\_x)$ in some neighbourhood $U\_x$ of $x$, and $g\_x \cdot P = f$ holds on all of $U\_x$. By the identity theorem, the $g\_x$ glue to a holomorphic function $g$ on $\Omega$.


