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  "body": "\n## Proof\n\nWe deduce the lemma from the generalized Euclid's lemma.\n\n1. Let $p$ be a [[prime number|0SN0MAEKYX]] and let $a$, $b$ be integers with $p \\mid ab$. If $p \\mid a$, we are done. So assume $p \\nmid a$.\n2. Let $g = \\gcd(p, a)$ be the [[greatest common divisor|02WF878BH1]]. It is defined as the largest common divisor, because $p \\neq 0$.\n3. The number $1$ is a common divisor of $p$ and $a$, so $g \\ge 1$.\n4. The number $g$ is a positive [[divisor|096PW7Q014]] of $p$. Since $p$ is prime, $g = 1$ or $g = p$.\n5. If $g = p$, then $p \\mid a$, which is false. So $g = 1$, and $p$ and $a$ are [[coprime|0GFKP8F8AG]].\n6. By the [[generalized Euclid's lemma|03BAREW1Z5]], $p \\mid ab$ and $\\gcd(p, a) = 1$ give $p \\mid b$.\n7. So $p \\mid a$ or $p \\mid b$.\n\n## Idea\n\nA prime has only two positive divisors, so a prime that does not divide $a$ has nothing in common with $a$. Then the coprime case of the lemma forces $p$ to divide $b$.\n",
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