Well-definedness of zeros and poles of rational functions in algebraic geometry

My question is about the well-definedness of zeros and poles of rational functions following Vakil's Rising Sea (November 18, 2017 Edition).

On a regular codimension $$1$$ point $$p$$, we define the number of zeros resp. poles of an element $$f \in \operatorname{Quot}(\mathcal{O}_{X,p})$$ via the discrete valuation given by the DVR $$\mathcal{O}_{X,p}$$ (p. 353). In algebraic geometry, one then usually considers some rational function $$s$$ and tries to analyze the number of zeros and poles of $$s$$. I'm confused in how this makes sense, i.e. why is $$s$$ naturally an element in $$\mathcal{O}_{X,p}$$?

Let me also recall quickly the definition of rational functions in Vakil's book. On p. 171 we define a rational function on a scheme $$X$$ as an equivalence class of pairs $$(U, s \in \Gamma(U, \mathcal{O}_X))$$ where $$\operatorname{Ass}(X) \subseteq U$$. Two pairs $$(U,s)$$ and $$(U',s')$$ are said to be equivalent if $$s|_{U \cap U'} = s'|_{U \cap U'}$$. I'm guessing that my confusion stems from an insufficient understanding of associated points.

So to summarize: Given a rational function $$s \in \Gamma(U,\mathcal{O}_X)$$ with $$\operatorname{Ass}(X) \subseteq U$$, why is $$s$$ naturally an element of $$\operatorname{Quot}(\mathcal{O}_{X,p})$$ for all regular codimension $$1$$ points $$p$$?

If $$p\in X$$ is a regular codimension one point, then it is in the closure of exactly one associated point $$\eta$$ of $$X$$ (ref), and in fact that point is the generic point of the unique irreducible component it is on. As the reduced locus is open in a noetherian scheme (ref), we can pick an integral affine open neighborhood $$\operatorname{Spec} A\subset X$$ of $$p$$. Then $$\mathcal{O}_{X,p}$$ is a localization of $$A$$, and $$\mathcal{O}_{X,\eta}$$ is the quotient field of $$A$$, so $$\operatorname{Frac}(\mathcal{O}_{X,p})=\mathcal{O}_{X,\eta}$$.

Now suppose $$U\subset X$$ contains all the associated points of $$X$$. Let $$p\in X$$ be a regular codimension one point and $$\eta$$ the unique associated point of $$X$$ which is a generalization of $$p$$. Then $$s\in\mathcal{O}_X(U)$$ maps to $$\mathcal{O}_{X,\eta}$$ by restriction, and as $$\mathcal{O}_{X,\eta}=\operatorname{Frac}(\mathcal{O}_{X,p})$$ we've explained what you were interested in.

Let me note a couple of small imprecisions in your question, just in case these were contributing to some of your troubles.

... I'm confused in how this makes sense, i.e. why is $$s$$ naturally an element in $$\mathcal{O}_{X,p}$$?

You stated this incorrectly here, but correctly after the line break: $$s$$ is naturally an element of $$\operatorname{Frac}(\mathcal{O}_{X,p})$$, not $$\mathcal{O}_{X,p}$$.

... Given a rational function $$s \in \Gamma(U,\mathcal{O}_X)$$ ...

It would be more precise to say "a rational function represented by $$s\in\Gamma(U,\mathcal{O}_X)$$".

• Excellent answer. Mar 23 '21 at 13:29