Primitive, Irreducible and Local Idempotents

Q: Is every primitive idempotent local in a finite-dimensional algebra?

Yes. If A is a finite-dimensional algebra over a field then eAe is a finite-dimensional algebra with no nontrivial idempotents, hence — being semiprimary — a local ring. So over such algebras the distinction between primitive and local collapses. It reappears as soon as the algebra is infinite-dimensional or the base is not a field.

Q: Why is the corner ring eRe rather than the subring generated by e?

Because eRe is precisely the set of elements acting as endomorphisms of the summand eR: an element r with er = r = re multiplies eR into itself and commutes with the R-action on the correct side. The subring generated by e inside R carries almost no information.

Q: Does a minimal right ideal always come from an idempotent?

No. Brauer's Lemma (10.22) says a minimal right ideal I equals eR for a right irreducible idempotent e exactly when I^2 ≠ 0. If I^2 = 0 then I lies in the radical and contains no nonzero idempotent at all; the strictly upper triangular matrices in the 2 × 2 triangular ring are such an I.

Executive Summary

An idempotent $e = e^{2}$ splits $R$ as a right module: $R = e R \oplus (1 - e) R$ . How far the splitting can be pushed is controlled entirely by one ring, the corner $e R e$ . Three successively stronger conditions on that corner give the three grades of idempotent used throughout the structure theory: $e R e$ has no nontrivial idempotents (primitive), $e R e$ is local (local), $e R e$ is a division ring (right irreducible).

Primitivity and locality are left-right symmetric; irreducibility is not. The single most useful result on this page is $(21.18)$ : $e$ is local exactly when $e R$ has a unique maximal submodule, namely $e J$ with $J = rad R$ — the fact that makes $e R \mapsto e R / e J$ a well-defined passage from principal indecomposables to simple modules.

3Grades of idempotent

$e R e$ The single test object

$e R / e J$ Simple quotient when local

StrictBoth implications

Overview

Let $R$ be a ring with identity and $e = e^{2} \in R$ , $f = 1 - e$ . The Peirce decompositions give

R = e R \oplus f R (right ideals), R = R e \oplus R f (left ideals),

(21.1)-(21.2)

Every direct decomposition of $R_{R}$ into two summands arises this way, from a unique idempotent.

Refining a decomposition means refining the idempotent: writing $e = α + β$ with $α, β$ nonzero orthogonal idempotents is the same as splitting $e R = α R \oplus β R$ . An idempotent that admits no such refinement is primitive. Requiring more — that the summand $e R$ be not merely indecomposable but strongly indecomposable, or actually simple — produces the two finer grades.

e

right irreducible

e

local

e

primitive

Neither implication reverses. The identity of $ℤ$ is primitive but not local; the idempotent $(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ in the ring of upper triangular $2 \times 2$ matrices over a field is local but not right irreducible. Both are computed in full below.

Learning Objectives

Read primitivity, locality and irreducibility of $e$ off the corner ring $e R e$ .
Prove that $e R$ minimal forces $e R e$ to be a division ring, and recover the converse over a semiprime ring.
Prove the equivalence $(21.18)$ : $e$ local $iff$ $\bar{e}$ right irreducible in $R / rad R$ $iff$ $e R / e J$ simple.
Exhibit an idempotent that is left irreducible but not right irreducible.
Use $M e \neq 0$ to detect $e R / e J$ among the composition factors of a finite-length module.
Explain why local idempotents, not primitive ones, index the simple modules of a semiperfect ring.

Definitions

Definition(21.8)Primitive idempotent

A nonzero idempotent $e \in R$ is primitive if any of the following equivalent conditions holds: $e R$ is indecomposable as a right $R$ -module; $R e$ is indecomposable as a left $R$ -module; the ring $e R e$ has no idempotents other than $0$ and its identity $e$ ; $e$ is not the sum of two nonzero orthogonal idempotents of $R$ .

Definition(21.9)Local idempotent

An idempotent $e \in R$ is local if the corner ring $e R e$ is a local ring; equivalently $e R$ is strongly indecomposable as a right $R$ -module, equivalently $R e$ is strongly indecomposable as a left $R$ -module. Since a local ring has no idempotents but $0$ and $1$ , a local idempotent is primitive (and in particular nonzero, because the zero ring is not local).

Definition(21.15)Right and left irreducible idempotents

A nonzero idempotent $e \in R$ is right irreducible if $e R$ is a minimal right ideal of $R$ , i.e. a simple right $R$ -module; it is left irreducible if $R e$ is a minimal left ideal. Unlike primitivity and locality, this notion genuinely depends on the side.

$e R e$: The corner ring at $e$ : the set of $r$ with $e r = r = r e$ , a ring with identity $e$ (usually not a subring containing $1$ ).
Orthogonal: Idempotents $α, β$ with $α β = β α = 0$ ; then $α + β$ is again idempotent.
Full idempotent: An idempotent $e$ with $R e R = R$ . Fullness is unrelated to primitivity: the identity is full and is often not primitive.
$J$: Throughout this page, $J = rad R$ , the Jacobson radical, and $\bar{R} = R / J$ , $\bar{e} = e + J$ .
Principal indecomposable: A right module of the form $e R$ with $e$ a primitive idempotent; a direct summand of $R_{R}$ that cannot be split further.

Rings have an identity and modules are unital. Simple and minimal are used interchangeably: a minimal right ideal is exactly a simple submodule of $R_{R}$ .

Core Concepts

Everything happens in the corner

The bridge is an isomorphism of rings, obtained by evaluating an endomorphism at $e$ :

{End}_{R} (e R) ≅ e R e, θ ⟼ θ (e) .

(21.7)

Well defined because $θ (e) = θ (e \cdot e) = θ (e) e$ and $θ (e) \in e R$ , so $θ (e) \in e R e$ .

A module is indecomposable precisely when its endomorphism ring has no nontrivial idempotents, and strongly indecomposable precisely when that endomorphism ring is local. Feeding $(21.7)$ into those two statements produces the equivalences in $(21.8)$ and $(21.9)$ at once. The third grade is Schur's Lemma: if $e R$ is simple then ${End}_{R} (e R)$ is a division ring.

Why irreducibility is the one-sided notion

Primitivity and locality are conditions on the abstract ring $e R e$ , and $e R e$ does not know which side we started on — replacing $R$ by $R^{op}$ replaces $e R e$ by $(e R e)^{op}$ , which is local exactly when $e R e$ is. Irreducibility is a condition on the size of $e R$ inside $R$ , and $e R$ and $R e$ can have very different sizes. Over a semiprime ring the discrepancy disappears, because there $e R e$ being a division ring is equivalent to irreducibility on either side.

The radical of a corner

Passing to $R / J$ is what converts the coarse condition (locality) into the sharp one (irreducibility). The tool is the computation of the radical of a corner ring.

rad (e R e) = J \cap e R e = e J e, e R e / rad (e R e) ≅ \bar{e} \bar{R} \bar{e} .

(21.10)

The corner of the quotient is the quotient of the corner. This is the identity that makes $(21.18)$ work.

So $e R e$ is local $iff$ $e R e / rad (e R e)$ is a division ring $iff$ $\bar{e} \bar{R} \bar{e}$ is a division ring. Since $\bar{R}$ is semiprimitive, hence semiprime, the last condition is exactly irreducibility of $\bar{e}$ in $\bar{R}$ .

Key Results

Proposition(21.16)Irreducible idempotents and division rings

Let $e$ be a nonzero idempotent of a ring $R$ .

If $e$ is right irreducible, then $e R e$ is a division ring.
Conversely, if $R$ is a semiprime ring and $e R e$ is a division ring, then $e$ is right irreducible.

Proof

(1) If $e R$ is a minimal right ideal it is a simple right $R$ -module, so Schur's Lemma makes ${End}_{R} (e R)$ a division ring; by $(21.7)$ that ring is $e R e$ .

(2) Assume $R$ semiprime and $e R e$ a division ring. It suffices to show that every nonzero element of $e R$ generates all of $e R$ . Take $0 \neq e r \in e R$ . Semiprimeness says that $a R a = 0$ forces $a = 0$ , so $e r R e r \neq 0$ : there is $s \in R$ with $(e r s e) r = e r s e r \neq 0$ , whence $e r s e \neq 0$ . Now $e r s e$ is a nonzero element of the division ring $e R e$ , so it has an inverse $e t e \in e R e$ with $(e r s e) (e t e) = e$ . Therefore $e \in e r R$ , giving $e R \subseteq e r R \subseteq e R$ and hence $e r R = e R$ . Thus $e R$ has no proper nonzero submodule.

Corollary(21.17)Comparisons

A right (or left) irreducible idempotent is always local, hence primitive — a division ring is a local ring.
If $R$ is semiprime, an idempotent is right irreducible if and only if it is left irreducible; both are equivalent to $e R e$ being a division ring.
If $R$ is semisimple, an idempotent is right irreducible if and only if it is local, if and only if it is primitive.

In (3) the extra input is that over a semisimple ring $e R$ is a semisimple module, and a semisimple indecomposable module is simple.

Proposition(21.18)Local idempotents modulo the radical

Let $e$ be a nonzero idempotent of $R$ , let $J = rad R$ and $\bar{R} = R / J$ . The following are equivalent:

$e$ is a local idempotent of $R$ ;
$\bar{e}$ is a right irreducible idempotent of $\bar{R}$ ;
$\bar{e}$ is a left irreducible idempotent of $\bar{R}$ ;
$e R / e J$ is a simple right $R$ -module;
$e J$ is the unique maximal submodule of $e R$ .

Proof

First, $\bar{e} \neq 0$ : an idempotent lying in $rad R$ is zero, since $1 - e$ would then be a unit annihilating $e$ . So all five statements concern a nonzero idempotent of $\bar{R}$ .

**(2) $\Leftrightarrow$ (3).** $\bar{R}$ is semiprimitive, hence semiprime, so $(21.17) (2)$ applies.

**(1) $\Leftrightarrow$ (2).** By $(21.10)$ , $\bar{e} \bar{R} \bar{e} ≅ e R e / rad (e R e)$ . A ring is local exactly when its quotient by its radical is a division ring, so $e R e$ is local iff $\bar{e} \bar{R} \bar{e}$ is a division ring. Since $\bar{R}$ is semiprime, $(21.16)$ turns the latter into right irreducibility of $\bar{e}$ .

**(2) $\Leftrightarrow$ (4).** The surjection $e R \to \bar{e} \bar{R}$ has kernel $e R \cap J = e J$ , so $e R / e J ≅ \bar{e} \bar{R}$ as right $R$ -modules, the right-hand side carrying the $R$ -action through $R \to \bar{R}$ . Submodules correspond, so one is simple iff the other is.

**(4) $\Rightarrow$ (5).** Let $I \subseteq e R$ be a submodule with $I not \subseteq e J$ . Its image in the simple module $e R / e J$ is nonzero, hence everything, so $e R = I + e J = I + (e R) J$ . The module $e R$ is cyclic, hence finitely generated, so Nakayama's Lemma gives $I = e R$ . Thus every proper submodule lies in $e J$ , and $e J$ is proper because $e \notin e J \subseteq J$ . **(5) $\Rightarrow$ (4)** is immediate.

Proposition(21.19)Detecting a composition factor

Let $e \in R$ be a local idempotent, $J = rad R$ , and let $M$ be a right $R$ -module of finite composition length. Then $M$ has a composition factor isomorphic to the simple module $e R / e J$ if and only if $M e \neq 0$ , if and only if ${Hom}_{R} (e R, M) \neq 0$ .

Proof

Take a composition series $M = M_{0} \supseteq M_{1} \supseteq \dots \supseteq M_{n} = 0$ and suppose $M e \neq 0$ . If $M_{i} e \subseteq M_{i + 1}$ for every $i$ , then iterating and using $e = e^{n}$ gives $M e = M e^{n} \subseteq M_{n} = 0$ , a contradiction. So some factor $V = M_{i} / M_{i + 1}$ satisfies $V e \neq 0$ .

Choose $v \in V$ with $v e \neq 0$ . Since $V$ is simple, $v e R = V$ , so $λ : e R \to V$ , $λ (e r) = v e r$ , is a well-defined surjective $R$ -homomorphism. Its kernel is a maximal submodule of $e R$ , and by $(21.18) (5)$ the only one is $e J$ . Hence $V ≅ e R / e J$ .

Conversely, if some $M_{i} / M_{i + 1} ≅ e R / e J$ then $(M_{i} / M_{i + 1}) e \neq 0$ , because $e \notin e J$ ; so $M_{i} e not \subseteq M_{i + 1}$ and in particular $M e \supseteq M_{i} e \neq 0$ . The last equivalence is the natural isomorphism ${Hom}_{R} (e R, M) ≅ M e$ , $θ \mapsto θ (e)$ .

Remark(10.22)Brauer's Lemma

A minimal right ideal $I$ of $R$ is of the form $e R$ for a right irreducible idempotent $e$ if and only if $I^{2} \neq 0$ . This is where irreducible idempotents come from in practice: minimal right ideals either square to zero — and then live inside the radical — or are generated by an idempotent.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Translate to the corner

Any question about the summand $e R$ becomes a question about the ring $e R e$ via ${End}_{R} (e R) ≅ e R e$ . Indecomposable, strongly indecomposable and simple correspond to three familiar ring conditions.

Move 2

Kill the radical, then use semiprimeness

Conditions that are awkward over $R$ become clean over $\bar{R} = R / rad R$ , which is semiprime. Transport with $(21.10)$ , argue in $\bar{R}$ , then lift the conclusion back through Nakayama.

Move 3

Use idempotency as a filter

The identity $e = e^{n}$ turns a chain of inclusions $M_{i} e \subseteq M_{i + 1}$ into $M e = 0$ . Multiplying by an idempotent is a projection, so it either kills a layer or survives to the bottom.

Move 2 is the reason the theory of semiperfect rings works at all: there, idempotents also lift back from $\bar{R}$ to $R$ , so the correspondence between local idempotents of $R$ and irreducible ones of $\bar{R}$ becomes a bijection on conjugacy classes, and $e R \mapsto e R / e J$ becomes a bijection between principal indecomposables and simple modules.

Worked Example

Upper triangular matrices: local but not right irreducible

Let $k$ be a field and $R = {(\begin{matrix} a & b \\ 0 & c \end{matrix}) : a, b, c \in k}$ . Then $rad R$ is the set of strictly upper triangular matrices and $(rad R)^{2} = 0$ , so $R$ is not semisimple. Put $e = (\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ and $f = 1 - e = (\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})$ .

R e = {(\begin{matrix} a & 0 \\ 0 & 0 \end{matrix})}, e R = {(\begin{matrix} a & b \\ 0 & 0 \end{matrix})}, e R e = {(\begin{matrix} a & 0 \\ 0 & 0 \end{matrix})} ≅ k .

(E.1)

Since $\dim_{k} R e = 1$ , the left ideal $R e$ is minimal and $e$ is left irreducible. But $\dim_{k} e R = 2$ and $e R \supseteq rad R ⊋ 0$ , with $rad R$ a right ideal, so $e R$ is not minimal and $e$ is not right irreducible. Nevertheless $e R e ≅ k$ is a division ring, so $e$ is a local idempotent, hence primitive.

Check $(21.18)$ on $e$ : here $e J = {(\begin{matrix} 0 & b \\ 0 & 0 \end{matrix})}$ is one-dimensional, so $e R / e J$ is one-dimensional and simple, and $e J$ is visibly the unique maximal submodule of the two-dimensional module $e R$ . The two simple right $R$ -modules are $e R / e J$ and $f R$ , and $R = e R \oplus f R$ is the decomposition of $R_{R}$ into principal indecomposables.

Matrix rings: all three grades coincide

Let $R = M_{n} (k)$ with $k$ a division ring and let $e = E_{11}$ be a matrix unit. Then $e r e = r_{11} e$ for $r = (r_{i j})$ , so $e R e ≅ k$ is a division ring; $e R$ is the set of matrices supported in the first row, a minimal right ideal of $k$ -dimension $n$ . Hence $e$ is right irreducible, left irreducible, local and primitive simultaneously — as $(21.17) (3)$ predicts, because $M_{n} (k)$ is semisimple. Note $e$ is also full: $R e R = R$ .

Primitive but not local

Take $R = ℤ$ and $e = 1$ . Then $e R = ℤ$ is indecomposable as a $ℤ$ -module, so $e$ is primitive; but $e R e = ℤ$ is not a local ring, so $e$ is not local. Replacing $ℤ$ by $ℤ_{(p)}$ makes $e = 1$ local — $e R e = ℤ_{(p)}$ is local — yet still not irreducible, since $ℤ_{(p)}$ is not a minimal ideal of itself.

Process and Workflow

Given a nonzero idempotent $e$ , which grade does it have?

Compute $e R e$ firstIt is a ring with identity

e

. Look for idempotents in it: if there are none besides

0

and

e

, then

e

is primitive; if

e R e

is local,

e

is local; if

e R e

is a division ring,

e

is local and, over a semiprime ring, irreducible.

$e R e$ hard to compute?Compute

rad (e R e) = e J e

instead, using

(21.10)

, and test whether

e R e / e J e

is a division ring. Equivalently test whether

e R / e J

is a simple module.

Need irreducibility?Measure

e R

directly — over a finite-dimensional algebra, compare

\dim_{k} e R

with the dimensions of the known simple modules. Irreducibility is a size condition and cannot be read off

e R e

unless

R

is semiprime.

Working modulo the radical?Locality of

e

R

is exactly irreducibility of

\bar{e}

R / rad R

, by

(21.18)

; if idempotents lift modulo

rad R

, the two settings carry the same information.

Comparison and Classification

The three grades side by side
Grade	Condition on $e R e$	Condition on $e R$	Side-neutral?
Primitive $(21.8)$	no idempotents but $0, e$	indecomposable	yes
Local $(21.9)$	local ring	strongly indecomposable	yes
Right irreducible $(21.15)$	division ring (if $R$ semiprime)	simple	no

Which grade does the idempotent have?
	Primitive	Local	Right irred.	Left irred.
$E_{11}$ in $M_{n} (k)$ , $k$ a division ring	yes	yes	yes	yes
$E_{11}$ in upper triangular $2 \times 2$ over a field	yes	yes	no	yes
$E_{22}$ in upper triangular $2 \times 2$ over a field	yes	yes	yes	no
$1$ in $ℤ_{(p)}$	yes	yes	no	no
$1$ in $ℤ$	yes	no	no	no
$1$ in $k \times k$	no	no	no	no

Which grade does the idempotent have?

The table collapses under hypotheses: over a semiprime ring the last two columns agree; over a semisimple ring all four agree; over a von Neumann regular ring, primitive, irreducible on either side and $e R e$ a division ring are all equivalent.

Relationship Map

$e$ right irreducible — $e R$ simple
- implies
  - $e R e$ is a division ring $(21.16) (1)$
  - $e$ is local $(21.17) (1)$
  - $e$ is primitive
- is implied by, when $R$ is semiprime
  - $e R e$ a division ring $(21.16) (2)$
  - $e$ left irreducible $(21.17) (2)$
- is implied by, when $R$ is semisimple
  - $e$ primitive $(21.17) (3)$

The vertical structure is best seen as a filtration by hypotheses on the ambient ring: the stronger the ring, the more the grades collapse.

Arbitrary ringirreducible

\Rightarrow

local

\Rightarrow

primitive, all strict

Semiprimeright irreducible

\Leftrightarrow

left irreducible

\Leftrightarrow

e R e

a division ring

Semisimpleprimitive

\Leftrightarrow

local

\Leftrightarrow

irreducible

Simple artinian $M_{n} (D)$ all primitive idempotents are conjugate;

e R

is the unique simple module

Applications and Industry Use

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

Representation theory

Projective indecomposables

For a finite-dimensional algebra $A$ , decomposing $1$ into primitive orthogonal idempotents produces the projective indecomposable modules $e_{i} A$ ; $(21.18)$ then identifies their simple tops $e_{i} A / e_{i} J$ , which is how the Cartan matrix and the decomposition matrix are indexed.

Quivers

Vertices are idempotents

For a path algebra $k Q$ modulo an admissible ideal, the vertices of $Q$ are exactly a complete set of primitive orthogonal idempotents, and arrows $i \to j$ correspond to a basis of $e_{i} (J / J^{2}) e_{j}$ . Reconstructing the quiver from the algebra is idempotent bookkeeping.

Computer algebra

Splitting an algebra

GAP, Magma and Sage decompose a finite-dimensional algebra by computing the radical, splitting the semisimple quotient, then lifting a complete set of primitive orthogonal idempotents. $(21.18)$ is the correctness statement behind the lift.

Coding theory

Idempotent generators

Cyclic codes of length $n$ over $𝔽_{q}$ with $\gcd (n, q) = 1$ are ideals of the semisimple ring $𝔽_{q} [x] / (x^{n} - 1)$ , each generated by a unique idempotent; the minimal codes are exactly those generated by primitive idempotents, and $(21.17) (3)$ says primitive, local and irreducible agree there.

The honest description is that these notions are infrastructure for decomposition. They are what one computes with when a module or an algebra has to be broken into indecomposable pieces and the pieces then have to be named.

Standards and Notation

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

Corner ring

e R e

; some authors write

R_{e}

{End}_{R} (e R)

PrimitiveUniversal. Occasionally minimal idempotent in the ordering

e' \leq e iff e e' = e' e = e'

Local idempotentLam's term; Anderson–Fuller say local for the module

e R

and use primitive with local endomorphism ring

IrreducibleAlso minimal idempotent in the older literature — a genuine clash with the ordering sense above

Principal indecomposable

e R

with

e

primitive; over an artinian algebra these are the projective indecomposables

GAP / MagmaPrimitiveIdempotents, CentralIdempotents, IdempotentLift in the algebra packages

Failure Modes and Common Mistakes

Do not conclude from $e R e$ being a division ring that $e R$ is simple: that needs $R$ semiprime. In the triangular example $e R e ≅ k$ while $e R$ has length two.
Do not assume $e R ≅ f R$ follows from $e R e ≅ f R f$ ; the corners can agree while the modules differ.
Do not forget $e \neq 0$ in the definitions of primitive and right irreducible; $0$ satisfies the corner conditions vacuously since $0 R 0$ is the zero ring.
A minimal right ideal need not be $e R$ for any idempotent — only those with $I^{2} \neq 0$ are, by Brauer's Lemma.

Best Practices

State the side. Write right irreducible, never just irreducible, unless $R$ has been assumed semiprime.
When you need Krull–Schmidt, verify locality of the endomorphism ring, not merely indecomposability.
Compute $e R / e J$ early: for a local idempotent it is the unique simple quotient of $e R$ and is the natural label for that principal indecomposable.
Use $M e \neq 0$ as a cheap test for the occurrence of a composition factor; it is a single multiplication, whereas building a composition series is not.
Record whether $e$ is full separately — fullness controls the ideal correspondence between $R$ and $e R e$ and has nothing to do with the grades on this page.

Quick Reference

Corner isomorphism

{End}_{R} (e R) ≅ e R e

Primitive

e R e

has no idempotent but

0, e

Local

e R e

is a local ring

Right irreducible

e R

is a minimal right ideal

Radical of a corner

rad (e R e) = e J e = J \cap e R e

Local criterion

e

local

iff

e R / e J

simple

iff

e J

the unique maximal submodule

Factor test

e R / e J

occurs in

M

iff

M e \neq 0

Hom formula

{Hom}_{R} (e R, M) ≅ M e

Where each statement lives
Statement	Hypotheses	Reference
$e$ right irreducible $\Rightarrow$ $e R e$ a division ring	$e \neq 0$ idempotent	(21.16)(1)
$e R e$ a division ring $\Rightarrow$ $e$ right irreducible	$R$ semiprime	(21.16)(2)
Right irreducible $\Leftrightarrow$ left irreducible	$R$ semiprime	(21.17)(2)
Primitive $\Leftrightarrow$ local $\Leftrightarrow$ irreducible	$R$ semisimple	(21.17)(3)
$e$ local $\Leftrightarrow$ $\bar{e}$ irreducible in $R / rad R$	$e \neq 0$ idempotent	(21.18)
$e R / e J$ a factor of $M$ $\Leftrightarrow$ $M e \neq 0$	$e$ local, $M$ of finite length	(21.19)
Minimal right ideal $I = e R$	$I^{2} \neq 0$	(10.22)

Frequently Asked Questions

Why introduce local idempotents at all when primitive idempotents already give indecomposable summands?

Because indecomposability alone does not give uniqueness of decompositions. The Krull–Schmidt–Azumaya theorem requires each summand to have a local endomorphism ring, and ${End}_{R} (e R) ≅ e R e$ , so the right hypothesis on the idempotent is exactly locality. Over $ℤ$ the identity is primitive but not local, and indeed direct-sum cancellation can fail over general noetherian rings.

Is every primitive idempotent local in a finite-dimensional algebra?

Yes. If $A$ is a finite-dimensional algebra over a field then $e A e$ is a finite-dimensional algebra with no nontrivial idempotents, hence — being semiprimary — a local ring. So over such algebras the distinction between primitive and local collapses. It reappears as soon as the algebra is infinite-dimensional or the base is not a field.

Can an idempotent be right irreducible without being left irreducible?

Yes, and the smallest example is the ring of upper triangular $2 \times 2$ matrices over a field: $E_{22}$ generates a one-dimensional right ideal but a two-dimensional left ideal. The phenomenon disappears over semiprime rings, where both conditions are equivalent to $e R e$ being a division ring.

What exactly does

e R / e J

do for me?

For a local idempotent it is the unique simple quotient — the top — of the principal indecomposable $e R$ . Over a semiperfect ring the assignment $e R \mapsto e R / e J$ is a bijection from the isomorphism classes of principal indecomposable modules to the isomorphism classes of simple modules, and $(21.19)$ lets you detect it inside any finite-length module by a single multiplication.

Why is the corner ring

e R e

rather than the subring generated by

e

Because $e R e$ is precisely the set of elements acting as endomorphisms of the summand $e R$ : an element $r$ with $e r = r = r e$ multiplies $e R$ into itself and commutes with the $R$ -action on the correct side. The subring generated by $e$ inside $R$ carries almost no information.

Does a minimal right ideal always come from an idempotent?

No. Brauer's Lemma $(10.22)$ says a minimal right ideal $I$ equals $e R$ for a right irreducible idempotent $e$ exactly when $I^{2} \neq 0$ . If $I^{2} = 0$ then $I$ lies in the radical and contains no nonzero idempotent at all; the strictly upper triangular matrices in the $2 \times 2$ triangular ring are such an $I$ .

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §21, results (21.8), (21.9), (21.15)–(21.19).
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §§21–27.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapters III–IV.
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

Show that two idempotents $e, f$ satisfy $e R ≅ f R$ if and only if $e = a b$ and $f = b a$ for some $a \in e R f$ , $b \in f R e$ .
Give an example of an indecomposable module whose endomorphism ring is not local, and explain how Krull–Schmidt fails for it.
Prove that in a von Neumann regular ring an idempotent is primitive if and only if $e R e$ is a division ring.
How do the primitive idempotents of $k G$ behave as $char k$ passes from zero to a prime dividing $| G |$ ?
Work out the primitive idempotents of $𝔽_{2} [x] / (x^{7} - 1)$ and identify the corresponding minimal cyclic codes.
Describe how a complete set of primitive orthogonal idempotents is computed algorithmically for a finite-dimensional algebra given by structure constants.
Under what conditions is a primitive idempotent of $R$ still primitive in $M_{n} (R)$ ?

Primitive and Local Idempotents

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Everything happens in the corner

Why irreducibility is the one-sided notion

The radical of a corner

Key Results

Proof Techniques and Method

Translate to the corner

Kill the radical, then use semiprimeness

Use idempotency as a filter

Worked Example

Upper triangular matrices: local but not right irreducible

Matrix rings: all three grades coincide

Primitive but not local

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Projective indecomposables

Vertices are idempotents

Splitting an algebra

Idempotent generators

Standards and Notation

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Everything happens in the corner

Why irreducibility is the one-sided notion

The radical of a corner

06Key Results

07Proof Techniques and Method

Translate to the corner

Kill the radical, then use semiprimeness

Use idempotency as a filter

08Worked Example

Upper triangular matrices: local but not right irreducible

Matrix rings: all three grades coincide

Primitive but not local

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Projective indecomposables

Vertices are idempotents

Splitting an algebra

Idempotent generators

13Standards and Notation

14Failure Modes and Common Mistakes

15Best Practices

16Quick Reference

17Frequently Asked Questions

18Related KEVOS Topics

19References

20AI 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

Applications and Industry Use

Standards and Notation

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions