# The Zariski topology of a variety is determined by its principal Cartier divisors

Let $$X$$ be a smooth separated integral variety over an algebraically closed field.

Question:

Is it true that a basis for the Zariski topology is given by the family $$\mathcal{U}=\{X\setminus \mathrm{Supp}(\mathrm{div}(f))\mid f\in K(X)\}$$ where $$\mathrm{div}(f)$$ is the principal divisor defined by $$f$$?

Some ideas that I have had

• If $$X=\mathrm{Spec}(A)$$ is an affine integral variety this is true because for $$f\in A$$ we have that $$X\setminus\mathrm{div}(f)$$ is equal to the principal open set $$D(f)$$ and these generate the topology.

• The question is equivalent to the following:

For each prime divisor $$D$$ on $$X$$ and a point $$x$$ outside $$D$$ there is a rational function $$f$$ vanishing on $$D$$ and invertible on $$x$$.

Indeed, if $$\mathcal{U}$$ is a basis then we can find a function $$f$$ such that $$x\notin \mathrm{div}(f)$$ (i.e $$f$$ is invertible at $$x$$) and $$D$$ is a component of $$\mathrm{div}(f)$$. Then either $$f$$ or $$f^{-1}$$ works.

Conversely, consider an open set $$U$$ and $$x\in U$$. The complement $$F=X\setminus U$$ is a closed set with some irreducible components $$F_i$$. Take a function $$h_i\in \mathcal{O}_{X,x}\setminus \mathcal{O}_{X,F_i}$$. Then divisor of poles $$\mathrm{div}_{\infty}(h_i)$$ contains $$F_i$$ but not $$x$$. Let $$D=\bigcup \mathrm{div}_{\infty}(h_i)$$. By the hypothesis, for each component $$D_j$$ of $$D$$ there is a function $$f_j$$ vanishing on $$D_j$$ but invertible on $$x$$. Then, $$f=\prod_j f_j^{n_j}$$ vanish on $$D$$ and is invertible on $$x$$ for some appropriate choice of $$n_j\in \mathbb{Z}$$.

• If $$X$$ is a quasiprojective variety this is also true. We can see this using the idea in the previous point. Given a prime divisor $$D$$ and a point $$x$$ not in $$D$$ take an ample divisor $$H$$ different from $$D$$ and not containing $$x$$. Then for some $$n$$ big $$nH-D$$ is very ample (because the ample cone is open) then it is base-point free and so it has a section $$f$$ not vanishing at $$x$$ but vanishing at $$D$$.

I think this question is related to the concept of divisorial variety (cf. the remark after definition 3.1 in Divisorial Varieties, Borelli, 1963). In some sense, a variety satisfying this is like a principally divisorial variety.

If this is not true for smoothness, it would be still nice if we have the result for something less restrictive than quasi-projective. Something that contains smooth proper toric varieties for example.

Let $$U$$ be a an affine open set of $$X$$ containing $$x$$. If $$U$$ contains $$D$$, we choose $$f$$ as in your first bullet point vanishing on $$D$$ but not on $$x$$. If $$U$$ does not contain $$D$$, there must exist some $$f$$ on $$U$$ with a pole at $$D$$. If $$f$$ vanishes at $$x$$, add $$1$$ to $$f$$.
Why must there exist some $$f$$ on $$U$$ with a pole at $$D$$? Otherwise, we get a map $$\operatorname{Spec} R \to U$$ where $$R$$ is the local ring of $$X$$ at $$D$$, which consists of all rational functions on $$X$$ without a pole at $$D$$. This means we have two different maps $$\operatorname{Spec} R \to X$$, both compatible with $$\operatorname{Spec} K \to X$$, contradicting separatedness.