Perfect Rings with Simple Quotient and Commutative Perfect Rings

Q: Why is the converse direction of (23.23) proved with modules instead of matrices?

Because right T-nilpotency of M_n(rad k) does not follow entrywise from right T-nilpotency of rad k. A product of matrices mixes entries along every index path, and the vanishing index for each path depends on the path. The criterion (23.16)(3) replaces the sequence condition by the requirement that every nonzero left module have nonzero annihilator submodule, and that requirement transfers along the equivalence between M_n(k)-modules and k-modules without any bookkeeping.

Q: Is there a version of (23.23) for perfect rings with several simple factors?

Not in the same explicit form. If R/rad R ≅ _i M_n_i(D_i) with m > 1 factors, the corresponding centrally primitive idempotents need not lift to central idempotents of R, so R need not decompose as a product matching the quotient. The theory of blocks and basic rings in §25 is the substitute; for the ring-level classification, only the simple and the commutative cases are clean.

Q: How does one recover n and k from R?

Decompose the right regular module R_R into indecomposable summands; by the simplicity of R/rad R they are all isomorphic to a single strongly indecomposable module M, and n is their number while k ≅ End(M_R). Krull-Schmidt-type uniqueness for semiperfect rings makes both invariants well defined.

Q: Is a local ring automatically perfect?

No. Local means only that rad k is the set of non-units, which makes k/rad k a division ring and k semiperfect. Perfectness additionally requires rad k to be T-nilpotent, which fails for k[[x]], for Z_(p), and for every local domain that is not a division ring.

Executive Summary

Wedderburn-Artin classifies semisimple rings as finite products of matrix rings over division rings. Theorems $(23.23)$ and $(23.24)$ extend that classification one level outwards, to two families of perfect rings where the answer is still completely explicit.

When the semisimple quotient is simple, a right perfect ring is exactly $M_{n} (k)$ for a local ring $k$ whose maximal ideal is right T-nilpotent, with $n$ and $k$ uniquely determined. When the ring is commutative, perfect means exactly a finite product of local rings with T-nilpotent maximal ideals. Both statements are the perfect-ring refinements of the semiperfect classifications $(23.10)$ and $(23.11)$ : the ring shape is unchanged, and only the T-nilpotency clause is added.

(23.23)Simple quotient

(23.24)Commutative case

$M_{n} (k)$ Normal form

Unique

n

and

k

Overview

For a general perfect ring $R$ , the quotient $R / rad R$ is a finite product of simple artinian rings, and the associated centrally primitive idempotents need not lift to central idempotents of $R$ . That is the obstruction to a global structure theorem, and it disappears in exactly two situations: when the product has one factor, and when the ring is commutative.

R right perfect, R / rad R simple iff R ≅ M_{n} (k), k local, rad k right T-nilpotent

(23.23)

The semiperfect version $(23.10)$ is the same statement without the T-nilpotency clause on $rad k$ .

The content is therefore concentrated in one technical point: how does T-nilpotency travel between $rad k$ and $rad M_{n} (k) = M_{n} (rad k)$ ? Downwards it is easy — restrict a sequence to the $(1, 1)$ corner. Upwards it is not, because a sequence of $n \times n$ matrices does not decompose into $n^{2}$ independent scalar sequences. The proof avoids the issue entirely by using the module criterion $(23.16)$ instead of the definition.

The commutative statement $(23.24)$ then follows from the semiperfect classification with almost no extra work, because T-nilpotency passes between a finite product and its factors in both directions. The related page Semiperfect Rings with Simple Quotient and Commutative Semiperfect Rings carries the unrefined versions.

Learning Objectives

State $(23.23)$ with the T-nilpotency hypothesis on the correct side.
Prove that right T-nilpotency of $M_{n} (rad k)$ forces right T-nilpotency of $rad k$ .
Use the criterion ${ann}_{N} (J) \neq 0$ and the column-module description of $M_{n} (k)$ -modules for the converse.
State and prove $(23.24)$ for commutative rings.
Show that T-nilpotency is inherited by, and reconstructed from, the factors of a finite direct product.
Classify concrete rings such as $M_{3} (ℤ / 9)$ , $M_{2} (ℤ_{(p)})$ and $ℤ / 12$ .

Definitions

Definition—The data in the classification

A ring $k$ is local if $k / rad k$ is a division ring; then $rad k$ is the unique maximal left ideal and the unique maximal right ideal, and consists precisely of the non-units. The classification $(23.23)$ requires in addition that this ideal be right T-nilpotent: every sequence $a_{1}, a_{2}, \dots \in rad k$ has $a_{n} \dots a_{2} a_{1} = 0$ for some $n$ .

$rad M_{n} (k)$: $= M_{n} (rad k)$ for every ring $k$ and every $n \geq 1$ ; this is the Morita invariance of the Jacobson radical.
$E_{i j}$: The matrix units of $M_{n} (k)$ , satisfying $E_{i j} E_{k l} = δ_{j k} E_{i l}$ and $\sum_{i} E_{i i} = 1$ .
$A^{n}$: The column module of a left $k$ -module $A$ ; every left $M_{n} (k)$ -module is isomorphic to $A^{n}$ with $A = E_{11} N$ .
${ann}_{N} (J)$: ${x \in N : J x = 0}$ . By $(23.16)$ , a right ideal $J$ is right T-nilpotent precisely when this is nonzero for every nonzero left module $N$ .
Simple ring: No two-sided ideals other than $0$ and itself. A semisimple simple ring is $M_{n} (D)$ for a division ring $D$ .

In the commutative setting left and right T-nilpotency coincide, so the side qualifiers may be dropped from (23.24) but not from (23.23).

Core Concepts

Why the simple quotient case is the tractable one

Write $1 = e_{1} + \dots + e_{m}$ as a sum of orthogonal local idempotents, which is possible for any semiperfect ring by $(23.6)$ . If $R / rad R$ is simple artinian, all the simple right modules ${\bar{e}}_{i} \bar{R}$ are isomorphic, hence so are the $e_{i} R$ ; the right regular module is then $M \oplus \dots \oplus M$ for a single strongly indecomposable $M$ , and $R ≅ End (M^{m}) ≅ M_{m} (k)$ with $k = End (M)$ local. That is $(23.10)$ , and perfectness adds nothing to the shape — only a condition on $rad k$ .

Transporting T-nilpotency across Morita equivalence

The definition of T-nilpotency is stated in terms of elements and is awkward under Morita equivalence; the module criterion $(23.16) (3)$ is stated in terms of module categories and is not. Since $M_{n} (k) -Mod$ and $k -Mod$ are equivalent via $N \mapsto E_{11} N$ and $A \mapsto A^{n}$ , the criterion transports for free. This is the reusable idea of the section.

Element form

$a_{n} \dots a_{1} = 0$

Immediate to check on a corner, useless for building matrices out of scalars.

Right module form

$M J = M \Rightarrow M = 0$

The unrestricted Nakayama lemma; the form used to produce projective covers.

Left module form

${ann}_{N} (J) \neq 0$ for $N \neq 0$

The form that transports along a Morita equivalence, and the one used in the proof of $(23.23)$ .

Commutativity kills the obstruction

For commutative $R$ the decomposition $1 = e_{1} + \dots + e_{m}$ into orthogonal local idempotents is automatically a decomposition into central idempotents, so $R ≅ e_{1} R \times \dots \times e_{m} R$ with each $e_{i} R = e_{i} R e_{i}$ local. Nothing needs to be lifted, and the classification is complete.

Key Results

Theorem(23.23)Right perfect rings with simple quotient

For a ring $R$ the following are equivalent:

$R$ is right perfect and $R / rad R$ is simple;
$R ≅ M_{n} (k)$ for some $n \geq 1$ and some local ring $k$ whose maximal ideal $rad k$ is right T-nilpotent.

When these hold, $n$ is uniquely determined and $k$ is unique up to isomorphism; moreover $R$ is indecomposable as a ring.

Proof

**(2) $\Rightarrow$ (1).** Let $R = M_{n} (k)$ with $k$ local. Then $rad R = M_{n} (rad k)$ and $R / rad R ≅ M_{n} (k / rad k)$ is a matrix ring over a division ring, hence simple artinian. It remains to prove that $M_{n} (rad k)$ is right T-nilpotent, and rather than manipulate matrix sequences we verify criterion (3) of $(23.16)$ .

Let $N$ be a nonzero left $R$ -module and put $A = E_{11} N$ , a left $k$ -module. If $A = 0$ then, since $E_{i i} = E_{i 1} E_{11} E_{1 i}$ , we get $E_{i i} N = 0$ for all $i$ and therefore $N = (\sum_{i} E_{i i}) N = 0$ ; so $A \neq 0$ . As $rad k$ is right T-nilpotent, $(23.16)$ applied over $k$ gives some $0 \neq a \in A$ with $(rad k) a = 0$ . Under the standard isomorphism $N ≅ A^{n}$ consider the column $v = (a, a, \dots, a)^{t} \neq 0$ . For $r = (r_{i j}) \in M_{n} (rad k)$ the $i$ -th entry of $r v$ is $\sum_{j} r_{i j} a = 0$ . Hence ${ann}_{N} (rad R) \neq 0$ for every $N \neq 0$ , and $(23.16)$ returns that $rad R$ is right T-nilpotent. So $R$ is right perfect.

**(1) $\Rightarrow$ (2).** A right perfect ring is semiperfect $(23.19)$ , so $(23.10)$ applies: $R ≅ M_{n} (k)$ for some local ring $k$ , with $n$ and $k$ uniquely determined and $R$ indecomposable. It remains to see that $rad k$ is right T-nilpotent. Let $a_{1}, a_{2}, \dots \in rad k$ and set $b_{i} = a_{i} E_{11} \in M_{n} (rad k) = rad R$ . Since $E_{11} E_{11} = E_{11}$ ,

b_{n} b_{n - 1} \dots b_{1} = (a_{n} a_{n - 1} \dots a_{1}) E_{11} .

Right T-nilpotency of $rad R$ gives $n$ with the left-hand side zero, and comparing $(1, 1)$ entries yields $a_{n} \dots a_{1} = 0$ .

Corollary—Matrix rings preserve perfectness

For any ring $k$ and any $n \geq 1$ , $M_{n} (k)$ is right perfect if and only if $k$ is right perfect. The same holds with left in place of right, and with semiperfect in place of perfect $(23.9)$ .

Proof

$rad M_{n} (k) = M_{n} (rad k)$ and $M_{n} (k) / rad M_{n} (k) ≅ M_{n} (k / rad k)$ , which is semisimple exactly when $k / rad k$ is. For the radical condition, run the two arguments of the previous proof with $rad k$ in place of the maximal ideal: neither used locality of $k$ , only the module criterion $(23.16)$ in one direction and the corner computation $b_{i} = a_{i} E_{11}$ in the other.

Theorem(23.24)Commutative perfect rings

A commutative ring $R$ is perfect if and only if it is a finite direct product of local rings each of whose maximal ideals is T-nilpotent.

Proof

( $\Rightarrow$ ) A perfect ring is semiperfect $(23.19)$ , so by the commutative semiperfect classification $(23.11)$ we may write $R ≅ R_{1} \times \dots \times R_{m}$ with each $R_{i}$ local. Then $rad R = rad R_{1} \times \dots \times rad R_{m}$ . Given a sequence $a_{1}, a_{2}, \dots \in rad R_{i}$ , the elements ${\tilde{a}}_{j} = (0, \dots, a_{j}, \dots, 0)$ lie in $rad R$ , and a vanishing product ${\tilde{a}}_{n} \dots {\tilde{a}}_{1} = 0$ forces $a_{n} \dots a_{1} = 0$ in the $i$ -th coordinate. So each $rad R_{i}$ is T-nilpotent.

( $\Leftarrow$ ) Each factor $R_{i}$ is local with T-nilpotent maximal ideal, so $R_{i} / rad R_{i}$ is a field and $R_{i}$ is perfect by $(23.18)$ ; commutativity makes the two sides agree. For the product, $R / rad R ≅ \prod_{i} R_{i} / rad R_{i}$ is a finite product of fields, hence semisimple. Given a sequence $(a_{j}^{(1)}, \dots, a_{j}^{(m)})_{j \geq 1}$ in $rad R$ , choose $n_{i}$ with $a_{n_{i}}^{(i)} \dots a_{1}^{(i)} = 0$ for each $i$ ; products only get shorter-lived, so $n = \max_{i} n_{i}$ annihilates every coordinate simultaneously and the product of the first $n$ terms is $0$ . Hence $rad R$ is T-nilpotent and $R$ is perfect.

Remark—The finiteness of the product is essential

The step $n = \max_{i} n_{i}$ is the only place where finiteness is used, and it is indispensable. An infinite product of local rings with T-nilpotent maximal ideals — for instance $\prod_{i \geq 1} k [x] / (x^{i})$ — has unbounded nilpotence indices, and the product ring is not even semilocal.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Three moves carry both theorems.

Corner restriction. To push a property from $M_{n} (k)$ down to $k$ , embed a sequence of $k$ via $a \mapsto a E_{11}$ and read the answer off the $(1, 1)$ entry. This works for T-nilpotency, nilness and nilpotence alike.
Change of criterion before change of ring. Do not transport a definition stated with elements; first replace it by an equivalent statement about the module category, then transport. The proof of $(2) \Rightarrow (1)$ in $(23.23)$ is the model.
Reduce to the semiperfect classification. Both theorems obtain the shape of the ring from $(23.10)$ or $(23.11)$ and then verify only the extra radical condition. Perfectness never has to be re-derived from scratch.

Worked Example

A finite matrix ring

Take $k = ℤ / 9 ℤ$ and $R = M_{3} (k)$ . The ring $k$ is local with $rad k = 3 ℤ / 9 ℤ$ and $(rad k)^{2} = 0$ , so $rad k$ is nilpotent, hence T-nilpotent on both sides. Therefore

rad R = M_{3} (3 ℤ / 9 ℤ), (rad R)^{2} = M_{3} (9 ℤ / 9 ℤ) = 0, R / rad R ≅ M_{3} (𝔽_{3}) .

(E.1)

$R$ is semiprimary, hence perfect, and its radical quotient is simple — the hypotheses of $(23.23)$ with $n = 3$ .

The uniqueness clause says that $n = 3$ and $k = ℤ / 9$ are recoverable from $R$ alone: $n$ is the number of indecomposable summands of $R_{R}$ and $k$ is the endomorphism ring of any one of them.

Where the T-nilpotency clause bites

Take $k = ℤ_{(p)}$ , the localisation of $ℤ$ at the prime $p$ , and $R = M_{2} (k)$ . Then $k$ is local with $rad k = p ℤ_{(p)}$ , so $R$ is semiperfect with

R / rad R ≅ M_{2} (𝔽_{p}) simple, yet p^{n} \neq 0 for all n .

(E.2)

$rad k$ is not nil, so not T-nilpotent on either side; $R$ satisfies $(23.10)$ but not $(23.23)$ .

So $R$ is semiperfect with simple radical quotient and is not perfect on either side. The same conclusion holds for $M_{n} (k [[x]])$ . This is exactly why $(23.23)$ must carry the T-nilpotency hypothesis explicitly rather than deriving it from the shape $M_{n} (k)$ .

A one-sided instance

Let $k = k_{0} \cdot 1 + J$ be the local ring of $(23.22)$ : finitely supported strictly upper triangular $ℕ \times ℕ$ matrices over a field $k_{0}$ , adjoined to the scalars. Then $rad k = J$ is right T-nilpotent but not left T-nilpotent, so $R = M_{2} (k)$ is right perfect with $R / rad R ≅ M_{2} (k_{0})$ simple, and is not left perfect. Both hypotheses of $(23.23)$ are one-sided for a reason.

The commutative classification in action

$ℤ / 12 ℤ ≅ ℤ / 4 ℤ \times ℤ / 3 ℤ$ : two local factors with maximal ideals $2 ℤ / 4 ℤ$ (square zero) and $0$ . Both are nilpotent, hence T-nilpotent, so $ℤ / 12$ is perfect — as it must be, being finite and therefore artinian.

A non-artinian instance: let $A = k [t_{1}, t_{2}, \dots] / (t_{i} t_{j} (i \neq j), t_{i}^{i + 1})$ , a local ring whose maximal ideal is T-nilpotent but not nilpotent. Then $A \times ℤ / 4 ℤ$ has no chain condition on ideals, yet $(23.24)$ certifies it as perfect. By contrast $ℤ_{(p)} \times ℤ / 4 ℤ$ is semiperfect and not perfect, because the first factor's maximal ideal is not nil.

Frameworks and Models

The two classifications sit inside a single ladder of structure theorems, each obtained from the one above by weakening the condition on the radical.

Structure theorems by radical condition — shape of the ring is constant; only the radical hypothesis changes
- $rad R = 0$
  - Wedderburn-Artin: $R ≅ \prod_{i} M_{n_{i}} (D_{i})$
  - simple case: $R ≅ M_{n} (D)$ , $D$ a division ring
- idempotents lift, no radical condition
  - semiperfect with simple quotient: $R ≅ M_{n} (k)$ , $k$ local $(23.10)$
  - commutative semiperfect: finite product of local rings $(23.11)$
- $rad R$ right T-nilpotent
  - right perfect with simple quotient: $R ≅ M_{n} (k)$ , $rad k$ right T-nilpotent $(23.23)$
  - commutative perfect: finite product of local rings with T-nilpotent maximal ideal $(23.24)$
- $rad R$ nilpotent
  - semiprimary with simple quotient: $R ≅ M_{n} (k)$ , $rad k$ nilpotent
  - commutative artinian: finite product of artinian local rings $(23.12)$

Reading down the ladder, the normal form $M_{n} (k)$ never changes. Every theorem in this family is a statement about which local rings $k$ are admissible, and the answer is always "those whose maximal ideal satisfies the corresponding nilpotence condition".

Process and Workflow

How do I identify a given ring against $(23.23)$ and $(23.24)$ ?

$R$ is commutativeDecompose

1

into orthogonal primitive idempotents. If there are infinitely many,

R

is not even semiperfect. Otherwise check each factor: local with T-nilpotent maximal ideal gives perfect by

(23.24)

$R / rad R$ is simpleWrite

R ≅ M_{n} (k)

using

(23.10)

, identify

k

as the endomorphism ring of an indecomposable summand of

R_{R}

, and test

rad k

for right T-nilpotency.

$R / rad R$ has several simple factorsNo normal form is available: the centrally primitive idempotents of

R / rad R

need not lift centrally. Fall back on Bass's Theorem P and treat the ring as a whole.

$R$ is a matrix ring over somethingUse the corollary:

M_{n} (k)

is right perfect exactly when

k

is. Reduce to the coefficient ring before doing any work.

For finite rings every branch terminates immediately: a finite ring is artinian, hence semiprimary, hence perfect, and $(23.23)$ reduces to the classical statement that a finite ring with simple radical quotient is a matrix ring over a finite local ring.

Comparison and Classification

Refining the semiperfect classifications
Hypothesis on $R$	Normal form	Condition on the local ring $k$	Lam
Semiperfect, $R / rad R$ simple	$M_{n} (k)$	$k$ local, no further condition	(23.10)
Right perfect, $R / rad R$ simple	$M_{n} (k)$	$rad k$ right T-nilpotent	(23.23)
Semiprimary, $R / rad R$ simple	$M_{n} (k)$	$rad k$ nilpotent	(23.19) plus (23.10)
Commutative semiperfect	$\prod_{i = 1}^{m} R_{i}$	each $R_{i}$ local	(23.11)
Commutative perfect	$\prod_{i = 1}^{m} R_{i}$	each $rad R_{i}$ T-nilpotent	(23.24)
Commutative artinian	$\prod_{i = 1}^{m} R_{i}$	each $R_{i}$ artinian local	(23.12)

Test cases against the two classifications
	Semiperfect	Right perfect	Perfect	Semiprimary
$M_{3} (ℤ / 9 ℤ)$	yes	yes	yes	yes
$M_{2} (ℤ_{(p)})$	yes	no	no	no
$M_{2} (k [[x]])$	yes	no	no	no
$M_{2}$ of the ring of $(23.22)$	yes	yes	no	no
$ℤ / 12 ℤ$	yes	yes	yes	yes
$A \times ℤ / 4 ℤ$ , $A$ as in the example	yes	yes	yes	no
$\prod_{i \geq 1} k [x] / (x^{i})$	no	no	no	no

Test cases against the two classifications

Failure Modes and Common Mistakes

Do not expect a normal form when $R / rad R$ has more than one simple factor; the centrally primitive idempotents may fail to lift centrally, which is why $(23.23)$ is restricted to the simple case.
Do not confuse " $R / rad R$ simple" with " $R$ simple". A ring with simple radical quotient is usually far from simple — $M_{3} (ℤ / 9)$ has the proper two-sided ideal $M_{3} (3 ℤ / 9)$ .
Do not try to prove $M_{n} (rad k)$ right T-nilpotent by manipulating matrix entries; use the module criterion $(23.16)$ .

Quick Reference

(23.23)right perfect

+

simple quotient

iff M_{n} (k)

k

local,

rad k

right T-nilpotent

(23.24)commutative perfect

iff

finite product of local rings with T-nilpotent maximal ideal

Uniqueness

n

and

k

are determined by

R

;

R

is indecomposable

Radical

rad M_{n} (k) = M_{n} (rad k)

Morita

M_{n} (k)

right perfect

iff k

right perfect

Key toolcriterion

(23.16)

{ann}_{N} (J) \neq 0

for all

N \neq 0

Semiperfect versions

(23.10)

and

(23.11)

, without the T-nilpotency clause

Failure case

M_{2} (ℤ_{(p)})

: semiperfect, simple quotient, not perfect

Checklist for applying the classifications
Step	What to verify	Reference
1	$R / rad R$ semisimple	(23.18)
2	Is that quotient simple, or is $R$ commutative?	(23.10), (23.11)
3	Extract $n$ and the local ring $k$ , or the local factors $R_{i}$	(23.6), (23.10)
4	Test $rad k$ for right T-nilpotency	(23.13), (23.16)
5	Conclude right perfect, perfect, or neither	(23.23), (23.24)

Frequently Asked Questions

Why is the converse direction of

(23.23)

proved with modules instead of matrices?

Because right T-nilpotency of $M_{n} (rad k)$ does not follow entrywise from right T-nilpotency of $rad k$ . A product of matrices mixes entries along every index path, and the vanishing index for each path depends on the path. The criterion $(23.16) (3)$ replaces the sequence condition by the requirement that every nonzero left module have nonzero annihilator submodule, and that requirement transfers along the equivalence between $M_{n} (k)$ -modules and $k$ -modules without any bookkeeping.

Is there a version of

(23.23)

for perfect rings with several simple factors?

Not in the same explicit form. If $R / rad R ≅ \prod_{i} M_{n_{i}} (D_{i})$ with $m > 1$ factors, the corresponding centrally primitive idempotents need not lift to central idempotents of $R$ , so $R$ need not decompose as a product matching the quotient. The theory of blocks and basic rings in §25 is the substitute; for the ring-level classification, only the simple and the commutative cases are clean.

Does

(23.24)

cover all commutative artinian rings?

Yes. A commutative artinian ring is a finite product of artinian local rings, and an artinian local ring has nilpotent maximal ideal, which is T-nilpotent. The extra generality of $(23.24)$ lies in allowing T-nilpotent but non-nilpotent maximal ideals, which produces perfect commutative rings with no chain condition at all.

How does one recover

n

and

k

from

R

Decompose the right regular module $R_{R}$ into indecomposable summands; by the simplicity of $R / rad R$ they are all isomorphic to a single strongly indecomposable module $M$ , and $n$ is their number while $k ≅ End (M_{R})$ . Krull-Schmidt-type uniqueness for semiperfect rings makes both invariants well defined.

Is a local ring automatically perfect?

No. Local means only that $rad k$ is the set of non-units, which makes $k / rad k$ a division ring and $k$ semiperfect. Perfectness additionally requires $rad k$ to be T-nilpotent, which fails for $k [[x]]$ , for $ℤ_{(p)}$ , and for every local domain that is not a division ring.

What replaces

(23.24)

for noncommutative perfect rings?

Nothing as sharp. One has Bass's Theorem P as a characterisation, the decomposition $1 = e_{1} + \dots + e_{m}$ into orthogonal local idempotents from $(23.6)$ , and the block theory of §25. But there is no finite list of building blocks: the local rings with one-sided T-nilpotent radical are already an unclassifiable family.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §23, (23.10)–(23.12) and (23.23)–(23.24) (pp. 349–357).
H. Bass, “Finitistic dimension and a homological generalization of semi-primary rings”, Transactions of the American Mathematical Society 95 (1960), 466–488.
N. Jacobson, Basic Algebra II, 2nd edition, W. H. Freeman, 1989, Chapter 3 (modules over matrix rings and Morita theory).
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §§27–28.
H. Matsumura, Commutative Ring Theory, Cambridge Studies in Advanced Mathematics 8, Cambridge University Press, 1986, §8 (structure of artinian and semilocal rings).

AI Suggested Questions

Prove that every left $M_{n} (k)$ -module is isomorphic to the column module over $A = E_{11} N$ .
Give a local ring whose maximal ideal is T-nilpotent but not nilpotent and which is not commutative.
Work out the block decomposition of a perfect ring whose radical quotient has three simple factors.
Show that perfectness is preserved under Morita equivalence, directly from Bass's Theorem P.
Which local rings arise as endomorphism rings of indecomposable projective modules over a perfect ring?
Compare $(23.23)$ with the classification of semiprimary rings with simple radical quotient.
Does an infinite product of perfect rings ever remain perfect, and under what restriction on the factors?

Special Perfect Rings

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Why the simple quotient case is the tractable one

Transporting T-nilpotency across Morita equivalence

$a_{n} \dots a_{1} = 0$

$M J = M \Rightarrow M = 0$

${ann}_{N} (J) \neq 0$ for $N \neq 0$

Commutativity kills the obstruction

Key Results

Proof Techniques and Method

Worked Example

A finite matrix ring

Where the T-nilpotency clause bites

A one-sided instance

The commutative classification in action

Frameworks and Models

Process and Workflow

Comparison and Classification

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 simple quotient case is the tractable one

Transporting T-nilpotency across Morita equivalence

an⋯a1=0

MJ=M⇒M=0

annN(J)≠0 for N≠0

Commutativity kills the obstruction

06Key Results

07Proof Techniques and Method

08Worked Example

A finite matrix ring

Where the T-nilpotency clause bites

A one-sided instance

The commutative classification in action

09Frameworks and Models

10Process and Workflow

11Comparison and Classification

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

$a_{n} \dots a_{1} = 0$

$M J = M \Rightarrow M = 0$

${ann}_{N} (J) \neq 0$ for $N \neq 0$

Key Results

Proof Techniques and Method

Worked Example

Frameworks and Models

Process and Workflow

Comparison and Classification

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions