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  "body": "\n## Definition\n\nLet $L/K$ be a [[finite extension|0NRYPM9Y27]] and let $\\mathrm{Tr}_{L/K}$ be the [[field trace|73XV4RP5PT]].\nThe *trace form* of $L/K$ is the map\n$$B \\colon L \\times L \\to K, \\qquad B(x, y) = \\mathrm{Tr}_{L/K}(xy).$$\n\n## Remarks\n\nSince $\\mathrm{Tr}_{L/K}$ is $K$-linear and multiplication is commutative and distributive, $B$ is a [[symmetric bilinear form|0744CN4MDA]] on the $K$-vector space $L$.\nIt is [[nondegenerate|0KAN330PAH]] if and only if $L/K$ is separable.\nIts quadratic form is $x \\mapsto \\mathrm{Tr}_{L/K}(x^2)$.\n\nExample: $L = \\Q(i)$ over $K = \\Q$.\nCheck: $(a + bi)(c + di) = (ac - bd) + (ad + bc)i$, so $B(a + bi, c + di) = 2(ac - bd)$; in the basis $1, i$ the matrix of $B$ is $\\begin{pmatrix} 2 \u0026 0 \\\\ 0 \u0026 -2 \\end{pmatrix}$, which is invertible.\n\nNon-example of nondegeneracy: let $K = \\F_p(t)$ and $L = K(s)$ with $s^p = t$.\nThe trace of every element of $L$ is $0$, so the trace form of $L/K$ is identically zero and degenerate, as expected for an inseparable extension.\n",
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