---
tag: QSK2J5HEER
kind: statement
role: theorem
name: Lenagan's theorem
msc: [16N40, 16P60, 16P40]
uses: [0B1261B5ED, 0CD87SFM2N, NZ1G9WRWFB, C2H26BYWHE, 3G2JQEYBEK, YQNZTQGP1X, 0NZ8JA25WK, PP11DV3ZMQ, QHD4KJYFSA]
about: [NZ1G9WRWFB, C2H26BYWHE, 3G2JQEYBEK]
sources:
  - url: https://en.wikipedia.org/wiki/Nil_ideal
---

## Statement

Let $R$ be a [[ring|0B1261B5ED]] with identity.
For 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.
One sets $\operatorname{Kdim} 0 = -1$.
Let $\alpha \ge 0$ be an ordinal, and consider descending chains $M_0 \supseteq M_1 \supseteq \cdots$ of submodules of $M$.
Then $\operatorname{Kdim} M \le \alpha$ means: in every such chain, all but finitely many factors $M_i / M_{i+1}$ have Krull dimension less than $\alpha$.
If such ordinals exist, $\operatorname{Kdim} M$ is the least of them; otherwise $M$ has no Krull dimension.
The ring $R$ has right Krull dimension if $\operatorname{Kdim} R_R \le \alpha$ for some ordinal $\alpha$.

Assume that $R$ has right Krull dimension.
Then the [[upper nilradical|3G2JQEYBEK]] of $R$ is a [[nilpotent ideal|C2H26BYWHE]].
Equivalently, every [[nil ideal|NZ1G9WRWFB]] of $R$, that is, every [[two-sided ideal|0CD87SFM2N]] whose elements are all nilpotent, is nilpotent.

## Remarks

The nonzero modules of Krull dimension $0$ are exactly the nonzero [[Artinian modules|YQNZTQGP1X]].
Every right Noetherian ring has right Krull dimension.
So the theorem contains the two-sided case of [[Levitzki's theorem|QHD4KJYFSA]]: in a right Noetherian ring every nil ideal is nilpotent.
It also covers right Artinian rings, where the nil ideals lie in the [[Jacobson radical|7HWZ06SRMF]], which is nilpotent.

An outline of the proof: one first shows that the [[prime radical|9AVE43G8RT]] $N$ of $R$ is nilpotent.
The quotient $R/N$ is semiprime and still has right Krull dimension, which makes it a right [[Goldie ring|4AY3RMCTA3]].
A semiprime right Goldie ring has no nonzero nil ideals, so every nil ideal of $R$ lies in $N$.

The hypothesis cannot be dropped.
Let $k$ 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 $x_i$.
The 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.
It is not nilpotent, since $\bar x_n^{\,n-1} \ne 0$ lies in $I^{n-1}$ for every $n$.
This ring has no Krull dimension.

## Explanations

### L1 human

An element is nilpotent when some power of it is zero.
An ideal can consist of nilpotent elements and still have no single power that kills all of its elements at once.
Lenagan's theorem says that this cannot happen in rings that satisfy a mild finiteness condition, the existence of Krull dimension.
In such rings, "every element dies at some power" upgrades to "every product of a fixed number of elements of the ideal is zero".

### L2 human

Krull dimension of a module measures how far the module is from being Artinian, by a deviation of its lattice of submodules.
Rings with right Krull dimension include all right Noetherian and all right Artinian rings, and many rings that are neither.
The theorem extends the classical facts that nil ideals are nilpotent in right Artinian rings and, by Levitzki's theorem, in right Noetherian rings.
It 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.
The 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.