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  "body": "\n## Definition\n\nLet $K$ be a [[commutative ring|0GYGHKX834]].\nLet $A$ be a [[$K$-module|09VDS2W1E8]] with a direct sum decomposition $A = A_0 \\oplus A_1$.\nThen $A$ with a $K$-bilinear product $A \\times A \\to A$ is a *superalgebra* over $K$ if\n$$A_i A_j \\subseteq A_{i + j}$$\nfor all $i, j \\in \\Z/2\\Z$, with indices added modulo $2$.\nElements of $A_0$ are *even*, elements of $A_1$ are *odd*, and a nonzero element of $A_0 \\cup A_1$ is *homogeneous* of parity $\\abs{a} \\in \\{0, 1\\}$.\nUnless stated otherwise the product is associative with a unit.\n\n## Remarks\n\nA superalgebra is *supercommutative* if $ab = (-1)^{\\abs{a}\\abs{b}} ba$ for all homogeneous $a, b$.\nExample: the exterior algebra $\\Lambda(V)$ of a vector space $V$, with $A_0$ the sum of the even powers $\\Lambda^{2m} V$ and $A_1$ the sum of the odd powers.\nThe product maps $\\Lambda^p V \\times \\Lambda^r V$ into $\\Lambda^{p+r} V$, and $p + r$ has the parity of the sum, so the condition holds; moreover $\\Lambda(V)$ is supercommutative.\nExample: the matrix ring $M_2(K)$, with $A_0$ the diagonal matrices and $A_1$ the off-diagonal ones, is a superalgebra that is not supercommutative.\nNon-example: $K[x]$ with $A_0 = K$ and $A_1 = xK[x]$ fails the condition, since $x \\in A_1$ but $x \\cdot x = x^2 \\notin A_0$.\nEquivalently, a superalgebra is a monoid object in the category of $\\Z/2\\Z$-graded modules, with the tensor product of graded modules.\n",
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