Let X be a locally Noetherian scheme and let f be a rational function on X (i.e. t

herbariak1

herbariak1

Answered question

2022-05-28

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 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 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 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 & Explanation

Bruce Bridges

Bruce Bridges

Beginner2022-05-29Added 13 answers

Here's how I think this works. Take a codimension 1 point p. The irreducible components containing p correspond to minimal primes of O X , p and there's only one of these. Let's say the generic point for it is η. Then there is an inclusion O X , p O X , η which is just taking the fraction field. Your U has to contain η, 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.

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