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  "body": "\n## Definition\n\nLet $K$ be a field with a [[valuation|SVY882HKSJ]] $v \\colon K \\to \\R \\cup \\set{\\infty}$.\nLet $f = a_0 + a_1 X + \\cdots + a_n X^n \\in K[X]$ with $n \\ge 1$ and $a_0 a_n \\ne 0$.\nLet $S = \\set{(i, v(a_i)) : 0 \\le i \\le n,\\ a_i \\ne 0} \\subseteq \\R^2$.\nThe *Newton polygon* of $f$ is the lower boundary of the [[convex hull|0TCHAS04WK]] of $S$.\nIt is the graph of the largest convex function $\\varphi \\colon [0, n] \\to \\R$ with $\\varphi(i) \\le v(a_i)$ for every $(i, v(a_i)) \\in S$.\n\nIt is a broken line from $(0, v(a_0))$ to $(n, v(a_n))$ whose vertices lie in $S$ and whose slopes increase from left to right.\n\n## Remarks\n\nMain property: let $L$ be a splitting field of $f$ over $K$ and $w$ an extension of $v$ to $L$.\nIf the Newton polygon has a segment of slope $s$ and horizontal length $\\ell$, then exactly $\\ell$ roots of $f$ in $L$ have $w$-value $-s$, counted with multiplicity.\nExample: over $\\Q_p$, take $f = X^2 + X + p$.\nThe points are $(0, 1)$, $(1, 0)$, $(2, 0)$, and the lower boundary has a segment of slope $-1$ from $(0, 1)$ to $(1, 0)$ and a segment of slope $0$ from $(1, 0)$ to $(2, 0)$.\nSo one root has valuation $1$ and one has valuation $0$; this agrees with the product of the roots being $p$ and their sum being $-1$.\nNon-example: for $f = 1 + pX + X^2$ the points are $(0, 0)$, $(1, 1)$, $(2, 0)$.\nThe broken line through all three points is not the Newton polygon, because it is not the graph of a convex function.\nThe point $(1, 1)$ lies above the segment from $(0, 0)$ to $(2, 0)$, and this segment is the whole Newton polygon.\nSo both roots have valuation $0$.\n",
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