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  "body": "\n## Definition\n\nLet $F$ be a [[finite field|0ZXZ82FTX2]] with $q$ elements.\nLet $\\chi, \\lambda \\colon F^\\times \\to \\C^\\times$ be multiplicative characters, that is, [[group homomorphisms|0JYTG5K9Y4]], each extended to $F$ by the value $0$ at $0$.\nThe *Jacobi sum* of $\\chi$ and $\\lambda$ is\n$$J(\\chi, \\lambda) = \\sum_{a \\in F} \\chi(a) \\lambda(1 - a).$$\n\n## Remarks\n\nSome books set the trivial character equal to $1$ at $0$; this changes $J$ only when $\\chi$ or $\\lambda$ is trivial.\nIf $\\chi$, $\\lambda$ and $\\chi\\lambda$ are nontrivial, then $J(\\chi, \\lambda) = g(\\chi)g(\\lambda)/g(\\chi\\lambda)$, where $g$ is the [[Gauss sum|EQ6A3ZN48K]] with a fixed nontrivial additive character, and so $\\abs{J(\\chi, \\lambda)} = \\sqrt{q}$.\n\nExample: let $q = 5$ and $\\chi = \\lambda$ the [[Legendre symbol|0JTC045NNZ]] modulo $5$.\nThe terms with $a = 0$ and $a = 1$ vanish.\nFor $a = 2, 3, 4$ the products $\\chi(a)\\chi(1-a)$ are $(-1)(1) = -1$, $(-1)(-1) = 1$ and $(1)(-1) = -1$.\nSo $J(\\chi, \\chi) = -1$; here $\\chi^2$ is trivial and the general rule gives $J(\\chi, \\chi^{-1}) = -\\chi(-1) = -1$.\n\nNon-example: the sum $\\sum_{a} \\chi(a) \\lambda(a)$ is not a Jacobi sum, since the second argument is $a$ and not $1 - a$.\nBy orthogonality it equals $q - 1$ if $\\chi\\lambda$ is trivial and $0$ otherwise.\n",
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