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  "body": "\n## Statement\n\nLet $A$ be a [[Noetherian|0MVH4Y1CRZ]] [[integral domain|09S3PFBEPM]] of [[Krull dimension|MVGBS3J1P7]] $1$.\nLet $K$ be the [[field of fractions|0VF11ZTTMD]] of $A$ and let $L$ be a finite field extension of $K$.\nLet $B$ be a ring with $A \\subseteq B \\subseteq L$.\nThen $B$ is Noetherian and has Krull dimension at most $1$.\nMoreover, for every nonzero ideal $J$ of $B$, the quotient $B/J$ has finite [[length|CZSCNZH8VY]] as an $A$-module.\n\n## Remarks\n\nThe general case reduces to the case $L = K$.\nThere the key step is the bound $\\ell_A(B/aB) \\le \\ell_A(A/aA)$ for each nonzero $a \\in A$, where $\\ell_A$ is the length; the right side is finite because $A$ has dimension $1$.\nEach nonzero ideal $J$ of $B$ contains a nonzero element $a$ of $A$, so $J/aB$ has finite length, and $J$ is finitely generated.\nA main application: the [[integral closure|G0V86SJBQW]] of $A$ in $L$ is a Noetherian integrally closed domain of dimension at most $1$, that is, a Dedekind domain or a field.\nFor example, with $A = \\Z$ and $L$ a number field, this shows that the ring of integers of $L$ is Noetherian.\nThe ring $B$ need not be a finitely generated $A$-module, so the result is not a consequence of [[Hilbert's basis theorem|0YKA9J3XEE]].\n\n## Explanations\n\n### L1 human\n\nStart with a nice ring of dimension one, such as the integers.\nTake any ring that sits between it and a finite extension of its fraction field.\nThe theorem says that this new ring is still Noetherian and still has dimension at most one.\nSo rings of integers in number fields keep the finiteness of $\\Z$.\n\n### L2 human\n\nThe theorem is the standard route to the following fact.\nThe integral closure of a one-dimensional Noetherian domain in a finite extension of its fraction field is a Dedekind domain, even when it is not a finite module.\nIt is special to dimension $1$: for any Noetherian domain, the [[Mori–Nagata theorem|J8YYYP5CDN]] only gives a Krull domain, and the integral closure can fail to be Noetherian.\nThe proof uses [[lengths of modules|CZSCNZH8VY]] and the fact that $A/aA$ is Artinian for $a \\ne 0$.\n",
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