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  "body": "\n## Definition\n\nLet $k$ be a [[field|04C5ZBG1WT]], let $S = k[x_1, \\ldots, x_n]$ be the [[polynomial ring|0SDC934R52]], and fix a [[monomial order|E874NARWPV]] on $S$.\nFor nonzero $h \\in S$, let $\\operatorname{LT}(h)$ be its leading term.\nLet $f \\in S$ and let $(g_1, \\ldots, g_s)$ be an ordered list of nonzero elements of $S$.\nThe *multivariate division algorithm* computes $q_1, \\ldots, q_s, r \\in S$ as follows.\n\n1. Start with $q_1 = \\cdots = q_s = 0$, $r = 0$ and $p = f$.\n2. While $p \\ne 0$: if some $\\operatorname{LT}(g_i)$ divides $\\operatorname{LT}(p)$, take the least such $i$, and replace $q_i$ by $q_i + \\operatorname{LT}(p)/\\operatorname{LT}(g_i)$ and $p$ by $p - (\\operatorname{LT}(p)/\\operatorname{LT}(g_i)) \\, g_i$.\n3. Otherwise, replace $r$ by $r + \\operatorname{LT}(p)$ and $p$ by $p - \\operatorname{LT}(p)$.\n4. When $p = 0$, return $q_1, \\ldots, q_s$ and the *remainder* $r$.\n\n## Remarks\n\nThe algorithm stops, because each step makes the leading monomial of $p$ strictly smaller, or makes $p$ zero, and a monomial order is a well-order.\nThe output satisfies $f = q_1 g_1 + \\cdots + q_s g_s + r$.\nNo monomial of $r$ is divisible by any $\\operatorname{LT}(g_i)$, and the leading monomial of $q_i g_i$ is at most that of $f$ whenever $q_i g_i \\ne 0$.\n\nExample: in $k[x, y]$ with the lexicographic order $x \u003e y$, divide $f = x^2 y + x y^2 + y^2$ by $(g_1, g_2) = (xy - 1, \\; y^2 - 1)$.\nThe steps give $q_1 = x + y$, $q_2 = 1$ and $r = x + y + 1$; check: $(x + y)(xy - 1) + (y^2 - 1) + x + y + 1 = x^2 y + x y^2 + y^2$.\nWith the list in the order $(y^2 - 1, \\; xy - 1)$, the algorithm gives the quotients $x + 1$ for $y^2 - 1$ and $x$ for $xy - 1$, and the remainder $2x + 1$.\nSo in general the remainder depends on the order of the list.\nFor a [[Gröbner basis|C83KPT47R0]], the remainder does not depend on the order, and it is zero exactly when $f$ lies in the ideal $(g_1, \\ldots, g_s)$.\n",
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