Bass’s Theorem P: Characterisations of Perfect Rings

Q: Why does a right-handed condition correspond to a left-handed chain condition?

Because right T-nilpotency of rad R has two faces. Theorem (23.16) shows it is equivalent to an unrestricted Nakayama lemma for right modules and, simultaneously, to the statement that ann_N(rad R) ≠ 0 for every nonzero left module N. The products a_n … a_2 a_1 that appear in the definition read naturally as a descending chain Ra_1 Ra_2a_1 … of left ideals, which is where the left-handed chain condition comes from.

Q: Which implication is hard, and why?

The hard one is right perfect DCC on principal left ideals. T-nilpotency controls products along one chosen sequence; a chain condition must control all chains simultaneously, and there is no direct combinatorial passage between the two. Bass's route goes through the module category: over a right perfect ring every flat right module is projective, and that fact is strong enough to produce the chain condition.

Q: Is a right perfect ring left perfect if it happens to be commutative?

Yes, trivially: in a commutative ring the products a_1 … a_n and a_n … a_1 agree, so left and right T-nilpotency coincide. The asymmetry in Theorem P is invisible in the commutative case, which is why the commutative classification (23.24) is so clean.

Q: Does condition (2) alone imply R/rad R is semisimple?

Yes — that is precisely the content of the chain (2) (3) (4) (1). The semisimplicity is extracted at the last step, from the prohibition on infinite orthogonal idempotent families together with the fact that a nil radical lifts idempotents. Only the T-nilpotency half of (1) follows from (2) by an elementary argument.

Executive Summary

Bass's Theorem P $(23.20)$ is the structural core of the theory of perfect rings. It converts the definition — semisimple quotient plus right T-nilpotent radical — into three conditions of a completely different character: a descending chain condition on principal left ideals, a descending chain condition on cyclic submodules of arbitrary left modules, and a joint condition on idempotents and socles.

The side switch is not a misprint. Right perfectness is a statement about the category of right $R$ -modules; the equivalent chain conditions live on the left. Both facets come from the same source, Theorem $(23.16)$ , which characterises right T-nilpotency of a right ideal by a right-module condition and a left-module condition simultaneously.

(23.20)Lam reference

7Equivalent conditions

LeftSide of the DCC

1960Bass

Overview

Right artinian rings are right perfect, as recorded in Perfect Rings and Semiprimary Rings. Bass's theorem says that perfectness is itself a chain condition, just a weaker one: instead of DCC on all left ideals, DCC on the principal ones.

R right perfect iff R satisfies DCC on principal left ideals R a_{1} \supseteq R a_{2} a_{1} \supseteq R a_{3} a_{2} a_{1} \supseteq \dots

(23.20)

The displayed chain is the one produced by an arbitrary sequence in $R$ ; stationarity of all such chains is the content of condition (2).

One direction is elementary and is proved in full below: a stationary chain of that shape, together with the fact that $1 - r a$ is a unit whenever $a \in rad R$ , forces $a_{n} \dots a_{1} = 0$ . The other direction — right perfect implies DCC on principal left ideals — is the deep one, and Lam defers it to §24, where it is extracted from the homological characterisation that every flat right module over a right perfect ring is projective.

The theorem also explains the name. Nothing in conditions (2)–(4) mentions right modules, yet the class they define is precisely the one whose right module category has projective covers and satisfies flat implies projective. The homological facts, not the chain conditions, fix the terminology.

Learning Objectives

State Bass's Theorem P with all four of Lam's conditions and the correct side on each.
Prove that DCC on principal left ideals implies $rad R$ is right T-nilpotent.
Prove the implications (2) $\Rightarrow$ (3) $\Rightarrow$ (4) and sketch (4) $\Rightarrow$ (1).
Construct the strictly descending chain attached to an infinite orthogonal family of idempotents.
Quote the Bjork and Jonah supplements and say exactly what they add.
Apply the theorem to decide perfectness for concrete rings.

Definitions

Right perfect: $R / rad R$ is semisimple and $rad R$ is right T-nilpotent, i.e. every sequence $a_{1}, a_{2}, \dots \in rad R$ has $a_{n} \dots a_{2} a_{1} = 0$ for some $n$ . Definition $(23.18)$ .
DCC on principal left ideals: Every chain $R b_{1} \supseteq R b_{2} \supseteq \dots$ of principal left ideals stabilises. Equivalently, $R$ has no strictly descending sequence of cyclic submodules of $_{R} R$ .
${ann}_{N} (J)$: For a left module $N$ and a right ideal $J$ , the submodule ${x \in N : J x = 0}$ . It is nonzero for every $N \neq 0$ exactly when $J$ is right T-nilpotent, by $(23.16)$ .
$soc (N)$: The socle: the sum of all simple submodules of $N$ . Condition (4) of $(23.20)$ asks that $soc (N) \neq 0$ for every $N \neq 0$ .
Infinite orthogonal family: Idempotents $e_{1}, e_{2}, \dots \neq 0$ with $e_{i} e_{j} = 0$ for $i \neq j$ . Such a family exists in $End (V)$ for $V$ of infinite dimension, and in any infinite Boolean ring.

Modules are unital and rings have an identity. Chains are indexed by the natural numbers throughout; DCC for countable chains is equivalent to DCC for arbitrary families here, since a non-stationary family yields a non-stationary sequence.

Core Concepts

The engine: the general Nakayama lemma

Everything in the proof rests on Theorem $(23.16)$ , developed on the T-Nilpotency page. For a right ideal $J \subseteq R$ , the following are equivalent: $J$ is right T-nilpotent; $M J = M \Rightarrow M = 0$ for every right $R$ -module $M$ ; and ${ann}_{N} (J) = 0 \Rightarrow N = 0$ for every left $R$ -module $N$ .

The second condition is Nakayama's lemma $(4.22)$ with the finite generation hypothesis deleted; the third is its left-handed shadow. A hypothesis that is simultaneously about right modules and about left modules is exactly what a theorem with a side switch needs.

Where each of the four conditions bites

Condition (1)

Right perfect

Semisimple quotient plus right T-nilpotent radical. The definition; the form used when quoting homological consequences.

Condition (2)

DCC on principal left ideals

The cheapest condition to falsify. One strictly descending chain of principal left ideals kills right perfectness outright.

Condition (3)

DCC on cyclic submodules of every left module

Formally stronger than (2) — take the module to be $R R$ — and the version that survives base change to module categories.

Condition (4)

No infinite orthogonal idempotents, and every nonzero left module has a simple submodule

Two separate finiteness statements: one bounds the idempotent structure, the other supplies socles. Together they rebuild both halves of the definition.

How (4) reconstructs the definition

The socle half of (4) gives T-nilpotency: a simple submodule of $N$ is annihilated by $rad R$ , so ${ann}_{N} (rad R) \neq 0$ for every $N \neq 0$ , and $(23.16)$ applies. The idempotent half gives semisimplicity of $R / rad R$ : a non-semisimple $R / rad R$ can be split off repeatedly, producing an infinite orthogonal family that lifts back to $R$ because the radical is nil.

Key Results

Lemma—The easy half of (2) implies (1)

Let $R$ be any ring satisfying DCC on principal left ideals. Then $rad R$ is right T-nilpotent.

Proof

Let $a_{1}, a_{2}, a_{3}, \dots \in rad R$ and put $b_{n} = a_{n} a_{n - 1} \dots a_{1}$ . Since $b_{n + 1} = a_{n + 1} b_{n} \in R b_{n}$ , the principal left ideals descend:

R b_{1} \supseteq R b_{2} \supseteq R b_{3} \supseteq \dots

By hypothesis the chain is stationary, say $R b_{n} = R b_{n + 1}$ . Write $b_{n} = r b_{n + 1} = r a_{n + 1} b_{n}$ for some $r \in R$ , so that $(1 - r a_{n + 1}) b_{n} = 0$ . Now $a_{n + 1} \in rad R$ and $rad R$ is a two-sided ideal, so $r a_{n + 1} \in rad R$ and $1 - r a_{n + 1} \in U (R)$ by the characterisation $(4.1)$ of the radical. Multiplying by its inverse gives $b_{n} = a_{n} \dots a_{1} = 0$ , which is precisely right T-nilpotency.

Theorem(23.20)Bass's Theorem P

For any ring $R$ with identity the following conditions are equivalent.

$R$ is right perfect, i.e. $R / rad R$ is semisimple and $rad R$ is right T-nilpotent.
$R$ satisfies DCC on principal left ideals.
Every left $R$ -module $N$ satisfies DCC on cyclic submodules.
$R$ contains no infinite set of nonzero orthogonal idempotents, and every nonzero left $R$ -module contains a simple submodule.

Proof

**(2) $\Rightarrow$ (3).** Any descending chain of cyclic submodules of a left module $N$ may be written as $R x \supseteq R a_{1} x \supseteq R a_{2} a_{1} x \supseteq \dots$ : if $R y \subseteq R x$ then $y = a x$ for some $a \in R$ . The corresponding chain $R \supseteq R a_{1} \supseteq R a_{2} a_{1} \supseteq \dots$ of principal left ideals of $R$ is stationary by (2), and stationarity there forces stationarity of the chain in $N$ , since $R a_{n + 1} a_{n} \dots a_{1} = R a_{n} \dots a_{1}$ implies the corresponding equality after applying the map $r \mapsto r x$ .

**(3) $\Rightarrow$ (4).** Let $N \neq 0$ be a left module. Among the nonzero cyclic submodules of $N$ , condition (3) provides a minimal one, $R x$ ; minimality forces $R x$ to be simple, so $N$ has a simple submodule. Next suppose, for a contradiction, that $e_{1}, e_{2}, \dots$ are nonzero orthogonal idempotents. Each $f_{n} : = 1 - e_{1} - \dots - e_{n}$ is idempotent, and

(1 - e_{n + 1}) f_{n} = f_{n} - e_{n + 1} = f_{n + 1},

using $e_{n + 1} e_{i} = 0$ for $i \leq n$ . Hence $R f_{1} \supseteq R f_{2} \supseteq \dots$ . The inclusions are strict: if $f_{n} = r f_{n + 1}$ for some $r \in R$ , right multiplication by $e_{n + 1}$ gives $f_{n} e_{n + 1} = e_{n + 1}$ on the left, while $r f_{n + 1} e_{n + 1} = r (e_{n + 1} - e_{n + 1}) = 0$ on the right, forcing $e_{n + 1} = 0$ . This strictly descending chain of cyclic submodules of $_{R} R$ contradicts (3).

**(4) $\Rightarrow$ (1).** Let $N \neq 0$ be a left $R$ -module. By (4) it has a simple submodule $N_{0}$ , and $rad R$ annihilates every simple left module, so ${ann}_{N} (rad R) \supseteq N_{0} \neq 0$ . By criterion (3) of $(23.16)$ , $rad R$ is right T-nilpotent; in particular it is nil.

It remains to show $S : = R / rad R$ is semisimple as a left $R$ -module. Two facts are available: every nonzero $R$ -submodule of $_{R} S$ contains a simple submodule, by (4); and every simple $R$ -submodule of $_{R} S$ is a direct summand, because $S$ is a semiprime ring $(10.23)$ . If $_{R} S$ were not semisimple, iterating these two facts as in the proof of $(4.14)$ yields decompositions $S = A_{1} \oplus B_{1}$ , $B_{1} = A_{2} \oplus B_{2}, \dots$ with each $A_{i}$ simple, and the associated projections form an infinite orthogonal family of nonzero idempotents of $S$ . Since $rad R$ is nil, such a family lifts to an infinite orthogonal family of nonzero idempotents of $R$ by $(21.25)$ and $(21.28)$ — contradicting (4). Hence $S$ is semisimple and $R$ is right perfect.

**(1) $\Rightarrow$ (2)** is the deep implication and is not proved here; it is obtained in §24 from the homological characterisation of right perfect rings, and is the subject of Flat Implies Projective.

Corollary(23.21)Right DCC forces left DCC on principal ideals

If $R$ satisfies DCC on right ideals — that is, $R$ is right artinian — then $R$ satisfies DCC on principal left ideals.

Proof

A right artinian ring is semiprimary, hence perfect, hence right perfect $(23.19)$ . Apply the implication (1) $\Rightarrow$ (2) of $(23.20)$ . Note that the conclusion is genuinely one-sided in flavour: a right artinian ring need not be left artinian, so DCC on all left ideals cannot be concluded.

Remark—Three further characterisations

Lam records three supplements to the list. Bjork proved that for any left module over any ring, DCC on cyclic submodules is equivalent to DCC on finitely generated submodules; this adds

(5) every left $R$ -module satisfies DCC on finitely generated submodules;
(6) $R$ satisfies DCC on finitely generated left ideals.

Jonah added a genuinely surprising item, an ascending chain condition on the other side:

(7) $R$ satisfies ACC on principal right ideals.

Conditions (5)–(7) are each equivalent to right perfectness. The proofs are more technical and are not reproduced in §23.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Stationary chain plus a unit

From $R b_{n} = R b_{n + 1}$ one gets $(1 - r a) b_{n} = 0$ ; if $a$ lies in the radical the coefficient is a unit and $b_{n} = 0$ . This single trick converts every DCC hypothesis on principal ideals into a vanishing statement, and it is the standard replacement for Nakayama when finite generation is unavailable.

Move 2

Idempotents make strict chains

An infinite orthogonal family ${e_{i}}$ always produces the strictly descending chain $R (1 - e_{1}) ⊋ R (1 - e_{1} - e_{2}) ⊋ \dots$ . Strictness is checked by multiplying on the right by the next idempotent. Use this whenever a finiteness hypothesis must be contradicted.

Move 3

Minimal cyclic means simple

A cyclic submodule minimal among nonzero cyclic submodules is simple, because every nonzero submodule of $R x$ contains a nonzero cyclic one. This is how a chain condition manufactures socles.

The reason $(1) \Rightarrow (2)$ resists these methods is that it must produce a chain condition out of a purely sequential hypothesis. T-nilpotency controls products along a fixed sequence, whereas DCC must handle every possible refinement at once; bridging that gap is what the homological argument of §24 achieves.

Worked Example

Certifying that a ring is not right perfect

Take $R = k [[x]]$ , $k$ a field. Set $a_{i} = x$ for all $i$ . Then $b_{n} = x^{n}$ and

R ⊋ R x ⊋ R x^{2} ⊋ R x^{3} ⊋ \dots

(E.1)

Strict because $x^{n} \notin (x^{n + 1})$ , by comparing orders of vanishing.

Condition (2) fails, so $R$ is not right perfect. It is local, hence semiperfect — the gap between the two classes is exactly this chain. The same computation in $ℤ_{(p)}$ , with $a_{i} = p$ , shows $ℤ_{(p)}$ is not perfect either.

A Boolean ring: instantiating the idempotent chain

Let $R$ be the ring of subsets of $ℕ$ that are finite or cofinite, with addition given by symmetric difference and multiplication by intersection. This is a commutative ring with identity $ℕ$ , every element is idempotent, and $rad R = 0$ because the radical of any ring contains no nonzero idempotent.

Put $e_{n} = {n}$ . These are nonzero orthogonal idempotents, so condition (4) fails and $R$ is not perfect on either side. The proof of (3) $\Rightarrow$ (4) predicts a strictly descending chain; here it is explicit, with $f_{n} = ℕ ∖ {1, \dots, n}$ :

R f_{1} ⊋ R f_{2} ⊋ R f_{3} ⊋ \dots, R f_{n} = {A \in R : A \subseteq f_{n}},

(E.2)

Strict because ${n + 1} \subseteq f_{n}$ but ${n + 1} not \subseteq f_{n + 1}$ .

Note that $R$ is semiprimitive and von Neumann regular yet fails every condition of $(23.20)$ : perfectness is a finiteness condition, not a radical condition.

A positive certificate

Let $R = M_{2} (ℤ / 8 ℤ)$ . Then $rad R = M_{2} (2 ℤ / 8 ℤ)$ with $(rad R)^{3} = 0$ , and $R / rad R ≅ M_{2} (𝔽_{2})$ is simple artinian. So $R$ is semiprimary, hence perfect $(23.19)$ , and Theorem P guarantees DCC on principal left ideals — visible directly here, since $R$ is a finite ring of $8^{4} = 4096$ elements.

Process and Workflow

Compute $rad R$ Identify the radical and the quotient

R / rad R

. If the quotient is not semisimple, stop:

R

is not even semilocal, hence not perfect on either side.

Test the radical for nilnessIf

rad R

contains a non-nilpotent element,

R

is not perfect on either side, by

(23.14)

. This kills

k [[x]]

and

ℤ_{(p)}

immediately.

Look for infinite orthogonal idempotentsIf one exists, condition (4) fails and

R

is not right perfect. This is usually the fastest disqualifier for rings of infinite rank.

Try to build a bad sequenceSeek

a_{1}, a_{2}, \dots \in rad R

with

a_{n} \dots a_{1} \neq 0

for all

n

. Success proves failure of right perfectness; the same sequence read in the other order tests left perfectness.

Otherwise prove T-nilpotencyShow that any sequence eventually annihilates a suitable filtration, as in the infinite triangular example, or verify the criterion

{ann}_{N} (rad R) \neq 0

for all

N \neq 0

Which condition of $(23.20)$ should I verify?

Proving a ring IS right perfectUse (1) directly, or the module criterion (3) of

(23.16)

. Chain conditions are hard to establish from scratch; T-nilpotency usually has a structural proof.

Proving a ring is NOT right perfectUse (2) or the idempotent half of (4). A single explicit strictly descending chain settles the matter.

Transferring perfectness along a functorUse (3), which is stated for all left modules and therefore survives Morita equivalence without extra work.

Reasoning about right modulesUse (1) and the results of §24: projective covers exist for all right modules, and flat right modules are projective.

Comparison and Classification

The seven conditions, by side and by type
	Side used	Chain condition	Elementary to falsify	In Lam §23
(1) right perfect	right	no	no	yes
(2) DCC on principal left ideals	left	yes	yes	yes
(3) DCC on cyclic submodules of left modules	left	yes	partial	yes
(4) no infinite orthogonal idempotents plus socles	left	no	yes	yes
(5) DCC on f.g. submodules of left modules	left	yes	partial	cited
(6) DCC on f.g. left ideals	left	yes	yes	cited
(7) ACC on principal right ideals	right	yes	partial	cited

The seven conditions, by side and by type

Chain conditions, from strongest to weakest
Condition on $R$	Implies	Strictly weaker than	Class defined
DCC on all left ideals	DCC on f.g. left ideals	—	left artinian
DCC on f.g. left ideals	DCC on principal left ideals	DCC on all left ideals	right perfect
DCC on principal left ideals	$rad R$ right T-nilpotent	DCC on all left ideals	right perfect
$rad R$ nilpotent, quotient semisimple	perfect on both sides	one-sided artinian	semiprimary
$rad R$ nil, quotient semisimple	idempotents lift	right perfect	semiperfect

Applications and Industry Use

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

Homological algebra

Finitistic dimension

Bass introduced perfect rings while studying the finitistic global dimension of a ring. Theorem P is what allows the homological invariant to be recognised from a chain condition, and it remains the model for later "big module" characterisations of finiteness.

Module theory

Projective covers everywhere

Right perfect rings are exactly those over which every right module has a projective cover. This makes minimal projective resolutions available for arbitrary modules, not just finitely generated ones.

Representation theory

Infinite-dimensional algebras

Finite-dimensional algebras are semiprimary, so perfectness is automatic; the theorem matters for infinite-dimensional algebras and for endomorphism rings, where it decides whether the usual decomposition theory still applies.

Symbolic computation

Detecting finiteness

Condition (4) is a practical test in computer algebra: search for an infinite orthogonal idempotent family, which for a concretely presented ring is a finite computation on each candidate.

The honest summary is that Theorem P is infrastructure for module theory. Its downstream users are homological algebra and the representation theory of infinite-dimensional algebras; nothing outside algebra consumes it directly.

Computational Notes

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

For a finite-dimensional algebra $A$ over a field, perfectness is automatic — $rad A$ is nilpotent — so the interesting computation is the radical itself, at $O (n^{3})$ field operations by the trace-form method in characteristic $0$ .
For a finite ring given by its multiplication table, condition (4) is decidable by inspection: a finite ring cannot contain an infinite orthogonal family, and every nonzero module over a finite ring has a simple submodule.
For a finitely presented ring there is no algorithm: the word problem is undecidable, so neither $rad R$ nor the DCC on principal left ideals can be decided in general.
Falsification is often effective even when verification is not. Producing one strictly descending chain of principal left ideals is a finite certificate that a ring fails to be right perfect.

Failure Modes and Common Mistakes

Do not use (1) $\Rightarrow$ (2) as if it were elementary; it depends on the homological theory of §24. The converse direction is the cheap one.
Do not assume a ring with DCC on principal left ideals has DCC on principal right ideals; Jonah's condition (7) is an ascending chain condition on the right, not a descending one.
Do not forget that condition (3) quantifies over all left modules, including very large ones; verifying it for finitely generated modules only is not enough.

Quick Reference

Theorem Pright perfect

iff

DCC on principal left ideals

Module formevery left module has DCC on cyclic submodules

Finiteness formno infinite orthogonal idempotents, and all nonzero left modules have socle

Bjorkcyclic DCC

iff

finitely generated DCC

JonahACC on principal right ideals is also equivalent

Easy halfDCC on principal left ideals

\Rightarrow rad R

right T-nilpotent

Deep halfright perfect

\Rightarrow

DCC, proved via flat implies projective in §24

Corollaryright artinian

\Rightarrow

DCC on principal left ideals

(23.21)

Fast disqualifiers for right perfectness
Observed feature	Conclusion	Reason
$rad R$ has a non-nilpotent element	not perfect on either side	(23.14)
Infinite orthogonal idempotent family	not right perfect	(23.20)(4)
A nonzero left module with zero socle	not right perfect	(23.20)(4)
$R$ is a domain that is not a division ring	not right perfect	$R a ⊋ R a^{2} ⊋ \dots$
$R / rad R$ not semisimple	not perfect, not semiperfect	(23.18)

Frequently Asked Questions

Why does a right-handed condition correspond to a left-handed chain condition?

Because right T-nilpotency of $rad R$ has two faces. Theorem $(23.16)$ shows it is equivalent to an unrestricted Nakayama lemma for right modules and, simultaneously, to the statement that ${ann}_{N} (rad R) \neq 0$ for every nonzero left module $N$ . The products $a_{n} \dots a_{2} a_{1}$ that appear in the definition read naturally as a descending chain $R a_{1} \supseteq R a_{2} a_{1} \supseteq \dots$ of left ideals, which is where the left-handed chain condition comes from.

Which implication is hard, and why?

The hard one is right perfect $\Rightarrow$ DCC on principal left ideals. T-nilpotency controls products along one chosen sequence; a chain condition must control all chains simultaneously, and there is no direct combinatorial passage between the two. Bass's route goes through the module category: over a right perfect ring every flat right module is projective, and that fact is strong enough to produce the chain condition.

Is a right perfect ring left perfect if it happens to be commutative?

Yes, trivially: in a commutative ring the products $a_{1} \dots a_{n}$ and $a_{n} \dots a_{1}$ agree, so left and right T-nilpotency coincide. The asymmetry in Theorem P is invisible in the commutative case, which is why the commutative classification $(23.24)$ is so clean.

Does condition (2) alone imply

R / rad R

is semisimple?

Yes — that is precisely the content of the chain (2) $\Rightarrow$ (3) $\Rightarrow$ (4) $\Rightarrow$ (1). The semisimplicity is extracted at the last step, from the prohibition on infinite orthogonal idempotent families together with the fact that a nil radical lifts idempotents. Only the T-nilpotency half of (1) follows from (2) by an elementary argument.

How does Theorem P relate to the Hopkins-Levitzki theorem?

Hopkins-Levitzki says a semiprimary ring is left artinian if and only if it is left noetherian. Theorem P works one level down: it characterises the weaker perfect condition by a weaker chain condition, on principal rather than arbitrary one-sided ideals. Combining the two, a right perfect right noetherian ring is right artinian.

What is the practical value of Jonah's condition (7)?

It is the only characterisation on the list that is an ascending chain condition, and it lives on the same side as the word "right". For rings where ascending chains are easier to control than descending ones — for instance rings built from noetherian data — it can be the cheapest condition to check, though its proof is considerably more technical than the rest of the list.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §23, especially (23.16), (23.20)–(23.21) (pp. 352–356).
H. Bass, “Finitistic dimension and a homological generalization of semi-primary rings”, Transactions of the American Mathematical Society 95 (1960), 466–488.
J.-E. Bjork, “Rings satisfying a minimum condition on principal ideals”, Journal für die reine und angewandte Mathematik 236 (1969), 112–119.
D. Jonah, “Rings with the minimum condition for principal right ideals have the maximum condition for principal left ideals”, Mathematische Zeitschrift 113 (1970), 106–112.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §28.

AI Suggested Questions

Reconstruct the proof that over a right perfect ring every flat right module is projective.
Prove Bjork's theorem that DCC on cyclic submodules implies DCC on finitely generated submodules.
Give a ring satisfying ACC on principal right ideals that is not noetherian, and check it against Jonah's condition.
Show directly that a right perfect right noetherian ring is right artinian.
Which of the seven conditions in Theorem P are Morita invariant, and why?
Construct a nonzero module with zero socle over a ring that is semiperfect but not perfect.
How does Theorem P change if the ring is not assumed to have an identity?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

The engine: the general Nakayama lemma

Where each of the four conditions bites

Right perfect

DCC on principal left ideals

DCC on cyclic submodules of every left module

No infinite orthogonal idempotents, and every nonzero left module has a simple submodule

How (4) reconstructs the definition

06Key Results

07Proof Techniques and Method

Stationary chain plus a unit

Idempotents make strict chains

Minimal cyclic means simple

08Worked Example

Certifying that a ring is not right perfect

A Boolean ring: instantiating the idempotent chain

A positive certificate

09Process and Workflow

10Comparison and Classification

11Applications and Industry Use

Finitistic dimension

Projective covers everywhere

Infinite-dimensional algebras

Detecting finiteness

12Computational Notes

13Failure Modes and Common Mistakes

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

Process and Workflow

Comparison and Classification

Applications and Industry Use

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions