---
tag: QHD4KJYFSA
kind: statement
role: theorem
name: Levitzki's theorem
msc: [16N40, 16P40]
uses: [0670T59HGE, NZ1G9WRWFB, C2H26BYWHE, 5N2CJZ47D4, 0ZB2JG1M9C, HB4XDKJTW8, W29AA0266Z]
about: [NZ1G9WRWFB, C2H26BYWHE, 0670T59HGE]
sources:
  - url: https://en.wikipedia.org/wiki/Levitzky%27s_theorem
same_as:
  wikidata: Q6535741
  wikipedia: Levitzky's theorem
---

## Statement

Let $R$ be a right [[Noetherian ring|0670T59HGE]].
Then every [[nil|NZ1G9WRWFB]] one-sided ideal of $R$ is [[nilpotent|C2H26BYWHE]].
That 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 $n \ge 1$.

## 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|HB4XDKJTW8]] 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 $n$.

## Explanations

### L1 human

An ideal is nil if each of its elements becomes zero after raising it to some power.
It is nilpotent if one fixed power kills all products of its elements at once.
Nilpotent implies nil, but in general not the other way round.
Levitzki's theorem says that in a ring with the ascending chain condition on right ideals the two notions agree, even for one-sided ideals.

### L2 human

The theorem extends the classical fact that nil ideals of Artinian rings are nilpotent, since Artinian rings are Noetherian by the [[Hopkins–Levitzki theorem|W29AA0266Z]].
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.