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  "body": "\n## Statement\n\nLet $K$ be an [[algebraic number field|MWBDFS2KJ5]] with ring of integers $\\mathcal{O}_K$.\nLet $r_1$ be the number of field embeddings $K \\to \\R$ and $r_2$ the number of pairs of complex conjugate non-real embeddings $K \\to \\C$.\nLet $\\mu(K)$ be the set of [[roots of unity|0YV2CKP2AV]] in $K$.\nThen $\\mu(K)$ is a finite cyclic group, and the [[group of units|07C6ECNJET]] $\\mathcal{O}_K^\\times$ is finitely generated with\n$$\\mathcal{O}_K^\\times \\cong \\mu(K) \\times \\Z^{r_1 + r_2 - 1}.$$\n\n## Remarks\n\nThe proof uses the logarithmic embedding $u \\mapsto \\big(N_j \\log \\abs{\\sigma_j(u)}\\big)_j$ into $\\R^{r_1 + r_2}$, where $\\sigma_j$ runs over the real embeddings and one embedding from each complex pair, and $N_j$ is $1$ or $2$.\nIts kernel on $\\mathcal{O}_K^\\times$ is the finite group $\\mu(K)$, and its image lies in the hyperplane where the coordinates sum to $0$.\nThe hard part is that the image is a full [[lattice|WN9V7WSJKF]] in that hyperplane; this uses [[Minkowski's theorem|0K387S2PKQ]] to produce enough units.\n\nA set of $r = r_1 + r_2 - 1$ units that generate $\\mathcal{O}_K^\\times$ modulo $\\mu(K)$ is a system of *fundamental units*.\n\nExample: for $K = \\Q(\\sqrt{2})$, $r_1 = 2$ and $r_2 = 0$, so the rank is $1$.\nThe units of $\\Z[\\sqrt{2}]$ are exactly $\\pm (1 + \\sqrt{2})^k$ with $k \\in \\Z$.\nFor an imaginary quadratic field, $r_1 = 0$ and $r_2 = 1$, so the rank is $0$ and every unit is a root of unity; for $\\Q(i)$ the units are $\\pm 1, \\pm i$.\n\n## Explanations\n\n### L1 human\n\nThe units of a number field are the algebraic integers whose inverse is also an algebraic integer.\nThe theorem says that all of them are built from a finite list of basic units by multiplication, up to roots of unity.\nThe number of basic units is given by counting the real and complex embeddings of the field.\n\n### L2 human\n\nThe theorem is one of the two finiteness results at the start of algebraic number theory, with the finiteness of the class number.\nIn terms of [[free abelian groups|0WJX8MGG37]], $\\mathcal{O}_K^\\times / \\mu(K)$ is free of rank $r_1 + r_2 - 1$.\nThe covolume of the unit lattice is the [[regulator|RWK0HNF1NA]], which enters the [[class number formula|55XX5A7XZV]].\nFor real quadratic fields, the fundamental unit gives the solutions of Pell's equation.\n",
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