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  "body": "\n## Statement\n\nLet $R$ be a [[Dedekind domain|7NPY2CN5KB]] that is also a [[unique factorization domain|0W4X9G6263]].\nThen $R$ is a [[principal ideal domain|0CMD0BCH8T]].\n\n## Remarks\n\nLet $P$ be a nonzero [[prime ideal|00MHHZ9SNE]] and let $a \\in P$ be nonzero.\nThe element $a$ is not a unit, and one of its [[irreducible|0YMT0D2442]] factors $\\pi$ lies in $P$.\nIn a UFD, $\\pi$ is a [[prime element|01VH21EV6F]], so $(\\pi)$ is a nonzero prime ideal inside $P$; as nonzero primes of a Dedekind domain are maximal, $(\\pi) = P$.\nSo every nonzero prime is principal, and by the [[unique factorization of ideals|S6PHT75MEK]] every nonzero ideal is a product of principal ideals, hence principal.\n\nExample: $\\Z[\\sqrt{-5}]$ is a Dedekind domain, but $6 = 2 \\cdot 3 = (1 + \\sqrt{-5})(1 - \\sqrt{-5})$ gives two different factorizations into irreducibles, so it is not a UFD; consistent with this, $(2, 1 + \\sqrt{-5})$ is not principal.\nThe Dedekind hypothesis cannot be dropped: $k[x, y]$ for a field $k$ is a UFD, but the ideal $(x, y)$ is not principal.\n\n## Explanations\n\n### L1 human\n\nFor rings of algebraic integers, unique factorization of numbers and \"every ideal has one generator\" are the same property.\nOne direction holds in every ring: a principal ideal domain always has unique factorization.\nThis result gives the other direction for Dedekind domains.\nSo a ring of integers of a number field has unique factorization exactly when its ideals are all principal.\n\n### L2 human\n\nIn terms of the [[ideal class group|WQ46BD3ARY]], a Dedekind domain is a UFD exactly when its class group is trivial, and this is the same as being a PID.\nThe class group thus measures the failure of unique factorization of elements.\nFor example, the class group of $\\Z[\\sqrt{-5}]$ has order $2$.\nThe proof uses only that nonzero primes are maximal and that ideals factor into primes, so it is a good first exercise with the ideal factorization theorem.\n",
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