Indecomposable Rings and Connectedness

Q: How can a ring be indecomposable and still have lots of idempotents?

Because indecomposability constrains only the *central* idempotents. T_n(k) has a large supply of idempotents — every E_11 + E_12, for instance — but its centre is just the scalars, so it is indecomposable. The right slogan is that indecomposability is a property of Z(R), not of R.

Q: Why is the linkage relation defined with an auxiliary idempotent f rather than by eRe' ≠ 0?

Because eRe' ≠ 0 is not symmetric-looking enough to generate the right relation, and more importantly the condition that matters is that some *single* principal indecomposable fR maps nontrivially to both eR and e'R. In the artinian case that condition becomes *eR and e'R have a composition factor in common*, which is manifestly symmetric and is exactly what block theory needs.

Q: Is transitive?

Not in general, which is why is defined as the equivalence relation it generates. Two primitive idempotents can be joined only by a long walk through intermediate idempotents, and blocks are the connected components of that walk structure, not the classes of itself.

Q: Do blocks always exist?

No. An infinite product of fields has infinitely many centrally primitive idempotents and no finite orthogonal decomposition of 1 into them. Existence needs a hypothesis: a chain condition on ideals (22.2), or a decomposition of 1 into orthogonal primitive idempotents (22.5). Both hold for right artinian rings.

Q: What is the relation between blocks of R and blocks of R/rad R?

The semisimple quotient generally has at least as many blocks, since its blocks are its simple components and linkage in it is just isomorphism of idempotents. Passing to the quotient loses the maps eRf that lived in the radical, so linkage classes can split. Recovering the blocks of R from those of R/rad R requires lifting central idempotents, which is a separate and stronger statement.

Q: Is indecomposability preserved by standard constructions?

It is preserved by passing to matrix rings, because Z(M_n(R)) ≅ Z(R), and by passing to R[x] and R[[x]] for the same reason. It is not preserved by quotients: Z is indecomposable but Z/12 is not.

Executive Summary

Decomposing a ring as a direct product means splitting its identity into orthogonal central idempotents. A ring that admits no such splitting is indecomposable, and the indecomposable factors of a ring — its blocks — are as canonical as the prime factors of an integer: when they exist they are unique up to order.

The mechanism is a connectedness argument. Primitive idempotents $e, e'$ are linked when some principal indecomposable maps nontrivially into both $e R$ and $e' R$ ; linkage generates an equivalence relation, and $(22.5)$ says the class sums are precisely the centrally primitive idempotents. Blocks are the connected components of the linkage graph. Over a right artinian ring, linkage is the concrete condition that $e R$ and $e' R$ share a composition factor.

$Z (R)$ Where blocks live

1Blocks of an indecomposable ring

$\approx$ Linkage classes

(22.5)Blocks are linkage classes

Overview

Let $e$ be an idempotent of $R$ with complement $f = 1 - e$ . By $(21.5)$ , $e$ is central if and only if $e R f = f R e = 0$ , and in that case both Peirce summands in $R = e R \oplus f R$ are two-sided ideals, so

R ≅ e R \times f R as rings, with identities e and f .

(22.0)

Conversely, if $R = A \oplus B$ with $A, B$ ideals, write $1 = e + f$ with $e \in A$ , $f \in B$ ; then $e, f$ are orthogonal central idempotents and $A = e R$ , $B = f R$ .

So ring direct product decompositions and *orthogonal decompositions of $1$ into central idempotents* are the same data. A nonzero ring is indecomposable when neither exists nontrivially.

Indecomposability is genuinely weaker than the other irreducibility conditions in this collection. Simple $\Rightarrow$ indecomposable and local $\Rightarrow$ indecomposable, but not conversely; and for a commutative ring, indecomposable is exactly the statement that $Spec R$ is a connected topological space. That geometric reading is where the word connected in the title comes from, and the linkage relation below is its noncommutative replacement.

Learning Objectives

Translate between ring product decompositions and central idempotents.
State when a block decomposition exists and prove it is unique up to permutation.
Prove $(22.4)$ : a primitive idempotent lies in $c R$ or in $(1 - c) R$ for every central idempotent $c$ .
Prove that linkage is compatible with central idempotents $(22.3)$ .
Prove $(22.5)$ : linkage classes of primitive idempotents are exactly the blocks.
Use $(22.6)$ to compute blocks of a right artinian ring from composition factors.

Definitions

Definition(22.0a)Indecomposable ring

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

Definition(22.0b)Centrally primitive idempotent

A central idempotent $c \neq 0$ is centrally primitive in $R$ if $c$ cannot be written as $α + β$ with $α, β$ nonzero orthogonal central idempotents of $R$ . Any such decomposition automatically takes place inside $c R$ , because $α = (α + β) α = c α \in c R$ ; so $c$ is centrally primitive exactly when the ring $c R$ is indecomposable.

Definition(22.2a)Linkage

Let $E$ denote the set of primitive idempotents of $R \neq 0$ . For $e, e' \in E$ write $e \sim e'$ if there exists $f \in E$ with $e R f \neq 0 \neq e' R f$ . The relation $\sim$ is reflexive and symmetric; let $\approx$ be the equivalence relation it generates, so $e \approx e'$ means there is a chain $e \sim e_{1} \sim \dots \sim e_{m} \sim e'$ in $E$ . Idempotents with $e \approx e'$ are called linked.

Central idempotent: $e = e^{2} \in Z (R)$ ; equivalently $e R f = f R e = 0$ for $f = 1 - e$ , by (21.5).
Primitive idempotent: A nonzero idempotent that is not a sum of two nonzero orthogonal idempotents; equivalently $e R$ is an indecomposable right $R$ -module. Not required to be central.
Block: A ring direct factor $c_{i} R$ arising from a decomposition $1 = c_{1} + \dots + c_{r}$ into orthogonal centrally primitive idempotents.
${Hom}_{R} (f R, e R)$: Naturally isomorphic to $e R f$ by (21.6); so $e R f \neq 0$ says there is a nonzero map $f R \to e R$ .
Connected: For commutative $R$ : $Spec R$ is connected iff $R$ is indecomposable. The noncommutative analogue is connectivity of the linkage graph.

All rings have an identity. Ideal means two-sided ideal unless a side is named. The zero ring is excluded from being indecomposable, exactly as $1$ is excluded from being prime.

Core Concepts

Linkage is a statement about homomorphisms

The definition of $\sim$ looks like a computation with products of subsets, but $(21.6)$ makes it representation-theoretic:

{Hom}_{R} (f R, e R) ≅ e R f .

(22.2b)

So $e \sim e'$ says: some principal indecomposable $f R$ admits nonzero maps into both $e R$ and $e' R$ .

Two immediate consequences are worth recording. If $e$ and $e'$ are isomorphic idempotents in the sense of $(21.20)$ — that is, $e R ≅ e' R$ — then $e \sim e'$ . And if $e, f \in E$ with $e R f \neq 0$ , then $e \sim f$ directly, taking $f$ itself as the witness, since $f R f ∋ f \neq 0$ .

Why primitivity makes idempotents choose a side

A central idempotent $c$ cuts $R$ into $c R$ and $(1 - c) R$ . An arbitrary idempotent may straddle the two pieces, but a primitive one cannot: the decomposition $e = c e + (1 - c) e$ is a decomposition into orthogonal idempotents, so primitivity kills one of the terms. This is $(22.4)$ , and it is what makes the assignment *primitive idempotent $\mapsto$ block* well defined.

The connectedness picture

Form a graph whose vertices are the primitive idempotents in a fixed orthogonal decomposition of $1$ , with an edge $e - e'$ whenever $e \sim e'$ . Then $\approx$ is connectivity in this graph, and $(22.5)$ identifies the connected components with the blocks. A ring is indecomposable exactly when this graph is connected.

Orthogonal decomposition of

1

Linkage graph on

E

Connected componentsBlocks of

R

Key Results

Proposition(22.1)Uniqueness of a block decomposition

Suppose $1 = c_{1} + \dots + c_{r}$ where the $c_{i}$ are orthogonal centrally primitive idempotents of $R$ . Then:

every central idempotent of $R$ is the sum of a subset of ${c_{1}, \dots, c_{r}}$ ;
$c_{1}, \dots, c_{r}$ are the only centrally primitive idempotents of $R$ ; in particular any two distinct centrally primitive idempotents are orthogonal;
the decomposition $1 = c_{1} + \dots + c_{r}$ is unique up to a permutation of its terms.

Proof

Let $c$ be a central idempotent. For each $i$ , $c c_{i}$ is a central idempotent lying in $c_{i} R$ ; since $c_{i}$ is centrally primitive, the ring $c_{i} R$ is indecomposable, so its only central idempotents are $0$ and $c_{i}$ . Hence $c c_{i} \in {0, c_{i}}$ . Summing, $c = c (c_{1} + \dots + c_{r}) = \sum_{i : c c_{i} \neq 0} c_{i}$ , which is (1). Statements (2) and (3) follow: a centrally primitive $c$ is a sum of some $c_{i}$ , and being centrally primitive it must be a single one.

Proposition(22.2)Existence under a chain condition

Let $R$ be a ring whose two-sided ideals satisfy either the ascending or the descending chain condition — for instance $R$ left or right noetherian, or left or right artinian. Then $R$ has a block decomposition, and all conclusions of $(22.1)$ hold.

Proof sketch. Repeat, for the lattice of ideals, the standard argument that a module with a chain condition is a finite direct sum of indecomposables: if $R$ were not a finite direct sum of indecomposable ideals, one could split off a proper ideal repeatedly and build an infinite strictly monotone chain.

Counterexample—No block decomposition

Let $R = \prod_{i = 1}^{\infty} ℚ$ . The coordinate idempotents $ε_{i}$ are centrally primitive, and there are infinitely many of them. By $(22.1) (2)$ , if $1$ were a finite sum of orthogonal centrally primitive idempotents then only finitely many centrally primitive idempotents could exist. Hence $R$ has no block decomposition. Its ideals satisfy neither chain condition, as $(22.2)$ requires.

Lemma(22.4)A primitive idempotent lies in one piece

Let $c$ be a central idempotent of $R$ and $e \in E$ a primitive idempotent. Then either $e \in c R$ or $e \in (1 - c) R$ .

Proof

Write $e = c e + (1 - c) e$ . Because $c$ is central, $(c e)^{2} = c^{2} e^{2} = c e$ and $((1 - c) e)^{2} = (1 - c) e$ , and $c e \cdot (1 - c) e = c (1 - c) e^{2} = 0$ together with the symmetric product; so this is a decomposition of $e$ into orthogonal idempotents. Primitivity of $e$ forces one of them to vanish. If $c e = 0$ then $e = (1 - c) e \in (1 - c) R$ ; if $(1 - c) e = 0$ then $e = c e \in c R$ .

Lemma(22.3)Linkage respects central idempotents

Let $e, e' \in E$ be linked, $e \approx e'$ , and let $c$ be a central idempotent of $R$ . Then $e \in c R$ if and only if $e' \in c R$ .

Proof

By symmetry and by chaining along the definition of $\approx$ , it suffices to treat $e \sim e'$ and prove one implication. Fix $f \in E$ with $e R f \neq 0 \neq e' R f$ , and suppose $e \in c R$ , i.e. $c e = e$ .

Then $0 \neq e R f = (c e) R f = e R (c f)$ , using centrality of $c$ ; hence $c f \neq 0$ , so $f \notin (1 - c) R$ and $(22.4)$ gives $f \in c R$ , i.e. $c f = f$ .

Now $0 \neq e' R f = e' R (c f) = c e' R f$ , so $c e' \neq 0$ , and $(22.4)$ applied to $e'$ gives $e' \in c R$ .

Theorem(22.5)Blocks are the linkage classes

Suppose the identity of $R$ can be written as $1 = e_{1} + \dots + e_{n}$ with the $e_{i}$ orthogonal primitive idempotents. Then $1$ is a sum of orthogonal centrally primitive idempotents, so $R$ has a block decomposition. Moreover two primitive idempotents $e, e' \in E$ are linked if and only if they belong to the same block.

Proof

Construction. The $e_{i}$ are distinct elements of $E$ , and $\approx$ partitions ${e_{1}, \dots, e_{n}}$ into classes. Let $c_{1}, \dots, c_{r}$ be the class sums. Each $c_{i}$ is an idempotent (a sum of orthogonal idempotents), the $c_{i}$ are pairwise orthogonal, and $c_{1} + \dots + c_{r} = 1$ .

Centrality. If $e$ and $e''$ lie in different classes then $e R e'' = 0$ : otherwise $e \sim e''$ with witness $e''$ , putting them in the same class. Hence $c_{i} R c_{j} = 0$ for $i \neq j$ . For $a \in R$ ,

c_{i} a = c_{i} a (c_{1} + \dots + c_{r}) = c_{i} a c_{i} = (c_{1} + \dots + c_{r}) a c_{i} = a c_{i},

(22.5a)

so each $c_{i}$ is central.

Central primitivity. Let $c$ be a nonzero central idempotent of $c_{i} R$ ; we show $c = c_{i}$ . Write $c_{i} = e_{i_{1}} + \dots + e_{i_{m}}$ for its class. From $0 \neq c = c c_{i} = c (e_{i_{1}} + \dots + e_{i_{m}})$ we get $c e_{i_{j}} \neq 0$ for some $j$ , so $e_{i_{j}} \in c R$ by $(22.4)$ . All the $e_{i_{l}}$ are linked to $e_{i_{j}}$ , so $(22.3)$ puts every $e_{i_{l}}$ in $c R$ . Therefore $c c_{i} = c_{i}$ , and since $c \in c_{i} R$ gives $c = c c_{i}$ , we conclude $c = c_{i}$ . So $c_{i} R$ is indecomposable and $c_{i}$ is centrally primitive.

Blocks equal classes. Let $e \in E$ be arbitrary. By $(22.4)$ applied to each $c_{i}$ , $e$ lies in exactly one block $R_{i} = c_{i} R$ . Then $0 \neq e = e c_{i} = e (e_{i_{1}} + \dots + e_{i_{m}})$ , so $e e_{i_{j}} \neq 0$ for some $j$ , whence $e R e_{i_{j}} \neq 0$ and $e \sim e_{i_{j}}$ . Conversely $(22.3)$ shows that anything linked to $e_{i_{j}}$ lies in $c_{i} R$ . So the primitive idempotents of the block $R_{i}$ are precisely the members of the $i$ -th linkage class.

Theorem(22.6)The right artinian case

Let $R$ be a right artinian ring. Then $R$ has a unique block decomposition $R = R_{1} \oplus \dots \oplus R_{r}$ . For primitive idempotents $e, e' \in E$ one has $e \sim e'$ if and only if $e R$ and $e' R$ have a common composition factor. Consequently $e$ and $e'$ lie in the same block if and only if there are $e_{1} = e, e_{2}, \dots, e_{m} = e'$ in $E$ such that $e_{i} R$ and $e_{i + 1} R$ have a common composition factor for each $i < m$ .

Proof

$R_{R}$ has finite length, hence a Krull–Schmidt decomposition, so $1$ is a sum of orthogonal primitive idempotents and $(22.5)$ gives a block decomposition; uniqueness is $(22.1) (3)$ .

A right artinian ring is semiperfect, so every $f \in E$ is a local idempotent $(19.17)$ and $f R / f J$ is simple, $J = rad R$ . Every simple right $R$ -module occurs this way: given $V$ simple, choose an idempotent $\bar{x}$ of $R / J$ with $V ≅ \bar{x} (R / J)$ , lift it to an idempotent $f \in R$ , and $(21.18)$ shows $f \in E$ with $V ≅ f R / f J$ .

Now $e R f \neq 0$ says $(e R) f \neq 0$ , which by $(21.19)$ — applicable because $f$ is local and $e R$ has finite length — is equivalent to $e R$ having $f R / f J$ as a composition factor. So $e \sim e'$ , i.e. the existence of $f \in E$ with $e R f \neq 0 \neq e' R f$ , says exactly that $e R$ and $e' R$ share a composition factor. The final statement is $(22.5)$ rewritten with this description of $\sim$ .

Corollary—The semisimple case

If $R$ is semisimple then $e R$ is itself simple for every $e \in E$ , so sharing a composition factor means $e R ≅ e' R$ ; thus $e \sim e'$ if and only if $e ≅ e'$ as idempotents. The blocks are the simple components of the Wedderburn–Artin decomposition, and $R$ is indecomposable exactly when it is simple artinian.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Split by a central idempotent, then use primitivity

Multiplying by $c$ and $1 - c$ decomposes any element; when the element is a primitive idempotent, the decomposition must be trivial. This converts a global splitting of $R$ into a partition of $E$ .

Move 2

Sum a class to build a central idempotent

Centrality is proved, not assumed: $c_{i} R c_{j} = 0$ for distinct classes turns $c_{i} a = c_{i} a c_{i} = a c_{i}$ into an identity. Whenever a partition kills cross terms, the class sums are central.

Move 3

Read $e R f$ as a Hom group

$e R f ≅ {Hom}_{R} (f R, e R)$ turns an opaque product of subsets into a statement about maps between principal indecomposables, and then $(21.19)$ turns it into a statement about composition factors.

The pattern is worth naming: an equivalence relation defined by some object sees both of us is generated to a connectivity relation, and its classes then produce idempotents that are central for combinatorial reasons. The same argument shape reappears for blocks of group algebras, for connected components of quivers, and for the decomposition of a category into blocks.

Worked Example

Upper triangular matrices: many idempotents, one block

Let $k$ be a field and $R = T_{3} (k)$ , the upper triangular $3 \times 3$ matrices. The matrix units $e_{i} = E_{i i}$ for $i = 1, 2, 3$ are orthogonal primitive idempotents with $e_{1} + e_{2} + e_{3} = 1$ , and

e_{i} R e_{j} = {\begin{cases} k E_{i j} & i \leq j, \\ 0 & i > j . \end{cases}

(E.1)

So $e_{1} R e_{2} = k E_{12} \neq 0$ and $e_{2} R e_{2} \neq 0$ , giving $e_{1} \sim e_{2}$ ; likewise $e_{2} \sim e_{3}$ . All three are linked, the linkage graph is connected, and $(22.5)$ says $R$ has a single block: $T_{3} (k)$ is an indecomposable ring. This matches the direct computation $Z (T_{3} (k)) = k \cdot 1$ , whose only idempotents are $0$ and $1$ .

The composition-factor description of $(22.6)$ agrees. With $S_{1}, S_{2}, S_{3}$ the three one-dimensional simple modules, $e_{1} R$ has factors $S_{1}, S_{2}, S_{3}$ ; $e_{2} R$ has $S_{2}, S_{3}$ ; $e_{3} R = S_{3}$ . The Cartan matrix, with $(i, j)$ entry the multiplicity of $S_{j}$ in $e_{i} R$ , is

C = (\begin{matrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{matrix}),

(E.2)

No simultaneous row-and-column permutation makes $C$ block diagonal, which is the matrix form of connectivity.

A decomposable example with explicit arithmetic

Take $R = ℤ / 12 ℤ$ . Its idempotents are $0, 1, 4, 9$ : indeed $4^{2} = 16 \equiv 4$ and $9^{2} = 81 \equiv 9 (\mod 12)$ . They satisfy $4 + 9 = 13 \equiv 1$ and $4 \cdot 9 = 36 \equiv 0$ , so $1 = 4 + 9$ is an orthogonal decomposition into central idempotents and

ℤ / 12 ℤ ≅ 4 (ℤ / 12 ℤ) \times 9 (ℤ / 12 ℤ) ≅ ℤ / 3 ℤ \times ℤ / 4 ℤ .

(E.3)

The idempotent $4$ generates the copy of $ℤ / 3$ and $9$ the copy of $ℤ / 4$ .

Both factors are local rings, hence indecomposable, so $4$ and $9$ are centrally primitive and $(E . 3)$ is the block decomposition, unique by $(22.1) (3)$ . Two blocks, and correspondingly $Spec (ℤ / 12)$ is a two-point discrete space — disconnected, as the commutative dictionary predicts.

Process and Workflow

Decompose the identityWrite

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

with the

e_{i}

orthogonal primitive idempotents. This is available whenever

R_{R}

has a Krull–Schmidt decomposition, in particular for right artinian rings.

Build the linkage graphJoin

e_{i}

e_{j}

when

e_{i} R e_{j} \neq 0

, or, in the artinian case, when

e_{i} R

and

e_{j} R

share a composition factor. Reading the Cartan matrix is usually the fastest route.

Take connected componentsEach component gives a class sum

c_{i}

, and

(22.5)

certifies that

c_{i}

is a centrally primitive idempotent.

Read off the blocks

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

with each

c_{i} R

indecomposable, and by

(22.1)

this decomposition is the only one.

Comparison and Classification

Indecomposability against the neighbouring conditions
Ring	Indecomposable?	Simple?	Local?	Blocks
$D$ a division ring	yes	yes	yes	1
$M_{n} (D)$	yes	yes	no ( $n \geq 2$ )	1
$T_{n} (k)$ , $n \geq 2$	yes	no	no	1
$k [[x]]$	yes	no	yes	1
$ℤ$	yes	no	no	1
$ℤ / 12 ℤ$	no	no	no	2
$k \times M_{2} (k)$	no	no	no	2
$\prod_{i = 1}^{\infty} ℚ$	no	no	no	none exists

Which hypotheses give which conclusions
	Block decomposition exists	Unique up to order	Blocks are linkage classes	Linkage = shared composition factor
Arbitrary ring	no	yes, if it exists	no	no
Ideals satisfy ACC or DCC	yes	yes	partial	no
$1$ a sum of orthogonal primitive idempotents	yes	yes	yes	no
Right artinian	yes	yes	yes	yes
Semisimple	yes	yes	yes	yes, and linkage is isomorphism

Which hypotheses give which conclusions

The middle row is the one to notice: a chain condition on ideals delivers existence and uniqueness of blocks without saying anything about primitive idempotents, whereas $(22.5)$ delivers the combinatorial description without any chain condition. Right artinian rings satisfy both hypotheses, which is why the theory is complete there.

Relationship Map

Indecomposability is the weakest of the standard irreducibility conditions on a ring; each of the following implies it, and none of the reverse implications holds.

SimplePrimeIndecomposable

LocalNo nontrivial central idempotentsIndecomposable

For the first chain: a prime ring cannot contain nonzero orthogonal ideals $A, B$ with $A B = 0$ , and a nontrivial central idempotent would produce exactly that. $T_{2} (k)$ is indecomposable but not prime, and $ℤ$ is prime but not simple. For the second: in a local ring every idempotent is $0$ or $1$ , but $M_{2} (k)$ is indecomposable and far from local.

Applications and Industry Use

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

Modular representation theory

Blocks of a group algebra

For $k G$ with $char k = p$ , the block decomposition partitions the simple and projective modules into linkage classes. Brauer's theory of defect groups, and the whole local-global programme in representation theory, is organised block by block, and this is where the partition comes from.

Quiver algebras

Connected components

For a finite-dimensional algebra given by a quiver with relations, linkage of the vertex idempotents is connectivity of the underlying graph. An algebra is indecomposable exactly when its quiver is connected — the combinatorial statement that $(22.5)$ abstracts.

Algebraic geometry

Connectedness of a scheme

For commutative $R$ , idempotents correspond to clopen subsets of $Spec R$ , so indecomposability is connectedness. Non-central idempotents give the noncommutative analogue used in the theory of Azumaya algebras and Brauer groups.

Computer algebra

Splitting an algebra into blocks

Systems compute blocks of a finite-dimensional algebra by computing $Z (R)$ , splitting it into local factors, and reading the central primitive idempotents. Block structure is the first coarse invariant reported for a group algebra in GAP or Magma.

The honest summary is that block decomposition is a divide-and-conquer step. Nothing is proved about a block by decomposing; what is gained is that every subsequent question — modules, cohomology, characters — can be asked one block at a time, and blocks do not interact.

Failure Modes and Common Mistakes

Do not infer indecomposability of $R$ from indecomposability of $R_{R}$ as a module: $M_{2} (k)$ decomposes as a module but not as a ring.
Do not expect $(22.6)$ to hold without a chain condition; composition factors are not available in general, and $e R f \neq 0$ is then the only usable form of linkage.
Do not assume the block containing a simple module is determined by its dimension or its character alone; only the linkage class is.
Do not confuse a block of $R$ with a block of $R / rad R$ : the semisimple quotient generally has more blocks, since linkage collapses when the radical is killed.

Historical Notes and Lessons Learned

1907–1908Wedderburn's componentsFor semisimple algebras the decomposition into simple components is established; here blocks and simple components coincide, and no linkage theory is needed.
1935–1947Brauer's blocksBrauer introduces blocks of a group algebra in characteristic $p$ , distributing ordinary and modular characters into classes, and defines defect groups. The linkage relation is the abstract shadow of his character-theoretic partition.
1950sIdempotent-theoretic reformulationBlocks are recast as the centrally primitive idempotents of the algebra, making the theory available for arbitrary rings with enough idempotents rather than only group algebras.
1970s onwardsQuivers and connectednessGabriel's quiver formalism identifies blocks of a finite-dimensional algebra with connected components of its quiver, making the graph-theoretic reading of linkage explicit.

The methodological lesson is that a partition first observed among characters turned out to be a statement about central idempotents, and therefore about the ring alone. Once restated that way it generalised immediately beyond the setting in which it was found.

Quick Reference

Indecomposable

R \neq 0

, no central idempotent but

0, 1

Central test

e

central

iff

e R f = f R e = 0

f = 1 - e

Centrally primitive

c \neq 0

central,

c R

indecomposable

Linkage

e \sim e'

iff

e R f \neq 0 \neq e' R f

for some

f \in E

Hom form

e R f ≅ {Hom}_{R} (f R, e R)

Blocks

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

, unique up to order

Artinian test

e \sim e'

iff

e R

e' R

share a composition factor

Commutative caseIndecomposable

iff

Spec R

connected

Statements and their hypotheses
Statement	Hypotheses	Reference
Central idempotents are sums of the $c_{i}$	$1 = c_{1} + \dots + c_{r}$ centrally primitive orthogonal	(22.1)(1)
Block decomposition is unique up to order	one exists	(22.1)(3)
A block decomposition exists	ideals satisfy ACC or DCC	(22.2)
$e \in c R$ or $e \in (1 - c) R$	$e$ primitive, $c$ central idempotent	(22.4)
Linkage respects central idempotents	$e \approx e'$ , $c$ central idempotent	(22.3)
Blocks are linkage classes	$1$ a sum of orthogonal primitive idempotents	(22.5)
Linkage = shared composition factor	$R$ right artinian	(22.6)

Frequently Asked Questions

How can a ring be indecomposable and still have lots of idempotents?

Because indecomposability constrains only the central idempotents. $T_{n} (k)$ has a large supply of idempotents — every $E_{11} + λ E_{12}$ , for instance — but its centre is just the scalars, so it is indecomposable. The right slogan is that indecomposability is a property of $Z (R)$ , not of $R$ .

Why is the linkage relation defined with an auxiliary idempotent

f

rather than by

e R e' \neq 0

Because $e R e' \neq 0$ is not symmetric-looking enough to generate the right relation, and more importantly the condition that matters is that some single principal indecomposable $f R$ maps nontrivially to both $e R$ and $e' R$ . In the artinian case that condition becomes * $e R$ and $e' R$ have a composition factor in common*, which is manifestly symmetric and is exactly what block theory needs.

\sim

transitive?

Not in general, which is why $\approx$ is defined as the equivalence relation it generates. Two primitive idempotents can be joined only by a long walk through intermediate idempotents, and blocks are the connected components of that walk structure, not the classes of $\sim$ itself.

Do blocks always exist?

No. An infinite product of fields has infinitely many centrally primitive idempotents and no finite orthogonal decomposition of $1$ into them. Existence needs a hypothesis: a chain condition on ideals $(22.2)$ , or a decomposition of $1$ into orthogonal primitive idempotents $(22.5)$ . Both hold for right artinian rings.

What is the relation between blocks of

R

and blocks of

R / rad R

The semisimple quotient generally has at least as many blocks, since its blocks are its simple components and linkage in it is just isomorphism of idempotents. Passing to the quotient loses the maps $e R f$ that lived in the radical, so linkage classes can split. Recovering the blocks of $R$ from those of $R / rad R$ requires lifting central idempotents, which is a separate and stronger statement.

Is indecomposability preserved by standard constructions?

It is preserved by passing to matrix rings, because $Z (M_{n} (R)) ≅ Z (R)$ , and by passing to $R [x]$ and $R [[x]]$ for the same reason. It is not preserved by quotients: $ℤ$ is indecomposable but $ℤ / 12$ is not.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §22, results (22.1)–(22.6).
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981, §55 (blocks and central idempotents).
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.
R. Brauer, “On blocks of characters of groups of finite order I, II”, Proceedings of the National Academy of Sciences USA 32 (1946), 182–186 and 215–219.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Prove that $R$ is indecomposable if and only if $M_{n} (R)$ is, and identify where the centre calculation is used.
Show that idempotents of a commutative ring correspond bijectively to clopen subsets of its prime spectrum.
Compute the blocks of $k S_{3}$ for $k$ of characteristic $2$ , $3$ and $0$ , and compare the linkage graphs.
For a path algebra with relations, prove that blocks correspond to connected components of the quiver.
Give an example of a ring in which the relation $\sim$ on primitive idempotents fails to be transitive.
How does Brauer's first main theorem refine the block partition beyond what linkage alone provides?
Which conditions on a ring guarantee that central idempotents lift modulo the Jacobson radical?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Linkage is a statement about homomorphisms

Why primitivity makes idempotents choose a side

The connectedness picture

06Key Results

07Proof Techniques and Method

Split by a central idempotent, then use primitivity

Sum a class to build a central idempotent

Read eRf as a Hom group

08Worked Example

Upper triangular matrices: many idempotents, one block

A decomposable example with explicit arithmetic

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Blocks of a group algebra

Connected components

Connectedness of a scheme

Splitting an algebra into blocks

13Failure Modes and Common Mistakes

14Historical Notes and Lessons Learned

15Quick Reference

16Frequently Asked Questions

17Related KEVOS Topics

18References

19AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Read $e R f$ as a Hom group

Worked Example

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions