# Zeros and poles of rational functions on locally Noetherian schemes

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.

-

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.)

-
Thank you Adeel for your precise response. I did a little search before asking the question, including The Stacks Project and Fulton's Intersection Theory, but have not been able to fully convince myself that it wasn't possible without integrality conditions. I'll wait some time before accepting your answer just in case someone has something more to add. –  Nuno Apr 5 '13 at 14:42
I would imagine that the definition in the Stacks Project is the most general one... if there was a useful generalization to non-integral schemes I'm sure it would be in one of the standard references. Do you have a specific motivation, by the way? –  Adeel Apr 5 '13 at 20:53
No, not at all. I was reading the chapter on Weil divisors in Vakil's notes and at the very beginning he doesn't assume that the scheme is integral (or has integral stalks), so I was trying to adapt past definitions and exercises to this setting. But then we'll also require the scheme to be normal and everything goes smoothly. –  Nuno Apr 6 '13 at 14:08
The order of vanishing of an invertible rational function at a point $x\in X$ of codimension $1$ can be defined for any locally noetherian ring. Locally an invertible rational function is a quotient $a/b$ with $a, b$ regular elements of $O_{X,x}$. Then the order is given by the difference of the lenghts of $O_{X,x}/(a)$ and $O_{X,x}/(b)$. –  Cantlog Sep 30 '13 at 7:07
@Cantlog, one can identify $R(X)$ with $\mathrm{Frac}(\mathscr{O}_{X,x})$ only when $X$ is integral, see EGA I, section 7. –  Adeel Sep 30 '13 at 9:13
show 1 more comment

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.

-
That was what I first thought, but how can you take the germ of a rational function which domain of definition doesn't contain the point p? –  Nuno Apr 4 '13 at 16:56
Since we are talking about a pole we don't expect the rational function to be defined on it. The integral condition on stalks would make life easier, since then we would have an integral affine scheme Spec($A$) containing $p$ and it would be possible to look at the rational function as an element of K($A$) = K($\mathcal{O}_{X,p}$) and then use the discrete valuation. The question about zeros is easier and that's why I only briefly mentioned it at the end. –  Nuno Apr 4 '13 at 17:22
When $p \in U$, $f \in \mathcal{O}_X(U)$, then $f_p \in \mathcal{O}_{X,p}$. –  Martin Brandenburg Apr 4 '13 at 18:55
In this case we would have $\nu({f_p}) \ge 0$ so no pole. –  Nuno Apr 5 '13 at 14:45
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 $p$.