Levitzki's theorem

Remarks

The sum of all nilpotent ideals of a right Noetherian ring is itself a nilpotent ideal $N$, by the ascending chain condition. The ring $R/N$ has no nonzero nilpotent ideals, and it suffices to show that it has no nonzero nil one-sided ideals. In a nonzero nil one-sided ideal, choose a nonzero element $a$ whose right annihilator is maximal, which is possible by the chain condition. A short computation with this maximality gives $aRa = 0$, so $aR$ is a nonzero right ideal with $(aR)^2 = 0$, a contradiction.

Example: 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 $N^n = 0$.

The Noetherian hypothesis is needed. Let $k$ be a field and $R = k[x_1, x_2, \ldots]/(x_1^2, x_2^3, x_3^4, \ldots)$. The ideal $I$ generated by the images of all $x_i$ is nil, because $R$ is commutative and $I$ is generated by nilpotent elements. But $I$ is not nilpotent, because $x_n^n \ne 0$ in $R$, so $I^n \ne 0$ for every $n$.

Going deeper

For a student

The theorem extends the classical fact that nil ideals of Artinian rings are nilpotent, since Artinian rings are Noetherian by the Hopkins–Levitzki theorem. A typical consequence: the prime radical of a right Noetherian ring is nilpotent, and it is the largest nil ideal. So for right Noetherian rings the various nil radicals all coincide.