Zeros and poles of rational functions on locally Noetherian schemes

Question

Let $X$ be a locally Noetherian scheme and let $f$ be a rational function on $X$ (i.e. the equivalence class of a pair $(U,f)$, where $f \in \mathcal{O}_X(U)$ and $U$ contains the associated points of $X$, under obvious equivalence relation).

While reading Vakil's notes I wondered how could we define poles of such a rational function. After some thought I came up with the following definition: I'd say that a regular codimension one point $p$ is a pole if it's not in the domain of definition of $f$. If $X$ is also an integral scheme (or at least if all the stalks of $\mathcal{O}_X$ are integral domains, in which case we can cover $X$ with integral schemes), then this definition would coincide with the usual one, namely using the discrete valuation at $p$.

But there is something unnatural about my definition, since I was not able to relate the rational function with the discrete valuation on $\mathcal{O}_{X,p}$ and consequently was not able to determine the order of the pole. So I'd like to know if it's possible to define a meaningful notion of poles for rational functions on locally Noetherian schemes and how would it relate to my definition. By extension, consider the same question about zeros.

Answer 1

The order of vanishing of a rational function is only defined on locally noetherian, integral schemes, at points of codimension 1. The locally noetherian and codimension 1 conditions are to ensure that the order is finite, and that the maps $\mathrm{ord}_x : R(X) \to \mathbf{Z}$ are homomorphisms; see (Stacks, Comm. alg., Orders of vanishing). And the integrality assumption is so that rational functions correspond to elements of the fraction fields of stalks; otherwise as you see yourself the definition doesn't work at all.

In practice this doesn't really present a problem. For example when considering Weil divisors (1-codimensional cycles) associated to rational functions, if one has a Weil divisor associated to a rational function on a closed integral subscheme Z of X, one can consider the direct image of this cycle by the inclusion, to get a cycle on X. (This is how one defines rational equivalence of cycles.)

Answer 2

I don't understand the problem. A $1$-dimensional regular noetherian local ring is a discrete valuation domain (yes, it's automatically an integral domain), so you can take the discrete valuation of the germ $f_p \in \mathcal{O}_{X,p}$ as the definition of the order of $f$ at $p$. Actually this is the usual definition.

Answer 3

Here's how I think this works. Take a codimension $1$ point $p$. The irreducible components containing $p$ correspond to minimal primes of $\mathcal{O}_{X,p}$ and there's only one of these. Let's say the generic point for it is $\eta$. Then there is an inclusion $\mathcal{O}_{X,p} \to \mathcal{O}_{X,\eta}$ which is just taking the fraction field. Your $U$ has to contain $\eta$, so you can take the stalk of $f$ there and then find its valuation at $p$.

It's confusing. It's probably good to note that around any point you can find a Noetherian open subscheme, and then this is a finite disjoint union of Noetherian normal irreducible schemes, which is what you really like.