The Levitzki Radical and Locally Nilpotent Ideals

Q: Why introduce a third nil-type radical at all?

Because the upper nilradical is not known to absorb one-sided ideals, and the lower nilradical is often too small. The Levitzki radical is the largest radical in this family whose defining property is closed under the operations one actually performs — sums of one-sided ideals and generation of two-sided ideals — so theorems proved with it are unconditional.

Q: Is the Levitzki radical of R/L-rad R zero?

Yes. If M/L-rad R is a locally nilpotent ideal of the quotient, the finite-word argument used for sums shows M itself is locally nilpotent, hence contained in L-rad R. All four radicals on this page have that idempotence property.

Q: Does Levitzki's Theorem need both chain conditions?

No. Only ACC on right annihilators is used, and that is implied by right noetherianity. The stronger hypothesis appears in the statement because the conclusion — nilpotence rather than mere containment in Nil_* R — needs ACC on ideals to produce a maximal nilpotent ideal.

Q: Where exactly does Köthe's Conjecture enter this page?

Nowhere as an assumption, and that is the point. Every statement here is unconditional because *locally nilpotent* replaces *nil*. If Köthe's Conjecture were proved, the two middle terms of the chain would still differ — Golod's algebra separates them — but Nil^* R would acquire the one-sided closure properties that L-rad R already has.

Q: How does the twisted polynomial example avoid contradicting (10.18)?

Because A[x;] is not a polynomial ring in the sense of (10.18): the variable does not commute with the coefficients. The leading-coefficient identity becomes a^n(b) rather than ab, and it is precisely the shifting automorphism, together with van der Waerden's theorem on arithmetic progressions, that manufactures a locally nilpotent right ideal inside a prime ring.

Executive Summary

Nil is a property of individual elements and behaves badly for one-sided ideals; nilpotent is a uniform property and behaves well but is far too strong. Locally nilpotent — every finite subset generates a nilpotent subring — is the intermediate notion, and it is the one that works: locally nilpotent one-sided ideals may be added, may be expanded to two-sided ideals, and possess a largest member.

That largest member is the Levitzki radical. Two theorems make it and its neighbours usable. Utumi's Lemma pushes every nil one-sided ideal into ${Nil}_{*} R$ under ACC on right annihilators; Levitzki's Theorem specialises this to right noetherian rings, where all nil one-sided ideals turn out to be nilpotent and the four radicals partially collapse.

$L - rad R$ Largest locally nilpotent ideal

4Radicals in the chain

1939 / 1950Levitzki: proved / published

StrictAll three inclusions

Overview

For a one-sided ideal the implications $nilpotent \Rightarrow locally nilpotent \Rightarrow nil$ hold and neither converse does. The middle condition is finitary — it is a statement about finite subsets — and that is exactly what makes the standard closure arguments run.

{Nil}_{*} R \subseteq L - rad R \subseteq {Nil}^{*} R \subseteq rad R

(10.32)

The four radicals of a ring. The second inclusion holds because a locally nilpotent ideal is nil; the first because $L - rad R$ is a semiprime ideal; the third by $(4.11)$ .

The situation mirrors group theory exactly: the Hirsch–Plotkin radical of a group is the largest locally nilpotent normal subgroup, and it exists for the same finitary reason. The analogy is not decorative — the technique of proving closure by working with one finite subset at a time is identical.

Learning Objectives

State what it means for a subset of a ring to be locally nilpotent and check the two implications relating it to nil and nilpotent.
Prove Utumi's Lemma $(10.29)$ by choosing an element with maximal right annihilator.
Derive Levitzki's Theorem $(10.30)$ and identify ${Nil}_{*} R$ with the largest nilpotent one-sided ideal.
Prove $(10.31)$ : locally nilpotent one-sided ideals are closed under generation of two-sided ideals and under sums.
Prove that $L - rad R$ is a semiprime ideal and deduce the chain $(10.32)$ .
Describe a prime ring with nonzero Levitzki radical.

Definitions

Definition—Locally nilpotent

A subset $S \subseteq R$ is locally nilpotent if for every finite subset ${s_{1}, \dots, s_{n}} \subseteq S$ there is an integer $N = N (n)$ such that every product of $N$ elements drawn from ${s_{1}, \dots, s_{n}}$ — repetitions allowed — is zero. Equivalently, every subring without identity generated by finitely many elements of $S$ is nilpotent.

$L - rad R$: The Levitzki radical: the sum of all locally nilpotent ideals of $R$ . Some sources write $Levitzki (R)$ or $L (R)$ .
${ann}_{r} (a)$: ${x \in R : a x = 0}$ , a right ideal. ACC on right annihilators means the family of these, for $a$ ranging over $R$ , has no strictly increasing infinite chain.
Nil: Every element nilpotent, exponents unbounded.
Nilpotent: $𝔄^{n} = 0$ for one fixed $n$ .
Right noetherian: ACC on right ideals. It implies ACC on right annihilators, since annihilators are right ideals.

Right noetherian is a genuinely one-sided hypothesis in (10.30); the mirror statement for left noetherian rings follows by applying the theorem to the opposite ring.

Core Concepts

The maximal-annihilator technique

Utumi's argument is a template. Inside a set of elements that resists a conclusion, pick one whose right annihilator is as large as possible; then show that any counterexample to the conclusion produces an element with a strictly larger annihilator. The chain condition converts strictly larger into a contradiction.

Choose $a$ badlyAmong elements of the nil one-sided ideal that are not yet in

{Nil}_{*} R

, take

a

with

{ann}_{r} (a)

maximal.

Exploit nilpotence of $a x$ For

x \in R

with

a x \neq 0

, let

k

be minimal with

(a x)^{k} = 0

. Then

k \geq 2

Produce a new annihilating element

y = x (a x)^{k - 2}

satisfies

a y = (a x)^{k - 1} \neq 0

and

(a x a) y = (a x)^{k} = 0

, so

{ann}_{r} (a x a) ⊋ {ann}_{r} (a)

Conclude by maximalityHence

a x a \in {Nil}_{*} R

for all

x

; semiprimeness of

R / {Nil}_{*} R

then forces

a \in {Nil}_{*} R

, the desired contradiction.

Local nilpotence is closed under everything that matters

Two closure facts carry the definition. First, if $𝔄$ is a locally nilpotent left ideal then $𝔄 R$ is locally nilpotent: a finite subset of $𝔄 R$ is built from finitely many pairs $(a_{i j}, r_{i j})$ , and the finitely many products $r_{p q} a_{i j}$ lie in $𝔄$ , where a uniform bound exists. Second, a sum of two locally nilpotent ideals is locally nilpotent: reduce modulo the second one, and note that the set of products of a fixed length from a finite set is again finite.

Key Results

Lemma(10.29)Utumi

Let $R$ be a ring satisfying the ACC on right annihilators ${ann}_{r} (a) = {x \in R : a x = 0}$ , $a \in R$ . Then:

every nil one-sided ideal $𝔄$ of $R$ is contained in ${Nil}_{*} R$ ;
every nonzero nil right (resp. left) ideal $𝔅$ contains a nonzero nilpotent right (resp. left) ideal.

In particular, if $R$ is moreover semiprime, then every nil one-sided ideal of $R$ is zero — so Köthe's Conjecture holds for such rings.

Proof

(1), right ideals. Suppose $𝔄$ is a nil right ideal with $𝔄 not \subseteq {Nil}_{*} R$ . By ACC choose $a \in 𝔄 ∖ {Nil}_{*} R$ with ${ann}_{r} (a)$ maximal among ${{ann}_{r} (a') : a' \in 𝔄 ∖ {Nil}_{*} R}$ . Fix $x \in R$ ; we claim $a x a \in {Nil}_{*} R$ .

We may assume $a x \neq 0$ . Since $a x \in 𝔄$ is nilpotent, there is a least $k$ with $(a x)^{k} = 0$ , and $k \geq 2$ . Put $y = x (a x)^{k - 2}$ . Then $a y = (a x)^{k - 1} \neq 0$ , so $y \notin {ann}_{r} (a)$ , while $(a x a) y = (a x)^{2} (a x)^{k - 2} = (a x)^{k} = 0$ , so $y \in {ann}_{r} (a x a)$ . Since $a z = 0$ implies $(a x a) z = a x (a z) = 0$ , we have ${ann}_{r} (a) ⊊ {ann}_{r} (a x a)$ . As $a x a \in 𝔄$ , maximality forces $a x a \in {Nil}_{*} R$ .

Thus $\bar{a} \bar{R} \bar{a} = 0$ in the semiprime ring $R / {Nil}_{*} R$ , whence $\bar{a} = 0$ , i.e. $a \in {Nil}_{*} R$ — a contradiction. If instead $𝔄$ is a nil left ideal, then for each $a' \in 𝔄$ the right ideal $a' R$ is nil (because $R a'$ is nil and $(a' r)^{n + 1} = a' (r a')^{n} r$ ), so $a' R \subseteq {Nil}_{*} R$ by the case already proved, and $𝔄 \subseteq {Nil}_{*} R$ .

(2). Among the nonzero elements of $𝔅$ choose $b$ with ${ann}_{r} (b)$ maximal. It suffices to prove $b x b = 0$ for all $x \in R$ , since then $(b R)^{2} = b R b R = 0$ and $(R b)^{2} = R b R b = 0$ , and $b R$ , $R b$ are nonzero. If $𝔅$ is a right ideal, the computation in (1) applies verbatim: $b x b \neq 0$ would give ${ann}_{r} (b x b) ⊋ {ann}_{r} (b)$ with $b x b \in 𝔅$ nonzero, contradicting maximality.

If $𝔅$ is a left ideal and $b x b \neq 0$ , then $x b \in 𝔅$ is nilpotent and nonzero; take $k$ least with $(x b)^{k} = 0$ , so $k \geq 2$ and $(x b)^{k - 1} \neq 0$ . Now $(x b)^{k - 1} = (x b)^{k - 2} x b \in R b \subseteq 𝔅$ is a nonzero element of $𝔅$ , and ${ann}_{r} (b) \subseteq {ann}_{r} ((x b)^{k - 1})$ because $b z = 0$ gives $(x b)^{k - 1} z = (x b)^{k - 2} x (b z) = 0$ . The inclusion is strict: $x b$ annihilates $(x b)^{k - 1}$ on the right but not $b$ , since $b (x b) = b x b \neq 0$ . This contradicts the maximality of ${ann}_{r} (b)$ .

Finally, if $R$ is semiprime then ${Nil}_{*} R = 0$ , so (1) makes every nil one-sided ideal zero.

Theorem(10.30)Levitzki

Let $R$ be a right noetherian ring. Then every nil one-sided ideal $𝔄$ of $R$ is nilpotent. Moreover ${Nil}_{*} R = {Nil}^{*} R$ , and this common ideal is the largest nilpotent right ideal and also the largest nilpotent left ideal of $R$ .

Proof

Right annihilators are right ideals, so $R$ satisfies ACC on them and $(10.29) (1)$ applies: every nil one-sided ideal lies in ${Nil}_{*} R$ . It therefore suffices to prove that ${Nil}_{*} R$ is nilpotent.

The sum of two nilpotent ideals is nilpotent, since $𝔄^{m} = 𝔅^{n} = 0$ gives $(𝔄 + 𝔅)^{m + n} = 0$ . By ACC there is a maximal nilpotent ideal $N$ , and maximality plus this closure property show that $N$ contains every nilpotent ideal. Consequently $R / N$ has no nonzero nilpotent ideal — if $(𝔐 / N)^{k} = 0$ then $𝔐^{k} \subseteq N$ and $𝔐$ is nilpotent, so $𝔐 \subseteq N$ — hence $R / N$ is semiprime by $(10.16)$ and $N \supseteq {Nil}_{*} R$ .

Conversely $N$ is nilpotent, hence nil, hence $N \subseteq {Nil}_{*} R$ by $(10.29) (1)$ . So ${Nil}_{*} R = N$ is nilpotent. Any nil one-sided ideal is now contained in the nilpotent ideal $N$ and is therefore itself nilpotent. Finally ${Nil}^{*} R$ is a nil ideal, so ${Nil}^{*} R \subseteq {Nil}_{*} R$ , and the reverse inclusion always holds; and every nilpotent right (or left) ideal, being nil, lies in $N$ .

Proposition(10.31)Closure properties of locally nilpotent ideals

Let $𝔄, 𝔅$ be locally nilpotent one-sided ideals of a ring $R$ . Then $R 𝔄 R$ , $R 𝔅 R$ and $𝔄 + 𝔅$ are locally nilpotent.

Proof

Assume $𝔄$ is a left ideal, so $R 𝔄 R = 𝔄 R$ . Take a finite subset ${b_{1}, \dots, b_{n}} \subseteq 𝔄 R$ and write $b_{i} = \sum_{j} a_{i j} r_{i j}$ with $a_{i j} \in 𝔄$ , $r_{i j} \in R$ , finitely many terms in each sum. The set $S = {r_{p q} a_{i j}}$ is a finite subset of $𝔄$ , because $𝔄$ is a left ideal. Choose $N$ so that every product of $N$ elements of $S$ vanishes. Expanding a product of $N + 1$ of the $b_{i}$ gives a sum of terms of the form $a_{i_{1} j_{1}} (r_{i_{1} j_{1}} a_{i_{2} j_{2}}) \dots (r_{i_{N} j_{N}} a_{i_{N + 1} j_{N + 1}}) r_{i_{N + 1} j_{N + 1}}$ , each containing a product of $N$ elements of $S$ , hence zero. So $𝔄 R$ is locally nilpotent.

For sums, first note the two-sided case. Let $ℭ, 𝔇$ be locally nilpotent ideals and ${c_{1}, \dots, c_{n}} \subseteq ℭ + 𝔇$ finite. In $R / 𝔇$ the images lie in the locally nilpotent left ideal $(ℭ + 𝔇) / 𝔇$ , so there is $N$ with every product of $N$ of the $c_{i}$ lying in $𝔇$ . The set $P$ of all products of exactly $N$ elements from ${c_{1}, \dots, c_{n}}$ is finite — at most $n^{N}$ elements — and contained in $𝔇$ , so there is $M$ with every product of $M$ elements of $P$ equal to zero. Then every product of $N M$ of the $c_{i}$ vanishes.

For general one-sided $𝔄, 𝔅$ , apply the first paragraph to get locally nilpotent two-sided ideals $R 𝔄 R \supseteq 𝔄$ and $R 𝔅 R \supseteq 𝔅$ ; the previous paragraph makes $R 𝔄 R + R 𝔅 R$ locally nilpotent, and $𝔄 + 𝔅$ is a subset of it. Subsets of locally nilpotent sets are locally nilpotent.

Corollary(10.32)The Levitzki radical and the four-term chain

The sum of all locally nilpotent ideals of $R$ is locally nilpotent; call it $L - rad R$ . It is the largest locally nilpotent ideal of $R$ and contains every locally nilpotent one-sided ideal. Moreover

{Nil}_{*} R \subseteq L - rad R \subseteq {Nil}^{*} R \subseteq rad R,

and all three inclusions are strict for suitable rings.

Proof

Local nilpotence is a finitary condition: any finite subset of a sum of ideals lies in the sum of finitely many of them, which is locally nilpotent by $(10.31)$ and induction. Hence the total sum is locally nilpotent and is the largest such ideal; a locally nilpotent one-sided ideal $𝔄$ satisfies $𝔄 \subseteq R 𝔄 R \subseteq L - rad R$ .

$L - rad R$ is nil, so $L - rad R \subseteq {Nil}^{*} R$ , and ${Nil}^{*} R \subseteq rad R$ by $(4.11)$ . For the first inclusion it suffices to show $L - rad R$ is a semiprime ideal, since ${Nil}_{*} R$ is the smallest one. Let $𝔄$ be an ideal with $𝔄^{2} \subseteq L - rad R$ and let ${a_{1}, \dots, a_{n}} \subseteq 𝔄$ be finite. The $n^{2}$ products $a_{i} a_{j}$ lie in $L - rad R$ , so some $N$ annihilates all $N$ -fold products of them; grouping consecutive factors in pairs shows every product of $2 N$ of the $a_{i}$ is zero. Hence $𝔄$ is locally nilpotent and $𝔄 \subseteq L - rad R$ .

Proof Techniques and Method

The reusable moves behind the proofs above.

Move 1

Maximal annihilator

Pick the element whose right annihilator is largest in a suitable family; then show the obstruction manufactures a strictly larger annihilator. This is Utumi's contribution and it replaces a chain condition on ideals by one on annihilators.

Move 2

Count words, not elements

Products of a fixed length from a finite set form a finite set. Applying local nilpotence to that second finite set is what makes sums of locally nilpotent ideals locally nilpotent.

Move 3

Pair up factors

To show an ideal $𝔄$ with $𝔄^{2}$ locally nilpotent is itself locally nilpotent, group a long product of elements of $𝔄$ into consecutive pairs. The exponent doubles; nothing else changes.

Move 1 is the reason Levitzki's Theorem is stated for right noetherian rings: the hypothesis is used only through ACC on right annihilators, which is weaker. Stating the weaker hypothesis, as $(10.29)$ does, is what makes the lemma applicable to Goldie-type situations where full noetherianity is unavailable.

Worked Example

A prime ring with nonzero Levitzki radical

The first inclusion of $(10.32)$ is strict, and the standard witness — an example of J. Ram, reproduced by Lam — is a twisted polynomial ring. Fix a field $k$ and let $A$ be the $k$ -algebra with generators $t_{i}$ for $i \in ℤ$ and relations

t_{i_{1}} t_{i_{2}} t_{i_{3}} = 0 whenever i_{1} < i_{2} < i_{3} is an arithmetic progression in ℤ .

(E.1)

A $k$ -basis of $A$ consists of the monomials $t_{i_{1}} \dots t_{i_{r}}$ with $i_{1} < \dots < i_{r}$ containing no three-term arithmetic progression.

Let $σ$ be the $k$ -automorphism of $A$ with $σ (t_{i}) = t_{i + 1}$ and put $R = A [x; σ]$ , so that $x a = σ (a) x$ . Then:

** $R$ is prime.** Suppose $f R g = 0$ with $f, g \neq 0$ , leading coefficients $a, b$ and $\deg f = n$ . Testing against the multipliers $x^{j}$ and reading off top-degree coefficients gives $a σ^{n + j} (b) = 0$ for all $j \geq 0$ . Taking $j$ large shifts the indices in $b$ far to the right, so the monomials of $a$ and of $σ^{n + j} (b)$ are disjoint and their products contain no three-term progression — hence the product is a nonzero basis element, a contradiction.
** $L - rad R \neq 0$ .** The right ideal $t_{0} x R$ is locally nilpotent. A product of $N$ elements $t_{0} x f_{i}$ produces a monomial in the $t$ 's whose indices form an increasing sequence of partial sums $r (i_{1}), r (i_{1}) + r (i_{2}), \dots$ with gaps bounded by $m = \max_{i} r (i)$ ; by a theorem of van der Waerden type on sequences with bounded gaps, any sufficiently long such sequence contains a three-term arithmetic progression, so the product vanishes.
Consequently ${Nil}_{*} R = 0$ while $L - rad R \supseteq t_{0} x R \neq 0$ .

The bound on $N$ depends only on $m$ , which is exactly what local nilpotence requires: a uniform exponent per finite subset, not per element.

The other two strict inclusions

$L - rad R ⊊ {Nil}^{*} R$ : adjoin an identity to Golod's finitely generated infinite-dimensional nil algebra. Its augmentation ideal is nil, hence inside ${Nil}^{*}$ , but a finitely generated locally nilpotent algebra is nilpotent, so it cannot lie in $L - rad$ .
${Nil}^{*} R ⊊ rad R$ : any commutative domain with a nonzero maximal ideal, for instance $k [[x]]$ or $ℤ_{(p)}$ , has all three nil-type radicals zero and nonzero Jacobson radical.

Comparison and Classification

Nilpotence conditions on a one-sided ideal
Condition	Meaning	Closed under sums?	Generates a two-sided ideal of the same type?
Nilpotent	$𝔄^{n} = 0$ for a fixed $n$	yes	yes
Locally nilpotent	uniform exponent per finite subset	yes, $(10.31)$	yes, $(10.31)$
Nil	each element nilpotent	unknown — Köthe	unknown — Köthe
Quasi-regular	$1 - a$ invertible for all $a$	yes	yes, inside $rad R$

The four radicals across classes of rings
	${Nil}_{*} = L - rad$ ?	$L - rad = {Nil}^{*}$ ?	${Nil}^{*} = rad$ ?
Commutative	yes	yes	no
Left artinian	yes	yes	yes
Right noetherian	yes	yes	no
Algebraic algebra over a field	no	no	yes
Skew polynomial ring $A [x; σ]$	no	no	no
General ring	no	no	no

The four radicals across classes of rings

The row for algebraic algebras uses Amitsur's theorem $(4.19)$ : there $rad R$ is nil, so it coincides with ${Nil}^{*} R$ , while the lower two radicals can still be smaller.

Relationship Map

Nilpotent one-sided idealLocally nilpotent one-sided idealNil one-sided idealInside

rad R

Each radical collects the ideals at one level of this hierarchy: ${Nil}_{*} R$ is generated by the nilpotent behaviour visible to prime ideals, $L - rad R$ by local nilpotence, ${Nil}^{*} R$ by nilness, $rad R$ by quasi-regularity. The chain $(10.32)$ is the ordering of those four conditions.

How each radical is characterised
Radical	Defined as	Extremal property
${Nil}_{*} R$	intersection of all prime ideals	smallest semiprime ideal
$L - rad R$	sum of all locally nilpotent ideals	largest locally nilpotent ideal
${Nil}^{*} R$	sum of all nil ideals	largest nil ideal
$rad R$	intersection of all maximal left ideals	largest left ideal $𝔘$ with $1 + 𝔘 \subseteq U (R)$

Note the alternation: the outer two are intersections, the inner two are sums. That is not a coincidence — the lower nilradical is defined by a class of quotients, the middle two by a class of subobjects.

Applications and Industry Use

Applications here means where this structure is used — inside mathematics and in the engineering and computing disciplines that consume it.

Goldie theory

Orders in semisimple rings

The ACC on right annihilators is one half of the Goldie conditions. Utumi's Lemma is what guarantees that a semiprime Goldie ring has no nonzero nil one-sided ideals, which is the first step towards constructing its classical ring of quotients.

Noetherian ring theory

Bounded nilpotence

Levitzki's Theorem is used constantly in the theory of noetherian rings: it converts the qualitative hypothesis nil into the quantitative one nilpotent of some index, which is what induction on ideal filtrations requires.

Group theory

Hirsch–Plotkin radical

The same finitary construction produces the largest locally nilpotent normal subgroup of a group. Results transfer between the two settings, most visibly for group rings and for Burnside-type problems.

Symbolic computation

Structure of algebras

For finite-dimensional algebras all four radicals coincide and computer algebra systems report a single radical. The distinctions on this page become relevant only for infinite-dimensional or finitely presented algebras, where no algorithm exists.

The honest summary: the Levitzki radical is a technical device introduced because the upper nilradical is not known to behave. Its practical value is that theorems proved with it are unconditional, whereas the corresponding statements for ${Nil}^{*}$ are hostage to Köthe's Conjecture.

Failure Modes and Common Mistakes

Do not use $(10.31)$ for nil one-sided ideals; the corresponding statements there are exactly Köthe's Conjecture.
Do not assume $L - rad (R / L - rad R) = 0$ without checking; it is true, but it needs the same finitary argument rather than being formal.
Do not conflate the two Levitzki results on this page: $(10.30)$ is about noetherian rings and nilpotence, while the radical $L - rad$ is defined for all rings.
Do not expect the Levitzki radical to be computable; like the upper nilradical it is inaccessible for finitely presented algebras.

Historical Notes and Lessons Learned

1930Köthe's problemKöthe asks whether a ring without nonzero nil ideals can have a nonzero nil one-sided ideal, setting the agenda that the Levitzki radical later sidesteps.
1939Levitzki proves his theoremLevitzki establishes that nil one-sided ideals of a right noetherian ring are nilpotent.
1950Publication, at lastWar and circumstance delayed publication by eleven years; the paper appeared in a little-read journal and carried a small error. Divinsky's history records that the flaw persisted even after Jacobson included the theorem in his 1956 book.
1956–64The radical zooAmitsur, Baer, Levitzki and Brown–McCoy define competing radicals; the four-term chain $(10.32)$ becomes the standard comparison, and Golod's nil algebras show the middle inclusion is strict.
1958–60Goldie's theoremsThe ACC on right annihilators enters mainstream ring theory as one of the Goldie conditions, giving Utumi's argument a permanent home.
1970s onwardUtumi's argument becomes standardThe maximal-annihilator technique is now the textbook route to Levitzki's Theorem, replacing the original transfinite constructions with a two-paragraph proof.

The methodological lesson repeats one from the Jacobson radical: when a property behaves badly, do not weaken the theorems — find the nearby property that behaves well and prove the theorems for that. Locally nilpotent is that property, and the price is one extra quantifier in the definition.

Quick Reference

Locally nilpotentevery finite subset has a uniform vanishing exponent

Levitzki radical

L - rad R =

sum of all locally nilpotent ideals

Containsevery locally nilpotent one-sided ideal

Chain

{Nil}_{*} R \subseteq L - rad R \subseteq {Nil}^{*} R \subseteq rad R

UtumiACC on

{ann}_{r} (a)

\Rightarrow

nil one-sided ideals lie in

{Nil}_{*} R

Levitzki

R

right noetherian

\Rightarrow

nil one-sided ideals are nilpotent

Noetherian collapse

{Nil}_{*} R = L - rad R = {Nil}^{*} R

, nilpotent

StrictnessRam's

A [x; σ]

; Golod's algebra;

k [[x]]

Which hypothesis gives which conclusion about a nil one-sided ideal $𝔄$
Hypothesis on $R$	Conclusion	Reference
none	$𝔄 \subseteq rad R$	(4.11)
$𝔄$ two-sided	$𝔄 \subseteq {Nil}^{*} R$	(10.26)
ACC on right annihilators	$𝔄 \subseteq {Nil}_{*} R$	(10.29)(1)
ACC on right annihilators, $R$ semiprime	$𝔄 = 0$	(10.29)
right noetherian	$𝔄$ nilpotent	(10.30)
$𝔄$ locally nilpotent	$𝔄 \subseteq L - rad R$	(10.31)

Frequently Asked Questions

Why introduce a third nil-type radical at all?

Because the upper nilradical is not known to absorb one-sided ideals, and the lower nilradical is often too small. The Levitzki radical is the largest radical in this family whose defining property is closed under the operations one actually performs — sums of one-sided ideals and generation of two-sided ideals — so theorems proved with it are unconditional.

Is the Levitzki radical of

R / L - rad R

zero?

Yes. If $𝔐 / L - rad R$ is a locally nilpotent ideal of the quotient, the finite-word argument used for sums shows $𝔐$ itself is locally nilpotent, hence contained in $L - rad R$ . All four radicals on this page have that idempotence property.

Does Levitzki's Theorem need both chain conditions?

No. Only ACC on right annihilators is used, and that is implied by right noetherianity. The stronger hypothesis appears in the statement because the conclusion — nilpotence rather than mere containment in ${Nil}_{*} R$ — needs ACC on ideals to produce a maximal nilpotent ideal.

Where exactly does Köthe's Conjecture enter this page?

Nowhere as an assumption, and that is the point. Every statement here is unconditional because locally nilpotent replaces nil. If Köthe's Conjecture were proved, the two middle terms of the chain would still differ — Golod's algebra separates them — but ${Nil}^{*} R$ would acquire the one-sided closure properties that $L - rad R$ already has.

How does the twisted polynomial example avoid contradicting

(10.18)

Because $A [x; σ]$ is not a polynomial ring in the sense of $(10.18)$ : the variable does not commute with the coefficients. The leading-coefficient identity becomes $a σ^{n} (b)$ rather than $a b$ , and it is precisely the shifting automorphism, together with van der Waerden's theorem on arithmetic progressions, that manufactures a locally nilpotent right ideal inside a prime ring.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §10, (10.29)–(10.33).
J. Levitzki, “On multiplicative systems”, Compositio Mathematica 8 (1950), 76–80.
N. J. Divinsky, Rings and Radicals, University of Toronto Press, 1965, Chapter 1.
M. B. Nathanson, “Arithmetic progressions contained in sequences with bounded gaps”, Canadian Mathematical Bulletin 23 (1980), 491–493.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapter I.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Write out the proof that the Levitzki radical is a Morita invariant, or explain what is known about it.
How do the Goldie conditions interact with Utumi's Lemma in the proof of Goldie's theorem?
Give the details of van der Waerden's theorem in the form used for Ram's example: bounded gaps force arithmetic progressions.
Compare the Levitzki radical of a ring with the Hirsch–Plotkin radical of a group, including for group rings.
Is there a ring with ${Nil}_{*} R ⊊ L - rad R ⊊ {Nil}^{*} R ⊊ rad R$ , all inclusions strict simultaneously?
What is the Levitzki radical of a group algebra $k G$ for $G$ locally finite of characteristic $p$ ?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

The maximal-annihilator technique

Local nilpotence is closed under everything that matters

06Key Results

07Proof Techniques and Method

Maximal annihilator

Count words, not elements

Pair up factors

08Worked Example

A prime ring with nonzero Levitzki radical

The other two strict inclusions

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Orders in semisimple rings

Bounded nilpotence

Hirsch–Plotkin radical

Structure of algebras

12Failure Modes and Common Mistakes

13Historical Notes and Lessons Learned

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Comparison and Classification

Relationship Map

Applications and Industry Use

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions