Order codimension one rational function

Let $X$ be a normal locally Noetherian scheme (normal noetherian ring) and let $Z\subseteq X$ be an integral closed subscheme with generic point $\xi$ and $\dim {\mathcal{O}}_{X,\xi }=1$ (Integral schemes, Generic points of irreducible closed subsets). We call such a $Z$ a prime divisor also in this locally Noetherian setting. This extends the same codimension-one definition in Weil divisor normal noetherian scheme; it requires no quasi-compactness of $X$.

The local ring ${\mathcal{O}}_{X,\xi }$ is a discrete valuation ring. It is a Noetherian local ring, because $X$ is locally Noetherian; it is a domain with fraction field equal to the function field of the irreducible component containing $Z$, because $\xi$ lies in a unique irreducible component of the normal scheme $X$; it is integrally closed, by normality; and it has dimension equal to one, by the definition of a prime divisor. A one-dimensional Noetherian local integrally closed domain is a discrete valuation ring by the characterisation of discrete valuation rings, and Height-one localizations of normal Noetherian domains are DVRs is exactly this statement in global form (Equivalent characterizations of a DVR, Discrete valuation rings).

Let $v_{\xi }$ denote the discrete valuation of the fraction field $K=\mathrm{Frac}({\mathcal{O}}_{X,\xi })$ whose valuation ring is ${\mathcal{O}}_{X,\xi }$, normalised so that $v_{\xi }(\pi )=1$ for a uniformiser $\pi$ of ${\mathcal{O}}_{X,\xi }$ (Discrete valuations, The field of fractions $\mathrm{Frac}(D)=(D\setminus \{0\}{)}^{-1}D$ of an integral domain). Now let $f\in \mathrm{\Gamma }(X,K_X^{\times })$ be a global meromorphic unit (Sheaf total quotient rings). The generic point $\xi$ lies in a unique irreducible component $X_i$ of $X$; the sheaf $K_X$ restricts on the integral scheme $X_i$ to the constant sheaf with value the function field $K(X_i)$, and ${\mathcal{O}}_{X,\xi }={\mathcal{O}}_{X_i,\xi }$ has fraction field $K(X_i)$. The restriction of $f$ to $X_i$ is therefore an element of $K(X_i{)}^{\times }=K^{\times }$, written $f_{\xi }$, and the order of vanishing of $f$ along $Z$ is the integer ${\mathrm{ord}}_Z(f):=v_{\xi }(f_{\xi })\in \mathbb{Z}\mathrm{.}$ Since $v_{\xi }$ is a group homomorphism $K^{\times }\to \mathbb{Z}$ and $f_{\xi }$ depends only on the restriction of $f$ to $X_i$, this is well defined: ${\mathrm{ord}}_Z(1)=0$, ${\mathrm{ord}}_Z(fg)={\mathrm{ord}}_Z(f)+{\mathrm{ord}}_Z(g)$ and ${\mathrm{ord}}_Z(f^{-1})=-{\mathrm{ord}}_Z(f)$.

If $X$ is integral (Integral schemes) then $K_X$ is the constant sheaf with value $K(X)$, so a meromorphic unit is simply an element of $K(X{)}^{\times }$, and ${\mathrm{ord}}_Z(f)=v_{\xi }(f)$ for the element $f\in K(X{)}^{\times }=\mathrm{Frac}({\mathcal{O}}_{X,\xi }{)}^{\times }$. In this case ${\mathrm{ord}}_Z(f)\ge 0$ if and only if $f$ lies in ${\mathcal{O}}_{X,\xi }$, and ${\mathrm{ord}}_Z(f)=0$ if and only if $f$ is a unit of ${\mathcal{O}}_{X,\xi }$, since $v_{\xi }$ is the normalised valuation of a discrete valuation ring.

For $X=\mathrm{\varnothing }$ there are no prime divisors, so the order domain is empty. The meromorphic-unit group is trivial, with its unique identity; this does not give an order without a prime divisor.