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  "body": "\n## Proof\n\nLet $R$ be a [[Euclidean domain|0GJ2HRG95T]] with Euclidean function $\\delta \\colon R \\setminus \\set{0} \\to \\N$. Let $I$ be an [[ideal|0MYSA6MYNA]] of $R$.\n\n1. For $b \\in R$, the set $b R = \\set{b t : t \\in R}$ is an ideal that contains $b$. Every ideal containing $b$ contains $b R$. So the [[ideal generated by|030TZ0DQWT]] $b$ is $(b) = b R$.\n2. If $I = \\set{0}$, then $I = (0)$, so $I$ is [[principal|00A2EDFT9Q]].\n3. Now let $I \\neq \\set{0}$. The set $\\set{\\delta(c) : c \\in I, c \\neq 0}$ is a nonempty set of natural numbers.\n4. By the [[well-ordering principle|01BX15HSRR]], it has a least element. Choose $b \\in I$ with $b \\neq 0$ such that $\\delta(b)$ is this least element.\n5. Since $b \\in I$, step 1 gives $(b) \\subseteq I$.\n6. Let $a \\in I$. Since $b \\neq 0$, there are $q, r \\in R$ with $a = b q + r$, and $r = 0$ or $\\delta(r) \u003c \\delta(b)$.\n7. Since $I$ is an ideal, $b q \\in I$, so $r = a - b q \\in I$.\n8. If $r \\neq 0$, then $r$ is a nonzero element of $I$ with $\\delta(r) \u003c \\delta(b)$. This contradicts the choice of $b$. So $r = 0$.\n9. Hence $a = b q \\in (b)$. So $I \\subseteq (b)$, and with step 5, $I = (b)$.\n10. Every ideal of $R$ is principal, and $R$ is an integral domain. So $R$ is a [[principal ideal domain|0RZYDVZ475]].\n\n## Idea\n\nIn a nonzero ideal, pick a nonzero element $b$ with the smallest value of $\\delta$. Dividing any element of the ideal by $b$ leaves a remainder that lies in the ideal and is smaller than $b$. So the remainder must be zero, and $b$ generates the ideal.\n",
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  "name": "Proof that Euclidean domains are principal ideal domains",
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