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  "body": "\n## Definition\n\nLet $k$ be a [[field|04C5ZBG1WT]] and let $A$ be a finite-dimensional [[associative algebra|3CTW1P4HBP]] over $k$ with identity.\nThe algebra $A$ is a *Frobenius algebra* if there is a [[bilinear form|0SQZMXB6GP]] $\\sigma \\colon A \\times A \\to k$ such that:\n\n1. $\\sigma$ is [[nondegenerate|01NHFZF2SK]];\n2. $\\sigma(ab, c) = \\sigma(a, bc)$ for all $a, b, c \\in A$.\n\n## Remarks\n\nExample: the matrix algebra $M_n(k)$ with $\\sigma(a, b) = \\tr(ab)$.\nClause 2 holds because both sides equal $\\tr(abc)$.\nClause 1 holds because $\\tr(a E_{ji}) = a_{ij}$ for the matrix unit $E_{ji}$, so $\\sigma(a, \\cdot) = 0$ forces $a = 0$.\nAnother example is the group algebra $k[G]$ of a finite group $G$, with $\\sigma(a, b)$ the coefficient of the identity element of $G$ in $ab$.\nNon-example: $A = k[x, y]/(x^2, xy, y^2)$, with basis $1, x, y$, is not Frobenius.\nIf $\\sigma$ satisfies clause 2, then $\\sigma(a, b) = \\lambda(ab)$ for the linear form $\\lambda = \\sigma(1, \\cdot)$.\nThe plane spanned by $x$ and $y$ meets $\\ker \\lambda$ in some nonzero $u = \\alpha x + \\beta y$.\nSince $ub \\in k u$ for every $b \\in A$, $\\sigma(u, b) = 0$ for all $b$, and clause 1 fails.\nEquivalent forms: $A$ is Frobenius if and only if some linear form $\\lambda \\colon A \\to k$ has a kernel that contains no nonzero left ideal.\nThis holds if and only if $A \\cong \\Hom_k(A, k)$ as left $A$-modules.\nIf moreover $\\sigma$ is symmetric, $A$ is called a symmetric algebra.\n",
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