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  "body": "\n## Counterexample\n\nLet $k$ be a field and let $A = k[x_1, x_2, x_3, \\ldots]$ be the [[polynomial ring|0SDC934R52]] in countably many variables.\nSplit the variables into consecutive blocks $B_1, B_2, B_3, \\ldots$ where the block $B_i$ has exactly $i$ variables.\nLet $\\mathfrak{p}_i$ be the [[prime ideal|00MHHZ9SNE]] of $A$ generated by the variables of $B_i$.\nLet $S = A \\setminus \\bigcup_{i \\ge 1} \\mathfrak{p}_i$, a [[multiplicative set|AN9SKEZ3VY]], and let $R = S^{-1} A$ be the [[localization|4C1AE84N5N]].\nThen $R$ is a [[Noetherian ring|0670T59HGE]], but its [[Krull dimension|MVGBS3J1P7]] is infinite.\n\n## Remarks\n\nThe maximal ideals of $R$ are exactly the ideals $\\mathfrak{p}_i R$.\nThe localization of $R$ at $\\mathfrak{p}_i R$ is a localization of a polynomial ring in $i$ variables over the field generated by the other variables.\nSo it is Noetherian of dimension $i$.\nEach nonzero element of $R$ involves finitely many variables, so it lies in only finitely many maximal ideals.\nA ring whose localizations at maximal ideals are all Noetherian and whose nonzero elements lie in finitely many maximal ideals is Noetherian.\nThe [[height|87WH65FJB8]] of $\\mathfrak{p}_i R$ is $i$, so the dimension of $R$ is at least $i$ for every $i$.\nSo the statement \"a Noetherian ring has finite Krull dimension\" is false, although every Noetherian local ring has finite dimension.\n\n## Explanations\n\n### L1 human\n\nIn a Noetherian ring every ascending chain of ideals stops, so one might expect the length of chains of prime ideals to be bounded.\nEach single chain of primes is finite here, but the chains can be as long as one likes.\nThe ring is glued from local pieces of dimension $1, 2, 3, \\ldots$, each of them well behaved, and there is no common bound.\n",
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