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  "body": "\n## Statement\n\nLet $R$ be a [[unique factorisation domain|06YMXJMXYS]] and let $p \\in R$.\nThen $p$ is [[irreducible|052SDEPZH7]] if and only if $p$ is [[prime|0MNHXBSX3M]].\n\n## Remarks\n\n[[Prime elements are irreducible|0F0K8DB9TD]] in every integral domain. The other direction needs unique factorisation.\nThe proof: suppose $p$ is irreducible and $p c = a b$. Factor $a$, $b$ and $c$ into irreducibles. By uniqueness, $p$ is an associate of an irreducible factor of $a$ or of $b$.\nConversely, an integral domain in which every nonzero nonunit is a product of irreducibles, and every irreducible is prime, is a unique factorisation domain.\n\n## Explanations\n\n### L1 human\n\nIn the integers, Euclid's lemma says that a prime that divides a product divides a factor.\nIn a unique factorisation domain, the same thing is true for every irreducible element.\nIn $\\mathbb{Z}[\\sqrt{-5}]$ this fails, and so factorisation is not unique there.\n",
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