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  "body": "\n## Definition\n\nLet $R$ be a [[Dedekind domain|7NPY2CN5KB]] with field of fractions $K$.\nLet $I(R)$ be the set of nonzero [[fractional ideals|YB4JTB4WJN]] of $R$; under the product of fractional ideals it is an [[abelian group|082GHNPJ53]] with identity $R$.\nLet $P(R) = \\set{xR : x \\in K^{\\times}}$ be its subgroup of principal fractional ideals.\nThe *ideal class group* of $R$ is the [[quotient group|0KYX2N69J0]]\n$$\\operatorname{Cl}(R) = I(R) / P(R).$$\nFor a [[number field|MWBDFS2KJ5]] $K$ with ring of integers $\\mathcal{O}_K$, the class group of $K$ is $\\operatorname{Cl}(K) = \\operatorname{Cl}(\\mathcal{O}_K)$.\n\n## Remarks\n\nExample: $R = \\Z$ is a principal ideal domain, so every fractional ideal is $x\\Z$ with $x \\in \\Q^{\\times}$, $I(\\Z) = P(\\Z)$, and $\\operatorname{Cl}(\\Z)$ is trivial.\nExample: for $R = \\Z[\\sqrt{-5}]$, let $\\mathfrak{p} = (2, 1 + \\sqrt{-5})$.\nThen $\\mathfrak{p}^2$ is generated by $4$, $2 + 2\\sqrt{-5}$ and $(1 + \\sqrt{-5})^2 = -4 + 2\\sqrt{-5}$; all are multiples of $2$ and the ideal contains $4$ and $(2 + 2\\sqrt{-5}) - (-4 + 2\\sqrt{-5}) = 6$, hence $2$, so $\\mathfrak{p}^2 = (2)$.\nThe ideal $\\mathfrak{p}$ is not principal, since $a^2 + 5b^2 = 2$ has no integer solution; so its class has order $2$, and in fact $\\operatorname{Cl}(\\Z[\\sqrt{-5}]) \\cong \\Z/2\\Z$.\nNon-example: the nonzero integral ideals of $R$ alone do not form a group under multiplication.\nA proper ideal $\\mathfrak{a}$ has no integral ideal $\\mathfrak{b}$ with $\\mathfrak{a}\\mathfrak{b} = R$, since $\\mathfrak{a}\\mathfrak{b} \\subseteq \\mathfrak{a}$.\nThis is why fractional ideals are used.\nEquivalently, $\\operatorname{Cl}(R)$ is the set of nonzero ideals of $R$ modulo the relation $\\mathfrak{a} \\sim \\mathfrak{b}$ if $x\\mathfrak{a} = y\\mathfrak{b}$ for some nonzero $x, y \\in R$.\nThe class group is trivial if and only if $R$ is a principal ideal domain.\n",
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