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  "body": "\n## Proof\n\nLet $R$ be a [[Euclidean domain|0GJ2HRG95T]] and let $a, b \\in R$.\n\n1. Let $I = \\set{s a + t b : s, t \\in R}$. It contains $0 = 0 a + 0 b$, and also $a = 1 a + 0 b$ and $b = 0 a + 1 b$.\n2. For $s a + t b$ and $s' a + t' b$ in $I$ and $r \\in R$, we have $(s a + t b) - (s' a + t' b) = (s - s') a + (t - t') b \\in I$ and $r (s a + t b) = (r s) a + (r t) b \\in I$.\n3. By steps 1 and 2, $I$ is an [[ideal|0MYSA6MYNA]] of $R$.\n4. Since [[Euclidean domains are principal ideal domains|0QKVE1GD9A]], $I$ is a [[principal ideal|00A2EDFT9Q]]: $I = (d)$ for some $d \\in R$.\n5. The set $d R = \\set{d r : r \\in R}$ is an ideal that contains $d$. Every ideal containing $d$ contains $d R$. So the [[ideal generated by|030TZ0DQWT]] $d$ is $(d) = d R$.\n6. Since $d \\in (d) = I$, there are $s, t \\in R$ with $d = s a + t b$.\n7. Since $a \\in I = d R$, $d$ [[divides|04GNK7K4VP]] $a$. In the same way, $d \\mid b$.\n8. Let $c \\in R$ with $c \\mid a$ and $c \\mid b$, say $a = c a_1$ and $b = c b_1$. Then $d = s c a_1 + t c b_1 = c (s a_1 + t b_1)$, so $c \\mid d$.\n9. By steps 7 and 8, $d$ is a [[greatest common divisor|0F7KNJ1ZYN]] of $a$ and $b$. By step 6, $d = s a + t b$.\n\n## Idea\n\nThe combinations $s a + t b$ form an ideal. In a Euclidean domain every ideal is generated by one element $d$. This $d$ is itself a combination of $a$ and $b$, it divides both, and every common divisor of $a$ and $b$ divides every combination of them, hence divides $d$.\n",
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