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  "body": "\n## Statement\n\nLet $R$ be a [[ring|0B1261B5ED]] with identity.\nFor a right $R$-[[module|0NZ8JA25WK]] $M$, the [[Krull dimension|PP11DV3ZMQ]] $\\operatorname{Kdim} M$ (in the sense of Gabriel and Rentschler) is defined by transfinite recursion on ordinals.\nOne sets $\\operatorname{Kdim} 0 = -1$.\nLet $\\alpha \\ge 0$ be an ordinal, and consider descending chains $M_0 \\supseteq M_1 \\supseteq \\cdots$ of submodules of $M$.\nThen $\\operatorname{Kdim} M \\le \\alpha$ means: in every such chain, all but finitely many factors $M_i / M_{i+1}$ have Krull dimension less than $\\alpha$.\nIf such ordinals exist, $\\operatorname{Kdim} M$ is the least of them; otherwise $M$ has no Krull dimension.\nThe ring $R$ has right Krull dimension if $\\operatorname{Kdim} R_R \\le \\alpha$ for some ordinal $\\alpha$.\n\nAssume that $R$ has right Krull dimension.\nThen the [[upper nilradical|3G2JQEYBEK]] of $R$ is a [[nilpotent ideal|C2H26BYWHE]].\nEquivalently, every [[nil ideal|NZ1G9WRWFB]] of $R$, that is, every [[two-sided ideal|0CD87SFM2N]] whose elements are all nilpotent, is nilpotent.\n\n## Remarks\n\nThe nonzero modules of Krull dimension $0$ are exactly the nonzero [[Artinian modules|YQNZTQGP1X]].\nEvery right Noetherian ring has right Krull dimension.\nSo the theorem contains the two-sided case of [[Levitzki's theorem|QHD4KJYFSA]]: in a right Noetherian ring every nil ideal is nilpotent.\nIt also covers right Artinian rings, where the nil ideals lie in the [[Jacobson radical|7HWZ06SRMF]], which is nilpotent.\n\nAn outline of the proof: one first shows that the [[prime radical|9AVE43G8RT]] $N$ of $R$ is nilpotent.\nThe quotient $R/N$ is semiprime and still has right Krull dimension, which makes it a right [[Goldie ring|4AY3RMCTA3]].\nA semiprime right Goldie ring has no nonzero nil ideals, so every nil ideal of $R$ lies in $N$.\n\nThe hypothesis cannot be dropped.\nLet $k$ be a field and $R = k[x_1, x_2, \\dots] / (x_1, x_2^2, x_3^3, \\dots)$, with $\\bar x_i$ the image of $x_i$.\nThe ideal $I$ generated by all $\\bar x_i$ is nil, since each element of $I$ lies in an ideal generated by finitely many nilpotent elements of a commutative ring.\nIt is not nilpotent, since $\\bar x_n^{\\,n-1} \\ne 0$ lies in $I^{n-1}$ for every $n$.\nThis ring has no Krull dimension.\n\n## Explanations\n\n### L1 human\n\nAn element is nilpotent when some power of it is zero.\nAn ideal can consist of nilpotent elements and still have no single power that kills all of its elements at once.\nLenagan's theorem says that this cannot happen in rings that satisfy a mild finiteness condition, the existence of Krull dimension.\nIn such rings, \"every element dies at some power\" upgrades to \"every product of a fixed number of elements of the ideal is zero\".\n\n### L2 human\n\nKrull dimension of a module measures how far the module is from being Artinian, by a deviation of its lattice of submodules.\nRings with right Krull dimension include all right Noetherian and all right Artinian rings, and many rings that are neither.\nThe theorem extends the classical facts that nil ideals are nilpotent in right Artinian rings and, by Levitzki's theorem, in right Noetherian rings.\nIt is a typical use of Krull dimension: a statement proved for Noetherian rings by an ascending chain argument is carried over by induction on the deviation.\nThe commutative example $k[x_1, x_2, \\dots] / (x_1, x_2^2, x_3^3, \\dots)$ shows that some finiteness condition is needed even without noncommutativity.\n",
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