Indecomposable Modules: First Examples and Pathologies

Q: Which group algebras have finite representation type?

For k of characteristic p, exactly those kG whose Sylow p-subgroups are cyclic — Higman's theorem. In characteristic 0, or when p does not divide the group order, kG is semisimple and the indecomposables are the finitely many simple modules.

Executive Summary

For a semisimple algebra, indecomposable and simple mean the same thing and the module theory is finished. Outside that case the simple modules are only the top layer, and the objects that actually classify modules are the indecomposables: modules admitting no nontrivial direct sum decomposition.

Two examples set the scale. Over $k [x] / (x^{n})$ there are exactly $n$ indecomposables, given by Jordan blocks. Over the three-dimensional algebra $k [x, y] / (x^{2}, x y, y^{2})$ there is an indecomposable of every positive dimension, and — if $k$ is infinite — a family of two-dimensional indecomposables parametrised by $k$ itself. Classification is hopeless in general; the modern theory measures how hopeless.

$n$ Indecomposables over

k [x] / (x^{n})

$\leq 2$ Over any

2

-dimensional algebra

$\infty$ Over

k [x, y] / (x^{2}, x y, y^{2})

$| k |$ Two-dimensional indecomposables there

Overview

A nonzero module $M$ is indecomposable if $M = N \oplus N'$ with $N, N'$ submodules forces $N = 0$ or $N' = 0$ . Simple modules are indecomposable; the converse holds precisely when the module is semisimple. Over a finite-dimensional algebra every finite-dimensional module is a finite direct sum of indecomposables, and by Krull–Schmidt the summands are unique up to isomorphism and order — so classifying indecomposables classifies everything.

The examples below are chosen to mark the boundary. Dimension $2$ is always tame: at most two indecomposables. Dimension $3$ can already be wild in the informal sense — infinitely many indecomposables, in every dimension. The group-algebra instance is $k (C_{p} \times C_{p})$ in characteristic $p$ , which is why modular representation theory of non-cyclic $p$ -groups is hard.

This page gives first examples only. The general machinery — Krull–Schmidt, local endomorphism rings, principal indecomposables — is developed on Krull–Schmidt Theorem and Strongly Indecomposable Modules.

Learning Objectives

State the definition of indecomposability and contrast it with simplicity.
List the indecomposables over a two-dimensional algebra in each of the three cases for its defining quadratic.
Derive the classification over $k [x] / (x^{n})$ from the Jordan canonical form.
Construct the modules $M_{2 n + 1}$ and $M_{2 n}$ over $k [x, y] / (x^{2}, x y, y^{2})$ and prove indecomposability.
Prove that the family $M (a)$ consists of pairwise non-isomorphic two-dimensional indecomposables.
Define representation type and place the standard examples on either side of the divide.

Definitions

DefinitionIndecomposable module and representation type

A nonzero module $M$ over a ring $R$ is indecomposable if it is not the direct sum of two nonzero submodules. A finite-dimensional $k$ -algebra $R$ has finite representation type if it has only finitely many finite-dimensional indecomposable modules up to isomorphism, and infinite representation type otherwise.

ConstructionThe three-dimensional algebra

Let $R$ be the commutative $k$ -algebra with two generators $α, β$ subject to

α^{2} = β^{2} = α β = β α = 0, R = k [x, y] / (x^{2}, x y, y^{2})

${1, α, β}$ is a $k$ -basis, so $\dim_{k} R = 3$ ; $rad R = k α + k β$ with $(rad R)^{2} = 0$ .

An $R$ -module is thus a $k$ -vector space with two commuting operators $α, β$ whose pairwise products all vanish. There is a unique simple $R$ -module, namely $k$ with $α = β = 0$ .

Simple: No submodules other than $0$ and $M$ . Simple implies indecomposable; the converse fails as soon as $rad R \neq 0$ .
Indecomposable: No decomposition $M = N \oplus N'$ with both summands nonzero. Equivalently, $End (_{R} M)$ has no idempotents besides $0$ and $1$ .
$M_{i} = k [x] / (x^{i})$: The Jordan block modules over $k [x] / (x^{n})$ , for $1 \leq i \leq n$ .
$M (a)$: The two-dimensional module $k w \oplus k v$ over $R$ with $α v = w$ and $β v = a w$ , and $α, β$ killing $w$ .
Wild versus tame: Modern refinements of infinite representation type, measuring whether indecomposables come in one-parameter families or in families of arbitrary complexity.

All modules on this page are finite-dimensional over $k$ unless stated otherwise. Over a finite-dimensional algebra this is equivalent to being finitely generated.

Core Concepts

Two-dimensional algebras: three cases and no more

A two-dimensional $k$ -algebra $R$ is spanned by $1$ and one further element, so it is commutative and $R ≅ k [x] / (q (x))$ for a monic quadratic $q$ . The classification of its indecomposables follows the factorisation of $q$ :

Indecomposables over a two-dimensional algebra $k [x] / (q)$
Case	$R$	Indecomposable modules	Count
$q$ irreducible over $k$	a quadratic field extension	$_{R} R$ only	$1$
$q = (x - a) (x - b)$ , $a \neq b$	$k \times k$ , semisimple	the two simple modules	$2$
$q = (x - a)^{2}$	$≅ k [x] / (x^{2})$ , local	$k$ and $_{R} R$	$2$

In the first case $R$ is a field, so every module is an $R$ -vector space and the indecomposables are the one-dimensional ones. In the second $R$ is semisimple. In the third the classification comes from Jordan form, as below.

Jordan form is a classification of indecomposables

A finitely generated module over $R = k [x] / (x^{n})$ is a finite-dimensional $k$ -space equipped with an endomorphism $A$ with $A^{n} = 0$ . The Jordan canonical form for a nilpotent operator says such a pair is a direct sum of Jordan blocks with uniquely determined multiplicities, so

M ≅ r_{1} M_{1} \oplus r_{2} M_{2} \oplus \dots \oplus r_{n} M_{n}, M_{i} = k [x] / (x^{i})

The $r_{i}$ are uniquely determined; $M_{1}, \dots, M_{n}$ are exactly the indecomposables.

Key Results

Proposition(7.22)Indecomposables in every dimension

Let $R = k [x, y] / (x^{2}, x y, y^{2})$ with generators $α, β$ as above, and let $n \geq 1$ . Define $R$ -modules

$M_{2 n + 1} = U \oplus V$ as $k$ -spaces, where $U$ has basis $u_{0}, u_{1}, \dots, u_{n}$ and $V$ has basis $v_{1}, \dots, v_{n}$ , with $α (U) = β (U) = 0$ and $α (v_{i}) = u_{i}$ , $β (v_{i}) = u_{i - 1}$ for $1 \leq i \leq n$ ;
$M_{2 n} = W \oplus V$ , where $W$ has basis $w_{1}, \dots, w_{n}$ and $V$ has basis $v_{1}, \dots, v_{n}$ , with $α (W) = β (W) = 0$ , $α (v_{i}) = w_{i}$ for all $i$ , $β (v_{i}) = w_{i - 1}$ for $i \geq 2$ and $β (v_{1}) = 0$ .

These are $R$ -modules of $k$ -dimensions $2 n + 1$ and $2 n$ respectively, and both are indecomposable. Together with the one-dimensional simple module, $R$ therefore has an indecomposable module of every positive dimension.

Lemma(7.23)A subspace observation

In $M = M_{2 n + 1}$ , let $S$ be any nonzero $k$ -subspace of $V$ . Then $β S$ is not contained in $α S$ .

Proof

Let $m$ be the largest index with $S \subseteq k v_{m} + \dots + k v_{n}$ ; since $S \neq 0$ we have $1 \leq m \leq n$ , and by maximality there is $s = a_{m} v_{m} + \dots + a_{n} v_{n} \in S$ with $a_{m} \neq 0$ .

Then $β s = a_{m} u_{m - 1} + \dots + a_{n} u_{n - 1}$ has nonzero coefficient at $u_{m - 1}$ . But every element of $S$ is a combination of $v_{m}, \dots, v_{n}$ , so $α S \subseteq k u_{m} + \dots + k u_{n}$ , in which $u_{m - 1}$ does not appear. Hence $β s \notin α S$ .

Proofof (7.22) for

M = M_{2 n + 1}

Let $π : M \to V$ be the projection with respect to $M = U \oplus V$ . We claim that every nonzero submodule $N \subseteq M$ satisfies $\dim_{k} N \geq 1 + 2 \dim_{k} π (N)$ .

Put $S = π (N)$ . If $S = 0$ then $N$ is a nonzero subspace of $U$ and the inequality reads $\dim_{k} N \geq 1$ , which holds. So assume $S \neq 0$ . Because $α$ and $β$ annihilate $U$ , for any $x \in N$ we have $α x = α π (x)$ and $β x = β π (x)$ ; hence $α S = α N \subseteq N$ and $β S = β N \subseteq N$ , and both lie in $U$ .

Therefore $α S + β S \subseteq N \cap U$ , while $π$ induces an isomorphism $N / (N \cap U) ≅ S$ . Using $(7.23)$ , $β S$ is not contained in $α S$ , so $\dim_{k} (α S + β S) \geq \dim_{k} α S + 1 = \dim_{k} S + 1$ , the last equality because $α$ is injective on $V$ . Combining,

\dim_{k} N = \dim_{k} (N \cap U) + \dim_{k} S \geq (\dim_{k} S + 1) + \dim_{k} S = 1 + 2 \dim_{k} S

Now suppose $M = N \oplus N'$ with $N, N'$ nonzero submodules. Applying $π$ gives $S + S' = V$ with $S = π (N)$ , $S' = π (N')$ , so $\dim_{k} S + \dim_{k} S' \geq n$ . Adding the two inequalities,

2 n + 1 = \dim_{k} M = \dim_{k} N + \dim_{k} N' \geq 2 + 2 (\dim_{k} S + \dim_{k} S') \geq 2 + 2 n

a contradiction. Hence $M_{2 n + 1}$ is indecomposable.

Proofof (7.22) for

M = M_{2 n}

, sketch

Let $π : M \to V$ be the projection for $M = W \oplus V$ and suppose $M = N \oplus N'$ with both summands nonzero; set $S = π (N)$ , $S' = π (N')$ . If $β S$ is not contained in $α S$ , or $β S'$ is not contained in $α S'$ , or $\dim_{k} S + \dim_{k} S' > n$ , the dimension count above applies verbatim and gives a contradiction.

So assume $β S \subseteq α S$ , $β S' \subseteq α S'$ and $V = S \oplus S'$ . Both are nonzero: if say $S' = 0$ and $S = V$ , then $N' \subseteq W = α V = α S = α N \subseteq N$ , contradicting directness.

Hence $W = α V = α S \oplus α S'$ is a decomposition into nonzero subspaces. Let $λ$ be the $k$ -endomorphism of $W$ with $λ (w_{i}) = w_{i - 1}$ for $i \geq 2$ and $λ (w_{1}) = 0$ ; by construction $λ α = β$ on $V$ , so $λ (α S) = β S \subseteq α S$ and likewise for $S'$ . Thus $W$ decomposes as a direct sum of two nonzero $k [λ]$ -modules. But $k [λ] ≅ k [x] / (x^{n})$ and $W$ is free of rank one over it, hence indecomposable — a contradiction.

Proposition(7.24)A one-parameter family in dimension two

For $a \in k$ let $M (a) = k w \oplus k v$ with the $R$ -action $α w = 0$ , $α v = w$ , $β w = 0$ , $β v = a w$ ; in the ordered basis ${w, v}$ ,

α \mapsto (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}), β \mapsto (\begin{matrix} 0 & a \\ 0 & 0 \end{matrix})

Then $M (a)$ is an indecomposable $R$ -module, and $M (a)$ and $M (a')$ are isomorphic only if $a = a'$ . Consequently, if $k$ is infinite then $R$ has infinitely many pairwise non-isomorphic two-dimensional indecomposable modules, and $R$ has infinite representation type.

Proof

Indecomposability. On a one-dimensional $R$ -module, $α$ acts as a scalar $c$ with $c^{2} = 0$ , hence as zero, and similarly $β$ . If $M (a)$ were decomposable it would be a sum of two one-dimensional submodules, so $α$ would act as zero on $M (a)$ — contradicting $α v = w \neq 0$ .

Non-isomorphism. Write $M (a') = k w' \oplus k v'$ with $α w' = 0$ , $α v' = w'$ , $β w' = 0$ , $β v' = a' w'$ , and let $h : M (a) \to M (a')$ be an $R$ -isomorphism, say $h (w) = b w' + c v'$ and $h (v) = d w' + e v'$ .

From $h (α w) = α h (w)$ we get $0 = α (b w' + c v') = c w'$ , so $c = 0$ . From $h (α v) = α h (v)$ we get $h (w) = b w' = α (d w' + e v') = e w'$ , so $b = e$ . From $h (β v) = β h (v)$ we get $a b w' = β (d w' + e v') = e a' w'$ , that is $a b = e a' = b a'$ .

The matrix of $h$ in the bases ${w, v}$ and ${w', v'}$ is $(\begin{matrix} b & d \\ 0 & e \end{matrix})$ , which is invertible, so $b = e \neq 0$ . Cancelling $b$ in $a b = b a'$ gives $a = a'$ .

CorollaryElementary abelian

p

-groups

Let $k$ have characteristic $p$ and let $G = ⟨ f ⟩ \times ⟨ g ⟩$ be elementary abelian of order $p^{2}$ . Then

k G ≅ k [X, Y] / (X^{p} - 1, Y^{p} - 1) = k [X, Y] / ((X - 1)^{p}, (Y - 1)^{p}) ≅ k [x, y] / (x^{p}, y^{p})

Since $p \geq 2$ , the algebra $k [x, y] / (x^{2}, x y, y^{2})$ is a quotient of $k [x, y] / (x^{p}, y^{p})$ , so all the modules above are $k G$ -modules. Hence $k G$ has indecomposable modules in every positive dimension, and for infinite $k$ infinitely many two-dimensional ones, given by

f \mapsto (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix}), g \mapsto (\begin{matrix} 1 & a \\ 0 & 1 \end{matrix}), a \in k

Proof Techniques and Method

How these proofs work, and which move to reuse.

Indecomposability proofs come in three standard flavours; all three appear above.

Move 1

Dimension inequality

Show every nonzero submodule is larger than its projection forces it to be. Two summands then over-fill the module, and the contradiction is arithmetic.

Move 2

Idempotents in $End$

$M$ is indecomposable exactly when $End (_{R} M)$ has no idempotent other than $0$ and $1$ ; for finite-dimensional modules this holds precisely when the endomorphism ring is local.

Move 3

Transport to a simpler algebra

The $M_{2 n}$ argument converts a decomposition of an $R$ -module into a decomposition of a $k [λ]$ -module, where indecomposability is already known from Jordan form.

For distinguishing modules, the technique of $(7.24)$ is the one to keep: write a hypothetical isomorphism in coordinates, impose the module relations one generator at a time, and let invertibility force the parameters to agree. It is the standard way to prove a family is genuinely a family.

Worked Example

Decomposing a module over $k [x] / (x^{3})$

Let $M = k^{4}$ with $x$ acting by the nilpotent operator $N$ given by $N e_{1} = e_{2}$ , $N e_{2} = e_{3}$ , $N e_{3} = 0$ , $N e_{4} = 0$ . Then $rank N = 2$ , $rank N^{2} = 1$ , $N^{3} = 0$ .

The number of Jordan blocks is $\dim \ker N = 2$ , and the block sizes are read off from the ranks: sizes $3$ and $1$ . Hence

M ≅ k [x] / (x^{3}) \oplus k [x] / (x) = M_{3} \oplus M_{1}

and the decomposition is unique by Krull–Schmidt. Over this algebra there are exactly three indecomposables, so the classification is complete.

The module $M_{5}$ over $R = k [x, y] / (x^{2}, x y, y^{2})$

Take $n = 2$ : $U$ has basis $u_{0}, u_{1}, u_{2}$ and $V$ has basis $v_{1}, v_{2}$ , with

α v_{1} = u_{1}, α v_{2} = u_{2}, β v_{1} = u_{0}, β v_{2} = u_{1}, α (U) = β (U) = 0

Check the inequality of the proof on a sample submodule. Take $S = k v_{2} \subseteq V$ : then $α S = k u_{2}$ and $β S = k u_{1}$ , so $β S$ is not inside $α S$ , as $(7.23)$ predicts. The smallest submodule with $π (N) = S$ is $N = k v_{2} + k u_{1} + k u_{2}$ , of dimension $3 = 1 + 2 \cdot 1$ — the bound is attained.

If $M_{5} = N \oplus N'$ with both nonzero, then $\dim N + \dim N' \geq 2 + 2 (\dim S + \dim S') \geq 2 + 4 = 6 > 5$ . So $M_{5}$ is indecomposable.

Two members of the family $M (a)$

Compare $M (0)$ and $M (1)$ . In $M (0)$ the operator $β$ acts as zero, while in $M (1)$ it acts as the same nonzero map as $α$ . Any isomorphism would have to carry $\ker β$ to $\ker β$ ; in $M (0)$ that kernel is the whole module, in $M (1)$ it is one-dimensional. So $M (0)$ and $M (1)$ are not isomorphic, in agreement with $(7.24)$ . Note also $M (0) ≅ M_{2}$ in the earlier notation.

Frameworks and Models

Modern representation theory sorts finite-dimensional algebras over an algebraically closed field into three mutually exclusive classes.

Representation type
- Finite — finitely many indecomposables
  - $k [x] / (x^{n})$
  - path algebras of Dynkin quivers $A_{n}$ , $D_{n}$ , $E_{6}$ , $E_{7}$ , $E_{8}$ (Gabriel)
  - $k G$ with cyclic Sylow $p$ -subgroup (Higman)
- Tame — indecomposables in each dimension come in finitely many one-parameter families
  - $k [x, y] / (x^{2}, y^{2})$ and other special biserial algebras
  - $k G$ for $G$ with dihedral, semidihedral or quaternion Sylow $2$ -subgroups in characteristic $2$
- Wild — classification contains the classification of pairs of matrices up to simultaneous conjugacy
  - $k ⟨ x, y ⟩$ -type problems
  - most group algebras of non-cyclic $p$ -groups in characteristic $p$

Comparison and Classification

Representation type of standard algebras
Algebra	$\dim_{k}$	Indecomposables	Type
$k$	$1$	$1$	finite
$k \times k$	$2$	$2$	finite
quadratic field extension	$2$	$1$	finite
$k [x] / (x^{n})$	$n$	$n$	finite
$M_{n} (k)$	$n^{2}$	$1$	finite
$T_{2} (k)$ upper triangular	$3$	$3$	finite
$k [x, y] / (x^{2}, x y, y^{2})$	$3$	one in every dimension	infinite
$k [x, y] / (x^{r}, y^{s})$ , $r, s \geq 2$	$r s$	one in every dimension	infinite
$k G$ , $char k = p$ , Sylow $p$ cyclic	$\| G \|$	finitely many	finite
$k (C_{p} \times C_{p})$ , $char k = p$	$p^{2}$	one in every dimension	infinite

Simple, indecomposable, projective: which implications hold
	$R$ semisimple	$R$ local	$R$ arbitrary finite-dimensional
simple $\Rightarrow$ indecomposable	yes	yes	yes
indecomposable $\Rightarrow$ simple	yes	no	no
every module is a sum of simples	yes	no	no
finitely many indecomposables	yes	partial	no
Krull–Schmidt holds for finite-dimensional modules	yes	yes	yes

Simple, indecomposable, projective: which implications hold

Relationship Map

simpleindecomposabledirect summandarbitrary module

Reading right to left: every finite-dimensional module is a direct sum of indecomposables, uniquely by Krull–Schmidt. The simple modules are the tops of the finitely many principal indecomposables; the other indecomposables record the ways modules can be glued along the radical, and there may be a great many of them.

All finite-dimensional modulesclosed under sums, subquotients

Indecomposable modulesthe building blocks, by Krull–Schmidt

Principal indecomposablesthe direct summands of

_{R} R

; finitely many, in bijection with the simple modules

Simple modules

M_{1}, \dots, M_{r}

; the tops of the principal indecomposables

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

Why $p$ -groups are hard

Higman's theorem says $k G$ in characteristic $p$ has finite representation type exactly when the Sylow $p$ -subgroups are cyclic. The elementary abelian example above is the smallest obstruction, and it is why block theory works with invariants rather than classifications.

Quiver representations

Gabriel's theorem

Indecomposable representations of a quiver over an algebraically closed field are finite in number precisely for Dynkin quivers, where they biject with the positive roots. This turned representation type into Lie theory.

Linear algebra

Pairs of matrices

Classifying modules over $k ⟨ x, y ⟩$ -like algebras is classifying pairs of matrices up to simultaneous conjugacy — the standard benchmark for a wild problem, and provably as hard as any classification in linear algebra.

Computational algebra

Decomposition algorithms

Splitting a module into indecomposables is done by searching for idempotents in the endomorphism algebra. Krull–Schmidt guarantees the output is canonical, which is what makes such algorithms well posed.

The honest summary: this is where the structure theory stops being a classification and becomes a taxonomy. What downstream users take from it is the knowledge of which problems admit lists and which do not.

Failure Modes and Common Mistakes

Do not conclude from a one-parameter family that the algebra is tame — tameness is a bound on how many families occur in each dimension, not the existence of one.
Do not assume the indecomposables of a quotient algebra are among those of the original; they are, but only because modules over a quotient pull back, and the reverse inclusion is false.
Do not use dimension counts alone to prove indecomposability; the projection inequality of $(7.22)$ needs the specific structure of the action.

Historical Notes and Lessons Learned

1954Higman $k G$ in characteristic $p$ has finite representation type if and only if the Sylow $p$ -subgroups of $G$ are cyclic.
1968RoiterProof of the first Brauer–Thrall conjecture: an algebra of infinite representation type has indecomposable modules of arbitrarily large dimension. Auslander later extended this to artin algebras.
1972GabrielA connected quiver has finitely many indecomposable representations exactly when its underlying graph is a simply laced Dynkin diagram, and they correspond to the positive roots.
1970sNazarova–RoiterThe second Brauer–Thrall conjecture proved over algebraically closed fields: infinite type forces infinitely many indecomposables in each of infinitely many dimensions. Ringel extended this to perfect fields.
1977–1980DrozdThe tame–wild dichotomy: over an algebraically closed field every finite-dimensional algebra is of finite, tame or wild representation type, and no algebra is both tame and wild.

The lesson is one of expectation management. Wedderburn theory suggested that algebras could be classified by their modules; these examples showed that beyond the semisimple case the right question is not what are the indecomposables but how badly do they proliferate. The answer is now a structure theory in its own right, built on almost split sequences and Auslander–Reiten quivers.

Quick Reference

Indecomposable

M \neq 0

and

M = N \oplus N'

forces

N = 0

N' = 0

Equivalently

End (_{R} M)

has no idempotents besides

0, 1

$2$ -dimensional algebrasat most two indecomposables, in all three cases

$k [x] / (x^{n})$ exactly

n

indecomposables, the Jordan blocks

$k [x, y] / (x^{2}, x y, y^{2})$ one indecomposable in every positive dimension

$M (a)$

α v = w

β v = a w

; pairwise non-isomorphic

Group algebrasfinite type iff Sylow

p

-subgroups are cyclic

Krull–Schmidtdecomposition unique up to isomorphism and order

Brauer–Thrall Iinfinite type implies unbounded dimensions

Statement finder
Result	Content	Reference
Two-dimensional algebras	three cases, at most two indecomposables	before (7.22)
Jordan classification	$n$ indecomposables over $k [x] / (x^{n})$	before (7.22)
Every dimension occurs	$M_{2 n}$ and $M_{2 n + 1}$ are indecomposable	(7.22)
Key subspace lemma	$β S$ not inside $α S$	(7.23)
A family in dimension two	$M (a)$ indecomposable, pairwise non-isomorphic	(7.24)
Elementary abelian groups	$k (C_{p} \times C_{p})$ has infinite type	after (7.24)

Frequently Asked Questions

Why does the theory switch from simple to indecomposable modules?

Because for a non-semisimple algebra the simple modules do not generate the module category under direct sums. Every finite-dimensional module is still a direct sum of indecomposables, uniquely by Krull–Schmidt, so the indecomposables are the correct building blocks; the simple modules are only their tops.

Is a module with local endomorphism ring the same as an indecomposable one?

Local endomorphism ring implies indecomposable always. The converse holds for finite-dimensional modules over a finite-dimensional algebra, because the endomorphism ring is then a finite-dimensional algebra with no nontrivial idempotents, hence local. In general the converse fails.

How can a three-dimensional algebra be so complicated?

Because its modules encode a pair of linear maps with all products zero, and the simultaneous classification of two maps is already an infinite problem. Dimension of the algebra is a poor measure of the difficulty of its module category; the shape of the radical is a better one.

Does the family

M (a)

exhaust the two-dimensional indecomposables?

Up to isomorphism, the two-dimensional modules over this algebra correspond to pairs of nilpotent square-zero operators with zero products, and after normalisation those with $α \neq 0$ are the $M (a)$ . Modules with $α = 0$ and $β \neq 0$ give one further isomorphism class, and the rest are decomposable.

What is the significance of the Brauer–Thrall conjectures?

They assert that infinite representation type is not a mild condition: it forces unbounded dimensions (first conjecture, Roiter) and infinitely many indecomposables in infinitely many dimensions (second conjecture, Nazarova–Roiter and Ringel). Together they say there is no algebra with infinitely many indecomposables all of bounded size.

Which group algebras have finite representation type?

For $k$ of characteristic $p$ , exactly those $k G$ whose Sylow $p$ -subgroups are cyclic — Higman's theorem. In characteristic $0$ , or when $p$ does not divide the group order, $k G$ is semisimple and the indecomposables are the finitely many simple modules.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §7 (pp. 122–123).
R. S. Pierce, Associative Algebras, Graduate Texts in Mathematics 88, Springer-Verlag, 1982, Chapters 5–6.
M. Auslander, I. Reiten and S. O. Smalø, Representation Theory of Artin Algebras, Cambridge Studies in Advanced Mathematics 36, Cambridge University Press, 1995.
D. G. Higman, “Indecomposable representations at characteristic $p$ ”, Duke Mathematical Journal 21 (1954), 377–381.
P. Gabriel, “Unzerlegbare Darstellungen I”, Manuscripta Mathematica 6 (1972), 71–103.
D. J. Benson, Representations and Cohomology I, Cambridge Studies in Advanced Mathematics 30, Cambridge University Press, 1991, Chapter 4.

AI Suggested Questions

Prove Higman's theorem that $k G$ has finite representation type if and only if the Sylow $p$ -subgroups are cyclic.
List all indecomposable representations of the quiver $A_{3}$ and match them with the positive roots.
Show that the two-dimensional modules over $k [x, y] / (x^{2}, x y, y^{2})$ are classified by a point of the projective line over $k$ .
What are the Auslander–Reiten quiver and almost split sequences, and how do they organise indecomposables?
Give a proof of Krull–Schmidt for finite-dimensional modules using local endomorphism rings.
Explain Drozd's tame–wild dichotomy and give an algebra of tame but not finite type.
How does one compute a decomposition into indecomposables algorithmically over a finite field?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Two-dimensional algebras: three cases and no more

Jordan form is a classification of indecomposables

06Key Results

07Proof Techniques and Method

Dimension inequality

Idempotents in End

Transport to a simpler algebra

08Worked Example

Decomposing a module over k[x]/(x3)

The module M5 over R=k[x,y]/(x2,xy,y2)

Two members of the family M(a)

09Frameworks and Models

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Why p-groups are hard