Central Idempotents and Ring Direct Decompositions

Q: Why insist on central idempotents when ordinary idempotents already decompose the ring?

An arbitrary idempotent e decomposes R as a right module, R = eR ⊕ (1-e)R, but eR is then only a right ideal and the decomposition carries no ring structure. Centrality is exactly the condition making both summands two-sided ideals, so that multiplication is componentwise and R ≅ eR × (1-e)R as rings.

Q: How do I know I have found all the central idempotents?

Once you exhibit a decomposition 1 = c_1 + … + c_r into orthogonal centrally primitive idempotents, (22.1) tells you the complete list: the 2^r partial sums, and nothing else. Before that, the reliable method is to compute the centre and find its idempotents.

Q: What happens to modules under a block decomposition?

Every R-module M splits as M = c_1M ⊕ … ⊕ c_rM, with c_iM a module over the block c_iR on which the other blocks act as zero. The module category of R is therefore the product of the module categories of the blocks, which is why representation theory is done one block at a time.

Executive Summary

Every way of writing a ring $R$ as a direct product of two rings is encoded by a single element: a central idempotent $c$ , with the two factors being $c R$ and $(1 - c) R$ . The dictionary is exact and cheap to use, and it turns a question about ring decompositions into a question about one distinguished set of elements of the centre $Z (R)$ .

The content of §22 is what happens when the process is pushed to the end. If $1$ can be written as a sum of finitely many orthogonal centrally primitive idempotents, then that expression is unique up to order, every central idempotent is a partial sum of it, and $R$ is a direct product of indecomposable rings in exactly one way. A chain condition on the ideals of $R$ guarantees this happens.

$c = c^{2} \in Z (R)$ The splitting datum

$2^{r}$ Central idempotents when there are

r

blocks

UniqueDecomposition into blocks

ACC or DCCSufficient condition on ideals

Overview

Let $R$ be a ring with identity and let $e = e^{2} \in R$ with complementary idempotent $f = 1 - e$ . The Peirce decomposition writes $R = e R e \oplus e R f \oplus f R e \oplus f R f$ as abelian groups. The idempotent $e$ is central precisely when the two off-diagonal corners vanish, $e R f = f R e = 0$ ; in that case the coarser decomposition $R = e R \oplus f R$ has both summands two-sided ideals, and multiplication is componentwise.

e central iff e R f = f R e = 0 iff R = e R \times f R as rings, f = 1 - e

(22.0)

The summand $e R$ is a ring in its own right, with identity element $e$ rather than $1$ .

The converse direction is the reason the dictionary is useful: a direct sum decomposition $R = A \oplus B$ into two-sided ideals produces the idempotent, by splitting $1 = e + f$ with $e \in A$ and $f \in B$ . Nothing is lost in either direction, so ring-theoretic decompositions and central idempotents are the same data.

Two warnings frame the rest of this collection. First, $R$ can be riddled with idempotents and still be indecomposable: $M_{n} (D)$ has an abundance of them but a field as its centre. Second, finest decompositions need not exist at all — an infinite direct product of fields has no finite one — which is exactly why a chain condition is imposed. The pages The Block Decomposition of a Ring and Indecomposable Rings and Connectedness take these two points further.

Learning Objectives

State the equivalence between central idempotents, ideal direct sums, and ring product decompositions.
Recognise when a central idempotent $c$ is centrally primitive by testing $c R$ for indecomposability.
Prove that every central idempotent is a partial sum of a given decomposition of $1$ into centrally primitive idempotents.
Deduce uniqueness of the block decomposition up to a permutation of the factors.
Apply the chain condition criterion $(22.2)$ to a noetherian or artinian ring.
Compute all central idempotents of $ℤ / 12 ℤ$ and identify the corresponding factors.

Definitions

Definition(22.0)Indecomposable ring, centrally primitive idempotent

A ring $R \neq 0$ is indecomposable if it is not the direct sum of two nonzero two-sided ideals; equivalently, if its only central idempotents are $0$ and $1$ .

A central idempotent $c \in R$ is centrally primitive if $c \neq 0$ and $c$ cannot be written as $c = α + β$ with $α, β$ nonzero orthogonal central idempotents of $R$ .

$Z (R)$: The centre of $R$ . Every central idempotent lies here, and idempotents of $Z (R)$ are exactly the central idempotents of $R$ .
Orthogonal: Idempotents $e, f$ with $e f = f e = 0$ . For central idempotents $e f = f e$ automatically, so orthogonality is the single condition $e f = 0$ .
$c R$ as a ring: For $c$ a central idempotent, $c R$ is closed under multiplication and $c$ acts on it as an identity; it is a ring in its own right, and simultaneously an ideal of $R$ .
Primitive idempotent: An idempotent $e \neq 0$ that is not a sum of two nonzero orthogonal idempotents — no centrality required. This is a different and weaker-sounding but incomparable condition; see the pitfalls below.
Block: A summand $c_{i} R$ arising from a decomposition of $1$ into orthogonal centrally primitive idempotents.

Rings have an identity and are not assumed commutative. The zero ring is excluded from the definition of indecomposable, exactly as $1$ is excluded from the primes.

Core Concepts

From an idempotent to a product, and back

Suppose $c$ is a central idempotent and $d = 1 - c$ . Then $c d = c - c^{2} = 0$ , so $c R$ and $d R$ meet in $0$ , and $r = c r + d r$ for every $r$ , so $R = c R \oplus d R$ . Both summands are two-sided ideals because $c$ is central, and $(c r) (c s) = c (r s)$ shows $c R$ is closed under multiplication with $c$ as identity. Multiplication in $R$ is componentwise, so $R ≅ c R \times d R$ as rings.

Conversely let $R = A \oplus B$ with $A, B$ ideals, and write $1 = e + f$ with $e \in A$ , $f \in B$ . For $a \in A$ we get $a = a e + a f$ ; but $a f \in A \cap B = 0$ , so $a = a e$ , and symmetrically $a = e a$ . Thus $e$ is an identity for $A$ ; taking $a = e$ gives $e^{2} = e$ . For arbitrary $r \in R$ write $r = a + b$ ; then $e r = e a + e b = a = a e + b e = r e$ , using $e b \in A \cap B = 0$ . Hence $e$ is central, $A = e R$ , and likewise for $f$ and $B$ .

R = A \oplus B

(ideals)

1 = e + f

e, f

central orthogonal idempotents

R ≅ e R \times f R

Decompositions of c happen inside cR

This small observation is what makes the theory finite and local. Let $c$ be a central idempotent and suppose $c = α + β$ with $α, β$ orthogonal central idempotents of $R$ . Then

α = α^{2} = (α + β) α = c α \in c R,

(22.0a)

and symmetrically $β = c β \in c R$ .

So any splitting of $c$ already lives in the ring $c R$ . Because central idempotents of $R$ lying in $c R$ are precisely the central idempotents of the ring $c R$ — centrality transfers both ways, since $c$ is central and acts as the identity of $c R$ — we obtain the working criterion: ** $c$ is centrally primitive in $R$ if and only if the ring $c R$ is indecomposable.**

The Boolean algebra of central idempotents

Write $S$ for the set of central idempotents of $R$ . It is closed under multiplication, and under the operation $e \lor e' = e + e' - e e'$ . With these two operations and complement $e \mapsto 1 - e$ , $S$ becomes a Boolean ring: every element satisfies $x^{2} = x$ , and $S$ is the Boolean algebra of "clopen pieces" of $R$ . Centrally primitive idempotents are exactly its atoms, and a block decomposition exists precisely when $S$ is a finite Boolean algebra with $1$ the join of its atoms.

Key Results

Proposition(22.1)Uniqueness of a centrally primitive decomposition

Let $R$ be a ring and suppose that $1 = c_{1} + \dots + c_{r}$ where $c_{1}, \dots, c_{r}$ are pairwise orthogonal centrally primitive idempotents of $R$ . Then:

every central idempotent $c \in R$ equals $\sum_{i \in T} c_{i}$ for a unique subset $T \subseteq {1, \dots, r}$ ;
$c_{1}, \dots, c_{r}$ are the only centrally primitive idempotents of $R$ ; in particular any two distinct centrally primitive idempotents of $R$ are orthogonal;
the decomposition $1 = c_{1} + \dots + c_{r}$ is unique up to a permutation of its summands.

Proof

(1). Let $c$ be a central idempotent and fix $i$ . The element $c c_{i}$ is idempotent, since $c$ and $c_{i}$ commute, and it is central in the ring $c_{i} R$ : for $x \in c_{i} R$ we have $(c c_{i}) x = c x = x c = x (c c_{i})$ , because $c_{i}$ acts as the identity of $c_{i} R$ . Since $c_{i}$ is centrally primitive, $c_{i} R$ is indecomposable, so its only central idempotents are $0$ and $c_{i}$ . Hence $c c_{i} \in {0, c_{i}}$ for each $i$ . Now

c = c \cdot 1 = c (c_{1} + \dots + c_{r}) = \sum_{i : c c_{i} \neq 0} c_{i} .

(22.1a)

Uniqueness of $T$ follows on multiplying $\sum_{i \in T} c_{i}$ by $c_{j}$ : the product is $c_{j}$ if $j \in T$ and $0$ otherwise, so $T$ is recovered from $c$ .

(2). Let $c$ be centrally primitive. By (1), $c = \sum_{i \in T} c_{i}$ with $T \neq \emptyset$ since $c \neq 0$ . If $| T | \geq 2$ , pick $j \in T$ and split $c = c_{j} + \sum_{i \in T ∖ {j}} c_{i}$ into two nonzero orthogonal central idempotents, contradicting central primitivity. So $| T | = 1$ and $c = c_{j}$ .

(3). If $1 = d_{1} + \dots + d_{s}$ is another such decomposition, then each $d_{j}$ is centrally primitive, hence equals some $c_{i}$ by (2); distinct $d_{j}$ give distinct $c_{i}$ by orthogonality, and summing shows the two index sets coincide. So $s = r$ and the families agree up to order.

Corollary(22.1a)Existence criterion

A ring $R \neq 0$ is a finite direct product of indecomposable rings if and only if $1 \in R$ is a sum of finitely many pairwise orthogonal centrally primitive idempotents. When this holds, the factors and the idempotents are uniquely determined, and one writes $R = c_{1} R \oplus \dots \oplus c_{r} R$ for the block decomposition of $R$ .

Counterexample(22.1b)Block decompositions need not exist

Let $R = \prod_{n \geq 1} ℚ$ , the full direct product of countably many copies of $ℚ$ . The elements $ε_{n}$ (the identity in slot $n$ , zero elsewhere) form an infinite family of pairwise orthogonal centrally primitive idempotents. No finite subfamily sums to $1$ , and by $(22.1) (2)$ any decomposition of $1$ into orthogonal centrally primitive idempotents would have to use all of them. So $R$ has no block decomposition. The failure is a failure of finiteness: the ideals of $R$ satisfy neither chain condition.

Proposition(22.2)Chain conditions force a block decomposition

Let $R \neq 0$ be a ring whose two-sided ideals satisfy either the ascending chain condition or the descending chain condition — for instance, $R$ right noetherian, left noetherian, right artinian or left artinian. Then $1$ is a sum of finitely many orthogonal centrally primitive idempotents, so $R$ has a block decomposition, and all conclusions of $(22.1)$ apply.

Proof

Assume the DCC on ideals. Call a nonzero central idempotent $c$ good if it is a finite sum of orthogonal centrally primitive idempotents, and suppose some central idempotent is not good. Among the ideals $c R$ with $c$ a nonzero central idempotent that is not good, choose one minimal, say $c R$ .

Then $c$ is not centrally primitive (otherwise $c$ is good, being its own one-term decomposition), so $c = a + b$ with $a, b$ nonzero orthogonal central idempotents. Both lie in $c R$ by $(22.0 a)$ , so $a R \subseteq c R$ , and the inclusion is strict: $b$ lies in $c R$ but not in $a R$ , since $b = a x$ would give $b = a b = a^{2} x = a x = b$ together with $a b = 0$ , forcing $b = 0$ . Likewise $b R ⊊ c R$ . By minimality $a$ and $b$ are good, and concatenating their decompositions gives one for $c$ : the pieces coming from $a$ and from $b$ are orthogonal because they lie in $a R$ and $b R$ respectively and $a b = 0$ . This contradicts the choice of $c$ . Hence every nonzero central idempotent, $1$ included, is good.

Assume instead the ACC on ideals, and suppose $1$ is not good. A nonzero central idempotent that is not good is in particular not centrally primitive, so it splits as $a + b$ with $a, b$ nonzero orthogonal central idempotents, and at least one of the two pieces is again not good — otherwise concatenation would make the original good. Starting from $b_{0} = 1$ and applying this repeatedly gives nonzero central idempotents $a_{1}, a_{2}, \dots$ with $b_{n - 1} = a_{n} + b_{n}$ and $b_{n}$ never good. The $a_{n}$ are pairwise orthogonal, since $a_{n + 1} \in b_{n} R$ and $a_{n} b_{n} = 0$ , so the sums below are direct and each inclusion is strict:

a_{1} R ⊊ a_{1} R \oplus a_{2} R ⊊ a_{1} R \oplus a_{2} R \oplus a_{3} R ⊊ \dots

(22.2a)

This is an infinite strictly ascending chain of two-sided ideals, contradicting the ACC. Hence $1$ is good.

Remark(22.2b)What the chain condition is really for

Only chain conditions on two-sided ideals are needed, which is much weaker than a chain condition on one-sided ideals. Commutative noetherian rings, artinian rings, and finite-dimensional algebras all qualify. Rings that fail badly, such as infinite products, fail exactly here.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Multiply by the pieces of 1

Given $1 = c_{1} + \dots + c_{r}$ , the identity $c = \sum_{i} c c_{i}$ turns a global statement about $c$ into $r$ local statements inside the rings $c_{i} R$ . Every part of $(22.1)$ is this move.

Move 2

Localise centrality

A central idempotent of $R$ lying in $c R$ is a central idempotent of the ring $c R$ , and conversely when $c$ is central. Centrality is therefore not lost when passing to a direct factor, which is what makes indecomposability of $c R$ the right test.

Move 3

Minimal counterexample under DCC

To prove every object of a class decomposes, take a minimal offender and split it; the two smaller pieces decompose by minimality, and reassembling contradicts the choice. The same skeleton proves the Krull–Schmidt existence statement for modules.

The third move is worth isolating because it is the only place a hypothesis enters. Results $(22.0)$ and $(22.1)$ are hypothesis-free bookkeeping; all the mathematical content about existence sits in $(22.2)$ .

Worked Example

All central idempotents of $ℤ / 12 ℤ$

Take $R = ℤ / 12 ℤ$ , which is commutative, so every idempotent is central. Solving $x^{2} \equiv x (\mod 12)$ , that is $12 ∣ x (x - 1)$ , gives $x \in {0, 1, 4, 9}$ : indeed $4^{2} = 16 = 4$ and $9^{2} = 81 = 9$ in $ℤ / 12$ . Four central idempotents, so by the count $2^{r}$ we expect $r = 2$ blocks.

1 = 4 + 9, 4 \cdot 9 = 36 = 0 in ℤ / 12 ℤ .

(E.1)

An orthogonal decomposition of the identity.

The two factors are $4 R = {0, 4, 8}$ with identity $4$ , and $9 R = {0, 3, 6, 9}$ with identity $9$ . As rings, $4 R ≅ 𝔽_{3}$ and $9 R ≅ ℤ / 4 ℤ$ — check the identities: $4 \cdot 8 = 32 = 8$ and $9 \cdot 3 = 27 = 3$ in $ℤ / 12$ , as required. Both factors are indecomposable, the first because it is a field, the second because it is local. So $4$ and $9$ are centrally primitive and

ℤ / 12 ℤ ≅ ℤ / 4 ℤ \times ℤ / 3 ℤ

(E.2)

The Chinese Remainder Theorem, read as a block decomposition.

Every central idempotent is a partial sum of ${4, 9}$ , as $(22.1) (1)$ predicts: $\emptyset \mapsto 0$ , ${4} \mapsto 4$ , ${9} \mapsto 9$ , ${4, 9} \mapsto 1$ .

Many idempotents, no decomposition

Now let $R = M_{3} (ℚ)$ . It contains a great many idempotents — every projection onto a subspace of $ℚ^{3}$ along a complement — but $Z (R) = ℚ \cdot I_{3}$ is a field, whose only idempotents are $0$ and $I_{3}$ . Hence $R$ is indecomposable: the abundance of idempotents produces module decompositions of $R_{R}$ , not ring decompositions.

Comparison and Classification

Central idempotents and blocks in familiar rings
Ring	Centre	Central idempotents	Blocks
Division ring $D$	a field	$0, 1$	$1$
$M_{n} (D)$	$Z (D)$	$0, 1$	$1$
Upper triangular $T_{n} (k)$ , $k$ a division ring	$Z (k)$	$0, 1$	$1$
Any local ring	local	$0, 1$	$1$
$ℤ / 12 ℤ$	itself	$0, 4, 9, 1$	$2$
$M_{2} (ℚ) \times ℚ [x] / (x^{2})$	$ℚ \times ℚ [x] / (x^{2})$	four of them	$2$
Semisimple ring with $n$ simple components	$n$ fields	$2^{n}$	$n$
$\prod_{n \geq 1} ℚ$	itself	one per subset of $ℕ$	none exists

The table makes the two independent phenomena visible. Rows one to four are indecomposable for very different internal reasons; the last row is not indecomposable at all, yet still has no block decomposition.

Relationship Map

Central idempotent $c$ of $R$ — equivalently an idempotent of $Z (R)$
- gives
  - an ideal direct sum $R = c R \oplus (1 - c) R$
  - a ring isomorphism $R ≅ c R \times (1 - c) R$
  - a splitting of every $R$ -module $M = c M \oplus (1 - c) M$
- is centrally primitive when
  - $c R$ is an indecomposable ring
  - $c$ is an atom of the Boolean algebra of central idempotents
- exists in abundance when
  - $R$ is semisimple: $2^{n}$ of them
  - $R$ is a finite product of local rings

The module-splitting bullet deserves emphasis. Because $c$ is central, $c M$ is a submodule for every $R$ -module $M$ , and $M = c M \oplus (1 - c) M$ . So a block decomposition of $R$ decomposes the entire module category into a product of the module categories of the blocks — this is the categorical content of the theory, and it is why representation theorists work one block at a time.

Chain condition on ideals

1

= sum of orthogonal centrally primitive idempotentsBlock decomposition existsUniqueness by

(22.1)

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

Working one block at a time

Because a central idempotent splits every module, the representation theory of a finite group over a field of characteristic $p$ decomposes into independent problems, one per block of $k G$ . Brauer's theory of blocks, defect groups and defect zero all begin here.

Coding theory

Idempotent generators of cyclic codes

A cyclic code of length $n$ over $𝔽_{q}$ with $\gcd (n, q) = 1$ is an ideal of $𝔽_{q} [x] / (x^{n} - 1)$ , hence generated by an idempotent. The primitive idempotents of that commutative ring are the block idempotents, and they give the minimal cyclic codes into which every cyclic code decomposes.

Symbolic computation

Splitting an algebra before working with it

Computer algebra systems decompose a finite-dimensional algebra into blocks before doing anything else, because linear algebra of size $\dim R$ is replaced by independent problems of size $\dim c_{i} R$ . The block idempotents are computed from the centre.

Commutative algebra

Connected components of a scheme

For commutative $R$ , idempotents correspond to clopen subsets of $Spec R$ , and a block decomposition is a decomposition into connected components. The finiteness supplied by $(22.2)$ is the statement that a noetherian scheme has finitely many connected components.

The honest summary: this is infrastructure. Central idempotents are almost never the object of study; they are the first thing you compute so that the object of study breaks into independent pieces.

Standards and Notation

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

Block idempotent

c_{i}

e_{i}

; in group representation theory usually

e_{B}

for the block

B

Blocks

R = R_{1} \oplus \dots \oplus R_{r}

with

R_{i} = c_{i} R

; some authors write

B_{0}

for the principal block

Centre

Z (R)

here;

C

in Lam's §22,

Cent (R)

elsewhere

OrthogonalityFor central idempotents

e f = f e

is automatic, so many sources write only

e f = 0

GAPCentralIdempotentsOfAlgebra, PrimitiveIdempotentsOfAlgebra

Magma / SageCentralIdempotents, A.central_orthogonal_idempotents()

Computational Notes

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

For a finite-dimensional algebra $R$ over a field $k$ given by structure constants with $\dim_{k} R = n$ , the standard route to the block decomposition is:

Compute $Z (R)$ as the nullspace of the linear map $x \mapsto (x b_{i} - b_{i} x)_{i}$ for a basis $b_{1}, \dots, b_{n}$ ; this is one nullspace computation on an $n^{2} \times n$ matrix.
Compute $rad Z (R)$ ; since $Z (R)$ is commutative and finite-dimensional, $Z (R) / rad Z (R)$ is a product of fields.
Find the primitive idempotents of that product of fields — a factorisation problem over $k$ — and lift them through the nilpotent ideal $rad Z (R)$ by Newton iteration, using $e \mapsto 3 e^{2} - 2 e^{3}$ .
The lifted idempotents are the centrally primitive idempotents of $R$ , and the blocks are $c_{i} R$ .

Step 3 is the only step whose cost is not polynomial linear algebra: it needs factorisation of polynomials over $k$ , which is fast over finite fields and over $ℚ$ but is exactly where the difficulty concentrates over general fields. The Newton iteration doubles the precision each round, so $O (\log m)$ steps suffice when $(rad Z (R))^{m} = 0$ .

Failure Modes and Common Mistakes

Do not assume a direct sum decomposition into right ideals gives central idempotents; it gives orthogonal idempotents summing to $1$ , which is a much weaker structure. Centrality requires the summands to be two-sided ideals.
Do not confuse $c R$ , a ring with identity $c$ , with a unital subring of $R$ : it does not contain $1$ unless $c = 1$ .
Two distinct block decompositions cannot coexist, but a ring may admit many decompositions into non-primitive factors — $(22.1) (3)$ is uniqueness of the finest one only.

Quick Reference

Central idempotent

c^{2} = c

c r = r c

for all

r

; equivalently

c R f = f R c = 0

with

f = 1 - c

Splitting

R = c R \oplus (1 - c) R

, both ideals,

R ≅ c R \times (1 - c) R

Indecomposable

R \neq 0

with only central idempotents

0, 1

Centrally primitive

c \neq 0

central idempotent with

c R

indecomposable

Uniqueness

(22.1)

: partial sums, only centrally primitive idempotents, unique up to order

Existence

(22.2)

: ACC or DCC on two-sided ideals suffices

Counting

r

blocks

\Rightarrow

exactly

2^{r}

central idempotents

Failure case

\prod_{n} ℚ

: infinitely many blocks-to-be, no decomposition

Which statement to cite
Need	Statement	Hypotheses
Split $R$ using an element	$R ≅ c R \times (1 - c) R$	$c$ a central idempotent
Recognise the finest splitting	$c$ centrally primitive iff $c R$ indecomposable	none
List all central idempotents	$(22.1) (1)$	$1$ a finite sum of orthogonal centrally primitive idempotents
Uniqueness of blocks	$(22.1) (3)$	same as above
Existence of blocks	$(22.2)$	ACC or DCC on two-sided ideals

Frequently Asked Questions

Why insist on central idempotents when ordinary idempotents already decompose the ring?

An arbitrary idempotent $e$ decomposes $R$ as a right module, $R = e R \oplus (1 - e) R$ , but $e R$ is then only a right ideal and the decomposition carries no ring structure. Centrality is exactly the condition making both summands two-sided ideals, so that multiplication is componentwise and $R ≅ e R \times (1 - e) R$ as rings.

Is a ring with no nontrivial idempotents the same as an indecomposable ring?

No, it is strictly stronger. Indecomposability only forbids nontrivial idempotents in the centre. $M_{n} (D)$ for $n \geq 2$ has many idempotents yet is indecomposable, since its centre is a field.

Does a block decomposition always exist?

No. An infinite direct product of fields has none. A chain condition on two-sided ideals — ACC or DCC, so in particular one-sided noetherian or artinian — is enough $(22.2)$ , and it is the hypothesis used in practice.

How do I know I have found all the central idempotents?

Once you exhibit a decomposition $1 = c_{1} + \dots + c_{r}$ into orthogonal centrally primitive idempotents, $(22.1)$ tells you the complete list: the $2^{r}$ partial sums, and nothing else. Before that, the reliable method is to compute the centre and find its idempotents.

Do the blocks of

R

correspond to the blocks of

R / rad R

In general no, and this is the main subtlety of §22. Upper triangular $n \times n$ matrices over a field form an indecomposable ring whose radical quotient is a product of $n$ copies of the field, with $2^{n}$ central idempotents. Only quotients by the square of a nilpotent ideal preserve central idempotents; see the page on lifting central idempotents.

What happens to modules under a block decomposition?

Every $R$ -module $M$ splits as $M = c_{1} M \oplus \dots \oplus c_{r} M$ , with $c_{i} M$ a module over the block $c_{i} R$ on which the other blocks act as zero. The module category of $R$ is therefore the product of the module categories of the blocks, which is why representation theory is done one block at a time.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §22 (pp. 336–344), especially (22.1) and (22.2).
T. Y. Lam, A First Course in Noncommutative Rings, §21, for the Peirce decomposition and the criterion that an idempotent is central if and only if its off-diagonal Peirce corners vanish.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §7 and §27.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 1, on idempotents and Peirce decompositions.
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981, on central idempotents and block decompositions of algebras.

AI Suggested Questions

Show that the central idempotents of $R$ form a Boolean ring under $e \lor f = e + f - e f$ and ordinary multiplication.
Prove that two central idempotents $e$ and $f$ are isomorphic as idempotents if and only if $e = f$ .
For which commutative rings does the set of idempotents form a finite Boolean algebra, and how does this relate to connectedness of the spectrum?
Give an example of a ring whose two-sided ideals satisfy the DCC but whose one-sided ideals do not.
How are the central idempotents of $M_{n} (R)$ related to those of $R$ , and why does the matrix construction not create new blocks?
Work out the block decomposition of $ℤ / n ℤ$ for a general $n$ and relate it to the prime factorisation.
What is the analogue of a central idempotent for a ring without identity, and how much of $(22.1)$ survives?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

From an idempotent to a product, and back

Decompositions of c happen inside cR

The Boolean algebra of central idempotents

06Key Results

07Proof Techniques and Method

Multiply by the pieces of 1

Localise centrality

Minimal counterexample under DCC

08Worked Example

All central idempotents of ℤ/12ℤ

Many idempotents, no decomposition

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Working one block at a time

Idempotent generators of cyclic codes

Splitting an algebra before working with it

Connected components of a scheme

12Standards and Notation

13Computational Notes

14Failure Modes and Common Mistakes

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

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Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

All central idempotents of $ℤ / 12 ℤ$

Comparison and Classification

Relationship Map

Applications and Industry Use

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions