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  "body": "\n## Statement\n\nLet $R$ be an [[integral domain|09S3PFBEPM]] and let $p \\in R$ be a [[prime element|0MNHXBSX3M]].\nThen $p$ is [[irreducible|052SDEPZH7]].\n\n## Remarks\n\nThe proof: let $p = a b$. Then $p \\mid a b$, so $p \\mid a$, say. Write $a = p c$. Then $p = p c b$, and [[cancellation|0GJM24EXP9]] gives $c b = 1$, so $b$ is a unit.\nThe converse is false in general: in $\\mathbb{Z}[\\sqrt{-5}]$, [[$2$ is irreducible but not prime|0CHE3J7QMX]].\nIn a [[unique factorization domain|06YMXJMXYS]], [[irreducible elements are prime|0QRCFZKXH8]].\n\n## Explanations\n\n### L1 human\n\nSuppose that a prime $p$ factors as $a b$.\nThen $p$ divides one factor, so that factor is at least as large as $p$ for divisibility.\nThe other factor must then be a unit.\n",
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