Ring of rational functions ideal generators There is an affine variety $X\subset \mathbb{A}^n$ with its ring of rational functions which is  the quotient ring of $\mathbb{k}[X]$ (each $f\in \mathbb{k}(X)$ has a form $\frac{p}{q}$ where $q$ does not divide zero). For each $f\in \mathbb{k}(X)$ we can build an ideal $I_f=\{\phi\in \mathbb{k}[X]: f\phi\in \mathbb{k}[X]\}$. Is it true that $I_f$ is generated by all possible denominators of $f$?
So, we build the set of pairs $(p_i,q_i)$ such that $f$ is represented as $\frac{p_i}{q_i}$. It is obvious that $q_if\in \mathbb{k}[X]$ and also each sum $\sum \psi_i q_i$ is in $I_f$. I'd like to understand how to represent each $\phi \in I_f$ like this. Could you give me a little hint?

I am not good with algebraic geometry but here is my try)
If you prove the result for irreducible variety everything will follow (because $k[X]$ is direct sum of algebras of irreducible components and everything else also decomposes into direct sum)
Now, for irreducible $X$ if $\phi \in I_f$ then $f$ has the form $\frac{\phi f}{\phi}$ because there is no zero divisors. The result follows
P.S. It is generated by denominators, not equal to the set of all denominators, because $I_f$ is a direct sum of corresponding ideals on the component, not the union.
