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text: '[summary: While there are two distinct but closely related notions of "[5m2 prime]" and "[5m1 irreducible]", the two ideas are actually the same in a certain class of [3gq ring]. This result is basically what enables the [-fundamental_theorem_of_arithmetic].]\n\nLet $R$ be a [3gq ring] which is a [principal_ideal_domain PID], and let $r \\not = 0$ be an element of $R$.\nThen $r$ is [5m1 irreducible] if and only if $r$ is [5m2 prime].\n\nIn fact, it is easier to prove a stronger statement: that the following are equivalent.\n%%note:Every proof known to the author is of this shape either implicitly or explicitly, but when it's explicit, it should be clearer what is going on.%%\n\n1. $r$ is irreducible. \n2. $r$ is prime.\n3. The generated [ideal_ring_theory ideal] $\\langle r \\rangle$ is [maximal_ideal maximal] in $R$.\n\n[todo: motivate the third bullet point]\n\n# Proof\n\n## $2 \\Rightarrow 1$\nA proof that "prime implies irreducible" appears on the page for [5m1 irreducibility].\n\n## $3 \\Rightarrow 2$\nWe wish to show that if $\\langle r \\rangle$ is maximal, then it is prime.\n(Indeed, $r$ is [5m2 prime] if and only if its generated ideal is [prime_ideal prime].)\n\nAn ideal $I$ is maximal if and only if the [quotient_ring quotient] $R/I$ is a [481 field]. ([ideal_maximal_iff_quotient_is_field Proof.])\n\nAn ideal $I$ is [prime_ideal prime] if and only if the quotient $R/I$ is an [-5md]. ([ideal_prime_iff_quotient_is_integral_domain Proof.])\n\nAll fields are integral domains. (A proof of this appears on [5md the page on integral domains].)\n\nHence maximal ideals are prime.\n\n## $1 \\Rightarrow 3$\nLet $r$ be irreducible; then in particular it is not invertible, so $\\langle r \\rangle$ isn't simply the whole ring.\n\nTo show that $\\langle r \\rangle$ is maximal, we need to show that if it is contained in any larger ideal then that ideal is the whole ring.\n\nSuppose $\\langle r \\rangle$ is contained in the larger ideal $J$, then.\nBecause we are in a principal ideal domain, $J = \\langle a \\rangle$, say, for some $a$, and so $r = a c$ for some $c$.\nIt will be enough to show that $a$ is invertible, because then $\\langle a \\rangle$ would be the entire ring.\n\nBut $r$ is irreducible, so one of $a$ and $c$ is invertible; if $a$ is invertible then we are done, so suppose $c$ is invertible.\n\nThen $a = r c^{-1}$.\nWe have supposed that $J$ is indeed larger than $\\langle r \\rangle$: that there is $j \\in J$ which is not in $\\langle r \\rangle$.\nSince $j \\in J = \\langle a \\rangle$, we can find $d$ (say) such that $j = a d$; so $j = r c^{-1} d$ and hence $j \\in \\langle r \\rangle$, which is a contradiction.',
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