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  "body": "\n## Proof\n\nLet $K$ be a [[field|0QA8CHH8MG]].\n\n1. The field $K$ is an [[integral domain|09S3PFBEPM]]: $1 \\neq 0$, and if $a b = 0$ with $a \\neq 0$, then $b = a^{-1} a b = 0$.\n2. By [[the theorem on polynomial rings|0B5Z1EC66J]], $K[x]$ is an integral domain.\n3. For nonzero $h \\in K[x]$, the [[degree|0DPF6WRZW6]] $\\deg h$ is an integer at least $0$. So $\\delta(h) = \\deg h$ defines a function $\\delta \\colon K[x] \\setminus \\set{0} \\to \\N$.\n4. Let $f, g \\in K[x]$ with $g \\neq 0$. Let $c$ be the [[leading coefficient|0P05RJC5XA]] of $g$, so $c \\neq 0$ and $c$ has an inverse $c^{-1}$ in $K$.\n5. Put $g_1 = c^{-1} g$. Its coefficients are those of $g$ times $c^{-1}$. Since $c^{-1} a = 0$ only for $a = 0$, the polynomial $g_1$ has degree $\\deg g$ and leading coefficient $c^{-1} c = 1$.\n6. So $g_1$ is [[monic|0BDX4MBKT4]]. By the [[division algorithm|0MES3KS0ZC]] in $K[x]$, there are $q_1, r \\in K[x]$ with $f = q_1 g_1 + r$ and $\\deg r \u003c \\deg g_1 = \\deg g$.\n7. Put $q = c^{-1} q_1$. Then $q g = q_1 c^{-1} g = q_1 g_1$, so $f = q g + r$.\n8. If $r \\neq 0$, then step 6 gives $\\delta(r) = \\deg r \u003c \\deg g = \\delta(g)$. So $r = 0$ or $\\delta(r) \u003c \\delta(g)$.\n9. By steps 2, 3 and 8, $K[x]$ is a [[Euclidean domain|0GJ2HRG95T]] with the degree as Euclidean function.\n\n## Idea\n\nOver a field, every nonzero polynomial becomes monic after we divide it by its leading coefficient. Long division by a monic polynomial always works, so we can divide by any nonzero polynomial and get a remainder of smaller degree.\n",
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