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  "body": "\n## Definition\n\nLet $R$ be a [[commutative ring|0310REJJY8]].\nAn $R$-module $P$ is *invertible* if $P$ is a [[finitely generated|05QY1VEBYD]] [[projective module|BFFA03GE3V]] of rank one.\nThis means that for every [[maximal ideal|0PV74TR5FP]] $\\mathfrak{m}$ of $R$, the [[localization|X6JD4AG973]] $P_{\\mathfrak{m}}$ is isomorphic to $R_{\\mathfrak{m}}$.\nThe *Picard group* $\\operatorname{Pic}(R)$ is the set of isomorphism classes $[P]$ of invertible $R$-modules, with the operation\n$$[P] \\cdot [Q] = [P \\otimes_R Q],$$\nwhere $\\otimes_R$ is the [[tensor product|F0HQXQ0HCD]].\nThe identity element is $[R]$, and the inverse of $[P]$ is $[\\Hom_R(P, R)]$.\n\n## Remarks\n\nThe operation is well defined, associative and commutative because the tensor product is so up to natural isomorphism.\nFor invertible $P$, the evaluation map $P \\otimes_R \\Hom_R(P, R) \\to R$ is an isomorphism, since it is one after localization at every maximal ideal.\nAn $R$-module $P$ is invertible exactly when there is a module $Q$ with $P \\otimes_R Q \\cong R$.\nIf $R$ is an integral domain, $\\operatorname{Pic}(R)$ is isomorphic to the group of [[invertible fractional ideals|E190TAV2S6]] modulo the principal fractional ideals.\n\nExample: for $R = \\Z[\\sqrt{-5}]$, the ideal $P = (2, 1 + \\sqrt{-5})$ is an invertible module that is not free, while $P \\otimes_R P \\cong P^2 = (2) \\cong R$; in fact $\\operatorname{Pic}(R)$ has order $2$.\nIf $R$ is a [[principal ideal domain|0CMD0BCH8T]] or a [[local ring|MZ02BKPHJA]], then $\\operatorname{Pic}(R)$ is trivial.\n\nNon-example: the module $R^2$ is finitely generated and projective, but $(R^2)_{\\mathfrak{m}} \\cong R_{\\mathfrak{m}}^2$ is not isomorphic to $R_{\\mathfrak{m}}$ for $R \\ne 0$; so $R^2$ is not invertible and has no class in $\\operatorname{Pic}(R)$.\n",
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