Counterexamples Catalogue | KEVOS® Mathematics

Q: Is there a single ring that refutes most of these implications at once?

End(V_k) for V infinite-dimensional over a division ring comes closest. It is left primitive but not simple, von Neumann regular but not semisimple, semiprimitive with no chain condition on either side, and not Dedekind-finite, so left-invertible fails to mean invertible. What it cannot do is break sidedness — it is symmetric enough that one needs a triangular ring for that.

Q: Does the failure of Krull–Schmidt over Z[-5] contradict (19.21)?

No. (19.21) requires each indecomposable summand to have a *local* endomorphism ring. For the ideal I = (2, 1+-5) we have End_R(I) ≅ R, which is not local, so the theorem does not apply. Over a left artinian ring, by contrast, indecomposable finitely generated modules do have local endomorphism rings, and Krull–Schmidt holds.

Q: Are Gardam's units relevant to the zero-divisor conjecture?

Not directly. Gardam's counterexample is to Kaplansky's *unit* conjecture, over the field of two elements, for the torsion-free group usually called the Promislow or Hantzsche–Wendt group. That group is virtually free abelian, and its group algebra is still known to be a domain, so the zero-divisor conjecture survives intact for it and remains open in general.

Executive Summary

Every hypothesis in this subject is there because some ring violates the theorem without it. This page is the register of those rings. A dozen objects do almost all the work: the triangular rings of $(1.22)$ – $(1.24)$ , the power series ring $k [[x]]$ , the endomorphism ring of an infinite-dimensional vector space, the Weyl algebra, the group algebra $𝔽_{p} C_{p}$ , the quaternions $ℍ$ , and Bass's infinite-matrix ring of $(23.22)$ .

Each entry below states the implication that fails, the witness, and the calculation. Where the claim is not known to be false — Köthe's conjecture is the important case — that is recorded too, since an absent counterexample is itself information.

17Failed implications

5Construction families

$(1.24)$ The workhorse

1Open (Köthe)

Overview

Noncommutative ring theory has an unusually high density of near-misses. Statements that are true for commutative rings, or true under a chain condition, or true on one side, fail in general — and the failures are not pathological curiosities but the reason the theory is organised as it is.

Three themes account for most of them. Sidedness: a left hypothesis rarely implies its right analogue. Finiteness: dropping a chain condition breaks nilpotence and composition series. Local versus global: a property holding for each element need not hold for the ideal it generates.

Read alongside Left–Right Symmetry, which classifies which notions are side-neutral, and Ring Class Hierarchy, whose strict containments are certified by the witnesses collected here.

Learning Objectives

Produce a ring that is left artinian but not right artinian, and verify both halves.
Explain why $rad (k [[x]])$ is not nil and why that does not contradict $(4.12)$ .
Exhibit a nil ideal that is not nilpotent, and say which hypothesis of $(4.13)$ it violates.
Give a group algebra that is not semisimple and locate the failure in Maschke's hypotheses.
Construct a failure of Krull–Schmidt and identify the missing hypothesis of $(19.21)$ .
Recognise which of these failures are theorems and which are open questions.

Definitions

Five constructions generate almost every witness on this page.

Triangular ring $A = (\begin{matrix} R & M \\ 0 & S \end{matrix})$: For rings $R, S$ and an $(R, S)$ -bimodule $M$ . By $(1.22)$ , $A$ is left noetherian (resp. artinian) iff $R$ and $S$ are and $_{R} M$ is; right noetherian (resp. artinian) iff $R$ and $S$ are and $M_{S}$ is. Choosing $M$ small on one side and large on the other manufactures sidedness failures to order.
$End (V_{k})$ , $\dim V_{k}$ infinite: For $V$ a right vector space over a division ring $k$ . Left primitive, von Neumann regular, semiprimitive, but neither noetherian nor artinian, not simple, and not Dedekind-finite.
Power series and localisations: $k [[x]]$ and $ℤ_{(p)}$ are commutative local domains: the radical is the maximal ideal, nonzero and containing no nilpotent element at all.
Group algebras in the modular case: $k G$ with $char k = p$ dividing $| G |$ . Never semisimple; $𝔽_{p} C_{p} ≅ 𝔽_{p} [u] / (u^{p})$ is the smallest instance.
Free and skew constructions: $k ⟨ x, y ⟩$ , $k [x; σ]$ with $σ$ not surjective, and the Weyl algebra $A_{1} (k)$ supply simple rings without chain conditions and one-sided noetherian rings.

Throughout, rings have an identity and modules are unital; ‘artinian’ and ‘noetherian’ are always qualified by a side.

Core Concepts

What a good counterexample has to do

A witness must satisfy every hypothesis of the claimed implication, not merely most. The commonest defective counterexample fails silently because it violates an unstated standing assumption — that the ring has an identity, that modules are unital, or that a chain condition is on the correct side.

Minimality is worth having

$ℤ$ is a better witness than an exotic construction for semiprimitive does not imply semisimple, because its radical and its lack of DCC are both immediate. The same applies to $ℤ / 6 ℤ$ for the failure of idempotent lifting and to $ℍ$ for the failure of root counting. Reach for exotic constructions only when the elementary ones cannot satisfy the hypotheses.

Sidedness is manufactured, not found

Theorem $(1.22)$ turns the construction of one-sided witnesses into bookkeeping: choose $M$ finitely generated on one side and not on the other. With $S \subseteq R$ fields and $\dim_{S} R$ infinite, $(1.24)$ gives left artinian and left noetherian but neither right condition. With $S = ℤ$ , $R = ℚ$ , $(1.23)$ gives left noetherian, not right noetherian, and neither artinian.

Choose

R

S

Choose bimodule

M

Read off

_{R} M

Read off

M_{S}

Key Results

The principal entries, each with the hypotheses that make the failure genuine. The first result is not itself a counterexample but the machine that produces several of them.

Theorem(1.22)Chain conditions for a triangular ring

Let $R$ and $S$ be rings, let $M$ be an $(R, S)$ -bimodule, and let $A = (\begin{matrix} R & M \\ 0 & S \end{matrix})$ with matrix addition and multiplication. Then $A$ is left noetherian if and only if $R$ and $S$ are left noetherian and $M$ is noetherian as a left $R$ -module; $A$ is right noetherian if and only if $R$ and $S$ are right noetherian and $M$ is noetherian as a right $S$ -module. The same two statements hold with noetherian replaced throughout by artinian.

The point is that the two sides are governed by two genuinely different modules, $_{R} M$ and $M_{S}$ , and nothing forces them to behave alike.

Counterexample(1.24)Left artinian does not imply right artinian

Let $S \subseteq R$ be fields with $\dim_{S} R$ infinite — for instance $S = ℚ$ , $R = ℝ$ — and let

A = (\begin{matrix} R & R \\ 0 & S \end{matrix}) = {(\begin{matrix} a & m \\ 0 & s \end{matrix}) : a, m \in ℝ, s \in ℚ} .

(C.1)

Then $A$ is left artinian and left noetherian, and is neither right artinian nor right noetherian.

Proof

Left side. Put $N = (\begin{matrix} 0 & ℝ \\ 0 & ℚ \end{matrix})$ and $P = (\begin{matrix} 0 & ℝ \\ 0 & 0 \end{matrix})$ . A direct multiplication check shows both are left ideals, and $A ⊋ N ⊋ P ⊋ 0$ .

The factors are $A / N ≅ ℝ$ with $A$ acting through the entry $a$ ; $N / P ≅ ℚ$ with $A$ acting through $s$ ; and $P ≅ ℝ$ with $A$ acting through $a$ . Each is a one-dimensional vector space over a field acting through a surjection from $A$ , hence a simple left $A$ -module. So $_{A} A$ has a composition series of length $3$ , and by $(1.19)$ it is both artinian and noetherian.

Right side. For a $ℚ$ -subspace $V \subseteq ℝ$ , the set $(\begin{matrix} 0 & V \\ 0 & 0 \end{matrix})$ is a right ideal, because

(\begin{matrix} 0 & v \\ 0 & 0 \end{matrix}) (\begin{matrix} a & m \\ 0 & s \end{matrix}) = (\begin{matrix} 0 & v s \\ 0 & 0 \end{matrix}), s \in ℚ .

(C.2)

Since $\dim_{ℚ} ℝ$ is infinite, choose a $ℚ$ -independent sequence $α_{1}, α_{2}, \dots$ and set $V_{n} = {span}_{ℚ} {α_{n}, α_{n + 1}, \dots}$ and $W_{n} = {span}_{ℚ} {α_{1}, \dots, α_{n}}$ . The corresponding right ideals form a strictly descending chain and a strictly ascending chain respectively, so $A$ is neither right artinian nor right noetherian.

Note where the asymmetry lives: in $(C . 2)$ the right action multiplies $v$ by a rational scalar, whereas the left action multiplies by a real one. The bimodule $M = ℝ$ is one-dimensional as a left $ℝ$ -module and infinite-dimensional as a right $ℚ$ -module, which is exactly the criterion of $(1.22)$ .

Counterexample—The Jacobson radical need not be nil

$R = k [[x]]$ for a field $k$ is a commutative local domain with $rad R = (x) \neq 0$ , and $R$ has no nonzero nilpotent element whatsoever. The same holds for $R = ℤ_{(p)}$ with $rad R = p ℤ_{(p)}$ . This does not contradict $(4.12)$ : neither ring is artinian on either side, since $(x) ⊋ (x^{2}) ⊋ \dots$ never terminates.

Counterexample—A nil ideal that is not nilpotent

Let $k$ be a field and

A = k [x_{1}, x_{2}, x_{3}, \dots] / (x_{1}^{2}, x_{2}^{3}, x_{3}^{4}, \dots),

(C.3)

and let $N = (x_{1}, x_{2}, \dots)$ be the ideal generated by the images of the variables. Every element of $N$ involves finitely many variables, each nilpotent, and $A$ is commutative, so every element of $N$ is nilpotent: $N$ is nil. But $x_{n}^{n} \neq 0$ in $A$ , so $N^{n} \neq 0$ for every $n$ and $N$ is not nilpotent. Consistent with $(4.13)$ , which requires $A$ to be left artinian — and $A$ is not.

Counterexample(19.21)Krull–Schmidt fails without local endomorphism rings

Let $R = ℤ [\sqrt{- 5}]$ , a Dedekind domain with class number $2$ , and let $I = (2, 1 + \sqrt{- 5})$ , a non-principal ideal with $I^{2} = (2)$ . Steinitz's theorem for Dedekind domains gives $I \oplus J ≅ R \oplus I J$ for fractional ideals $I, J$ , so

I \oplus I ≅ R \oplus I^{2} = R \oplus (2) ≅ R \oplus R,

(C.4)

while $I not ≅ R$ because $I$ is not principal. Rank-one torsion-free modules over a domain are indecomposable, so a finitely generated module here has two genuinely different decompositions into indecomposables. The hypothesis that fails in $(19.21)$ is that the endomorphism rings be local: ${End}_{R} (I) ≅ R$ , which is a domain with two maximal ideals above $2$ and $3$ — not local.

Counterexample(23.22)Right perfect does not imply left perfect, or semiprimary

Let $k$ be a field, let $J$ be the set of $ℕ \times ℕ$ matrices over $k$ with finitely many nonzero entries, all strictly above the diagonal, and put $R = k \cdot 1 + J$ . Then $J$ is an ideal with $R / J ≅ k$ , so $R$ is local and $J = rad R$ .

$J$ is right T-nilpotent: given $a_{1}, a_{2}, \dots \in J$ , the matrix $a_{1}$ annihilates all but finitely many basis vectors, and each further factor lowers the surviving span, so $a_{n} \dots a_{2} a_{1} = 0$ for large $n$ . Hence $R$ is right perfect. But $J$ is not left T-nilpotent: with $a_{i} = E_{i, i + 1}$ one has $a_{1} a_{2} \dots a_{n} = E_{1, n + 1} \neq 0$ for every $n$ . And $J$ is not nilpotent, so $R$ is right perfect without being semiprimary.

Counterexample—Nilpotent elements need not form an ideal

In $M_{2} (k)$ the matrix units $E_{12}$ and $E_{21}$ both square to zero, but $(E_{12} + E_{21})^{2} = I \neq 0$ . Indeed $rad M_{2} (k) = 0$ while nilpotent elements abound. In a commutative ring the nilpotents do form an ideal; noncommutatively they need not, which is the whole reason the nilradicals of §10 must be defined by ideals rather than by elements.

Proof Techniques and Method

How these witnesses are built, and which construction to reach for.

Factory 1

Triangular rings

Break sidedness. $(1.22)$ reduces every chain condition on $(\begin{matrix} R & M \\ 0 & S \end{matrix})$ to conditions on $_{R} M$ and $M_{S}$ separately, so any asymmetry in the bimodule becomes an asymmetry in the ring.

Factory 2

Infinite-dimensional endomorphism rings

Break finiteness. $End (V_{k})$ is primitive and von Neumann regular but has no chain conditions, is not simple, and satisfies $R ≅ R \oplus R$ as left modules, killing Dedekind finiteness and invariant basis number in one object.

Factory 3

Complete local rings

Break nilpotence. $k [[x]]$ and $ℤ_{(p)}$ have a large radical with no nilpotents at all, refuting every claim that reads radical, therefore nil.

Factory 4

Modular group algebras

Break semisimplicity. $k G$ with $char k ∣ | G |$ is never semisimple, and $𝔽_{p} C_{p} ≅ 𝔽_{p} [u] / (u^{p})$ is local with $rad = (u)$ of nilpotency index exactly $p$ .

Factory 5

Skew and free constructions

Break commutative intuition. $A_{1} (k)$ in characteristic zero is simple but not artinian; $k [x; σ]$ with $σ$ not surjective is left but not right noetherian $(1.25)$ ; free algebras are left primitive $(11.23)$ , $(11.27)$ .

When none of the five suffices, the next move is a transfinite or limit construction: Bergman's left primitive ring that is not right primitive, Bass's ring of $(23.22)$ , and Smoktunowicz's nil ring with non-nil polynomial ring are all of this type. They are hard to build and easy to quote.

Worked Example

The triangular ring in full

Take $A = (\begin{matrix} ℝ & ℝ \\ 0 & ℚ \end{matrix})$ from $(C . 1)$ and compute everything.

Radical. The ideal $P = (\begin{matrix} 0 & ℝ \\ 0 & 0 \end{matrix})$ satisfies $P^{2} = 0$ , and

A / P ≅ ℝ \times ℚ,

(C.5)

a product of two fields, hence semisimple; so $rad A = P$ .

Simple modules. Exactly two, up to isomorphism: $S_{1} = ℝ$ with $(\begin{matrix} a & m \\ 0 & s \end{matrix})$ acting as $a$ , and $S_{2} = ℚ$ acting as $s$ . Their annihilators intersect in $P$ , confirming $(4.2)$ .

Verification against the unit criterion. For $m \in ℝ$ ,

{(\begin{matrix} 1 & m \\ 0 & 1 \end{matrix})}^{- 1} = (\begin{matrix} 1 & - m \\ 0 & 1 \end{matrix}),

(C.6)

so $1 + rad A \subseteq U (A)$ as $(4.5)$ requires.

Chain conditions. Left: composition length $3$ , so left artinian and left noetherian. Right: $(\begin{matrix} 0 & V_{n} \\ 0 & 0 \end{matrix})$ for $V_{n} = {span}_{ℚ} {α_{n}, α_{n + 1}, \dots}$ is a strictly descending chain of right ideals; the spans of initial segments give a strictly ascending one.

A second computation: roots over the quaternions

In $ℍ [x]$ the polynomial $x^{2} + 1$ of degree $2$ has infinitely many roots: any $q = b i + c j + d k$ with $b^{2} + c^{2} + d^{2} = 1$ satisfies $q^{2} = - 1$ . The set of roots is a whole conjugacy class — a two-sphere — which is precisely the phenomenon analysed in Vanishing Polynomials and the Niven–Jacobson Theorem.

Comparison and Classification

Failed implications and their witnesses
False implication	Witness	Why it fails
left artinian ⟹ right artinian	$(\begin{matrix} ℝ & ℝ \\ 0 & ℚ \end{matrix})$ , $(1.24)$	$M$ is $1$ -dimensional on the left, infinite on the right
left noetherian ⟹ right noetherian	$(\begin{matrix} ℚ & ℚ \\ 0 & ℤ \end{matrix})$ , $(1.23)$ ; also $ℤ ⟨ x, y ⟩ / (y^{2}, y x)$ , $(1.26)$	$ℚ$ is not a finitely generated $ℤ$ -module
left noetherian ⟹ left artinian	$ℤ$ , $k [x]$	no DCC on ideals
$rad R$ is nil	$k [[x]]$ , $ℤ_{(p)}$	domain with nonzero radical
nil ideal ⟹ nilpotent ideal	$k [x_{1}, x_{2}, \dots] / (x_{1}^{2}, x_{2}^{3}, \dots)$	unbounded nilpotency indices
nilpotent elements form an ideal	$M_{2} (k)$	$(E_{12} + E_{21})^{2} = I$
semiprimitive ⟹ semisimple	$ℤ$	no DCC; needs $(4.14)$
prime ⟹ primitive	$ℤ$	commutative primitive means field, $(11.8)$
primitive ⟹ simple	$End (V_{k})$ , $\dim V_{k}$ infinite	finite-rank maps form a proper ideal
simple ⟹ artinian	Weyl algebra $A_{1} (k)$ , $char k = 0$	simple noetherian domain of infinite dimension
left primitive ⟹ right primitive	Bergman's ring (1965); also Jategaonkar's	primitivity is genuinely one-sided
von Neumann regular ⟹ semisimple	$End (V_{k})$ , $\dim V_{k}$ infinite	$(4.25)$ needs a chain condition as well
left-invertible ⟹ invertible	$End (V_{k})$ , shift operator	not Dedekind-finite
idempotents lift modulo any ideal	$ℤ \to ℤ / 6 ℤ$	$(6)$ is not nil; contrast $(21.28)$
Krull–Schmidt for f.g. modules	$ℤ [\sqrt{- 5}]$ , $I \oplus I ≅ R \oplus R$	endomorphism rings not local
semiperfect ⟹ perfect	$ℤ_{(p)}$	$p ℤ_{(p)}$ is not T-nilpotent
right perfect ⟹ left perfect	Bass's ring, $(23.22)$	T-nilpotency is one-sided
$k G$ semisimple for all finite $G$	$𝔽_{p} C_{p} ≅ 𝔽_{p} [u] / (u^{p})$	$\| G \|$ not invertible; Maschke $(6.1)$ fails
$\deg f = n$ bounds the roots of $f$	$x^{2} + 1$ over $ℍ$	roots form a conjugacy class
$rad (R [T]) = (rad R) [T]$	$R = k [[x]]$	$rad R [T] = 0$ since $R$ is a domain
$rad A$ is the radical of the trace form	$A = 𝔽_{p} C_{p}$	trace form vanishes identically in characteristic $p$
$rad (A \otimes_{k} K) = 0$ when $rad A = 0$	$A = K = 𝔽_{p} (t^{1 / p})$ over $k = 𝔽_{p} (t)$	inseparable extension

Properties of the four workhorse rings
	$k [[x]]$	$End (V_{k})$	$(\begin{matrix} ℝ & ℝ \\ 0 & ℚ \end{matrix})$	$A_{1} (k)$ , char $0$
Left artinian	no	no	yes	no
Right artinian	no	no	no	no
Left noetherian	yes	no	yes	yes
$rad R = 0$	no	yes	no	yes
Simple	no	no	no	yes
Left primitive	no	yes	no	yes
Domain	yes	no	no	yes
Dedekind-finite	yes	no	yes	yes

Properties of the four workhorse rings

Relationship Map

The implications below are the true ones; each arrow that is absent from this chain is absent because of a witness above.

semisimpleleft artiniansemiprimaryright perfectsemiperfectsemilocal

No arrow reverses. Reversals fail by ℤ_(p), Bass's ring (23.22), the triangular ring, and 𝔽_p C_p respectively.

Failures of sidedness — one side holds, the other does not
- artinian: $(1.24)$
- noetherian: $(1.23)$ , $(1.25)$ , $(1.26)$
- primitive: Bergman 1965
- perfect: $(23.22)$
Failures of finiteness — a chain condition is silently needed
- nil not nilpotent: $(4.13)$ needs left artinian
- radical not nil: $(4.12)$ needs left artinian
- semiprimitive not semisimple: $(4.14)$ needs left artinian
- regular not semisimple: $(4.25)$ needs left noetherian
Failures of elementwise reasoning — a property of elements does not pass to ideals
- nilpotent elements do not form an ideal
- nil one-sided ideals may not sum to a nil ideal — Köthe, open
- left-invertible elements need not be invertible
Failures of base change — the property is not stable under an extension
- radical grows under inseparable field extension
- $rad R [T]$ is not $(rad R) [T]$
- simplicity is not preserved by tensoring with a non-splitting field

Failure Modes and Common Mistakes

Do not use $ℤ / 6 ℤ$ to refute idempotents lift modulo nil ideals: $(6) \subseteq ℤ$ is not nil, so $(21.28)$ never applied.
Do not cite $M_{2} (k)$ as a ring with a nonzero nil ideal; it has none, and its nilpotent elements are a red herring.
Do not claim $k [[x]]$ refutes $(4.12)$ — it is not artinian, so the theorem was never in play.
Do not assume a witness for one side is automatically a witness for the other; transpose it through $R^{op}$ explicitly.

Best Practices

State the implication being refuted in full, with every hypothesis, before naming the witness.
Verify the witness satisfies each hypothesis explicitly; the verification is usually shorter than the search.
Prefer commutative witnesses where they exist: $ℤ$ , $k [[x]]$ and $ℤ [\sqrt{- 5}]$ settle four of the entries above.
Record which hypothesis the witness violates — that is what makes the catalogue useful rather than merely negative.
When you cannot find a witness, check whether the statement is a known open problem before assuming it is a theorem.

Historical Notes and Lessons Learned

1930Köthe's questionKöthe asks whether a ring with no nonzero nil ideal can have a nonzero nil one-sided ideal. Nearly a century later there is still no example and no proof.
1937Mal'cevConstructs a domain that cannot be embedded in a division ring, refuting the naive noncommutative analogue of the field of fractions.
1960BassIntroduces perfect rings and exhibits a right perfect ring that is not left perfect and not semiprimary, the ring reproduced in (23.22).
1964Golod and ShafarevichConstruct finitely generated nil algebras that are infinite-dimensional, settling the Kurosh problem negatively and supplying a supply of nil-but-not-nilpotent objects.
1965BergmanA left primitive ring that is not right primitive, confirming that primitivity is genuinely one-sided; Jategaonkar later adds further examples.
2000SmoktunowiczA nil ring whose polynomial ring is not nil, refuting Amitsur's conjecture; two years later, a simple nil ring.
2021GardamA nontrivial unit in the group algebra of a torsion-free group over the field of two elements, refuting Kaplansky's unit conjecture after seventy years.

The pattern is instructive: the counterexamples that mattered most were not found by inspection but by construction — free algebras, transfinite matrix constructions, and computer search in Gardam's case. Where inspection fails, the honest position is that the question is open.

Quick Reference

Sidedness

(\begin{matrix} ℝ & ℝ \\ 0 & ℚ \end{matrix})

— left artinian, not right artinian

Radical not nil

k [[x]]

rad = (x)

Nil not nilpotent

k [x_{i}] / (x_{i}^{i + 1})

Semiprimitive not semisimple

ℤ

Prime not primitive

ℤ

Primitive not simple

End (V_{k})

\dim V_{k}

infinite

Simple not artinian

A_{1} (k)

, characteristic

0

No Krull–Schmidt

ℤ [\sqrt{- 5}]

No idempotent lifting

ℤ \to ℤ / 6 ℤ

Perfect one side onlyBass's ring

(23.22)

Not semisimple group algebra

𝔽_{p} C_{p}

Too many roots

x^{2} + 1

over

ℍ

Which hypothesis each witness attacks
Hypothesis dropped	Immediate consequence	Witness
left DCC	$rad R$ need not be nilpotent or even nil	$k [[x]]$
left DCC	nil ideals need not be nilpotent	$k [x_{1}, x_{2}, \dots] / (x_{i}^{i + 1})$
local endomorphism rings	decompositions are not unique	$ℤ [\sqrt{- 5}]$
$\| G \|$ invertible in $k$	the group algebra is not semisimple	$𝔽_{p} C_{p}$
nil ideal	idempotents need not lift	$ℤ \to ℤ / 6 ℤ$
commutativity of coefficients	root counting fails	$ℍ [x]$
symmetry of the bimodule	chain conditions become one-sided	$(1.22)$ – $(1.24)$

Frequently Asked Questions

Is there a single ring that refutes most of these implications at once?

$End (V_{k})$ for $V$ infinite-dimensional over a division ring comes closest. It is left primitive but not simple, von Neumann regular but not semisimple, semiprimitive with no chain condition on either side, and not Dedekind-finite, so left-invertible fails to mean invertible. What it cannot do is break sidedness — it is symmetric enough that one needs a triangular ring for that.

Why is there no counterexample to Köthe's conjecture in this catalogue?

Because none is known. The conjecture has been open since about 1930 and is proved for right noetherian rings $(10.30)$ , for algebras algebraic over a field $(4.19)$ , for algebras of dimension less than the cardinality of the base field $(4.20)$ , and for PI-algebras. A counterexample would have to avoid all of those classes.

Does the failure of Krull–Schmidt over

ℤ [\sqrt{- 5}]

contradict

(19.21)

No. $(19.21)$ requires each indecomposable summand to have a local endomorphism ring. For the ideal $I = (2, 1 + \sqrt{- 5})$ we have ${End}_{R} (I) ≅ R$ , which is not local, so the theorem does not apply. Over a left artinian ring, by contrast, indecomposable finitely generated modules do have local endomorphism rings, and Krull–Schmidt holds.

How small can a ring be and still be left artinian but not right artinian?

It cannot be finite: a finite ring satisfies both chain conditions trivially. It cannot be commutative either. Some infinite-dimensional asymmetry in a bimodule is unavoidable, and $(1.24)$ is close to the cheapest realisation — a $3$ -step composition series on the left over a ring built from two fields.

Are Gardam's units relevant to the zero-divisor conjecture?

Not directly. Gardam's counterexample is to Kaplansky's unit conjecture, over the field of two elements, for the torsion-free group usually called the Promislow or Hantzsche–Wendt group. That group is virtually free abelian, and its group algebra is still known to be a domain, so the zero-divisor conjecture survives intact for it and remains open in general.

Which of these witnesses can a computer algebra system verify?

The finite-dimensional ones. GAP, Magma and Sage will compute $rad (𝔽_{p} C_{p})$ , the radical of a triangular algebra over a field, and the class group computation behind the $ℤ [\sqrt{- 5}]$ example. The infinite-dimensional witnesses — $k [[x]]$ , $End (V_{k})$ , Bass's ring — must be argued by hand.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, especially §1 (1.22)–(1.26), §4 (4.12)–(4.14), §11, §19 and §23 (23.22).
T. Y. Lam, Exercises in Classical Ring Theory, 2nd edition, Problem Books in Mathematics, Springer, 2003.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964.
H. Bass, “Finitistic dimension and a homological generalization of semi-primary rings”, Transactions of the American Mathematical Society 95 (1960), 466–488.
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.

AI Suggested Questions

Give the full verification that Bergman's 1965 ring is left primitive but not right primitive.
Construct a ring that is left perfect but not right perfect, dual to the example in (23.22).
What is the smallest dimension of a finite-dimensional algebra whose radical is not detected by the trace form?
Sketch Smoktunowicz's construction of a nil ring whose polynomial ring is not nil.
Which of the failures listed here become true if the ring is assumed to satisfy a polynomial identity?
Find a Dedekind domain with class number 3 and describe the corresponding Krull-Schmidt failure.
Are there analogues of these counterexamples for rings without an identity element, and which ones change?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

What a good counterexample has to do

Minimality is worth having

Sidedness is manufactured, not found

06Key Results

07Proof Techniques and Method

Triangular rings

Infinite-dimensional endomorphism rings

Complete local rings

Modular group algebras

Skew and free constructions

08Worked Example

The triangular ring in full

A second computation: roots over the quaternions

09Comparison and Classification

10Relationship Map

11Failure Modes and Common Mistakes

12Best Practices

13Historical Notes and Lessons Learned

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

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