Right Perfect but Not Left Perfect: A Counterexample

Q: Why does finite support matter so much?

Two reasons. It makes every element of J nilpotent, which gives J ⊆ rad R and hence locality. And it supplies the starting index n in the T-nilpotency proof: because a_1 kills e_j for all large j, the whole column space is pushed into a finite stage V_n of the filtration at the first step, after which each further factor drops the index by one.

Q: Is the opposite ring also interesting?

It is the same example read backwards. R^op may be realised as the ring of finitely supported strictly *lower* triangular matrices adjoined to k; it is left perfect and not right perfect. Together the pair shows neither class contains the other.

Q: Could a commutative ring do this?

No. In a commutative ring the products a_1 a_2 … a_n and a_n … a_2 a_1 coincide, so left and right T-nilpotency are the same condition and left perfect equals right perfect equals perfect. Any separating example must be noncommutative, and (23.24) classifies the commutative perfect rings without any side qualifier.

Q: How is this ring related to the triangular matrix algebras T_n(k)?

It is the union, or direct limit, of the algebras k 1 + n_n where n_n is the strictly upper triangular part of T_n(k), embedded in the top left corner. Each stage is a finite-dimensional algebra with nilpotent radical of index n, hence semiprimary and perfect; the index grows without bound in the limit, which is precisely how a nilpotent radical degenerates into one that is only T-nilpotent on one side.

Q: What fails on the left in module-theoretic terms?

By (23.16) applied on the other side, left T-nilpotency of J would be equivalent to the implication JN = N N = 0 for all left modules N. Since J is not left T-nilpotent, some nonzero left module satisfies JN = N; equivalently, some flat left R-module fails to be projective, and some right R-module fails to have a projective cover.

Executive Summary

Almost every class of rings in this collection is left-right symmetric: semisimple, semiprimary, semiperfect, semilocal, and the Jacobson radical itself. Perfect is the exception, and $(23.22)$ is the example that proves it.

The ring is small enough to compute with: take the field $k$ , let $J$ be the set of $ℕ \times ℕ$ matrices over $k$ with finitely many nonzero entries, all of them strictly above the diagonal, and set $R = k \cdot 1 + J$ . Then $R$ is local with $rad R = J$ and $R / J ≅ k$ ; $J$ is right T-nilpotent but not left T-nilpotent; so $R$ is right perfect and not left perfect.

(23.22)Lam reference

LocalRing type

Right onlyPerfect on which side

$E_{1, n + 1}$ Witness to failure

Overview

Definition $(23.18)$ makes right perfectness depend on right T-nilpotency of $rad R$ : every sequence $a_{1}, a_{2}, \dots$ in the radical must satisfy $a_{n} \dots a_{2} a_{1} = 0$ for some $n$ . Left perfectness asks the same of the products taken in the opposite order, $a_{1} a_{2} \dots a_{n}$ .

Nothing forces these to agree. Multiplication in a noncommutative ring can be arranged so that products accumulating on the left die out while products accumulating on the right do not, and infinite matrix units are the cleanest device for arranging it: $E_{i, i + 1}$ multiplied on the right by $E_{i + 1, i + 2}$ lengthens the gap, while multiplication on the left annihilates.

E_{i j} E_{k l} = {\begin{cases} E_{i l} & j = k, \\ 0 & j \neq k, \end{cases} E_{1, 2} E_{2, 3} \dots E_{n, n + 1} = E_{1, n + 1} \neq 0 .

(23.22a)

The displayed product is never zero, however long the chain: the radical is not left T-nilpotent.

The ring appears in Catalogue of Counterexamples and in Left-Right Symmetry: What Transfers and What Does Not. It is also the standard witness that the two homological conditions of §24 — every flat right module projective, every flat left module projective — are independent.

Learning Objectives

Define $R = k \cdot 1 + J$ precisely and verify that $J$ is a two-sided ideal with $R / J ≅ k$ .
Prove that $J$ is right T-nilpotent by the filtration $V_{1} \subseteq V_{2} \subseteq \dots$ of the column space.
Show that $rad R = J$ and hence that $R$ is local and right perfect.
Exhibit the sequence $a_{i} = E_{i, i + 1}$ and conclude that $R$ is not left perfect.
Write down the strictly descending chain of principal right ideals that Bass's Theorem P predicts.
State which of semiperfect, perfect, semiprimary and artinian $R$ satisfies.

Definitions

Construction(23.22)The ring

R = k \cdot 1 + J

Let $k$ be a field and index rows and columns by $ℕ = {1, 2, 3, \dots}$ . Put

J = {a = (a_{i j}) : a_{i j} \in k, a_{i j} = 0 unless i < j, only finitely many a_{i j} \neq 0},

so $J$ is the $k$ -span of the matrix units $E_{i j}$ with $i < j$ . Let $1$ denote the infinite identity matrix $diag (1, 1, 1, \dots)$ and set $R = k \cdot 1 + J$ , a $k$ -algebra of countably infinite dimension.

$V$: The column space $V = ⨁_{i \geq 1} k e_{i}$ , on which $R$ acts on the left by matrix multiplication. The action is faithful.
$V_{n}$: The finite-dimensional subspace $k e_{1} \oplus \dots \oplus k e_{n}$ , with $V_{0} = 0$ . These form an exhaustive filtration of $V$ .
$E_{i j}$: The matrix unit; as an operator on $V$ it sends $e_{j} \mapsto e_{i}$ and kills every other basis vector.
Right perfect: $R / rad R$ semisimple and $rad R$ right T-nilpotent, Definition $(23.18)$ .

Any division ring may replace the field k throughout; nothing in the argument uses commutativity of the coefficients.

Core Concepts

Why the two sides behave differently

Read every element of $J$ as an operator on the column space $V$ . A strictly upper triangular matrix sends $e_{j}$ into $span (e_{1}, \dots, e_{j - 1})$ , so it lowers the filtration index by at least one:

V_{n}

V_{n - 1}

V_{n - 2}

\dots

V_{0} = 0

Composing operators on the left therefore drives everything down the filtration and, because each element of $J$ has finite support, the starting point is a fixed $V_{n}$ . That is right T-nilpotency. Composing on the right does the opposite: it lets the operator reach further out along the basis, and $E_{1, 2}$ followed on the right by $E_{2, 3}, E_{3, 4}, \dots$ simply migrates the surviving entry to $E_{1, n + 1}$ .

What the asymmetry costs and does not cost

$J$ is nil: every element has finite support, so it lies in a finite strictly upper triangular block and is nilpotent. Consequently idempotents lift modulo $J$ , and since $R / J ≅ k$ is a division ring, $R$ is local — in particular semiperfect. Semiperfectness sees nothing of the asymmetry, which is exactly why it is a two-sided notion.

Locally nilpotent, but not uniformly

By $(23.15)$ a right T-nilpotent one-sided ideal lies in the lower nilradical and is locally nilpotent. That is visible here: any finite set of elements of $J$ is supported in a finite square block ${1, \dots, N}^{2}$ , and the strictly upper triangular $N \times N$ matrices form a nilpotent algebra of index $N$ . The index grows with the block, which is precisely why $J$ itself is not nilpotent.

Key Results

Counterexample(23.22)A right perfect ring that is not left perfect

Let $k$ be a field, $J$ the set of finitely supported strictly upper triangular $ℕ \times ℕ$ matrices over $k$ , and $R = k \cdot 1 + J$ . Then:

$R$ is a ring and $J$ is a two-sided ideal with $R / J ≅ k$ ;
$J$ is right T-nilpotent, hence nil, hence $J = rad R$ and $R$ is a local ring;
$R$ is right perfect;
$J$ is not left T-nilpotent, so $R$ is not left perfect.

Proof

(1). If $a, b \in J$ have supports inside ${1, \dots, N}^{2}$ then so does $a b$ , and $(a b)_{i j} = \sum_{m} a_{i m} b_{m j}$ is nonzero only when $i < m < j$ ; so $J$ is closed under multiplication and $J R = R J = J$ because $R = k \cdot 1 + J$ . Thus $J$ is a two-sided ideal and $R / J ≅ k$ via $λ \cdot 1 + a \mapsto λ$ .

(2). Let $a_{1}, a_{2}, a_{3}, \dots \in J$ and let $R$ act on $V = ⨁_{i \geq 1} k e_{i}$ by matrix multiplication. Every $a \in J$ satisfies $a (V_{m}) \subseteq V_{m - 1}$ , since $a e_{j}$ is a combination of $e_{i}$ with $i < j$ . Because $a_{1}$ has finite support there is $n$ with $a_{1} e_{j} = 0$ for all $j > n$ , that is $a_{1} (V) = a_{1} (V_{n}) \subseteq V_{n - 1}$ . Iterating,

a_{2} a_{1} (V) \subseteq a_{2} (V_{n - 1}) \subseteq V_{n - 2}, \dots, a_{n} \dots a_{2} a_{1} (V) \subseteq V_{0} = 0 .

The action of $R$ on $V$ is faithful, so $a_{n} \dots a_{2} a_{1} = 0$ and $J$ is right T-nilpotent. By $(23.14)$ it is nil, so $J \subseteq rad R$ ; and $R / J ≅ k$ is a division ring, so $J$ is a maximal (left and right) ideal and $rad R = J$ . A ring whose quotient by its radical is a division ring is local.

(3). $R / rad R ≅ k$ is semisimple and $rad R = J$ is right T-nilpotent by (2), which is Definition $(23.18)$ .

(4). Take $a_{i} = E_{i, i + 1} \in J$ . Since $E_{i j} E_{k l} = δ_{j k} E_{i l}$ , an induction gives $a_{1} a_{2} \dots a_{n} = E_{1, 2} E_{2, 3} \dots E_{n, n + 1} = E_{1, n + 1} \neq 0$ for every $n$ . So no $n$ works for this sequence and $J$ is not left T-nilpotent. Since $rad R = J$ , $R$ fails the left half of $(23.18)$ and is not left perfect.

Corollary—Consequences

The classes of left perfect and right perfect rings are distinct; neither contains the other, since $R^{op}$ is left perfect and not right perfect.
T-nilpotency is a genuinely one-sided property of a two-sided ideal.
A right perfect ring need not be semiprimary: $J^{n} \neq 0$ for all $n$ .
By Bass's Theorem P, $R$ satisfies DCC on principal left ideals but fails DCC on principal right ideals.
By the homological characterisations of §24, every flat right $R$ -module is projective, but some flat left $R$ -module is not projective.

Proposition—The failing chain, explicitly

With $R$ as above, $E_{1, n} R = {span}_{k} {E_{1, m} : m \geq n}$ for every $n \geq 2$ , so

E_{1, 2} R ⊋ E_{1, 3} R ⊋ E_{1, 4} R ⊋ \dots

is a strictly descending chain of principal right ideals.

Proof

For $λ \in k$ and $b \in J$ we have $E_{1, n} (λ \cdot 1 + b) = λ E_{1, n} + \sum_{m} b_{n m} E_{1, m}$ , and $b_{n m} \neq 0$ forces $m > n$ . So $E_{1, n} R \subseteq span {E_{1, m} : m \geq n}$ ; conversely $E_{1, n} E_{n, m} = E_{1, m}$ realises every $m > n$ , giving equality. The inclusions are strict because $E_{1, n + 1} \in E_{1, n} R$ but $E_{1, n} \notin E_{1, n + 1} R$ .

Worked Example

Small computations in $R$

Write $a = E_{1, 2} + 3 E_{2, 5}$ and $b = E_{2, 3}$ with $k = ℚ$ . Then

a b = E_{1, 2} E_{2, 3} + 3 E_{2, 5} E_{2, 3} = E_{1, 3}, b a = E_{2, 3} E_{1, 2} + 3 E_{2, 3} E_{2, 5} = 0 .

(E.1)

Order matters completely: $a b \neq 0$ while $b a = 0$ .

The unit group is transparent as well. For $λ \neq 0$ and $a \in J$ with $a^{N} = 0$ ,

(λ \cdot 1 + a)^{- 1} = λ^{- 1} (1 - λ^{- 1} a + λ^{- 2} a^{2} - \dots + (- 1)^{N - 1} λ^{- (N - 1)} a^{N - 1}),

(E.2)

A finite geometric series, since every element of $J$ is nilpotent. So $U (R) = R ∖ J$ , confirming that $R$ is local.

A sequence that dies, and a sequence that does not

Take $a_{i} = E_{i, i + 1}$ for all $i$ . Reading the products in the two possible orders:

a_{1} a_{2} \dots a_{n} = E_{1, n + 1} \neq 0 for all n, a_{2} a_{1} = E_{2, 3} E_{1, 2} = 0 .

(E.3)

Right T-nilpotency is witnessed at $n = 2$ ; left T-nilpotency never occurs.

For a less degenerate right-handed test take $a_{i} = E_{1, i + 1}$ for all $i$ . Then $a_{2} a_{1} = E_{1, 3} E_{1, 2} = 0$ again, since the column index $3$ of the left factor does not match the row index $1$ of the right factor. The general proof shows this is unavoidable: after at most $n$ steps the composite operator has pushed every basis vector out of the filtration.

What the module theory looks like

$V = ⨁_{i \geq 1} k e_{i}$ is a left $R$ -module. Its socle is ${ann}_{V} (J) = k e_{1}$ , which is nonzero — as Bass's Theorem P demands of every nonzero left module over a right perfect ring. On the right side no such guarantee exists, and indeed the failure of DCC on principal right ideals in the chain $(E_{1, n} R)_{n \geq 2}$ is the concrete symptom.

Comparison and Classification

What $R = k \cdot 1 + J$ satisfies
	Left version	Right version	Symmetric class?
Perfect	no	yes	no
Semiperfect	yes	yes	yes
Semilocal	yes	yes	yes
Semiprimary	no	no	yes
Artinian	no	no	no
Noetherian	no	no	no
DCC on principal one-sided ideals	yes	no	no
Every flat module projective	no	yes	no

What $R = k \cdot 1 + J$ satisfies

Comparison with neighbouring examples
Ring	$rad$	Right perfect?	Left perfect?
$k \cdot 1 + J$ of $(23.22)$	$J$ , nil, not nilpotent	yes	no
Its opposite ring	$J^{op}$	no	yes
$T_{n} (k)$ , upper triangular $n \times n$	strictly upper triangular, nilpotent	yes	yes
$k [[x]]$	$(x)$ , not nil	no	no
$M_{2} (ℤ / 8 ℤ)$	$M_{2} (2 ℤ / 8 ℤ)$ , nilpotent	yes	yes

Relationship Map

The example sits at a precise point in the hierarchy: inside semiperfect, inside right perfect, outside perfect, outside semiprimary.

$R = k \cdot 1 + J$ — local, infinite-dimensional over $k$
- belongs to
  - local rings, hence semiperfect $(23.1)$
  - right perfect rings $(23.18)$
  - rings with nil radical
  - rings whose radical is locally nilpotent $(23.15)$
- does not belong to
  - left perfect rings
  - semiprimary rings
  - one-sided artinian rings
  - one-sided noetherian rings
- witnesses
  - T-nilpotency is one-sided
  - perfectness is one-sided
  - flat implies projective is one-sided

J

right T-nilpotent

R

right perfect

R

semiperfect

R

semilocal

Design Considerations

Design considerations here means the choices made when modelling a problem with these algebraic structures.

Fix the side once. Decide at the outset whether your modules are left or right modules and keep the perfectness hypothesis on the matching side. Mixing sides silently is how the majority of errors with perfect rings arise.
Test with the opposite ring. Every statement about right perfect rings yields a statement about left perfect rings by passing to $R^{op}$ ; if a claimed theorem is not stable under that translation, one of the two sides has been misassigned.
Choose the concrete model. The same abstract ring can be presented as finitely supported infinite matrices or as a direct limit of triangular matrix algebras $T_{n} (k)$ along the maps that place a block in the top left corner. The matrix picture makes T-nilpotency visible; the direct-limit picture makes local nilpotence visible.
Expect asymmetry only in the radical. The semisimple quotient of this ring is a field, so all the one-sidedness lives in $J$ . When designing a counterexample of this type, put the asymmetry in the radical and keep the quotient as simple as possible.

Standards and Notation

Standards here covers notation, symbol and markup standards, and reference implementations, rather than material or design codes.

Matrix units

E_{i j}

, with

E_{i j} E_{k l} = δ_{j k} E_{i l}

Index conventionrows and columns indexed by

ℕ

;

i < j

means strictly above the diagonal

Alternative namethe ring is sometimes written

k + {Str}_{\infty} (k)

or as

varinjlim T_{n} (k)

Opposite ring

R^{op}

is the analogous ring of strictly lower triangular matrices

ImplementationsGAP and Magma handle only finite matrix algebras; this ring must be modelled as a direct limit of the

T_{n} (k)

Failure Modes and Common Mistakes

Do not write $E_{i, i + 1} E_{i + 1, i + 2} = 0$ ; the product is $E_{i, i + 2}$ . It is the reversed product that vanishes.
Do not claim $R$ is artinian or noetherian on either side; it is neither, and none of the perfect ring theory requires it to be.
Do not assume the failure of left perfectness can be seen from a single element. Every element of $J$ is nilpotent; the obstruction is visible only along an infinite sequence.

Quick Reference

The ring

R = k \cdot 1 + J

J

finitely supported strictly upper triangular

Radical

rad R = J

, nil and locally nilpotent, not nilpotent

Quotient

R / rad R ≅ k

, so

R

is local

Right T-nilpotentyes — filtration argument on

V_{n}

Left T-nilpotentno —

a_{1} \dots a_{n} = E_{1, n + 1} \neq 0

Perfectright perfect only

Chain conditionsDCC on principal left ideals holds; on principal right ideals it fails

Units

U (R) = R ∖ J

, inverses given by a finite geometric series

The three separations this one ring provides
Claim refuted	Witness	Where used
T-nilpotency is left-right symmetric	$a_{i} = E_{i, i + 1}$	(23.13), (23.14)
Right perfect implies left perfect	$rad R = J$	(23.18), (23.22)
Right perfect implies semiprimary	$E_{1, n + 1} \in J^{n}$	(23.19)
Perfectness is detectable elementwise	every element of $J$ is nilpotent	(23.13)

Frequently Asked Questions

Why does finite support matter so much?

Two reasons. It makes every element of $J$ nilpotent, which gives $J \subseteq rad R$ and hence locality. And it supplies the starting index $n$ in the T-nilpotency proof: because $a_{1}$ kills $e_{j}$ for all large $j$ , the whole column space is pushed into a finite stage $V_{n}$ of the filtration at the first step, after which each further factor drops the index by one.

Is the opposite ring also interesting?

It is the same example read backwards. $R^{op}$ may be realised as the ring of finitely supported strictly lower triangular matrices adjoined to $k$ ; it is left perfect and not right perfect. Together the pair shows neither class contains the other.

Could a commutative ring do this?

No. In a commutative ring the products $a_{1} a_{2} \dots a_{n}$ and $a_{n} \dots a_{2} a_{1}$ coincide, so left and right T-nilpotency are the same condition and left perfect equals right perfect equals perfect. Any separating example must be noncommutative, and $(23.24)$ classifies the commutative perfect rings without any side qualifier.

Does this ring have a chain condition of any kind?

It satisfies DCC on principal left ideals — that is exactly what Bass's Theorem P extracts from right perfectness — and, by Jonah's condition, ACC on principal right ideals. It has no DCC on principal right ideals, no ACC or DCC on arbitrary one-sided ideals, and is neither left nor right noetherian.

How is this ring related to the triangular matrix algebras

T_{n} (k)

It is the union, or direct limit, of the algebras $k \cdot 1 + 𝔫_{n}$ where $𝔫_{n}$ is the strictly upper triangular part of $T_{n} (k)$ , embedded in the top left corner. Each stage is a finite-dimensional algebra with nilpotent radical of index $n$ , hence semiprimary and perfect; the index grows without bound in the limit, which is precisely how a nilpotent radical degenerates into one that is only T-nilpotent on one side.

What fails on the left in module-theoretic terms?

By $(23.16)$ applied on the other side, left T-nilpotency of $J$ would be equivalent to the implication $J N = N \Rightarrow N = 0$ for all left modules $N$ . Since $J$ is not left T-nilpotent, some nonzero left module satisfies $J N = N$ ; equivalently, some flat left $R$ -module fails to be projective, and some right $R$ -module fails to have a projective cover.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §23, (23.13)–(23.22) (pp. 351–357).
H. Bass, “Finitistic dimension and a homological generalization of semi-primary rings”, Transactions of the American Mathematical Society 95 (1960), 466–488.
T. Y. Lam, Exercises in Classical Ring Theory, 2nd edition, Problem Books in Mathematics, Springer, 2003, exercises for §23.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §28.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, §2.7.

AI Suggested Questions

Construct explicitly a nonzero left $R$ -module $N$ with $J N = N$ for the ring of $(23.22)$ .
Describe all the finitely generated projective modules over $k \cdot 1 + J$ .
Does $k \cdot 1 + J$ satisfy ACC on principal right ideals, and how does Jonah's theorem apply?
Give a right perfect ring that is not left perfect and whose semisimple quotient is not a division ring.
What happens to the example if the index set $ℕ$ is replaced by $ℤ$ or by an arbitrary totally ordered set?
Compare this ring with the endomorphism ring of an infinite-dimensional vector space: which chain conditions do they share?
Show that a flat left module over this ring need not be projective, following Bass's argument in reverse.

One-Sided Perfect Rings

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Why the two sides behave differently

What the asymmetry costs and does not cost

Locally nilpotent, but not uniformly

Key Results

Worked Example

Small computations in $R$

A sequence that dies, and a sequence that does not

What the module theory looks like

Comparison and Classification

Relationship Map

Design Considerations

Standards and Notation

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why the two sides behave differently

What the asymmetry costs and does not cost

Locally nilpotent, but not uniformly

06Key Results

07Worked Example

Small computations in R

A sequence that dies, and a sequence that does not

What the module theory looks like

08Comparison and Classification

09Relationship Map

10Design Considerations

11Standards and Notation

12Failure Modes and Common Mistakes

13Quick Reference

14Frequently Asked Questions

15Related KEVOS Topics

16References

17AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Worked Example

Small computations in $R$

Comparison and Classification

Relationship Map

Design Considerations

Standards and Notation

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions