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  "body": "\n## Statement\n\nLet $R$ be a right [[Noetherian ring|0670T59HGE]].\nThen every [[nil|NZ1G9WRWFB]] one-sided ideal of $R$ is [[nilpotent|C2H26BYWHE]].\nThat is, if $I$ is a right or [[left ideal|5N2CJZ47D4]] of $R$ in which every element is [[nilpotent|0ZB2JG1M9C]], then $I^n = 0$ for some $n \\ge 1$.\n\n## Remarks\n\nThe sum of all nilpotent ideals of a right Noetherian ring is itself a nilpotent ideal $N$, by the ascending chain condition.\nThe ring $R/N$ has no nonzero nilpotent ideals, and it suffices to show that it has no nonzero nil one-sided ideals.\nIn a nonzero nil one-sided ideal, choose a nonzero element $a$ whose right [[annihilator|HB4XDKJTW8]] is maximal, which is possible by the chain condition.\nA short computation with this maximality gives $aRa = 0$, so $aR$ is a nonzero right ideal with $(aR)^2 = 0$, a contradiction.\n\nExample: in the ring of upper triangular $n \\times n$ matrices over a field, the strictly upper triangular matrices form a nil ideal $N$, and indeed $N^n = 0$.\n\nThe Noetherian hypothesis is needed.\nLet $k$ be a field and $R = k[x_1, x_2, \\ldots]/(x_1^2, x_2^3, x_3^4, \\ldots)$.\nThe ideal $I$ generated by the images of all $x_i$ is nil, because $R$ is commutative and $I$ is generated by nilpotent elements.\nBut $I$ is not nilpotent, because $x_n^n \\ne 0$ in $R$, so $I^n \\ne 0$ for every $n$.\n\n## Explanations\n\n### L1 human\n\nAn ideal is nil if each of its elements becomes zero after raising it to some power.\nIt is nilpotent if one fixed power kills all products of its elements at once.\nNilpotent implies nil, but in general not the other way round.\nLevitzki's theorem says that in a ring with the ascending chain condition on right ideals the two notions agree, even for one-sided ideals.\n\n### L2 human\n\nThe theorem extends the classical fact that nil ideals of Artinian rings are nilpotent, since Artinian rings are Noetherian by the [[Hopkins–Levitzki theorem|W29AA0266Z]].\nA typical consequence: the prime radical of a right Noetherian ring is nilpotent, and it is the largest nil ideal.\nSo for right Noetherian rings the various nil radicals all coincide.\n",
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