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  "body": "\n## Definition\n\nLet $K$ be a [[field|0QA8CHH8MG]] and let $\\Gamma$ be an [[abelian group|071DDFVG4J]] with a total order making it a [[linearly ordered group|1P8087YGYQ]].\nA *Hahn series* over $K$ with exponents in $\\Gamma$ is a map $f \\colon \\Gamma \\to K$, written $f = \\sum_{\\gamma \\in \\Gamma} f(\\gamma) t^\\gamma$, whose support $\\set{\\gamma \\in \\Gamma : f(\\gamma) \\ne 0}$ is [[well-ordered|0F1D6PK9RZ]].\nThe set of Hahn series is written $K((t^\\Gamma))$.\nAddition is pointwise, and multiplication is\n$$(fg)(\\gamma) = \\sum_{\\alpha + \\beta = \\gamma} f(\\alpha) g(\\beta).$$\n\n## Remarks\n\nFor each $\\gamma$ the sum defining $(fg)(\\gamma)$ has finitely many nonzero terms, and the support of $fg$ is again well-ordered.\nWith these operations $K((t^\\Gamma))$ is a field, as Hahn showed.\nExample: for $\\Gamma = \\Z$, the well-ordered subsets are the subsets bounded below, so $K((t^\\Z))$ is the [[field of formal Laurent series|2S4K8FA8QY]] $K((t))$.\nExample: for $\\Gamma = \\Q$, the series $\\sum_{n \\ge 1} t^{1 - 1/n}$ is a Hahn series, since its support $\\set{0, 1/2, 2/3, \\dots}$ is well-ordered.\nNon-example: for $\\Gamma = \\Z$, the formal sum $\\sum_{n \\ge 1} t^{-n}$ is not a Hahn series, since its support $\\set{-1, -2, \\dots}$ has no least element.\nThe map $v(f) = \\min \\set{\\gamma : f(\\gamma) \\ne 0}$ for $f \\ne 0$ is a [[valuation|SVY882HKSJ]] on $K((t^\\Gamma))$ with value group $\\Gamma$.\n",
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