---
tag: Y6D894R6HB
kind: definition
name: Divisor class group
msc: [13C20, 13F05]
defines: WN8PK6RQZR
uses: [CB9WJCN9BM, 87WH65FJB8, MGM3CNJW9V, 0YMSEZE1VH, 0YYWCNTPWR, 0VF11ZTTMD]
sources:
  - url: https://en.wikipedia.org/wiki/Fractional_ideal
---

## Definition

Let $A$ be a [[Krull domain|CB9WJCN9BM]] with [[field of fractions|0VF11ZTTMD]] $K$, and let $X$ be the set of prime ideals of $A$ of [[height|87WH65FJB8]] one.
For $P \in X$, the localization $A_P$ is a [[discrete valuation ring|MGM3CNJW9V]]; let $v_P \colon K^\times \to \Z$ be its valuation.

1. The *divisor group* $\operatorname{Div}(A)$ is the [[free abelian group|0YMSEZE1VH]] on the set $X$.
2. For $x \in K^\times$, the *principal divisor* of $x$ is $\operatorname{div}(x) = \sum_{P \in X} v_P(x) P$; the sum is finite because $A$ is a Krull domain.
3. The map $x \mapsto \operatorname{div}(x)$ is a group homomorphism $K^\times \to \operatorname{Div}(A)$; its image is the subgroup $\operatorname{Prin}(A)$ of principal divisors.
4. The *divisor class group* of $A$ is the [[quotient group|0YYWCNTPWR]] $\operatorname{Cl}(A) = \operatorname{Div}(A)/\operatorname{Prin}(A)$.

## Remarks

$\operatorname{Cl}(A) = 0$ if and only if $A$ is a [[unique factorisation domain|06YMXJMXYS]].
Example: for $A = \Z$, the height-one primes are $p\Z$, and the divisor $\sum n_p \, p\Z$ is $\operatorname{div}\left(\prod p^{n_p}\right)$, so every divisor is principal and $\operatorname{Cl}(\Z) = 0$.
Non-example of a principal divisor: in $A = \Z[\sqrt{-5}]$, the divisor $P$ with $P = (2, 1 + \sqrt{-5})$ is not principal, since no element of $A$ has norm $2$; but $2P = \operatorname{div}(2)$, because $P^2 = (2)$.
For a Dedekind domain, $\operatorname{Cl}(A)$ is the ideal class group, the group of nonzero fractional ideals modulo principal ones.
The notation $\operatorname{Cl}(A)$ is standard.