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  "body": "\n## Statement\n\nLet $R$ be a [[principal ideal domain|0RZYDVZ475]] and let $p \\in R$ with $p \\neq 0$.\nThen $p$ is [[irreducible|052SDEPZH7]] if and only if the principal ideal $(p)$ is a [[maximal ideal|09J6KGT65G]].\n\n## Remarks\n\nThe proof: an ideal $I$ with $(p) \\subseteq I$ is $(a)$ with $a \\mid p$. If $p$ is irreducible, then $a$ is a unit or an associate of $p$, so $I = R$ or $I = (p)$.\nA consequence is that $R/(p)$ is a [[field|0QA8CHH8MG]] for each irreducible $p$. For example, $\\mathbb{Q}[x]/(x^2 - 2)$ is a field.\nIn a principal ideal domain, every nonzero prime ideal is maximal.\n\n## Explanations\n\n### L1 human\n\nIn a principal ideal domain, ideals that contain $(p)$ come from the divisors of $p$.\nAn irreducible $p$ has only trivial divisors.\nSo no ideal fits strictly between $(p)$ and the whole ring.\n",
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  "name": "In a PID a nonzero element is irreducible iff it generates a maximal ideal",
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