Strongly Indecomposable Modules and Local Endomorphism Rings

Q: Is the endomorphism ring of an indecomposable module always a domain?

No. End(Z/p^2Z) ≅ Z/p^2Z has p p = 0, and the ring computed in (19.15) has m^2 = 0. Indecomposability forbids idempotents, not zero divisors; those are different conditions, and locality controls neither directly.

Q: Does strong indecomposability behave well under field extension?

Not in general. If M is a module over a k-algebra and K k is an extension field, M ⊗_k K can decompose even when M is strongly indecomposable — this is the difference between indecomposable and absolutely indecomposable modules. Strong indecomposability is preserved when the residue division ring of End(M) stays a division ring after extension, which holds when k is a splitting field.

Q: What is the relation between locality of End(M_R) and locality of R?

None in either direction. R = Z is not local yet Z/p^nZ has local endomorphism ring; conversely R = k[[x]] is local while the module R ⊕ R has endomorphism ring M_2(R), which is not local. The two conditions concern different objects.

Q: How do indecomposable injectives fit in?

They are always strongly indecomposable. For M indecomposable injective, the endomorphisms with essential kernel form an ideal, and every endomorphism outside it is an automorphism — so that ideal is the radical and the quotient is a division ring. This gives a second large supply of local rings, developed in Lam's sequel volume rather than in §19.

Q: Can a module have finite length but decompose in more than one way?

It can decompose into indecomposables in more than one *literal* way — different submodules — but the multiset of isomorphism types is uniquely determined, by (19.17) together with Krull–Schmidt–Azumaya. Uniqueness is up to isomorphism and reindexing, never up to equality of submodules.

Executive Summary

Indecomposability of a module $M$ is a statement about idempotents in $E = End (M_{R})$ : $M$ is indecomposable exactly when $E$ has no idempotent other than $0$ and $1$ . Strong indecomposability asks for more — that $E$ be a local ring. Since a local ring has no nontrivial idempotents, strong indecomposability implies indecomposability, and the converse fails.

The distinction is not pedantic. Every uniqueness theorem for direct sum decompositions — the Krull–Schmidt–Azumaya theorem above all — takes local endomorphism rings as its hypothesis, not indecomposability. Lam's $(19.17)$ is the result that makes the hypothesis checkable: for modules of finite composition length the two notions agree.

$e = e^{2}$ Detects decompositions

Local $E$ Strong indecomposability

$ℤ$ Indecomposable, not strongly

$𝔪^{n} = 0$ Finite length

n

Overview

Fix a ring $R$ and a nonzero right $R$ -module $M$ , and write $E = End (M_{R})$ for its endomorphism ring, acting on the left of $M$ . Every internal direct sum decomposition $M = A \oplus B$ produces the projection onto $A$ along $B$ , an idempotent of $E$ ; and every idempotent of $E$ produces such a decomposition. So decomposition theory is idempotent theory in $E$ .

M = A \oplus B ⟷ e \in E, e = e^{2}, A = e M, B = (1 - e) M

(19.12a)

The correspondence between decompositions of $M$ and idempotents of $E$ .

Reading this backwards: $M$ indecomposable means $E$ has only the trivial idempotents. That is a weak condition on a ring — every domain satisfies it — and it is too weak to control decompositions of larger modules built from $M$ . Requiring $E$ to be local is the strengthening that works.

The last phrasing is the operational one. It says that if $f + g$ is an automorphism of $M$ , then $f$ or $g$ already is — precisely the statement that drives the exchange argument in the proof of the Krull–Schmidt–Azumaya theorem.

Learning Objectives

Prove the correspondence between direct decompositions of $M$ and idempotents of $End (M_{R})$ .
State $(19.12)$ and deduce that strong indecomposability implies indecomposability.
Prove $(19.13)$ : simple modules are strongly indecomposable.
Work through $(19.14)$ : $ℤ$ , $ℤ / p^{n} ℤ$ and the Prüfer group as $ℤ$ -modules.
Compute the endomorphism ring in the modular example $(19.15)$ and verify locality.
State $(19.17)$ and explain via $(19.18)$ why one chain condition is not enough.

Definitions

Definition(19.12)Strongly indecomposable module

A nonzero right $R$ -module $M$ is strongly indecomposable if $End (M_{R})$ is a local ring.

By $(19.2) (c)$ a local ring has no nontrivial idempotents, so every strongly indecomposable module is indecomposable. The converse is false; $(19.14)$ supplies the standard counterexample.

$End (M_{R})$: The ring of $R$ -module endomorphisms of $M$ , with pointwise addition and composition as multiplication. It acts on the left of $M$ , making $M$ an $E$ - $R$ -bimodule.
Indecomposable: $M \neq 0$ and $M = A \oplus B$ with submodules $A, B$ forces $A = 0$ or $B = 0$ .
Finite length: $M$ has a composition series; equivalently $M$ satisfies both ACC and DCC on submodules. The number of factors is the composition length.
Completely primary: A local ring with nilpotent maximal ideal — the class into which $(19.17)$ places $End (M_{R})$ for $M$ indecomposable of finite length.

Lam works with right modules throughout §19 and lets endomorphisms act on the left; that convention is kept here so that composition and module multiplication do not collide.

Core Concepts

Idempotents and decompositions

Suppose $M = A \oplus B$ with $A, B$ submodules. Define $e : M \to M$ by $e (a + b) = a$ . This is $R$ -linear, satisfies $e^{2} = e$ , and is nonzero iff $A \neq 0$ , and different from $1$ iff $B \neq 0$ .

Conversely, given $e = e^{2} \in E$ , put $A = e M$ and $B = (1 - e) M$ . Both are submodules because $e$ is $R$ -linear. Every $m$ splits as $m = e (m) + (1 - e) (m)$ , so $M = A + B$ ; and if $x = e (y) = (1 - e) (z)$ then $e (x) = e^{2} (y) = e (y) = x$ while also $e (x) = (e - e^{2}) (z) = 0$ , so $x = 0$ . Hence $M = A \oplus B$ .

What locality buys

If $E$ is local, then for any finite family $f_{1}, \dots, f_{s} \in E$ with $f_{1} + \dots + f_{s} = 1$ , some $f_{j}$ is an automorphism — this is $(19.1) (6)$ applied to the unit $1$ . In the Krull–Schmidt argument the $f_{j}$ are the composites $α_{1} β_{j}$ of two families of projections; locality selects one of them as invertible, and that single choice launches the induction.

M

strongly indecomposable

End (M_{R})

localsum of endomorphisms is an automorphism only if a summand isunique decomposition

Where the two notions agree

Two classes of modules collapse the distinction. For modules of finite composition length this is $(19.17)$ , proved via the Fitting decomposition — see Fitting's Lemma and Endomorphism Rings of Modules of Finite Length. For injective modules it is a separate classical fact: an indecomposable injective module has local endomorphism ring, because a non-injective-splitting endomorphism must have essential kernel. Both proofs share the same shape: show that every non-automorphism is, in a suitable sense, small.

Key Results

Proposition(19.12a)Idempotent criterion for indecomposability

Let $M$ be a nonzero right $R$ -module and $E = End (M_{R})$ . Then $M$ is indecomposable if and only if the only idempotents of $E$ are $0$ and $1$ .

Proof

If $M = A \oplus B$ with $A, B \neq 0$ , the projection $e$ onto $A$ along $B$ is idempotent with $e M = A \neq 0$ and $(1 - e) M = B \neq 0$ , so $e \neq 0, 1$ .

Conversely let $e = e^{2} \in E$ with $e \neq 0, 1$ . As computed above, $M = e M \oplus (1 - e) M$ . If $e M = 0$ then $e = 0$ ; if $(1 - e) M = 0$ then $e = 1$ . Both are excluded, so both summands are nonzero and $M$ is decomposable.

Corollary(19.12b)Strong implies plain

Every strongly indecomposable module is indecomposable.

Indeed if $E = End (M_{R})$ is local, then by $(19.2) (c)$ it has no idempotents other than $0$ and $1$ , so $(19.12 a)$ applies. The implication is strict: see $(19.14)$ .

Proposition(19.13)Simple modules

Every simple right $R$ -module $M$ is strongly indecomposable. More precisely, $End (M_{R})$ is a division ring, hence local with zero radical.

Proof

This is Schur's Lemma. Let $0 \neq f \in End (M_{R})$ . Then $\ker f$ is a proper submodule of $M$ , hence $0$ by simplicity, so $f$ is injective; and $im f$ is a nonzero submodule, hence all of $M$ , so $f$ is surjective. Thus $f$ is bijective and its set-theoretic inverse is again $R$ -linear. Every nonzero element of $End (M_{R})$ is a unit, so it is a division ring, and a division ring is local by $(19.1) (3)$ .

Example(19.14)Three abelian groups

** $M_{1} = ℤ$ as a $ℤ$ -module.** Indecomposable, since any two nonzero subgroups of $ℤ$ meet nontrivially. But $End (M_{1}) ≅ ℤ$ , which is not local — $4$ and $- 3$ are non-units with $4 + (- 3) = 1 \in U (ℤ)$ , contradicting $(19.1) (6)$ . So $M_{1}$ is not strongly indecomposable.
** $M_{2} = ℤ / p^{n} ℤ$ , $p$ prime, $n \geq 1$ .** $End (M_{2}) ≅ ℤ / p^{n} ℤ$ , in which every non-unit is a multiple of $p$ and hence nilpotent; by $(19.3) (a)$ this ring is local. So $M_{2}$ is strongly indecomposable, and its endomorphism ring is completely primary.
** $M_{3} = ℤ (p^{\infty})$ , the group of all $p^{n}$ -th roots of unity.** Its endomorphism ring is the inverse limit of the rings $ℤ / p^{n} ℤ$ , that is, the ring $ℤ_{p}$ of $p$ -adic integers, which is local with maximal ideal $p ℤ_{p}$ . So $M_{3}$ is strongly indecomposable — but $p ℤ_{p}$ is not nilpotent, indeed not nil.

Example(19.15)A modular representation of an elementary abelian group

Let $k$ be a field of characteristic $p > 0$ and let $G = ⟨ x, y ⟩$ be elementary abelian of order $p^{2}$ . Let $V = e_{1} k \oplus e_{2} k \oplus e_{3} k$ be the $3$ -dimensional right $k G$ -module defined by

\begin{aligned} e_{1} x & = e_{1}, & e_{2} x & = e_{2}, & e_{3} x & = e_{1} + e_{3}, \\ e_{1} y & = e_{1}, & e_{2} y & = e_{2}, & e_{3} y & = e_{2} + e_{3} . \end{aligned}

(19.15)

Then $End (V_{k G})$ consists exactly of the matrices $(\begin{matrix} a & 0 & b \\ 0 & a & c \\ 0 & 0 & a \end{matrix})$ with $a, b, c \in k$ . This ring is commutative and local with maximal ideal of square zero, so $V$ is a strongly indecomposable $k G$ -module. The computation is carried out in full below.

Theorem(19.17)Finite length forces locality

Let $M$ be an indecomposable right $R$ -module of finite composition length $n$ . Then $E = End (M_{R})$ is a local ring, and its unique maximal ideal $𝔪 = rad E$ satisfies $𝔪^{n} = 0$ . In particular $M$ is strongly indecomposable and $E$ is completely primary.

The proof runs through the Fitting decomposition $M = \ker (f^{m}) \oplus im (f^{m})$ for large $m$ : indecomposability forces one summand to vanish, so every non-automorphism is nilpotent, and $(19.3) (a)$ applies. Nakayama's Lemma applied to $M$ as a left $E$ -module then gives $𝔪^{n} = 0$ . Full details are on Fitting's Lemma and Endomorphism Rings of Modules of Finite Length.

Remark(19.18)One chain condition is not enough

ACC alone fails. $M_{1} = ℤ$ over $ℤ$ is indecomposable and noetherian, but $End (M_{1}) ≅ ℤ$ is not local. So the conclusion * $E$ is local* already breaks.
DCC alone fails differently. $M_{3} = ℤ (p^{\infty})$ satisfies DCC but not ACC. Here $End (M_{3}) ≅ ℤ_{p}$ is local, so the first conclusion survives — but $ℤ_{p}$ is a domain and its maximal ideal $p ℤ_{p}$ is very far from nilpotent, so the second conclusion $𝔪^{n} = 0$ fails.

The two halves of $(19.17)$ therefore fail independently, and both chain conditions are genuinely needed.

Proof Techniques and Method

How these arguments work, and which move to reuse.

Translate to the endomorphism ringRestate the module-theoretic question as a question about

E = End (M_{R})

: decompositions become idempotents, isomorphisms become units, direct summands become one-sided ideals.

Identify the non-automorphismsAsk which

f \in E

fail to be invertible. If they are all nilpotent,

(19.3) (a)

gives locality immediately.

Use the module to bound the radicalView

M

as a left

E

-module. Nakayama's Lemma converts a statement about

rad E

into a statement about the filtration

M \supseteq 𝔪 M \supseteq 𝔪^{2} M \supseteq \dots

, which is controlled by the composition length.

Feed the result into decomposition theoryOnce locality is known, Krull–Schmidt–Azumaya applies and the isomorphism types of the summands are determined.

Is my indecomposable module $M$ strongly indecomposable?

$M$ is simpleYes, by Schur's Lemma

(19.13)

End (M_{R})

is a division ring.

$M$ has finite composition lengthYes, by

(19.17)

, and

End (M_{R})

is completely primary with

𝔪^{n} = 0

$M$ is injective and indecomposableYes — a separate classical result, with

rad End (M_{R})

the endomorphisms with essential kernel.

$M$ satisfies only ACCNo general conclusion.

ℤ

over

ℤ

is a counterexample, so you must compute

End (M_{R})

directly.

$M$ satisfies only DCCStill no general conclusion, though it often works out; the Prüfer group has local endomorphism ring but non-nilpotent radical.

The reusable move is the second step. Every non-automorphism is nilpotent is a strong hypothesis that is nonetheless easy to verify in the finite-length setting, and $(19.3) (a)$ turns it into locality with no further work.

Worked Example

Computing $End (V)$ in the modular example (19.15)

Keep $k$ of characteristic $p$ , $G = ⟨ x, y ⟩$ elementary abelian of order $p^{2}$ , and $V$ as above. Write a vector as $v = e_{1} α_{1} + e_{2} α_{2} + e_{3} α_{3}$ and record it by the column $(α_{1}, α_{2}, α_{3})$ .

Step 1 — the action matrices

From the defining relations, $v x = e_{1} (α_{1} + α_{3}) + e_{2} α_{2} + e_{3} α_{3}$ and $v y = e_{1} α_{1} + e_{2} (α_{2} + α_{3}) + e_{3} α_{3}$ . So the operators are

X = I + E_{13}, Y = I + E_{23},

(E.1)

$E_{i j}$ is the matrix unit with $1$ in position $(i, j)$ .

These commute, since $E_{13} E_{23} = E_{23} E_{13} = 0$ , and each has order dividing $p$ : $X^{p} = (I + E_{13})^{p} = I + p E_{13} = I$ because $E_{13}^{2} = 0$ and $char k = p$ . So $V$ really is a $k G$ -module for $G$ elementary abelian of order $p^{2}$ .

Step 2 — impose $k G$ -linearity

A $k$ -linear map with matrix $A$ is a $k G$ -endomorphism iff it commutes with $X$ and $Y$ , equivalently with $E_{13}$ and $E_{23}$ .

From $A E_{13} = E_{13} A$ : the left side has only its third column nonzero, equal to the first column of $A$ ; the right side has only its first row nonzero, equal to the third row of $A$ . Comparing entry by entry gives $a_{21} = a_{31} = a_{32} = 0$ and $a_{33} = a_{11}$ .

From $A E_{23} = E_{23} A$ similarly: the third column of the left side is the second column of $A$ , the second row of the right side is the third row of $A$ . This gives $a_{12} = a_{32} = 0$ and $a_{33} = a_{22}$ .

End (V_{k G}) = {(\begin{matrix} a & 0 & b \\ 0 & a & c \\ 0 & 0 & a \end{matrix}) : a, b, c \in k} = k \cdot I \oplus k E_{13} \oplus k E_{23} .

(E.2)

Step 3 — the ring is local

Put $N = b E_{13} + c E_{23}$ . Then $N^{2} = 0$ , because each of the four products $E_{13} E_{13}$ , $E_{13} E_{23}$ , $E_{23} E_{13}$ , $E_{23} E_{23}$ vanishes. Hence $𝔪 = k E_{13} \oplus k E_{23}$ is a two-sided ideal of square zero, the ring is commutative, and

End (V_{k G}) / 𝔪 ≅ k, 𝔪^{2} = 0, (a I + N)^{- 1} = a^{- 1} I - a^{- 2} N (a \neq 0) .

(E.3)

So the non-units are exactly $𝔪$ , an ideal, and $(19.1) (4)$ makes $End (V_{k G})$ a local ring — completely primary, of $k$ -dimension $3$ . Therefore $V$ is strongly indecomposable.

Note what this example is not: $V$ is not free over $k G$ , since $\dim_{k} k G = p^{2}$ never divides $3$ , and it is not simple. It is an honest indecomposable of intermediate size, which is why elementary abelian $p$ -groups are the standard testing ground in modular representation theory.

Process and Workflow

Write down $End (M_{R})$ explicitlyFor a module given by matrices, this is a linear algebra problem: find all matrices commuting with the generators of the acting algebra.

Locate the non-unitsIn finite dimension, non-units are the endomorphisms with nonzero kernel. Ask whether they form an additive subgroup.

Test for nilpotenceIf every non-unit is nilpotent,

(19.3) (a)

closes the argument at once.

Otherwise identify the residue ringCompute

End (M_{R}) / rad

and check whether it is a division ring. A quotient with two or more simple modules means

M

is decomposable.

Record the Loewy boundIf

M

has length

n

(19.17)

guarantees

𝔪^{n} = 0

; a computed nilpotency index exceeding that signals an arithmetic error.

Comparison and Classification

Indecomposable versus strongly indecomposable
Module	Ring	$End$	Indecomposable?	Strongly?
Any simple $M$	any $R$	division ring	yes	yes
$ℤ$	$ℤ$	$ℤ$	yes	no
$ℤ / p^{n} ℤ$	$ℤ$	$ℤ / p^{n} ℤ$	yes	yes
$ℤ (p^{\infty})$	$ℤ$	$ℤ_{p}$	yes	yes
$ℚ$	$ℤ$	$ℚ$	yes	yes
$V$ of $(19.15)$	$k G$ , $G$ elem. abelian	$k [u, v] / (u, v)^{2}$	yes	yes
$R$ itself, $R = k \times k$	$k \times k$	$k \times k$	no	no
Nonprincipal ideal $𝔄$ of a Dedekind domain	Dedekind $R$	$R$	yes	no unless $R$ is local

Which hypotheses deliver which conclusion
	$End$ local	$rad End$ nilpotent	Krull–Schmidt uniqueness applies
Simple module	yes	yes	yes
Indecomposable, finite length	yes	yes	yes
Indecomposable, ACC only	no	no	no
Indecomposable, DCC only	partial	no	partial
Indecomposable injective	yes	no	yes

Which hypotheses deliver which conclusion

Relationship Map

The classes nest, and the nesting is strict at every stage.

Nonzero modulesno condition

Indecomposable

End

has only trivial idempotents

Strongly indecomposable

End

is local

$End$ completely primary…and

rad End

is nilpotent — e.g. finite length

Simple

End

is a division ring

$ℤ$ over $ℤ$ separates the first two bands.
$ℤ (p^{\infty})$ over $ℤ$ separates the second and third: its endomorphism ring $ℤ_{p}$ is local but not completely primary.
$ℤ / p^{2} ℤ$ over $ℤ$ separates the third and fourth: its endomorphism ring is completely primary but not a division ring.
Restricted to modules of finite length, the first two bands coincide — that is exactly $(19.17)$ .

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 of finite groups. Classifying indecomposable $k G$ -modules in characteristic $p$ is the central problem of modular representation theory; strong indecomposability is automatic there, since $k G$ is finite-dimensional and all modules of interest have finite length. Finite, tame and wild representation type are classified by how the indecomposables behave.
Integral representations and lattices. Over an order in a semisimple algebra, lattices have finite length only after reduction; strong indecomposability of the reductions is what makes the genus-theoretic bookkeeping work.
Computer algebra. The Meataxe algorithm decides indecomposability of a module over a finite-dimensional algebra by searching the endomorphism ring for a nontrivial idempotent — in effect running $(19.12 a)$ as a procedure. Failing to find one certifies locality when the module has finite length.
Homological algebra and derived categories. Objects with local endomorphism rings are the ones for which the Krull–Remak–Schmidt property holds in an additive category, which is the standing hypothesis in Auslander–Reiten theory and in the theory of tilting objects.
Coding theory over rings. Decomposition of a module of codewords over a finite chain ring is unique because the summands have local endomorphism rings, which underwrites canonical generator matrices.

Failure Modes and Common Mistakes

Do not confuse $End (M_{R})$ with $End (_{R} M)$ ; the two are opposite rings in general, though locality is preserved either way since it is a self-opposite condition.
Do not assume a direct summand of a strongly indecomposable module is anything: the only summands are $0$ and $M$ .
Do not conclude that $M \oplus M$ has local endomorphism ring — it never does for $M \neq 0$ , since $End (M \oplus M) ≅ M_{2} (End M)$ has nontrivial idempotents.
Do not expect strong indecomposability to be preserved by scalar extension: an absolutely indecomposable module keeps it, but in general $M \otimes_{k} K$ can decompose.

Best Practices

Verify finite length before invoking $(19.17)$ — say explicitly which chain conditions hold.
When computing an endomorphism ring by hand, present it as scalars plus nilpotents; that decomposition immediately exhibits the radical and the residue division ring.
State whether your modules are left or right, and on which side endomorphisms act. Almost all sign and order errors in this area come from that convention.
Sanity-check any computed $End (M_{R})$ against $\dim_{k} End (M) \leq (\dim_{k} M)^{2}$ and against the predicted nilpotency bound.
Where a result needs only indecomposability, say so; where it needs locality, say so. Conflating them is the most common error in citing Krull–Schmidt.

Quick Reference

Definition

M \neq 0

with

End (M_{R})

local

Indecomposable

End (M_{R})

has only idempotents

0, 1

Implicationstrongly indecomposable

\Rightarrow

indecomposable, not conversely

Simple modulesalways strongly indecomposable (Schur)

Finite lengthindecomposable

\Rightarrow

strongly indecomposable,

(19.17)

Nilpotency

𝔪^{n} = 0

for

n

the composition length

Counterexample

ℤ

over

ℤ

: ACC, indecomposable,

End ≅ ℤ

not local

Usehypothesis of Krull–Schmidt–Azumaya

The results of this page
Reference	Statement
(19.12)	Definition: $End (M_{R})$ local
(19.13)	Simple $\Rightarrow$ strongly indecomposable
(19.14)	$ℤ$ , $ℤ / p^{n} ℤ$ , $ℤ (p^{\infty})$ compared
(19.15)	A $3$ -dimensional strongly indecomposable $k G$ -module
(19.17)	Finite length: indecomposable $\Rightarrow$ local $End$ , $𝔪^{n} = 0$
(19.18)	ACC alone or DCC alone is insufficient

Frequently Asked Questions

Why not simply require indecomposability in the Krull–Schmidt theorem?

Because the uniqueness statement is then false. Over a Dedekind domain with class number $2$ and a nonprincipal ideal $𝔄$ , one has $𝔄 \oplus 𝔄 ≅ R \oplus R$ with all four summands indecomposable but $𝔄 ncong R$ . The endomorphism ring of $𝔄$ is $R$ itself, which is not local — exactly the hypothesis that fails.

Is the endomorphism ring of an indecomposable module always a domain?

No. $End (ℤ / p^{2} ℤ) ≅ ℤ / p^{2} ℤ$ has $p \cdot p = 0$ , and the ring computed in $(19.15)$ has $𝔪^{2} = 0$ . Indecomposability forbids idempotents, not zero divisors; those are different conditions, and locality controls neither directly.

Does strong indecomposability behave well under field extension?

Not in general. If $M$ is a module over a $k$ -algebra and $K \supseteq k$ is an extension field, $M \otimes_{k} K$ can decompose even when $M$ is strongly indecomposable — this is the difference between indecomposable and absolutely indecomposable modules. Strong indecomposability is preserved when the residue division ring of $End (M)$ stays a division ring after extension, which holds when $k$ is a splitting field.

What is the relation between locality of

End (M_{R})

and locality of

R

None in either direction. $R = ℤ$ is not local yet $ℤ / p^{n} ℤ$ has local endomorphism ring; conversely $R = k [[x]]$ is local while the module $R \oplus R$ has endomorphism ring $M_{2} (R)$ , which is not local. The two conditions concern different objects.

How do indecomposable injectives fit in?

They are always strongly indecomposable. For $M$ indecomposable injective, the endomorphisms with essential kernel form an ideal, and every endomorphism outside it is an automorphism — so that ideal is the radical and the quotient is a division ring. This gives a second large supply of local rings, developed in Lam's sequel volume rather than in §19.

Can a module have finite length but decompose in more than one way?

It can decompose into indecomposables in more than one literal way — different submodules — but the multiset of isomorphism types is uniquely determined, by $(19.17)$ together with Krull–Schmidt–Azumaya. Uniqueness is up to isomorphism and reindexing, never up to equality of submodules.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §19 (pp. 293–310), especially (19.12)–(19.18).
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §12 and §25.
T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics 189, Springer-Verlag, 1999, §3 (injective modules and their endomorphism rings).
G. Azumaya, “Corrections and supplementaries to my paper concerning Krull–Remak–Schmidt's theorem”, Nagoya Mathematical Journal 1 (1950), 117–124.
D. J. Benson, Representations and Cohomology, Volume I, Cambridge Studies in Advanced Mathematics 30, Cambridge University Press, 1991, Chapter 1.
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981, §6.

AI Suggested Questions

Prove that an indecomposable injective module has local endomorphism ring, identifying the radical explicitly.
Compute the endomorphism ring of each indecomposable module over $k [x] / (x^{n})$ and confirm locality.
For the Klein four group in characteristic 2, classify the indecomposable modules of dimension at most 4 and their endomorphism rings.
Give a module that is indecomposable but not strongly indecomposable over a commutative noetherian local ring.
How does the Meataxe algorithm detect idempotents in an endomorphism ring, and what is its complexity?
Under what conditions on a field extension does an indecomposable module remain indecomposable after scalar extension?
Describe the endomorphism ring of a uniserial module and explain when it is local.

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Idempotents and decompositions

What locality buys

Where the two notions agree

06Key Results

07Proof Techniques and Method

08Worked Example

Computing End(V) in the modular example (19.15)

Step 1 — the action matrices

Step 2 — impose kG-linearity

Step 3 — the ring is local

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

13Failure Modes and Common Mistakes

14Best Practices

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

Computing $End (V)$ in the modular example (19.15)

Step 2 — impose $k G$ -linearity

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions