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  "body": "\n## Definition\n\nLet $R$ be a [[commutative ring|0310REJJY8]] and let $x = (x_1, \\ldots, x_n)$ be a sequence of elements of $R$.\nLet $e_1, \\ldots, e_n$ be the standard basis of the [[free module|0DZ9DKAKZR]] $R^n$.\nFor $0 \\le p \\le n$, let $K_p = \\Lambda^p R^n$ be its $p$-th [[exterior power|0MDMZ965VE]].\nIt is free with basis the elements $e_{i_1} \\wedge \\cdots \\wedge e_{i_p}$ for $i_1 \u003c \\cdots \u003c i_p$.\nThe *Koszul complex* $K_\\bullet(x)$ is\n$$0 \\to K_n \\xrightarrow{d_n} K_{n-1} \\to \\cdots \\to K_1 \\xrightarrow{d_1} K_0 \\to 0,$$\nwith\n$$d_p(e_{i_1} \\wedge \\cdots \\wedge e_{i_p}) = \\sum_{j=1}^{p} (-1)^{j+1} x_{i_j}\\, e_{i_1} \\wedge \\cdots \\wedge \\widehat{e_{i_j}} \\wedge \\cdots \\wedge e_{i_p},$$\nwhere the hat marks an omitted factor.\nFor an [[$R$-module|0NZ8JA25WK]] $M$, the Koszul complex of $x$ on $M$ is $K_\\bullet(x; M) = K_\\bullet(x) \\otimes_R M$.\n\n## Remarks\n\nOne checks that $d_{p-1} \\circ d_p = 0$, so $K_\\bullet(x)$ is a complex, and $H_0(K_\\bullet(x)) = R/(x_1, \\ldots, x_n)$.\nIf $x_1, \\ldots, x_n$ is a [[regular sequence|69R5R2FCS5]], then $H_p(K_\\bullet(x)) = 0$ for $p \u003e 0$, so the complex is a free resolution of $R/(x_1, \\ldots, x_n)$.\nExample: for $n = 2$, the complex is $0 \\to R \\to R^2 \\to R \\to 0$ with $e_1 \\wedge e_2 \\mapsto x_1 e_2 - x_2 e_1$ and $e_i \\mapsto x_i$; the composite sends $e_1 \\wedge e_2$ to $x_1 x_2 - x_2 x_1 = 0$.\nNon-example of exactness: for $R = K[t]$ and $x = (t, t)$, which is not a regular sequence, $H_1$ is the set of $(a, -a)$ modulo the multiples of $(-t, t)$, which is isomorphic to $R/(t) \\ne 0$.\n",
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