The Radicals of a Ring Compared

Q: Why are there four radicals rather than one?

Because *negligible* admits four inequivalent definitions once chain conditions are dropped. Under the descending chain condition all four coincide (10.27), which is why classical Wedderburn theory needed only one. The proliferation is the price of generality, and each radical is the correct one for a different class of questions.

Q: Is the Jacobson radical always nil?

No, and this is the single most common error. rad(k[[x]]) = (x) contains no nonzero nilpotent. The correct general statement is the reverse inclusion: every nil one-sided ideal lies in rad R (4.11). Nilness of rad R requires a hypothesis such as left artinian (4.12).

Q: What exactly does Köthe's conjecture assert?

That a ring with no nonzero nil two-sided ideal has no nonzero nil one-sided ideal. Equivalently, that the sum of two nil left ideals is always nil; equivalently, that I[T] ⊆ rad R[T] whenever I is a nil ideal. It has been verified for right noetherian rings, for algebraic algebras, for PI-algebras and for algebras of dimension smaller than the cardinality of the base field, and remains open in general.

Q: Why does the Levitzki radical exist as a separate object?

Because local nilpotence is a finitary property and therefore behaves well under the operations one needs: sums of locally nilpotent one-sided ideals and their two-sided closures are again locally nilpotent (10.31). The corresponding statements for nil one-sided ideals are exactly what Köthe's conjecture would supply. The Levitzki radical is the largest radical below Nil^* R with an unconditional one-sided theory.

Q: How do the radicals behave under matrix rings and polynomial rings?

Matrix rings are uniformly good: rad M_n(R) = M_n(rad R) and Nil_* M_n(R) = M_n(Nil_* R) (10.21), so all the radicals are Morita invariant. Polynomial rings are mixed: the lower nilradical extends cleanly by Amitsur–McCoy (10.19), the Jacobson radical of R[T] has the form N[T] for a nil ideal N by (5.10), and the upper nilradical case is open.

Q: Is the Brown–McCoy radical worth carrying?

Rarely, but it clarifies what rad R is not. It is the intersection of maximal two-sided ideals, always contains rad R, and coincides with it for commutative rings. For End(V_k) with _k V countably infinite the Jacobson radical is zero while the Brown–McCoy radical is the ideal of finite-rank endomorphisms.

Executive Summary

A ring has more than one radical because there is more than one way to say negligible. Four candidates are standard: negligible in the sense of the prime spectrum, in the sense of local nilpotence, in the sense of nilness, and in the sense of invertibility. They are nested, and every inclusion can be strict.

{Nil}_{*} R

Levitzki (R)

{Nil}^{*} R

rad R

The chain is $(10.32)$ . Chain conditions collapse it: in a left artinian ring all four agree $(10.27)$ , and in a right noetherian ring the first three agree $(10.30)$ . In general they do not, and the hardest unsolved question in the subject — Köthe's conjecture — asks whether the third radical already captures all one-sided nilness.

4Radicals compared

3Strict inclusions

$(10.32)$ The chain

1930Köthe's question

Overview

In a commutative ring the question does not arise: the nilradical is the intersection of the primes, is the largest nil ideal, and is locally nilpotent, so the first three radicals coincide $(10.27)$ . Only the Jacobson radical is genuinely different, and even then the difference is familiar — the nilradical of $ℤ_{(p)}$ is $0$ while its Jacobson radical is $p ℤ_{(p)}$ .

Noncommutatively the three nil-flavoured radicals separate, because the three ways of asserting nilness stop being equivalent. Nilpotent means one exponent works for everything. Locally nilpotent means one exponent works for each finite subset. Nil means one exponent works for each element. Each weakening is strict, and the corresponding radicals are strictly nested.

nilpotent ⟹ locally nilpotent ⟹ nil,

(R.1)

For a one-sided ideal. Neither implication reverses.

The individual radicals are developed on The Jacobson Radical, The Lower Nilradical, Upper Nilradical and Köthe's Conjecture and The Levitzki Radical. This page is the comparison.

Learning Objectives

Give the defining description of each of the four radicals in a single uniform format.
Prove ${Nil}_{*} R \subseteq {Nil}^{*} R \subseteq rad R$ and explain each step.
State the hypotheses under which the chain collapses, and by how much.
Name a ring separating each consecutive pair.
State Köthe's conjecture and two of its equivalent formulations.
Compute all four radicals for a concrete noncommutative ring.

Definitions

${Nil}_{*} R$: The lower nilradical, $\sqrt{(0)}$ : the intersection of all prime ideals of $R$ , equivalently the smallest semiprime ideal $(10.13)$ . Its elements are the strongly nilpotent elements.
$Levitzki (R)$: The Levitzki radical, written $L -rad R$ by Lam: the sum of all locally nilpotent ideals, which is itself locally nilpotent by $(10.31)$ and contains every locally nilpotent one-sided ideal.
${Nil}^{*} R$: The upper nilradical: the sum of all nil ideals, which is nil by $(10.25)$ , hence the largest nil ideal $(10.26)$ . Equivalently ${a \in R : (a) is nil}$ .
$rad R$: The Jacobson radical: the intersection of the maximal left ideals, equivalently the largest left ideal $𝔘$ with $1 + 𝔘 \subseteq U (R)$ .
Brown–McCoy radical: The intersection of all maximal two-sided ideals of $R$ . It always contains $rad R$ and coincides with it for commutative rings.

Definition(10.26), (10.31)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 ${s_{1}, \dots, s_{n}}$ vanishes; equivalently, every subring without identity generated by finitely many elements of $S$ is nilpotent.

All four radicals are two-sided ideals, and all four are semiprime ideals of R. None of the four is one-sided in its definition, but only the Jacobson radical has a natural one-sided description.

Core Concepts

Four notions of negligible

Each radical answers a different question
Radical	Negligible means	Built from	Detected by
${Nil}_{*} R$	Invisible to every prime quotient	Prime ideals	m-systems: $a \in {Nil}_{*} R$ iff every m-system containing $a$ meets $0$
$Levitzki (R)$	Uniformly nilpotent on finite sets	Locally nilpotent ideals	Finitely generated subrings being nilpotent
${Nil}^{*} R$	Every element nilpotent	Nil ideals	Element-by-element nilpotence
$rad R$	Invisible to every simple module	Maximal left ideals	$1 - x y z$ being a unit for all $x, z$

Why the one-sided problem is hard

${Nil}^{*} R$ is built from nil two-sided ideals only. The reason is $(10.25)$ : the sum of a nil left ideal and a nil ideal is a nil left ideal, so the sum of all nil ideals is nil — but whether the sum of two nil left ideals is nil is exactly what nobody knows. If it always were, the upper nilradical would contain every nil one-sided ideal, and this is Köthe's conjecture.

Key Results

Proposition(10.27)The outer inclusions

For any ring $R$ , ${Nil}_{*} R \subseteq {Nil}^{*} R \subseteq rad R$ . If $R$ is commutative, ${Nil}_{*} R = {Nil}^{*} R$ is the nilradical. If $R$ is left artinian, all three coincide with $rad R$ .

Proof

${Nil}_{*} R$ is a nil ideal — every element of $\sqrt{(0)}$ is nilpotent — hence contained in the largest nil ideal ${Nil}^{*} R$ . The second inclusion is $(4.11)$ : a nil one-sided ideal lies in $rad R$ , and ${Nil}^{*} R$ is nil. For the commutative case, the nilradical of a commutative ring is an ideal, is nil, and equals the intersection of the primes, so both radicals equal it.

Now suppose $R$ is left artinian. By $(4.12)$ , $rad R$ is nilpotent. The quotient $R / {Nil}_{*} R$ is semiprime, so it contains no nonzero nilpotent ideal; the image of the nilpotent ideal $rad R$ is therefore zero, giving $rad R \subseteq {Nil}_{*} R$ . Combined with the two inclusions already proved, all three radicals are equal.

{Nil}_{*} R \subseteq Levitzki (R) \subseteq {Nil}^{*} R \subseteq rad R

(10.32)

The full chain. The second inclusion holds because a locally nilpotent ideal is nil; the first because the Levitzki radical is a semiprime ideal, and every semiprime ideal contains the smallest one.

Theorem(10.30)Levitzki's Theorem

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

Proof

We prove the key step, that ${Nil}_{*} R$ is nilpotent; the passage from there to the full statement is $(10.29) (1)$ , Utumi's lemma, which says that under the ascending chain condition on right annihilators every nil one-sided ideal lies in ${Nil}_{*} R$ .

Since $R$ is right noetherian it satisfies the ACC on two-sided ideals, so the family of nilpotent ideals has a maximal member $N$ . First, $R / N$ has no nonzero nilpotent ideal: if $M \supseteq N$ with $(M / N)^{k} = 0$ then $M^{k} \subseteq N$ , and $N^{m} = 0$ gives $M^{k m} = 0$ , so $M$ is nilpotent and maximality forces $M = N$ . Hence $R / N$ is semiprime, so $N$ is a semiprime ideal and therefore $N \supseteq {Nil}_{*} R$ , the smallest semiprime ideal. Conversely $N$ is nilpotent, so $N$ is contained in every prime ideal and $N \subseteq {Nil}_{*} R$ . Thus ${Nil}_{*} R = N$ is nilpotent.

Theorem(10.19)Amitsur–McCoy

For any ring $R$ and any set $T$ of commuting indeterminates, ${Nil}_{*} (R [T]) = ({Nil}_{*} R) [T]$ .

Theorem(5.10)Amitsur's Theorem on

R [T]

Let $R$ be any ring, $S = R [T]$ , $J = rad S$ and $N = R \cap J$ . Then $N$ is a nil ideal of $R$ and $J = N [T]$ . In particular, if $R$ has no nonzero nil ideal then $R [T]$ is Jacobson semisimple.

The converse question — whether $I$ nil implies $I [T] \subseteq rad R [T]$ , which is $(5.12)$ — is open, and is equivalent to Köthe's conjecture.

Conjecture(10.28)Köthe

If ${Nil}^{*} R = 0$ then $R$ has no nonzero nil one-sided ideal. Two equivalent formulations: the sum of any two nil left ideals of any ring is nil; and for every nil ideal $I$ of every ring $R$ , $I [T] \subseteq rad R [T]$ .

The conjecture is known for right noetherian rings by $(10.30)$ , for algebras algebraic over a field by $(4.19)$ , for algebras $R$ over a field $k$ with $\dim_{k} R < | k |$ by Amitsur's theorem $(4.20)$ , and for PI-algebras. It is open in general.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Quotient by the candidate

To show $X \subseteq {Nil}_{*} R$ , show $R / {Nil}_{*} R$ is semiprime and that the image of $X$ is a nilpotent ideal there. Semiprimeness kills it.

Move 2

Maximal counterexample

Under a chain condition, choose an element maximal with respect to a failing property — Utumi's proof of $(10.29)$ maximises the right annihilator ${ann}_{r} (a)$ and derives a strictly larger one.

Move 3

Geometric series

$y$ nilpotent gives $(1 - x y)^{- 1} = 1 + x y + (x y)^{2} + \dots$ , a finite sum. This one line is the whole of $(4.11)$ and the reason nil implies radical.

Move 4

Finitary transfer

Local nilpotence is a property of finite subsets, so it passes to sums and to two-sided closures $(10.31)$ . Nilness is a property of single elements and does not.

Move 5

Adjoin indeterminates

Passing to $R [T]$ converts a nilness question into a radical question, which is how $(5.12)$ becomes a reformulation of Köthe.

Move 6

Semiprimeness as a certificate

To prove an ideal contains ${Nil}_{*} R$ it suffices to prove it is semiprime — the smallest semiprime ideal is ${Nil}_{*} R$ by definition. This is how $(10.32)$ gets its first inclusion.

Worked Example

All four radicals of a triangular ring

Fix a prime $p$ and let $ℤ_{(p)}$ denote the localisation of $ℤ$ at $p$ . Put

T = (\begin{matrix} ℤ_{(p)} & ℚ \\ 0 & ℚ \end{matrix}) = {(\begin{matrix} a & q \\ 0 & r \end{matrix}) : a \in ℤ_{(p)}, q, r \in ℚ} .

(E.1)

Let $N$ be the strictly upper triangular part, the set of matrices $(\begin{matrix} 0 & q \\ 0 & 0 \end{matrix})$ . It is a two-sided ideal and $N^{2} = 0$ , and the quotient is

T / N ≅ ℤ_{(p)} \times ℚ .

(E.2)

The three nil radicals

$T / N$ is a product of commutative domains, so it is reduced, hence semiprime with no nonzero nil ideal and no nonzero locally nilpotent ideal. Therefore each of the three lower radicals of $T$ is contained in $N$ ; and $N$ itself is nilpotent, hence contained in all three. So

{Nil}_{*} T = Levitzki (T) = {Nil}^{*} T = N .

(E.3)

The Jacobson radical

Since $N \subseteq rad T$ , the quotient formula $(4.6)$ gives $rad (T) / N = rad (T / N)$ . The radical of the product is the product of the radicals, $rad (ℤ_{(p)}) \times rad (ℚ) = p ℤ_{(p)} \times 0$ . Pulling back,

rad T = (\begin{matrix} p ℤ_{(p)} & ℚ \\ 0 & 0 \end{matrix}) ⊋ N = {Nil}^{*} T .

(E.4)

A clean noncommutative separation of the upper nilradical from the Jacobson radical.

Check the separation directly: the element $(\begin{matrix} p & 0 \\ 0 & 0 \end{matrix})$ lies in $rad T$ but is not nilpotent, since its $n$ -th power is $(\begin{matrix} p^{n} & 0 \\ 0 & 0 \end{matrix}) \neq 0$ . So $rad T$ is not even nil, let alone nilpotent, and $T$ is not artinian on either side.

Comparison and Classification

Separating examples for each inclusion
Strict inclusion	Witness	Values
${Nil}_{*} R ⊊ Levitzki (R)$	J. Ram's twisted polynomial ring $A [x; σ]$ over the $k$ -algebra $A$ generated by $t_{i}$ $(i \in ℤ)$ with $t_{i_{1}} t_{i_{2}} t_{i_{3}} = 0$ for arithmetic progressions $i_{1} < i_{2} < i_{3}$	$R$ is prime, so ${Nil}_{*} R = 0$ , yet the right ideal $t_{0} x R$ is locally nilpotent, so $Levitzki (R) \neq 0$
$Levitzki (R) ⊊ {Nil}^{*} R$	$R = k \oplus G$ , the unitalisation of a Golod finitely generated nil algebra $G$ that is not nilpotent	${Nil}^{*} R = G$ ; $G$ is finitely generated and not nilpotent, so it is not locally nilpotent and $Levitzki (R) ⊊ G$
${Nil}^{*} R ⊊ rad R$	Any commutative local domain with nonzero maximal ideal, e.g. $k [[x]]$ or $ℤ_{(p)}$	The three lower radicals are $0$ ; $rad R$ is the maximal ideal
$rad R ⊊$ Brown–McCoy	$End (V_{k})$ for $V$ of countably infinite dimension over a division ring $k$	$rad = 0$ since the ring is left primitive; the unique maximal two-sided ideal is the ideal of finite-rank endomorphisms

When do the radicals coincide?
	${Nil}_{*} = Levitzki$	$Levitzki = {Nil}^{*}$	${Nil}^{*} = rad$	All four equal
Commutative	yes	yes	no	no
Left artinian	yes	yes	yes	yes
Right noetherian	yes	yes	no	no
Algebraic algebra over a field	partial	partial	yes	partial
$\dim_{k} R < \| k \|$	partial	partial	yes	partial
Semiprime	yes	partial	partial	no
General ring	no	no	no	no

When do the radicals coincide?

In the algebraic and small-dimension rows the equality of the upper nilradical with the Jacobson radical is (4.19) and (4.20); the lower equalities then follow only when a further hypothesis such as a chain condition is present.

Values on familiar rings
Ring	${Nil}_{*}$	${Nil}^{*}$	$rad$
$ℤ$	$0$	$0$	$0$
$ℤ / 12 ℤ$	$(6)$	$(6)$	$(6)$
$k [[x]]$	$0$	$0$	$(x)$
$k [x]$ , $k$ a field	$0$	$0$	$0$
$T_{n} (k)$ upper triangular	strictly upper triangular	same	same
$M_{n} (R)$	$M_{n} ({Nil}_{*} R)$	$M_{n} ({Nil}^{*} R)$	$M_{n} (rad R)$
$R [T]$	$({Nil}_{*} R) [T]$	not known in general	$N [T]$ with $N = R \cap rad R [T]$ nil

Relationship Map

Semiprime ideals of $R$ — All four radicals are semiprime ideals; they are ordered by which quotient property they force
- ${Nil}_{*} R$
  - $R / {Nil}_{*} R$ is semiprime
  - Smallest semiprime ideal; contained in every prime
- $Levitzki (R)$
  - $R / Levitzki (R)$ has no nonzero locally nilpotent one-sided ideal
  - Behaves well under sums and two-sided closure $(10.31)$
- ${Nil}^{*} R$
  - $R / {Nil}^{*} R$ has no nonzero nil ideal
  - Whether it has no nonzero nil one-sided ideal is Köthe's conjecture
- $rad R$
  - $R / rad R$ is semiprimitive
  - Intersection of the left primitive ideals $(11.5)$

Each radical is idempotent in the sense that applying it to the quotient returns zero, and each satisfies $Rad (R / I) = Rad (R) / I$ for any ideal $I \subseteq Rad (R)$ . That common formal behaviour is what makes them radicals in the axiomatic sense of Kurosh and Amitsur.

R

R / {Nil}_{*} R

semiprime

R / {Nil}^{*} R

no nil ideals

R / rad R

semiprimitive

Computational Notes

Computational notes cover algorithms, cost and library behaviour rather than manufacturing process.

For a finite-dimensional algebra $A$ over a field, all four radicals coincide (such an algebra is artinian), and the single computation is the standard radical algorithm: a trace-form nullspace in characteristic $0$ , costing $O (n^{3})$ field operations for $\dim_{k} A = n$ , and the Friedl–Rónyai iteration in characteristic $p$ .
For a commutative noetherian ring presented by generators and relations, ${Nil}_{*} R = {Nil}^{*} R$ is the radical of the zero ideal and is computed by Gröbner-basis radical algorithms in Macaulay2 or Singular.
For a general finitely presented noncommutative ring none of the four radicals is computable: the word problem is already undecidable, so membership tests cannot exist uniformly.
The Levitzki radical has no direct algorithm even in favourable cases, because local nilpotence quantifies over all finite subsets. In practice it is computed only when a theorem identifies it with one of its neighbours.

Failure Modes and Common Mistakes

Do not use the phrase the radical without saying which one, especially in sources predating 1945, where it usually means the largest nilpotent ideal.
Do not assume ${Nil}^{*} (R [T]) = ({Nil}^{*} R) [T]$ : the analogous statement for the lower nilradical is the Amitsur–McCoy theorem $(10.19)$ , but the upper version is open and equivalent to Köthe.
Do not conflate semiprime with semiprimitive: the first says ${Nil}_{*} R = 0$ , the second says $rad R = 0$ , and the second is strictly stronger.

Historical Notes and Lessons Learned

1908–1927The nilpotent radicalWedderburn and Artin work with the largest nilpotent ideal. It suffices only because the chain conditions in force make all radicals coincide.
1930Köthe's questionKöthe asks whether a ring with no nonzero nil ideals can have a nonzero nil one-sided ideal. It is still open.
1939 (published 1950)Levitzki's TheoremNil one-sided ideals in a right noetherian ring are nilpotent. Publication was delayed by the war, and the proof reached wide circulation only through Jacobson's 1956 book.
1945Jacobson's radicalThe radical is redefined by its action on simple modules, freeing the theory from chain conditions and creating the gap between nilness and radicality.
1943–1956Baer, Amitsur, KuroshBaer's radical ideal paper isolates the prime radical; the upper nil and locally nilpotent radicals follow, and the axiomatic theory of radical classes in the sense of Kurosh and Amitsur takes shape.
1964Golod–ShafarevichA finitely generated nil algebra that is not nilpotent, settling the Kurosh problem negatively and separating nil from locally nilpotent.
laterUtumi's argumentA short annihilator-maximisation proof, reproduced by Lam as (10.29), that yields Levitzki's Theorem and extends Köthe's conjecture to all rings with the ascending chain condition on right annihilators.

The lesson is that the multiplicity of radicals is not a defect of the theory but a record of which finiteness hypotheses have been discarded. Under the descending chain condition there is only one radical; each weakening of that condition splits it further.

Quick Reference

Chain

{Nil}_{*} R \subseteq Levitzki (R) \subseteq {Nil}^{*} R \subseteq rad R

${Nil}_{*} R$ Intersection of the primes; smallest semiprime ideal

$Levitzki (R)$ Largest locally nilpotent ideal

${Nil}^{*} R$ Largest nil ideal

$rad R$ Intersection of maximal left ideals

Left artinianAll four equal and nilpotent

Right noetherianFirst three equal and nilpotent

(10.30)

CommutativeFirst three equal the nilradical

Köthe

{Nil}^{*} R = 0 \Rightarrow

no nonzero nil one-sided ideal — open

Which radical to use
Question	Radical	Reason
Is the ring a subdirect product of prime rings?	${Nil}_{*} R$	Zero exactly when $R$ is semiprime
Can I do induction on finitely many elements?	$Levitzki (R)$	Local nilpotence is finitary
Is every element nilpotent?	${Nil}^{*} R$	Largest nil ideal
Is the ring detected by its simple modules?	$rad R$	Zero exactly when $R$ is semiprimitive
Does $R$ have a maximal two-sided ideal missing my element?	Brown–McCoy	Intersection of maximal two-sided ideals

Frequently Asked Questions

Why are there four radicals rather than one?

Because negligible admits four inequivalent definitions once chain conditions are dropped. Under the descending chain condition all four coincide $(10.27)$ , which is why classical Wedderburn theory needed only one. The proliferation is the price of generality, and each radical is the correct one for a different class of questions.

Is the Jacobson radical always nil?

No, and this is the single most common error. $rad (k [[x]]) = (x)$ contains no nonzero nilpotent. The correct general statement is the reverse inclusion: every nil one-sided ideal lies in $rad R$ $(4.11)$ . Nilness of $rad R$ requires a hypothesis such as left artinian $(4.12)$ .

What exactly does Köthe's conjecture assert?

That a ring with no nonzero nil two-sided ideal has no nonzero nil one-sided ideal. Equivalently, that the sum of two nil left ideals is always nil; equivalently, that $I [T] \subseteq rad R [T]$ whenever $I$ is a nil ideal. It has been verified for right noetherian rings, for algebraic algebras, for PI-algebras and for algebras of dimension smaller than the cardinality of the base field, and remains open in general.

Why does the Levitzki radical exist as a separate object?

Because local nilpotence is a finitary property and therefore behaves well under the operations one needs: sums of locally nilpotent one-sided ideals and their two-sided closures are again locally nilpotent $(10.31)$ . The corresponding statements for nil one-sided ideals are exactly what Köthe's conjecture would supply. The Levitzki radical is the largest radical below ${Nil}^{*} R$ with an unconditional one-sided theory.

How do the radicals behave under matrix rings and polynomial rings?

Matrix rings are uniformly good: $rad M_{n} (R) = M_{n} (rad R)$ and ${Nil}_{*} M_{n} (R) = M_{n} ({Nil}_{*} R)$ $(10.21)$ , so all the radicals are Morita invariant. Polynomial rings are mixed: the lower nilradical extends cleanly by Amitsur–McCoy $(10.19)$ , the Jacobson radical of $R [T]$ has the form $N [T]$ for a nil ideal $N$ by $(5.10)$ , and the upper nilradical case is open.

Is the Brown–McCoy radical worth carrying?

Rarely, but it clarifies what $rad R$ is not. It is the intersection of maximal two-sided ideals, always contains $rad R$ , and coincides with it for commutative rings. For $End (V_{k})$ with $\dim_{k} V$ countably infinite the Jacobson radical is zero while the Brown–McCoy radical is the ideal of finite-rank endomorphisms.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §4 (pp. 50–69), §5 (pp. 70–81) and §10 (pp. 163–181), especially (10.25)–(10.32).
N. J. Divinsky, Rings and Radicals, Mathematical Expositions 14, University of Toronto Press, 1965.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapters I and X.
E. S. Golod, “On nil-algebras and residually finite p-groups”, Izvestiya Akademii Nauk SSSR, Seriya Matematicheskaya 28 (1964), 273–276; and E. S. Golod and I. R. Shafarevich, “On the class field tower”, same volume, 261–272.
S. A. Amitsur, “Radicals of polynomial rings”, Canadian Journal of Mathematics 8 (1956), 355–361.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Give a self-contained account of Golod's construction of a finitely generated nil algebra that is not nilpotent.
What is currently known about Köthe's conjecture for graded rings and for rings with involution?
Verify that each of the four radicals is a radical in the sense of Kurosh–Amitsur, and identify the corresponding radical class.
Is the Levitzki radical of $R [T]$ equal to $Levitzki (R) [T]$ ?
How do the four radicals compare for a group algebra $k G$ as the characteristic of $k$ and the group $G$ vary?
Work through Utumi's proof of $(10.29)$ and identify exactly where the ascending chain condition on right annihilators is used.
Which of the four radicals is preserved by faithfully flat descent, and which are not?

Radicals Compared

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Four notions of negligible

Why the one-sided problem is hard

Key Results

Proof Techniques and Method

Quotient by the candidate

Maximal counterexample

Geometric series

Finitary transfer

Adjoin indeterminates

Semiprimeness as a certificate

Worked Example

All four radicals of a triangular ring

The three nil radicals

The Jacobson radical

Comparison and Classification

Relationship Map

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Four notions of negligible

Why the one-sided problem is hard

06Key Results

07Proof Techniques and Method

Quotient by the candidate

Maximal counterexample

Geometric series

Finitary transfer

Adjoin indeterminates

Semiprimeness as a certificate

08Worked Example

All four radicals of a triangular ring

The three nil radicals

The Jacobson radical

09Comparison and Classification

10Relationship Map

11Computational Notes

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

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions