Open Problems: Köthe, Kaplansky and the Group Ring Conjectures

Q: Why is Köthe's conjecture so much harder than it looks?

Because it is equivalent to statements that already look impossible to attack. Krempa showed it is equivalent to the nilness of M_2(N) for every nil ring N; nil rings can be as wild as Golod–Shafarevich and Smoktunowicz have shown, and nobody can compute their matrix rings. Every proof strategy that works — annihilator chains, algebraicity, polynomial identities — imposes a finiteness that the general case lacks.

Q: Does Gardam's counterexample make the group-ring problems easier or harder?

Harder, in the sense that the strongest of the four is now known to be false, so the implications U R and U D can no longer be used as a route to the others. It also shows that the unique product property, not torsion-freeness, is the operative hypothesis, which narrows where a proof of the zero-divisor conjecture could come from.

Q: If kG has a nontrivial unit, must it have a zero divisor?

No — and this is exactly the point. Gardam's group P is torsion-free and elementary amenable, so kP is known to be a domain, hence has no zero divisors and no nontrivial idempotents, yet it has a nontrivial unit over F_2. The unit and zero-divisor questions are genuinely independent.

Q: Is the Levitzki radical a substitute for the upper nilradical?

For many purposes, yes. By (10.31) locally nilpotent one-sided ideals sum to locally nilpotent ideals, which is precisely the additivity that nil one-sided ideals are not known to have. That is why L-rad R exists as a well-defined largest locally nilpotent ideal while the analogous 'largest nil one-sided' object does not.

Q: Are there results in this collection that assume Köthe's conjecture?

None. Lam proves everything unconditionally and marks clearly where a positive answer would strengthen a statement — in particular around (10.27) and (10.28). Any result you meet elsewhere that asserts additivity of nil one-sided ideals should be checked for a hidden hypothesis.

Q: What would a counterexample to Köthe's conjecture have to look like?

It would be a ring R with Nil^* R = 0 and a nonzero nil left ideal. By the known cases it could not be one-sided noetherian, could not satisfy ACC on one-sided annihilators, could not be an algebraic algebra over a field, could not satisfy a polynomial identity, and could not be an algebra of dimension smaller than the cardinality of its base field. That is a very narrow corridor.

Executive Summary

Two of the oldest questions in ring theory are still open. Köthe's conjecture $(10.28)$ , asked around 1930, would say that nil one-sided ideals behave as well as nil two-sided ideals. Kaplansky's zero-divisor conjecture would say that a group algebra of a torsion-free group over a field is a domain. Both have resisted every general technique.

The neighbouring unit conjecture was refuted in 2021 by Giles Gardam, seventy years after it was posed, using a computer search over the group algebra of a torsion-free crystallographic group in characteristic $2$ . That is the only one of these problems to have been settled, and its fall did not disturb the others.

1930Köthe asked

OpenKöthe status

2021Unit conjecture refuted

4Group-ring problems

Overview

The open problems here are not isolated curiosities; each marks a place where a proof technique runs out. Köthe's conjecture marks the boundary of what can be done with nil ideals in the absence of a chain condition. The group-ring problems mark the boundary of what can be extracted from the combinatorics of supports in $k G$ .

Two things make Köthe's conjecture unusually stubborn. It is equivalent to a family of statements that look much weaker — that $M_{2} (N)$ is nil for every nil ring $N$ , for instance — so any proof must handle all of them at once. And the natural counterexample searches run into the deep constructions of Golod–Shafarevich and Smoktunowicz, which produce nil algebras exotic enough that nobody can compute their one-sided ideal structure.

See Upper Nilradical and Köthe's Conjecture and The Group Ring Problems for the developed treatments; this page collects them and records their status.

Learning Objectives

State $(10.28)$ and prove the equivalence of the formulations $(10.28 a)$ and $(10.28 b)$ .
Recall Utumi's Lemma $(10.29)$ and derive Levitzki's Theorem $(10.30)$ from it.
List the classes of rings in which Köthe's conjecture is a theorem.
State Problems U, R, D and J of $(6.16)$ – $(6.19)$ with their standing hypotheses on $k$ and $G$ .
Reproduce the implication diagram $(6.20)$ and say which arrows are easy and which are deep.
Describe the current status of each problem and what a solution would require.

Definitions

Nil one-sided ideal: A left or right ideal every element of which is nilpotent. Nilpotency indices may be unbounded; a nil ideal need not be nilpotent.
${Nil}^{*} R$: The upper nilradical: the sum of all nil two-sided ideals of $R$ . By $(10.25)$ this sum is itself nil, so ${Nil}^{*} R$ is the largest nil ideal, and ${Nil}^{*} (R / {Nil}^{*} R) = 0$ .
${Nil}_{*} R$: The lower nilradical, or Baer radical: the intersection of all prime ideals, equivalently $\sqrt{(0)}$ in the sense of $(10.13)$ . Always contained in ${Nil}^{*} R$ .
$L - rad R$: The Levitzki radical: the largest locally nilpotent ideal. It satisfies ${Nil}_{*} R \subseteq L - rad R \subseteq {Nil}^{*} R \subseteq rad R$ , which is $(10.32)$ together with $(10.27)$ .
Torsion-free group: A group with no element of finite order other than the identity. This hypothesis is necessary in Problem D: if $x$ has order $n > 1$ then $(x - 1) (x^{n - 1} + \dots + x + 1) = 0$ in $k G$ .
Trivial unit: An element $a g$ with $a \in U (k)$ and $g \in G$ . Every group ring has these; the unit problem asks whether it has any others.

Core Concepts

Why nil one-sided ideals are hard

Two-sided nil ideals add well: by $(10.25)$ , if $𝔄$ is a nil left ideal and $𝔅$ a nil ideal, then $𝔄 + 𝔅$ is nil. That is enough to make ${Nil}^{*} R$ a nil ideal. What is missing is the case where both summands are merely one-sided.

The obstruction is concrete. For $a, b$ nilpotent in a noncommutative ring, $a + b$ need not be nilpotent — $E_{12} + E_{21}$ in $M_{2} (k)$ squares to the identity. Nil left ideals are more constrained than arbitrary sets of nilpotents, but no argument has yet converted that constraint into additivity.

The group-ring problems as a combinatorial question

For $α, β \in k G$ nonzero, the support of $α β$ is contained in the product of the supports. Proving $α β \neq 0$ amounts to finding a group element with a unique factorisation as a product of a support element of $α$ and one of $β$ — no cancellation can then occur. Groups in which this is always possible are the unique product groups, and for them Problems U, R and D all have affirmative answers.

The catch is that not every torsion-free group is a unique product group. The Promislow group — a torsion-free crystallographic group containing $ℤ^{3}$ with index $4$ , also called the Hantzsche–Wendt group — is a counterexample, and it is precisely where Gardam found his unit.

Ordered groups: where the answers are known

If $G$ carries a total order compatible with multiplication, the highest and lowest terms of a product cannot cancel, and $(6.29)$ gives everything at once. Right-orderable groups, torsion-free nilpotent groups and free groups are all covered, which is why the problems are only difficult for groups far from being orderable.

Key Results

Conjecture(10.28)Köthe

If $R$ is a ring with ${Nil}^{*} R = 0$ , then $R$ has no nonzero nil one-sided ideal.

Two standard reformulations: $(10.28 a)$ every nil left or right ideal of any ring $R$ is contained in ${Nil}^{*} R$ ; and $(10.28 b)$ the sum of two nil left ideals of any ring is nil.

Proof

The three formulations are equivalent. First a lemma: for $x \in R$ , the left ideal $R x$ is nil if and only if the right ideal $x R$ is nil. Suppose $R x$ is nil and take $a \in R$ . Then $a x \in R x$ , so $(a x)^{n} = 0$ for some $n$ , and

(x a)^{n + 1} = x (a x)^{n} a = 0 .

(O.1)

So every element of $x R$ is nilpotent. The converse is symmetric.

*The sum $S$ of all nil left ideals is a two-sided ideal, and equals the sum of all nil right ideals.* Let $𝔄$ be a nil left ideal, $a \in 𝔄$ and $r \in R$ . Then $R a \subseteq 𝔄$ is nil, so by the lemma $a R$ is nil; hence $a r$ is nilpotent and $R a r$ is a nil left ideal. Therefore $𝔄 r \subseteq S$ and $S$ is a right ideal, so an ideal. Moreover $a R$ nil gives $a \in a R \subseteq S'$ , where $S'$ is the sum of all nil right ideals; hence $S \subseteq S'$ , and symmetrically $S' \subseteq S$ .

** $(10.28 b) \Rightarrow (10.28 a)$ .** Assume sums of two nil left ideals are nil. By induction every finite sum of nil left ideals is nil, and every element of $S$ lies in such a finite sum, so $S$ is nil. Being a nil two-sided ideal, $S \subseteq {Nil}^{*} R$ , which is $(10.28 a)$ .

** $(10.28 a) \Rightarrow (10.28)$ .** Immediate: if ${Nil}^{*} R = 0$ then every nil one-sided ideal is contained in $0$ .

** $(10.28) \Rightarrow (10.28 a)$ .** Let $R$ be arbitrary and $\bar{R} = R / {Nil}^{*} R$ . By $(10.25)$ the sum of nil ideals is nil, so ${Nil}^{*} R$ is a nil ideal and ${Nil}^{*} \bar{R} = 0$ . Apply $(10.28)$ to $\bar{R}$ : it has no nonzero nil one-sided ideal. The image of a nil left ideal $𝔄 \subseteq R$ is a nil left ideal of $\bar{R}$ , hence zero, so $𝔄 \subseteq {Nil}^{*} R$ .

** $(10.28 a) \Rightarrow (10.28 b)$ .** If both summands lie in the nil ideal ${Nil}^{*} R$ , their sum does too, and ${Nil}^{*} R$ is nil.

Lemma(10.29)Utumi

Let $R$ satisfy the ascending chain condition on right annihilators ${ann}_{r} (a) = {x \in R : a x = 0}$ , $a \in R$ . Then (1) every nil one-sided ideal of $R$ is contained in ${Nil}_{*} R$ ; and (2) every nonzero nil right (respectively left) ideal contains a nonzero nilpotent right (respectively left) ideal. In particular, if $R$ is also semiprime then every nil one-sided ideal is zero.

Theorem(10.30)Levitzki

Let $R$ be a right noetherian ring. Then every nil one-sided ideal of $R$ is nilpotent, ${Nil}_{*} R = {Nil}^{*} R$ , and this common ideal is the largest nilpotent right ideal and the largest nilpotent left ideal. Consequently Köthe's conjecture holds for right noetherian rings.

Proof

A right noetherian ring satisfies ACC on right annihilators, so $(10.29) (1)$ applies and every nil one-sided ideal lies in ${Nil}_{*} R$ . It therefore suffices to prove ${Nil}_{*} R$ nilpotent. Since $R$ is right noetherian, the set of nilpotent ideals has a maximal member $N$ . In $R / N$ there is no nonzero nilpotent ideal — if $𝔄 / N$ were one, $𝔄$ would be a nilpotent ideal strictly containing $N$ — so $R / N$ is semiprime and hence ${Nil}_{*} R \subseteq N$ . Combined with $N \subseteq {Nil}_{*} R$ , we get ${Nil}_{*} R = N$ , which is nilpotent.

Problem(6.16)–(6.19)The four group-ring problems

Let $k$ be a domain and $G$ a torsion-free group. Then:

Problem U $(6.16)$ : are all units of $k G$ trivial?
Problem R $(6.17)$ : is $k G$ reduced, that is, free of nonzero nilpotent elements?
Problem D $(6.18)$ : is $k G$ a domain?
Problem J $(6.19)$ : if $G \neq {1}$ , is $k G$ $J$ -semisimple?

The known implications $(6.20)$ are U $\Rightarrow$ R, U $\Rightarrow$ J, and R $\Leftrightarrow$ D. Of these, D $\Rightarrow$ R is trivial and R $\Rightarrow$ D is deep, resting on the $Δ$ -methods of $(6.22)$ – $(6.28)$ .

Proposition(6.21)Consequences of triviality of units

Let $k \neq 0$ be a ring and $G \neq {1}$ a group such that $A = k G$ has only trivial units. Then (1) if $k$ is reduced and $G$ has no element of order $2$ , $A$ is reduced; and (2) $A$ is $J$ -semisimple except in the single case $| k | = | G | = 2$ .

Note that neither part assumes $G$ torsion-free — the hypotheses are much weaker than in $(6.16)$ – $(6.19)$ , which is what makes the implications of $(6.20)$ usable.

Theorem(6.29)Ordered groups settle all four

Let $k$ be a domain and let $(G, <)$ be an ordered group, that is, a group with a total order satisfying $g < h \Rightarrow a g b < a h b$ for all $a, b$ . Then $A = k G$ has only trivial units and is a domain; if $G \neq {1}$ , $A$ is $J$ -semisimple.

The proof compares the largest and smallest support elements of a product, which cannot cancel. This settles Problems U, R, D and J for all orderable groups — a class including free groups, torsion-free nilpotent groups and free abelian groups.

Remark—Problem U has a negative answer

In 2021 Gardam exhibited a nontrivial unit in $𝔽_{2} P$ , where $P$ is the Promislow group: a torsion-free crystallographic group of Hirsch length $3$ that is not a unique product group. So Problem U fails in general, and with it Kaplansky's unit conjecture. Murray subsequently produced counterexamples over $𝔽_{p}$ for every prime $p$ .

This does not settle Problems R, D or J: the implications in $(6.20)$ run from U, not to it. Indeed $k P$ is still known to be a domain, since $P$ is torsion-free and elementary amenable.

Proof Techniques and Method

How these problems are attacked, and where each technique stops.

Technique 1

Annihilator chains

Utumi's argument $(10.29)$ picks an element with a maximal right annihilator and derives a contradiction. It converts a chain condition into control of nil one-sided ideals, and gives Köthe for right noetherian rings. It stops as soon as ACC on annihilators fails.

Technique 2

Algebraicity and cardinality

$(4.19)$ and $(4.20)$ give ${Nil}^{*} R = rad R$ for algebras algebraic over $k$ or of dimension smaller than $| k |$ . Any nil one-sided ideal lies in $rad R$ by $(4.11)$ , so Köthe follows. It stops for countably generated algebras over countable fields.

Technique 3

Support combinatorics

For group rings, control the support of a product. Orderability $(6.29)$ and the unique product property both make cancellation impossible at an extreme element. It stops at torsion-free groups that are not unique product groups.

Technique 4

$Δ$ -methods

Pass to the FC subgroup $Δ (G)$ of elements with finitely many conjugates, project onto $k Δ (G)$ , and use that $Δ (G)$ is abelian when torsion-free $(6.24)$ . This is how R $\Rightarrow$ D is proved $(6.28)$ .

Technique 5

K-theory and analysis

For the zero-divisor conjecture, Kropholler, Linnell and Moody handled torsion-free elementary amenable groups by K-theoretic means, and operator-algebraic methods handle groups satisfying Baum–Connes. These reach far beyond orderability but not to all torsion-free groups.

Technique 6

Machine search

Gardam's counterexample was found by encoding the unit equation over $𝔽_{2}$ as a satisfiability problem on a bounded support. This is the only technique on this list that has settled one of the problems.

Frameworks and Models

The equivalent forms of Köthe's conjecture are worth tabulating because a proof of any one is a proof of all.

Köthe's conjecture — ${Nil}^{*} R = 0 \Rightarrow$ no nonzero nil one-sided ideal
- Ideal-theoretic forms
  - every nil one-sided ideal lies in ${Nil}^{*} R$ , $(10.28 a)$
  - the sum of two nil left ideals is nil, $(10.28 b)$
- Matrix forms
  - $M_{2} (N)$ is nil for every nil ring $N$
  - ${Nil}^{*} (M_{n} (R)) = M_{n} ({Nil}^{*} R)$ for all $R$ and $n$
- Polynomial form (Krempa)
  - $N [x]$ is Jacobson radical for every nil ring $N$
- Known cases
  - right or left noetherian rings, $(10.30)$
  - rings with ACC on one-sided annihilators, $(10.29)$
  - algebras algebraic over a field, $(4.19)$
  - algebras with $\dim_{k} R < | k |$ , $(4.20)$
  - PI-algebras over any field

Process and Workflow

You need Köthe's conjecture for a particular ring. What can you use?

The ring is one-sided noetherianLevitzki's Theorem

(10.30)

gives it outright, and more: every nil one-sided ideal is nilpotent and

{Nil}_{*} R = {Nil}^{*} R

Only ACC on one-sided annihilatorsUtumi's Lemma

(10.29)

still applies and puts every nil one-sided ideal inside

{Nil}_{*} R

It is an algebra over a fieldIf it is algebraic over

k

, use

(4.19)

; if

\dim_{k} R < | k |

as cardinals, use

(4.20)

. Either gives

{Nil}^{*} R = rad R

, and

(4.11)

finishes the argument.

It satisfies a polynomial identityKöthe holds for PI-algebras over any field. For finitely generated PI-algebras Braun's theorem gives the stronger conclusion that the radical is nilpotent.

None of theseDo not assume the conjecture. Restate your result conditionally, or work with

{Nil}_{*} R

and

L - rad R

, which are unconditionally well behaved by

(10.31)

–

(10.32)

Fix the exact statementNote the standing hypotheses: a domain of coefficients, a torsion-free group, or a ring with an identity. Half the confusion in this area comes from statements quoted without them.

Locate it in the implication diagramDetermine whether the statement follows from another problem or implies it —

(6.20)

for group rings, the equivalence tree for Köthe.

Check the special classesOrderable, unique product, elementary amenable for groups; noetherian, algebraic, PI for rings. Most concrete cases fall into one.

If not, state your result conditionallyA theorem proved modulo Köthe's conjecture is a legitimate contribution provided the dependence is stated explicitly.

Comparison and Classification

Status of the principal open problems
Problem	Statement	Status	Known cases
Köthe $(10.28)$	${Nil}^{*} R = 0$ implies no nonzero nil one-sided ideal	Open since c. 1930	noetherian $(10.30)$ ; algebraic algebras $(4.19)$ ; $\dim_{k} R < \| k \|$ $(4.20)$ ; PI-algebras
Problem U $(6.16)$	$k G$ has only trivial units, $G$ torsion-free, $k$ a domain	False (Gardam, 2021)	ordered groups; unique product groups
Problem R $(6.17)$	$k G$ is reduced	Open	same, plus everything covered by D
Problem D $(6.18)$	$k G$ is a domain	Open	orderable; unique product; torsion-free elementary amenable
Problem J $(6.19)$	$k G$ is $J$ -semisimple for $G \neq {1}$	Open	ordered groups $(6.29)$ ; consequences of U where U holds
Semiprimitivity over $ℚ$	is $ℚ G$ $J$ -semisimple for every group $G$ ?	Open	nonalgebraic extensions of $ℚ$ , by Amitsur $(6.12)$
Semiprimitivity over $𝔽_{p}$	is $𝔽_{p} G$ $J$ -semisimple for every $p'$ -group $G$ ?	Open	nonalgebraic extensions of $𝔽_{p}$ , by Passman $(6.15)$
Idempotent conjecture	$k G$ has no idempotents besides $0, 1$ for $G$ torsion-free	Open	follows from D; known for large classes via Baum–Connes

Which technique reaches which problem
	Köthe	Problem D	Problem J	Unit conjecture
Annihilator chains	yes	no	no	no
Algebraicity / cardinality	yes	no	partial	no
Support combinatorics	no	yes	yes	yes
$Δ$ -methods	no	yes	partial	no
K-theory and operator algebras	no	yes	no	no
Machine search	no	no	no	yes

Which technique reaches which problem

Relationship Map

The group-ring implications of $(6.20)$ , with the strength of each arrow marked.

Problem UProblem RProblem D

U ⟹ R is Proposition (6.21)(1); R ⟹ D is the deep direction, proved via Δ-methods in (6.28); D ⟹ R is immediate. U also implies J by (6.21)(2), with the single exception |k| = |G| = 2.

The radical chain places Köthe's conjecture exactly. Every inclusion below is a theorem; Köthe's conjecture is the assertion that the third term absorbs all nil one-sided ideals, not merely all nil two-sided ones.

{Nil}_{*} R

L - rad R

{Nil}^{*} R

rad R

All groupsno problem is settled here

Torsion-freethe standing hypothesis of Problems U, R, D

Unique productU, R, D all affirmative

Right-orderableunique product holds

Orderableall four settled by

(6.29)

Torsion-free elementary amenable

k G

known to be a domain (Kropholler–Linnell–Moody); contains the Promislow group

P

Failure Modes and Common Mistakes

Do not drop torsion-freeness from Problems U, R and D: if $x \in G$ has order $n > 1$ then $(x - 1) (x^{n - 1} + \dots + 1) = 0$ , so $k G$ has zero divisors outright.
Do not forget the exception in $(6.21) (2)$ : when $| k | = | G | = 2$ , $k G$ has only trivial units but $rad (k G) \neq 0$ .
Do not conclude from $(6.12)$ and $(6.15)$ that the $J$ -semisimplicity problem is solved. Those theorems need $K$ nonalgebraic over the prime field; the prime-field case itself is the hard one.

Best Practices

When writing a result that would need Köthe's conjecture, restate it in terms of ${Nil}_{*} R$ or $L - rad R$ , both of which are unconditionally well behaved.
Quote the status of an open problem with a date; several statuses on this page have changed within living memory.
Verify that a group in your hands really is torsion-free, and say whether it is a unique product group — that single fact decides most group-ring questions.
Treat computer-assisted refutations as first-class mathematics but record the verification: Gardam's unit is checked by a finite multiplication in $𝔽_{2} P$ and is independently reproducible.
Do not weaken a conjecture silently when a special case is all you can prove; state the special case.

Historical Notes and Lessons Learned

c. 1930KötheGottfried Köthe conjectures that a ring with no nonzero nil ideal has no nonzero nil one-sided ideal, in the course of extending Wedderburn theory beyond chain conditions.
1939 / 1950LevitzkiLevitzki proves that in a right noetherian ring every nil one-sided ideal is nilpotent, giving the first substantial class where Köthe holds. Publication was delayed by the war until 1950.
1940s–50sKaplanskyKaplansky formulates the unit, zero-divisor and idempotent conjectures for group rings of torsion-free groups, and proves Köthe's conjecture for PI-algebras.
1964Golod and ShafarevichFinitely generated infinite-dimensional nil algebras are constructed, ending hopes that nil algebras are automatically tame and settling the Kurosh problem negatively.
1972KrempaKöthe's conjecture is shown equivalent to the statement that the polynomial ring over any nil ring is Jacobson radical, and to the nilness of 2 by 2 matrices over nil rings.
1988Kropholler, Linnell and MoodyThe zero-divisor conjecture is proved for torsion-free elementary amenable groups by K-theoretic methods, greatly enlarging the class of groups where Problem D is known.
2000–2002SmoktunowiczA nil ring whose polynomial ring is not nil refutes Amitsur's conjecture; a simple nil ring follows two years later. Köthe's conjecture survives both.
2021GardamA nontrivial unit is found in the group algebra of the Promislow group over the field of two elements, refuting Kaplansky's unit conjecture; Murray extends the construction to all positive characteristics.

The lesson of the last thirty years is that these problems fall to construction rather than to theory. Golod–Shafarevich, Smoktunowicz and Gardam all produced objects; no general structural insight has yet touched Köthe's conjecture, and its resolution is likely to look the same way.

Quick Reference

Köthe

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

no nonzero nil one-sided ideal — open

Equivalentsum of two nil left ideals is nil

Equivalent

M_{2} (N)

nil for every nil ring

N

Equivalent

N [x]

Jacobson radical for every nil ring

N

(Krempa)

Known fornoetherian

(10.30)

, algebraic algebras

(4.19)

, PI-algebras

Problem Urefuted 2021 (Gardam)

Problems R, D, Jopen; R and D are equivalent

Settled forordered groups

(6.29)

, unique product groups

Hardest case of J

ℚ G

and

𝔽_{p} G

over the prime fields

Never assumethat a nil one-sided ideal lies in

{Nil}^{*} R

Unconditional substitutes for the conjectural statements
Conjectural statement	Unconditional replacement	Reference
nil left ideals sum to a nil ideal	locally nilpotent one-sided ideals sum to a locally nilpotent ideal	$(10.31)$
${Nil}^{} (M_{n} (R)) = M_{n} ({Nil}^{} R)$	${Nil}_{} (M_{n} (R)) = M_{n} ({Nil}_{} R)$	$(10.21)$
nil one-sided ideals are nilpotent	true when R is right noetherian	$(10.30)$
${Nil}^{*} R = rad R$	true for algebras algebraic over $k$	$(4.19)$
$k G$ is a domain, $G$ torsion-free	true for $(G, <)$ ordered	$(6.29)$

Frequently Asked Questions

Why is Köthe's conjecture so much harder than it looks?

Because it is equivalent to statements that already look impossible to attack. Krempa showed it is equivalent to the nilness of $M_{2} (N)$ for every nil ring $N$ ; nil rings can be as wild as Golod–Shafarevich and Smoktunowicz have shown, and nobody can compute their matrix rings. Every proof strategy that works — annihilator chains, algebraicity, polynomial identities — imposes a finiteness that the general case lacks.

Does Gardam's counterexample make the group-ring problems easier or harder?

Harder, in the sense that the strongest of the four is now known to be false, so the implications U $\Rightarrow$ R and U $\Rightarrow$ D can no longer be used as a route to the others. It also shows that the unique product property, not torsion-freeness, is the operative hypothesis, which narrows where a proof of the zero-divisor conjecture could come from.

k G

has a nontrivial unit, must it have a zero divisor?

No — and this is exactly the point. Gardam's group $P$ is torsion-free and elementary amenable, so $k P$ is known to be a domain, hence has no zero divisors and no nontrivial idempotents, yet it has a nontrivial unit over $𝔽_{2}$ . The unit and zero-divisor questions are genuinely independent.

Is the Levitzki radical a substitute for the upper nilradical?

For many purposes, yes. By $(10.31)$ locally nilpotent one-sided ideals sum to locally nilpotent ideals, which is precisely the additivity that nil one-sided ideals are not known to have. That is why $L - rad R$ exists as a well-defined largest locally nilpotent ideal while the analogous 'largest nil one-sided' object does not.

Are there results in this collection that assume Köthe's conjecture?

None. Lam proves everything unconditionally and marks clearly where a positive answer would strengthen a statement — in particular around $(10.27)$ and $(10.28)$ . Any result you meet elsewhere that asserts additivity of nil one-sided ideals should be checked for a hidden hypothesis.

What would a counterexample to Köthe's conjecture have to look like?

It would be a ring $R$ with ${Nil}^{*} R = 0$ and a nonzero nil left ideal. By the known cases it could not be one-sided noetherian, could not satisfy ACC on one-sided annihilators, could not be an algebraic algebra over a field, could not satisfy a polynomial identity, and could not be an algebra of dimension smaller than the cardinality of its base field. That is a very narrow corridor.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §6 (pp. 82–106) and §10 (pp. 163–181).
D. S. Passman, The Algebraic Structure of Group Rings, Wiley-Interscience, 1977.
N. J. Divinsky, Rings and Radicals, University of Toronto Press, 1965.
A. Smoktunowicz, “Polynomial rings over nil rings need not be nil”, Journal of Algebra 233 (2000), 427–436.
G. Gardam, “A counterexample to the unit conjecture for group rings”, Annals of Mathematics 194 (2021), 967–979.
P. H. Kropholler, P. A. Linnell and J. A. Moody, “Applications of a new K-theoretic theorem to soluble group rings”, Proceedings of the American Mathematical Society 104 (1988), 675–684.

AI Suggested Questions

Write out Krempa's proof that Kothe's conjecture is equivalent to the nilness of 2 by 2 matrices over nil rings.
Which torsion-free groups are known not to be unique product groups, and how were they constructed?
Summarise the encoding Gardam used to search for a nontrivial unit and what made the search feasible.
For which classes of groups is the idempotent conjecture known, and how does Baum-Connes enter?
Is Kothe's conjecture known for graded rings or for monomial algebras, and what does the grading buy?
What is the current state of the semiprimitivity problem for rational group algebras?
Compare the Kurosh problem and Kothe's conjecture: why did one fall to Golod-Shafarevich and the other not?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why nil one-sided ideals are hard

The group-ring problems as a combinatorial question

Ordered groups: where the answers are known

06Key Results

07Proof Techniques and Method

Annihilator chains

Algebraicity and cardinality

Support combinatorics

Δ-methods

K-theory and analysis

Machine search

08Frameworks and Models

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Failure Modes and Common Mistakes

13Best Practices

14Historical Notes and Lessons Learned

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

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Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

$Δ$ -methods

Frameworks and Models

Process and Workflow

Comparison and Classification

Relationship Map

Failure Modes and Common Mistakes

Best Practices

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions