Principal weil divisor and class group

Assume the Axiom of Dependent Choice (The axiom of dependent choice: a relation in which every element is related to something admits an $\mathbb{N}$-indexed chain). Let $X$ be a normal Noetherian integral scheme (Weil divisor normal noetherian scheme, Integral schemes). Because $X$ is integral, ${\mathcal{K}}_X$ is the constant sheaf with value the function field $K(X)$ (Sheaf total quotient rings), so the global meromorphic units of $X$ are exactly the nonzero elements of $K(X)$: $\mathrm{\Gamma }(X,{\mathcal{K}}_X^{\times })=K(X{)}^{\times }\mathrm{.}$

For $f\in K(X{)}^{\times }$ define the principal Weil divisor ${\mathrm{div}}_W(f):={\sum }_Z{\mathrm{ord}}_Z(f)[Z]\in \mathrm{Div}(X),$ where $Z$ runs over the prime divisors of $X$ and ${\mathrm{ord}}_Z(f)\in \mathbb{Z}$ is the order of $f$ along $Z$ (Order codimension one rational function). This is a legitimate Weil divisor: its coefficients are integers, and its support is locally finite by A meromorphic unit has locally finite nonzero order support, a lemma whose only choice input is Dependent Choice through the finiteness of the minimal primes of a Noetherian ring; since $X$ is quasi-compact, the support is in fact finite. Here $\mathrm{Div}(X)$ is the group of Weil divisors of Weil divisor normal noetherian scheme.

The map ${\mathrm{div}}_W:K(X{)}^{\times }\to \mathrm{Div}(X)$ is a group homomorphism: for $f,g\in K(X{)}^{\times }$ and every prime divisor $Z$ the order is additive, ${\mathrm{ord}}_Z(fg)={\mathrm{ord}}_Z(f)+{\mathrm{ord}}_Z(g)$ and ${\mathrm{ord}}_Z(1)=0$ (Order codimension one rational function), so the coefficients of ${\mathrm{div}}_W(fg)$ and ${\mathrm{div}}_W(f)+{\mathrm{div}}_W(g)$ agree at every $Z$; hence ${\mathrm{div}}_W(fg)={\mathrm{div}}_W(f)+{\mathrm{div}}_W(g),$ and ${\mathrm{div}}_W(f^{-1})=-{\mathrm{div}}_W(f)$. In particular ${\mathrm{div}}_W(1)=0$, and a global regular unit $u\in \mathrm{\Gamma }(X,{\mathcal{O}}_X^{\times })$ has ${\mathrm{ord}}_Z(u)=0$ at every prime divisor, so ${\mathrm{div}}_W(u)=0$.

The image $P(X):=\{{\mathrm{div}}_W(f):f\in K(X{)}^{\times }\}\subseteq \mathrm{Div}(X)$ is a subgroup, because ${\mathrm{div}}_W$ is a group homomorphism and the image of a homomorphism is a subgroup (Monoid homomorphism and group homomorphism, Subgroup). The (Weil) divisor class group of $X$ is the quotient group $\mathrm{Cl}(X):=\mathrm{Div}(X)/P(X)=\mathrm{Div}(X)/\{{\mathrm{div}}_W(f):f\in K(X{)}^{\times }\}$ (The quotient group $G\mathrm{/}N$ and coset product $(gN)(hN)=ghN$). Thus $\mathrm{Cl}(X)$ is an abelian group, and two Weil divisors $D,D^{\mathrm{\prime }}$ have the same class in $\mathrm{Cl}(X)$ exactly when $D-D^{\mathrm{\prime }}={\mathrm{div}}_W(f)$ for some $f\in K(X{)}^{\times }$; one then says that $D$ and $D^{\mathrm{\prime }}$ are linearly equivalent as Weil divisors. This relation is an equivalence relation: it is reflexive via $f=1$, symmetric via ${\mathrm{div}}_W(f^{-1})=-{\mathrm{div}}_W(f)$, and transitive via the homomorphism property.

Three boundary cases are worth recording. First, an integral scheme is nonempty by definition, so the empty scheme is not an instance of this definition and no empty divisor group is being described. Second, if $X$ has no prime divisors (for instance $X=\mathrm{Spec}k$ for a field $k$), then $\mathrm{Div}(X)=0$, and $\mathrm{Cl}(X)=0$ as well: every nonzero function field element is a unit and the zero divisor is principal. Third, the constant function $f=1$ realises the zero class, so $\mathrm{Cl}(X)$ is the quotient by the subgroup generated by the divisors of the form ${\mathrm{div}}_W(f)$; no effectiveness hypothesis is imposed on the elements of $\mathrm{Div}(X)$ or on the divisors defining a class.