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  "body": "\n## Definition\n\nLet $A$ 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.\nFor $P \\in X$, the localization $A_P$ is a [[discrete valuation ring|MGM3CNJW9V]]; let $v_P \\colon K^\\times \\to \\Z$ be its valuation.\n\n1. The *divisor group* $\\operatorname{Div}(A)$ is the [[free abelian group|0YMSEZE1VH]] on the set $X$.\n2. 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.\n3. 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.\n4. The *divisor class group* of $A$ is the [[quotient group|0YYWCNTPWR]] $\\operatorname{Cl}(A) = \\operatorname{Div}(A)/\\operatorname{Prin}(A)$.\n\n## Remarks\n\n$\\operatorname{Cl}(A) = 0$ if and only if $A$ is a [[unique factorisation domain|06YMXJMXYS]].\nExample: 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 $\\operatorname{Cl}(\\Z) = 0$.\nNon-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 $P^2 = (2)$.\nFor a Dedekind domain, $\\operatorname{Cl}(A)$ is the ideal class group, the group of nonzero fractional ideals modulo principal ones.\nThe notation $\\operatorname{Cl}(A)$ is standard.\n",
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