Reduced expression for rational functions with a pole

Question

For a field $k$, let $V$ be an affine variety over $k$. Denote by $k(V)$ the function field of $V$, containing all rational functions $r:V\dashrightarrow \mathbb{A}_k^1$. My question is, if a rational function $f\in k(V)$ has a pole at $p\in V$, is there an expression $f=\frac{g}{h}$ where $g,h\in k[V]$ are regular functions, and $g(p)\neq 0$, $h(p)=0$?

When $V\subseteq \mathbb A_k^1$, this is clear, since if we have $f=\frac{g}{h}$ where $g(p)=h(p)=0$, we can simply reduce the expression of $g$ and $h$ and eliminate the factor $(x-p)$ until we get $f=\frac{g'}{h'}$ such that $g'(p)\neq 0$, $h'(p)=0$. But when $g,h$ are multivariate functions, I wonder how to get such a reduced expression?

Thank you very much for any help!

Answer 1

This is a community wiki answer consisting of the answer from the comments in order to mark this question as answered.

Consider the case when $V=\Bbb A^2$ with co-ordinate functions $x,y$ and $p$ defined by $x=y=0$. Then look at the rational function $r=x/y$, which you can not write as $g/h$ with $g(p)\neq0$. – Mohan