Basic Idempotents

Q: Why is a basic idempotent automatically full?

Because in the semisimple quotient each e_i sits in a different simple component and generates it as a two-sided ideal, so the e_i together generate all of R. Lifting is then a one-line application of Nakayama's Lemma to the cyclic module R/ReR. Irredundancy is not needed for fullness — completeness is.

Q: Can two different basic idempotents give non-isomorphic corner rings?

No, and that is the point of (25.6). Both corner rings are endomorphism rings of the module e_1R ⊕ … ⊕ e_rR, which is determined up to isomorphism by R alone.

Executive Summary

In a semiperfect ring $R$ the identity decomposes as $1 = e_{1} + \dots + e_{n}$ into orthogonal primitive idempotents, but the modules $e_{j} R$ repeat: each isomorphism type occurs with a multiplicity. A basic idempotent is a subsum that keeps exactly one representative of each type.

Lam's $(25.6)$ says the resulting idempotent $e$ is full — $R e R = R$ — and that the corner ring $e R e$ is determined up to isomorphism, independently of every choice made along the way, and is itself semiperfect. That corner ring is the basic ring of $R$ , the smallest ring carrying the same module theory.

$e = e_{1} + \dots + e_{r}$ One per isomorphism type

$R e R = R$ Always full

$e R e$ Unique up to isomorphism

(25.5)–(25.7)Lam's definition and results

Overview

The classification of principal indecomposables says that the projective right modules over a semiperfect ring are governed by a finite list $e_{1} R, \dots, e_{r} R$ . The regular module $R_{R}$ realises that list with multiplicities $n_{1}, \dots, n_{r}$ , and those multiplicities are exactly the matrix sizes in $\bar{R} = R / rad R ≅ \prod_{i} M_{n_{i}} (D_{i})$ .

Multiplicities carry no information about the module category — $M_{n} (R)$ and $R$ have equivalent module categories for every $n$ . Deleting them is the point of a basic idempotent.

R_{R} ≅ n_{1} (e_{1} R) \oplus \dots \oplus n_{r} (e_{r} R) ⟶ e R = e_{1} R \oplus \dots \oplus e_{r} R,

(25.5)

The passage from the regular module to a basic idempotent: keep the list, discard the multiplicities.

Right modules throughout. A basic idempotent defined by the right principal indecomposables also works on the left, because e and the corner ring eRe are two-sided objects; only the labelling of the list changes.

Learning Objectives

State the definition $(25.5)$ of a basic idempotent and of a basic ring.
Explain why a basic idempotent exists in every semiperfect ring.
Prove $R e R = R$ using the semisimple quotient and Nakayama's Lemma.
Prove that two basic idempotents give isomorphic corner rings.
Compute a basic idempotent of $M_{n} (k)$ , of a semisimple ring and of a product.
Explain why $e R e$ is itself a basic semiperfect ring.

Definitions

Definition(25.5)Basic idempotent, basic ring

Let $R$ be a semiperfect ring. An idempotent $e \in R$ is basic if it can be written $e = e_{1} + \dots + e_{r}$ with the $e_{i}$ mutually orthogonal primitive idempotents such that $e_{1} R, \dots, e_{r} R$ represent a complete set of isomorphism classes of principal indecomposable right $R$ -modules — each class occurring exactly once. A basic ring of $R$ is any ring of the form $e R e$ with $e$ a basic idempotent.

Existence is immediate: take any decomposition $1 = e_{1} + \dots + e_{n}$ into orthogonal primitive idempotents, which $(23.6)$ supplies, and keep one $e_{j}$ from each isomorphism class of the modules $e_{j} R$ . The number of summands is then $r$ , the number of simple right $R$ -modules.

Full idempotent: $R e R = R$ . Equivalently the right module $e R$ generates every right $R$ -module, i.e. $e R$ is a generator.
$e R e$: The corner ring, with identity $e$ . Its radical is $e (rad R) e = e R e \cap rad R$ , and $e R e / rad (e R e) ≅ \bar{e} \bar{R} \bar{e}$ .
Multiplicity $n_{i}$: The number of times $e_{i} R$ occurs in a decomposition of $R_{R}$ into principal indecomposables; equal to the matrix size in the $i$ -th Wedderburn factor of $\bar{R}$ .
Basic ring: $e R e$ for e basic. Determined up to isomorphism by R alone, by (25.6).

Core Concepts

Why fullness is the right property

For an arbitrary idempotent $e$ , the corner ring $e R e$ can be much poorer than $R$ : taking $e$ to be a primitive idempotent of $M_{n} (k)$ collapses everything to $k$ , which is fine, but taking $e$ to miss an isomorphism class of principal indecomposables loses a simple module for good. Fullness $R e R = R$ is exactly the condition that no information is lost, and the corner ring theory of §21 is stated in those terms.

e

basic

\bar{R} \bar{e} \bar{R} = \bar{R}

R e R + J = R

R e R = R

The chain is the proof of $(25.6)$ in miniature: verify fullness in the semisimple quotient, where it is a statement about simple rings, then remove the radical with Nakayama's Lemma.

The corner ring as an endomorphism ring

The identification $e R e ≅ {End}_{R} (e R)$ converts uniqueness of the basic ring into uniqueness of the module $e R$ . Two basic idempotents give modules $e R$ and $e' R$ that are isomorphic — both are the direct sum of one copy of each principal indecomposable — so their endomorphism rings are isomorphic. No compatibility between the two idempotents is needed.

Minimal progenerators

A finitely generated projective module is a generator precisely when every principal indecomposable occurs in it. So $e R$ for $e$ basic is a projective generator with all multiplicities equal to $1$ — the smallest one available. This is the reason $e R e$ deserves the name basic ring, and the reason it is Morita equivalent to $R$ .

Key Results

Proposition(25.6)Fullness and uniqueness of the basic ring

Let $R$ be a semiperfect ring and $e \in R$ a basic idempotent. Then:

$e$ is full, i.e. $R e R = R$ ;
if $e'$ is any other basic idempotent of $R$ then $e R e ≅ e' R e'$ as rings;
$e R e$ is a semiperfect ring.

Proof

Write $e = e_{1} + \dots + e_{r}$ as in $(25.5)$ , put $J = rad R$ and $\bar{R} = R / J ≅ S_{1} \times \dots \times S_{r}$ with each $S_{i}$ simple artinian. Each ${\bar{e}}_{i}$ is a primitive idempotent of $\bar{R}$ , so ${\bar{e}}_{i} \bar{R}$ is a simple right ideal and ${\bar{e}}_{i}$ lies in exactly one component; because $e_{1} R, \dots, e_{r} R$ is a complete irredundant list of principal indecomposables, the modules ${\bar{e}}_{i} \bar{R}$ exhaust the simple right $\bar{R}$ -modules without repetition, so after relabelling ${\bar{e}}_{i} \in S_{i}$ for each $i$ .

(1). The two-sided ideal $\bar{R} {\bar{e}}_{i} \bar{R}$ is a nonzero ideal of $\bar{R}$ contained in $S_{i}$ ; as $S_{i}$ is a simple ring, $\bar{R} {\bar{e}}_{i} \bar{R} = S_{i}$ . Summing over $i$ gives $\bar{R} \bar{e} \bar{R} = S_{1} \oplus \dots \oplus S_{r} = \bar{R}$ , that is, $R e R + J = R$ . The right $R$ -module $R / R e R$ is cyclic, hence finitely generated, and $(R / R e R) J = (J + R e R) / R e R = R / R e R$ ; Nakayama's Lemma forces $R / R e R = 0$ , i.e. $R e R = R$ .

(2). Let $e' = e'_{1} + \dots + e'_{r'}$ be a second basic idempotent. Both lists of principal indecomposables are complete and irredundant, so $r' = r$ and, after reindexing, $e_{i} R ≅ e'_{i} R$ for every $i$ . Hence $e R = ⨁_{i} e_{i} R ≅ ⨁_{i} e'_{i} R = e' R$ , and therefore

e R e ≅ {End}_{R} (e R) ≅ {End}_{R} (e' R) ≅ e' R e' .

(25.6)

(3). $e R$ is a finitely generated projective right $R$ -module, so ${End}_{R} (e R) ≅ e R e$ is semiperfect by $(25.3) (3)$ ; equivalently, $e R$ is a finite direct sum of strongly indecomposable modules and $(23.8)$ applies.

Corollary—A basic ring is basic

Let $R$ be semiperfect with basic idempotent $e$ and $B = e R e$ . Then $B / rad B$ is a finite direct product of division rings, so $1_{B} = e$ is a basic idempotent of $B$ and $B$ is its own basic ring.

Proof

By the corner ring theory, $rad B = e J e$ and $B / rad B ≅ \bar{e} \bar{R} \bar{e}$ . With the labelling of the previous proof, ${\bar{e}}_{i} \bar{R} {\bar{e}}_{j} = 0$ for $i \neq j$ because ${\bar{e}}_{i}$ and ${\bar{e}}_{j}$ lie in different simple components, while ${\bar{e}}_{i} S_{i} {\bar{e}}_{i} ≅ End ({\bar{e}}_{i} \bar{R})$ is a division ring since ${\bar{e}}_{i} \bar{R}$ is simple. Hence

B / rad B ≅ D_{1} \times \dots \times D_{r}, D_{i} = {\bar{e}}_{i} \bar{R} {\bar{e}}_{i} a division ring,

(E.0)

A finite product of division rings — exactly the criterion for a semiperfect ring to be basic.

Example(25.7)Three standard computations

$R = M_{n} (k)$ with $k$ local. There is a single principal indecomposable, so $e = E_{11}$ is basic and $e R e = k E_{11} ≅ k$ : the basic ring of a matrix ring over a local ring is that local ring.
$R$ commutative semiperfect. By $(23.11)$ , $R$ is a finite product of local rings, whose primitive idempotents give pairwise non-isomorphic principal indecomposables; hence $1$ is the only basic idempotent and $R$ is its own basic ring.
$R = \prod_{i = 1}^{t} M_{n_{i}} (D_{i})$ semisimple. Applying the first computation componentwise, any basic ring of $R$ is isomorphic to $D_{1} \times \dots \times D_{t}$ .

Remark—Basic idempotents are not unique

Only the ring $e R e$ is canonical. In $M_{2} (k)$ both $E_{11}$ and $E_{22}$ are basic, as is $(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ conjugated by any invertible matrix; all give corner rings isomorphic to $k$ . There is in general no canonical choice, and none is needed.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Prove ideal identities modulo the radical

An equality of ideals $𝔄 = R$ is proved by checking $𝔄 + J = R$ and applying Nakayama to the cyclic module $R / 𝔄$ . Downstairs the statement is about simple rings, where every nonzero ideal is everything.

Move 2

Replace a ring by an endomorphism ring

$e R e ≅ {End}_{R} (e R)$ turns a question about idempotents into a question about a module. Modules are easier to compare, because isomorphism of modules is exactly what the classification theorem controls.

Move 3

Compute corners componentwise

Over a product of simple rings, ${\bar{e}}_{i} \bar{R} {\bar{e}}_{j}$ vanishes unless $i = j$ , so a corner of a semisimple ring is a product of the diagonal corners. Each diagonal corner is a division ring when the idempotent is primitive.

Move 1 is the workhorse of the whole chapter. It is worth noticing that it needs finite generation of $R / 𝔄$ — free here, since the module is cyclic — and nothing else; no chain condition is used anywhere in $(25.6)$ .

Worked Example

A ring with repeated and unrepeated types

Let $k$ be a division ring and take

R = M_{2} (k) \times T_{2} (k), \dim_{k} R = 4 + 3 = 7,

(E.1)

$T_{2} (k)$ is the ring of upper triangular $2 \times 2$ matrices. Both factors are finite-dimensional, hence semiperfect, hence so is the product.

Write $f_{1} = (E_{11}, 0)$ , $f_{2} = (E_{22}, 0)$ , $f_{3} = (0, E_{11})$ , $f_{4} = (0, E_{22})$ . These are orthogonal primitive idempotents with $f_{1} + f_{2} + f_{3} + f_{4} = 1$ , so $R_{R} = f_{1} R \oplus f_{2} R \oplus f_{3} R \oplus f_{4} R$ with $k$ -dimensions $2, 2, 2, 1$ .

The types: $f_{1} R ≅ f_{2} R$ (the two rows of $M_{2} (k)$ are isomorphic modules), while $f_{3} R$ and $f_{4} R$ are non-isomorphic to each other and to $f_{1} R$ , since they are annihilated by the first factor. Hence $r = 3$ and the multiplicities are $n = (2, 1, 1)$ , matching

\bar{R} = R / rad R ≅ M_{2} (k) \times k \times k, rad R = 0 \times (\begin{matrix} 0 & k \\ 0 & 0 \end{matrix}) .

(E.2)

A basic idempotent is therefore $e = f_{1} + f_{3} + f_{4} = (E_{11}, 1)$ , and

e R e = E_{11} M_{2} (k) E_{11} \times T_{2} (k) ≅ k \times T_{2} (k), \dim_{k} e R e = 1 + 3 = 4 .

(E.3)

The multiplicity $2$ has been removed from the first factor; the second factor was already basic.

Checking fullness by hand

In the first coordinate, $M_{2} (k) E_{11} M_{2} (k)$ contains $E_{i 1} E_{11} E_{1 j} = E_{i j}$ for all $i, j$ , hence is all of $M_{2} (k)$ . In the second coordinate $e$ acts as the identity. So $R e R = R$ , as $(25.6)$ predicts. Note also $\bar{e} \bar{R} \bar{e} ≅ k \times k \times k$ , a product of division rings, confirming that $e R e$ is basic.

Process and Workflow

Split the identityWrite

1 = e_{1} + \dots + e_{n}

with orthogonal primitive idempotents. For a matrix presentation the diagonal matrix units usually work.

Group by isomorphism typeDecide which

e_{j} R

are isomorphic. Equivalently, compare

{\bar{e}}_{j} \bar{R}

inside the semisimple quotient, or test the criterion of

(21.20)

Select representativesChoose one idempotent per class and add them up. Any choice works; the sum is a basic idempotent.

Form the cornerCompute

e R e

. Verify

\bar{e} \bar{R} \bar{e}

is a product of division rings as a check that the selection was irredundant.

Is a given idempotent $e$ basic?

Check the quotientCompute

\bar{e} \bar{R} \bar{e}

. If it is a finite product of division rings and

\bar{R} \bar{e} \bar{R} = \bar{R}

, then

e

is basic.

Count summandsDecompose

e R

into principal indecomposables.

e

is basic exactly when every type occurs, and occurs once.

Full but not basicFullness alone is weaker:

e = 1

is always full, and so is any idempotent whose module

e R

contains every type with any multiplicities.

Comparison and Classification

Basic idempotents and basic rings in examples
Ring $R$	Types $r$	A basic idempotent	Basic ring $e R e$
Division ring $D$	1	$1$	$D$
$M_{n} (D)$	1	$E_{11}$	$D$
$M_{n} (k)$ , $k$ local	1	$E_{11}$	$k$
$\prod_{i = 1}^{t} M_{n_{i}} (D_{i})$	$t$	one $E_{11}$ per factor	$D_{1} \times \dots \times D_{t}$
$T_{n} (k)$ , $k$ a division ring	$n$	$1$	$T_{n} (k)$ itself
Commutative semiperfect	number of local factors	$1$	$R$ itself
$M_{2} (k) \times T_{2} (k)$	3	$(E_{11}, 1)$	$k \times T_{2} (k)$

Properties of an idempotent $e$ in a semiperfect ring
	$e$ primitive	$e$ basic	$e = 1$	$e$ central
$R e R = R$ automatic	no	yes	yes	no
$e R e$ semiperfect	yes	yes	yes	yes
$e R e$ local	yes	no	no	no
$e R e$ determined by $R$ alone	partial	yes	yes	no
$e R$ a generator of the module category	no	yes	yes	no

Properties of an idempotent $e$ in a semiperfect ring

The row * $e R e$ determined by $R$ alone* is marked part for primitive $e$ : the corner ring is determined once the isomorphism type of $e$ is fixed, but different primitive idempotents can give non-isomorphic local rings when $R$ has several types.

Relationship Map

Idempotents of a semiperfect ring $R$
- Primitive — $e R e$ local; $e R$ a principal indecomposable
  - sums of these give every idempotent up to isomorphism
  - not full in general
- Basic — one summand per isomorphism type
  - always full: $R e R = R$
  - $e R$ is a minimal progenerator
  - $e R e$ unique up to isomorphism, semiperfect and basic
- Central — gives block decompositions
  - $e R e = e R$ is a two-sided ideal and a ring
  - full only if $e = 1$

e

basic

e

fullideal lattices of

e R e

and

R

agreeblocks correspond

The final implication is the content of the basic ring theorem, which uses fullness through the corner ring correspondence $𝔄 \mapsto R 𝔄 R$ .

Design Considerations

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

Whether to pass to the basic ring at all. Do it when the question is categorical — modules, Ext groups, block structure. Do not do it when the question concerns $R$ itself as a ring, such as its centre's dimension, its cardinality, or a specific matrix realisation.
Which representatives to pick. Any selection works, so pick the ones with the simplest description: diagonal matrix units, vertex idempotents of a quiver, or the idempotents already used to present $R$ .
Left or right. The list of principal indecomposables can be indexed by either side; the resulting basic idempotents are basic for both, since $e R e$ does not know which side defined it.
Keeping the multiplicities. The data lost is exactly the tuple $(n_{1}, \dots, n_{r})$ of matrix sizes in $\bar{R}$ . Record it separately if the original ring must be reconstructed.

Failure Modes and Common Mistakes

Do not expect a canonical basic idempotent. Only the isomorphism class of $e R e$ is well defined.
Do not apply the construction to a ring that is merely semilocal: without idempotent lifting, the list of principal indecomposables may not exist.
Do not assume $\dim e R e$ divides $\dim R$ or bears any simple relation to it; the example above has $7$ and $4$ .
Do not confuse the multiplicity $n_{i}$ of a type in $R_{R}$ with the composition multiplicity of a simple module in a principal indecomposable.

Quick Reference

Hypothesis

R

semiperfect,

J = rad R

r

= number of simple right

R

-modules

Definition

e = e_{1} + \dots + e_{r}

orthogonal primitive, the

e_{i} R

a complete irredundant list

ExistenceTake a subsum of any decomposition

1 = e_{1} + \dots + e_{n}

Fullness

R e R = R

, proved via

\bar{R}

and Nakayama

Uniqueness

e R e ≅ {End}_{R} (e R)

is independent of the choice

Semiperfectness

e R e

is semiperfect, and is its own basic ring

Radical

rad (e R e) = e (rad R) e

;

e R e / rad (e R e) ≅ D_{1} \times \dots \times D_{r}

Trivial case

e = 1

is basic exactly when

R

is a basic ring

Statement locator
Fact	Statement	Reference
Definition	basic idempotent, basic ring $e R e$	(25.5)
Fullness	$e$ basic $\Rightarrow R e R = R$	(25.6)
Uniqueness	$e R e ≅ e' R e'$ for basic $e, e'$	(25.6)
Semiperfect corner	${End}_{R} (P)$ semiperfect for f.g. projective $P$	(25.3)(3), (23.8)
Corner radical	$rad (e R e) = e (rad R) e$	(21.10)
Ideal correspondence	$𝔄 \mapsto R 𝔄 R$ bijective for full $e$	(21.11)
Examples	$M_{n} (k)$ , commutative, semisimple	(25.7)

Frequently Asked Questions

Why is a basic idempotent automatically full?

Because in the semisimple quotient each ${\bar{e}}_{i}$ sits in a different simple component and generates it as a two-sided ideal, so the ${\bar{e}}_{i}$ together generate all of $\bar{R}$ . Lifting is then a one-line application of Nakayama's Lemma to the cyclic module $R / R e R$ . Irredundancy is not needed for fullness — completeness is.

Can two different basic idempotents give non-isomorphic corner rings?

No, and that is the point of $(25.6)$ . Both corner rings are endomorphism rings of the module $e_{1} R \oplus \dots \oplus e_{r} R$ , which is determined up to isomorphism by $R$ alone.

Is a basic idempotent unique up to conjugation by a unit?

Not in general. Two basic idempotents are built from isomorphic idempotents in the sense of $(21.20)$ , which gives an isomorphism of the corresponding modules but not automatically an inner automorphism of $R$ . Conjugacy holds in favourable cases, such as matrix rings over local rings, but should not be assumed.

What happens if I select two idempotents from the same isomorphism class?

The resulting idempotent is still full, and the corner ring is still semiperfect and Morita equivalent to $R$ , but it is no longer basic: its semisimple quotient acquires a matrix factor of size at least $2$ . The construction degrades gracefully, but the minimality is lost.

How do I recognise a basic ring without computing idempotents?

Reduce modulo the radical: a semiperfect ring is basic exactly when $R / rad R$ is a finite direct product of division rings. That criterion is proved on the Basic Rings and Morita Reduction page and needs no idempotent bookkeeping.

Does the basic idempotent depend on choosing right modules rather than left?

No. The number of isomorphism classes of principal indecomposables is the same on both sides — it is the number of simple components of $R / rad R$ — and an idempotent basic for the right theory is basic for the left theory as well.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §25 (pp. 374–376).
T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics 189, Springer-Verlag, 1999, §18 (Morita theory).
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §21 and §27.
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981, §6.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Prove that $e R$ is a generator for the category of right $R$ -modules if and only if $R e R = R$ .
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Give an example of a full idempotent that is not basic and describe how its corner ring differs.
Show that a semiperfect ring and its basic ring have the same number of blocks.
How does one recover $R$ from its basic ring together with the multiplicities $n_{1}, \dots, n_{r}$ ?
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01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why fullness is the right property

The corner ring as an endomorphism ring

Minimal progenerators

06Key Results

07Proof Techniques and Method

Prove ideal identities modulo the radical

Replace a ring by an endomorphism ring

Compute corners componentwise

08Worked Example

A ring with repeated and unrepeated types

Checking fullness by hand

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Design Considerations

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

Relationship Map

Design Considerations

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions