Fitting’s Lemma and Endomorphism Rings of Modules of Finite Length

Q: Why must one wait for both chains to stabilise?

The proof compares f^n with f^2n twice: directness of the sum uses (f^2n) = (f^n), and surjectivity uses im(f^2n) = im(f^n). Either alone gives only one half. In practice one takes n to be the composition length, which certainly works for both.

Q: Is the nilpotency bound m^n = 0 sharp?

Yes. For M = Z/p^nZ the composition length is n, End(M) ≅ Z/p^nZ, and m^n-1 = (p^n-1) ≠ 0. For other modules the true index is often much smaller — for the three-dimensional module of Lam's (19.15) it is 2 against a bound of 3.

Q: Where exactly does uniqueness fail without local endomorphism rings?

At the step (19.21a). One has 1_M_1 = _j (_1_j)|_M_1, and without locality none of the summands need be invertible — the identity can be a sum of non-units. Over a Dedekind domain with class number 2 that is exactly what happens, and A ⊕ A ≅ R ⊕ R with A R results.

Q: How does this relate to the Jordan canonical form?

Take R = k[x] acting on a finite-dimensional space V via a linear operator. Fitting's decomposition for f = x - is the splitting into the generalised -eigenspace and its complement; iterating gives the primary decomposition, and (19.22) says the resulting multiset of Jordan blocks is an invariant of the operator.

Executive Summary

Fitting's decomposition is the technical heart of §19. For $M$ of finite length and $f \in End (M_{R})$ , the kernel chain and the image chain both stabilise, and at the common stabilisation point $M$ splits as the direct sum of the two. Nothing about $R$ is assumed.

Applied to an indecomposable $M$ , one of the two summands must vanish, so every endomorphism is either an automorphism or nilpotent. That is precisely the hypothesis of $(19.3) (a)$ , and it makes $End (M_{R})$ a local ring with nilpotent radical. This single deduction supplies the hypothesis for the Krull–Schmidt–Azumaya theorem and characterises local right artinian rings.

$\ker \oplus im$ Fitting split

NilpotentEvery non-automorphism

$𝔪^{n} = 0$

n

= composition length

ACC or DCCEnough for existence

Overview

Let $R$ be an arbitrary ring, $M$ a right $R$ -module and $E = End (M_{R})$ , acting on the left. Any $f \in E$ generates two chains of submodules,

M \supseteq im (f) \supseteq im (f^{2}) \supseteq \dots, 0 \subseteq \ker (f) \subseteq \ker (f^{2}) \subseteq \dots

(19.16a)

Descending images and ascending kernels — the DCC controls the first, the ACC the second.

If $M$ has finite length both chains stabilise, and Fitting's theorem says the two stable pieces are complementary. The result is a decomposition manufactured by an arbitrary endomorphism, which is why it converts indecomposability — a statement about the absence of decompositions — into information about individual endomorphisms.

M

of finite length

M = \ker (f^{n}) \oplus im (f^{n})

M

indecomposable: one summand diesevery non-automorphism is nilpotent

End (M_{R})

is local

The name Fitting's Lemma is used both for the decomposition $(19.16)$ and for the corollary $(19.17)$ ; when precision matters, call the first the Fitting decomposition and the second the local endomorphism ring theorem.

Learning Objectives

Prove that both chains in $(19.16 a)$ stabilise and that the stable kernel and image are complementary.
Deduce that an endomorphism of an indecomposable finite length module is an automorphism or nilpotent.
Prove $(19.17)$ in both halves: locality, and $𝔪^{n} = 0$ for $n$ the composition length.
Prove $(19.19)$ : a nonzero right artinian ring is local iff it has no nontrivial idempotents.
Prove $(19.20)$ : ACC or DCC alone gives existence of a decomposition into indecomposables.
Locate the exact step of the Krull–Schmidt–Azumaya proof where locality is used.

Core Concepts

Why both chain conditions appear

The image chain descends and the kernel chain ascends, so stabilising both is exactly asking for DCC and ACC together — that is, finite length. Drop one and the theorem fails: for $f$ multiplication by $p$ on $ℤ$ , the kernels are all zero but the images $p^{i} ℤ$ never stabilise, and $ℤ \neq 0 \oplus p^{i} ℤ$ .

The two conclusions of (19.17) are independent

Locality of $E$ says the non-units form an ideal. Nilpotence of that ideal is a strictly stronger, quantitative statement, controlled by the composition length. The Prüfer group $ℤ (p^{\infty})$ shows they can come apart: it satisfies DCC, is indecomposable, and has $End ≅ ℤ_{p}$ — local, but with a maximal ideal that is not even nil.

Surjective endomorphisms of finite length modules

A fact used repeatedly below: if $N$ has finite length and $g : N \to N$ is surjective, then $g$ is an automorphism. Indeed $length (N) = length (\ker g) + length (im g) = length (\ker g) + length (N)$ , forcing $\ker g = 0$ . The dual statement — injective implies surjective — holds for the same reason.

Existence versus uniqueness

It is worth keeping the two halves of decomposition theory apart. Existence of a finite decomposition into indecomposables $(19.20)$ needs only one chain condition and no locality at all. Uniqueness $(19.21)$ needs local endomorphism rings and says nothing about existence. Finite length gives both, which is why $(19.22)$ is stated for it.

Key Results

Theorem(19.16)Fitting Decomposition Theorem

Let $R$ be any ring and $M$ a right $R$ -module of finite length. For every $f \in E = End (M_{R})$ there is an integer $n_{0}$ such that

M = \ker (f^{n}) \oplus im (f^{n}) for all n \geq n_{0} .

(19.16)

One may take $n_{0}$ to be any index at which both the image chain and the kernel chain of $(19.16 a)$ have stabilised; the composition length of $M$ always works.

Proof

Since $M$ has finite length it satisfies both chain conditions, so the descending chain of images and the ascending chain of kernels each stabilise. Choose $n$ with

im (f^{n}) = im (f^{n + 1}) = \dots, \ker (f^{n}) = \ker (f^{n + 1}) = \dots .

(19.16b)

The sum is direct. Let $a \in \ker (f^{n}) \cap im (f^{n})$ and write $a = f^{n} (b)$ for some $b \in M$ . Then $0 = f^{n} (a) = f^{2 n} (b)$ , so $b \in \ker (f^{2 n}) = \ker (f^{n})$ by stabilisation, whence $a = f^{n} (b) = 0$ .

The sum is everything. Let $c \in M$ . Then $f^{n} (c) \in im (f^{n}) = im (f^{2 n})$ , so $f^{n} (c) = f^{2 n} (d)$ for some $d \in M$ . Therefore $f^{n} (c - f^{n} (d)) = 0$ , and

c = {\underset{⏟}{(c - f^{n} (d))}}_{\in \ker (f^{n})} + {\underset{⏟}{f^{n} (d)}}_{\in im (f^{n})} .

(19.16c)

Both pieces are submodules because $f$ is $R$ -linear, so $M = \ker (f^{n}) \oplus im (f^{n})$ .

Corollary(19.16d)Dichotomy for indecomposable modules

Let $M$ be an indecomposable right $R$ -module of finite length and $f \in End (M_{R})$ . Then $f$ is either an automorphism of $M$ or nilpotent — and not both unless $M = 0$ .

Proof

Fix $n$ with $M = \ker (f^{n}) \oplus im (f^{n})$ . Indecomposability forces one summand to be zero.

If $\ker (f^{n}) = 0$ then $im (f^{n}) = M$ , so $f^{n}$ is bijective. Writing $u = (f^{n})^{- 1}$ , the elements $f^{n - 1} u$ and $u f^{n - 1}$ are respectively a right and a left inverse for $f$ , so $f$ is an automorphism.

Otherwise $\ker (f^{n}) \neq 0$ , so $im (f^{n}) = 0$ , i.e. $f^{n} = 0$ and $f$ is nilpotent.

Theorem(19.17)Endomorphism rings of finite length indecomposables

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

Proof

Locality. $M \neq 0$ gives $E \neq 0$ . By $(19.16 d)$ every element of $E ∖ U (E)$ is nilpotent, so $(19.3) (a)$ applies and $E$ is local. Since $R ∖ U (R) = rad R$ for a local ring, $𝔪 = rad E$ is exactly the set of nilpotent endomorphisms.

**Nilpotence of $𝔪$ .** Let $f_{1}, \dots, f_{n} \in 𝔪$ . Apply them to $M$ one at a time from the right: put $M_{0} = M$ and $M_{i} = f_{n + 1 - i} (M_{i - 1})$ , so that $M_{n} = (f_{1} f_{2} \dots f_{n}) (M)$ and

M = M_{0} \supseteq M_{1} \supseteq M_{2} \supseteq \dots \supseteq M_{n} .

(19.17a)

Suppose $M_{n} \neq 0$ ; we derive a contradiction. If some step failed to be strict, say $M_{i} = M_{i - 1}$ , then the map $g = f_{n + 1 - i}$ restricted to $M_{i - 1}$ would be a surjective endomorphism of the finite length module $M_{i - 1}$ , hence an automorphism of it. Then $g^{t} (M_{i - 1}) = M_{i - 1}$ for every $t$ ; but $g \in 𝔪$ is nilpotent, so $g^{t} = 0$ for large $t$ and $M_{i - 1} = 0$ , forcing $M_{n} = 0$ .

So all inclusions in $(19.17 a)$ are strict. Then $length (M_{i}) \leq n - i$ for each $i$ , and $M_{n} \neq 0$ gives $1 \leq length (M_{n}) \leq 0$ — absurd. Hence $M_{n} = 0$ , i.e. $(f_{1} \dots f_{n}) (M) = 0$ , and since $E$ acts faithfully on $M$ we get $f_{1} \dots f_{n} = 0$ . As $𝔪^{n}$ is generated by such products, $𝔪^{n} = 0$ .

Corollary(19.19)Local right artinian rings

A nonzero right artinian ring $R$ is local if and only if $R$ has no idempotents other than $0$ and $1$ .

Proof

Only if: immediate from $(19.2) (c)$ , and no chain condition is needed.

If: take $M = R_{R}$ , the right regular module. By the Hopkins–Levitzki theorem $(4.15)$ a right artinian ring is right noetherian, so $M$ has finite length. Its endomorphism ring, acting on the left, is $End (R_{R}) ≅ R$ via left multiplication. Idempotents of $R$ correspond to decompositions of $M$ , so the hypothesis says $M$ is indecomposable. By $(19.17)$ , $End (R_{R}) ≅ R$ is local — and moreover $rad R$ is nilpotent, so $R$ is completely primary.

Proposition(19.20)Existence of Krull–Schmidt decompositions

Let $R$ be any ring and $M$ a right $R$ -module whose submodules satisfy either the ACC or the DCC. Then $M$ is a finite direct sum of indecomposable submodules.

Proof

Call a submodule $N \subseteq M$ good if it is a finite direct sum of indecomposable submodules, and bad otherwise. Three observations: the zero module is good, being the empty direct sum; any indecomposable submodule is good; and if $N, N'$ are good with $N \cap N' = 0$ then $N \oplus N'$ is good, by concatenating the two decompositions.

Suppose $M$ is bad. Then $M \neq 0$ and $M$ is not indecomposable, so $M = M_{1} \oplus M_{1}'$ with $M_{1}, M_{1}' \neq 0$ . If both summands were good, so would $M$ be; hence one is bad, say $M_{1}$ .

Repeating the argument on $M_{1}$ gives $M_{1} = M_{2} \oplus M_{2}'$ with $M_{2}$ bad and $M_{2}' \neq 0$ , and so on indefinitely. This produces

M ⊋ M_{1} ⊋ M_{2} ⊋ \dots, 0 ⊊ M_{1}' ⊊ M_{1}' \oplus M_{2}' ⊊ M_{1}' \oplus M_{2}' \oplus M_{3}' ⊊ \dots

(19.20a)

The first chain is strictly descending because each $M_{i + 1}' \neq 0$ , and the second is strictly ascending for the same reason; the sums on the right are direct because $M_{i + 1}' \subseteq M_{i}$ while $M_{1}' \oplus \dots \oplus M_{i}'$ meets $M_{i}$ in $0$ . So $M$ satisfies neither ACC nor DCC, contradicting the hypothesis. Hence $M$ is good.

Theorem(19.21)Krull–Schmidt–Azumaya

Let $R$ be a ring and $M$ a right $R$ -module with two decompositions

M = M_{1} \oplus \dots \oplus M_{r} = N_{1} \oplus \dots \oplus N_{s},

(19.21)

where every $N_{j}$ is indecomposable and every $M_{i}$ is strongly indecomposable — that is, $End ((M_{i})_{R})$ is a local ring. Then $r = s$ , and after reindexing $M_{i} ≅ N_{i}$ for $1 \leq i \leq r$ .

Proof

We record only the step where locality is used; the full induction is carried out on The Krull–Schmidt Theorem. Let $α_{i}, β_{j} \in E = End (M_{R})$ be the projections attached to the two decompositions, so that $α_{1} + \dots + α_{r} = 1 = β_{1} + \dots + β_{s}$ .

Multiplying by $α_{1}$ gives $α_{1} = \sum_{j = 1}^{s} α_{1} β_{j}$ . Each $α_{1} β_{j}$ maps $M$ into $M_{1}$ , so restricting to $M_{1}$ yields an identity in $End ((M_{1})_{R})$ :

1_{M_{1}} = \sum_{j = 1}^{s} (α_{1} β_{j}) |_{M_{1}} .

(19.21a)

Here locality enters. $End ((M_{1})_{R})$ is local, so by $(19.1) (6)$ one of the summands is a unit; after reindexing, $(α_{1} β_{1}) |_{M_{1}}$ is an automorphism of $M_{1}$ . Consequently $β_{1} |_{M_{1}} : M_{1} \to N_{1}$ is a split monomorphism, and since $N_{1}$ is indecomposable it is an isomorphism. One then shows $M = M_{1} \oplus N_{2} \oplus \dots \oplus N_{s}$ and induct on $r$ .

Corollary(19.22)Krull–Schmidt Theorem

Let $M$ be a right $R$ -module of finite length. Then $M = M_{1} \oplus \dots \oplus M_{r}$ with each $M_{i}$ indecomposable; the integer $r$ is uniquely determined, and the sequence of isomorphism types of $M_{1}, \dots, M_{r}$ is unique up to permutation.

Proof. Existence is $(19.20)$ , since finite length gives both chain conditions. Each $M_{i}$ is indecomposable of finite length, hence strongly indecomposable by $(19.17)$ . Uniqueness is then $(19.21)$ .

Corollary(19.23)Finitely generated modules over artinian rings

Both conclusions of $(19.22)$ hold for every finitely generated right module over a right artinian ring $R$ — in particular for every finite-dimensional module over a finite-dimensional algebra over a field. The reason is Hopkins–Levitzki $(4.15)$ : over a right artinian ring, a finitely generated right module has a composition series.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Stabilise both chains at once

Pick a single $n$ working for images and kernels simultaneously; the two stabilisation conditions are then used against each other via $f^{2 n}$ .

Move 2

Length counts steps

A strictly descending chain of submodules of a length $n$ module has at most $n$ strict steps. Turning a nilpotency claim into a chain-length claim is the whole proof of $𝔪^{n} = 0$ .

Move 3

Surjective plus finite length equals bijective

This converts a failure of strictness into an automorphism, which then collides with nilpotence. It is the pivot of the second half of $(19.17)$ .

Move 4

Faithful action transfers bounds

$End (M)$ acts faithfully on $M$ , so $𝔪^{n} M = 0$ gives $𝔪^{n} = 0$ . Prove the module statement; the ring statement is free.

Move 5

Good versus bad, then contradict

For existence results, define the desired property, check it is closed under the relevant operations, and derive an infinite chain from a hypothetical counterexample.

Move 6

Split $1$ and pick a unit

Writing $1$ as a sum of endomorphisms and invoking $(19.1) (6)$ selects one summand as invertible. This is the only place locality is used in Krull–Schmidt–Azumaya.

Move 6 explains why the theorem's hypothesis is local endomorphism ring rather than indecomposable: indecomposability rules out idempotents, but it gives no way to select an invertible summand from a sum equal to $1$ .

Worked Example

A: the Fitting decomposition of multiplication by 2 on $ℤ / 12 ℤ$

Let $M = ℤ / 12 ℤ$ as a $ℤ$ -module — length $3$ , with composition factors $ℤ / 2, ℤ / 2, ℤ / 3$ — and let $f$ be multiplication by $2$ , so $f^{i}$ is multiplication by $2^{i}$ .

The two chains for $f =$ multiplication by 2 on $ℤ / 12 ℤ$
$i$	$\ker (f^{i})$	$\| \ker \|$	$im (f^{i})$	$\| im \|$
1	${0, 6}$	2	${0, 2, 4, 6, 8, 10}$	6
2	${0, 3, 6, 9}$	4	${0, 4, 8}$	3
3	${0, 3, 6, 9}$	4	${0, 4, 8}$	3
4	${0, 3, 6, 9}$	4	${0, 4, 8}$	3

Both chains stabilise at $i = 2$ : $\ker (f^{2}) = 3 ℤ / 12 ℤ$ and $im (f^{2}) = 4 ℤ / 12 ℤ$ . Check the theorem directly.

\ker (f^{2}) \cap im (f^{2}) = {0, 3, 6, 9} \cap {0, 4, 8} = {0}, 4 \cdot 3 = 12 = | M | .

(E.1)

ℤ / 12 ℤ = 3 ℤ / 12 ℤ \oplus 4 ℤ / 12 ℤ ≅ ℤ / 4 ℤ \oplus ℤ / 3 ℤ .

(E.2)

Fitting recovers the primary decomposition: the $2$ -part is the kernel, the $2$ -divisible part is the image.

This also demonstrates that $M$ is decomposable, consistent with $End (ℤ / 12 ℤ) ≅ ℤ / 12 ℤ$ being non-local: the element $4$ is a nontrivial idempotent, since $4^{2} = 16 = 4$ in $ℤ / 12$ .

B: the bound $𝔪^{n} = 0$ is sharp

Take $M = ℤ / p^{n} ℤ$ for a prime $p$ . Its submodules form a chain of length $n$ , so $M$ is indecomposable of composition length exactly $n$ , with all composition factors $ℤ / p ℤ$ .

E = End (ℤ / p^{n} ℤ) ≅ ℤ / p^{n} ℤ, 𝔪 = rad E = (p), 𝔪^{n} = (p^{n}) = 0 .

(E.3)

And $𝔪^{n - 1} = (p^{n - 1}) \neq 0$ , so the exponent in $(19.17)$ cannot be lowered in general. Every non-unit of $E$ is a multiple of $p$ and hence nilpotent, exactly as $(19.16 d)$ predicts, and $(19.3) (a)$ certifies locality.

For contrast, $ℤ / 12 ℤ$ is artinian with a nontrivial idempotent, and correspondingly not local: $(19.19)$ is an exact criterion, not a one-way implication.

Process and Workflow

Confirm finite lengthCheck ACC and DCC, or invoke Hopkins–Levitzki over a right artinian ring for a finitely generated module. Record the composition length

n

; it is the constant in every subsequent bound.

Decompose into indecomposablesExistence is

(19.20)

. In practice one finds idempotents in

End (M_{R})

and splits off summands until none remain.

Certify each summandBy

(19.17)

each indecomposable summand of finite length automatically has a local endomorphism ring — no further verification needed.

Apply uniqueness

(19.21)

now gives that the multiset of isomorphism types is an invariant of

M

. Only at this point may you speak of the indecomposable summands.

Extract numerical invariantsMultiplicities of indecomposables, Loewy lengths and the bound

𝔪^{n} = 0

are all now well defined and computable.

Comparison and Classification

What each hypothesis delivers
Hypothesis on $M_{R}$	Fitting split	Decomposition exists	$End$ local	Uniqueness
Finite length, indecomposable	yes	trivially	yes, $𝔪^{n} = 0$	yes
Finite length, general	yes	yes	for each summand	yes
ACC only	no	yes	not in general	no
DCC only	no	yes	not in general	no
Finitely generated over right artinian $R$	yes	yes	for each summand	yes
Finitely generated over a Dedekind domain	no	yes	no	no

The bottom row is the standard warning. Over a Dedekind domain $R$ with a nonprincipal ideal $𝔄$ , the Steinitz isomorphism gives $𝔄 \oplus 𝔄 ≅ R \oplus R$ when the class of $𝔄$ has order $2$ , with all four summands indecomposable and $𝔄 ncong R$ . ACC holds, existence holds, uniqueness fails — because $End (𝔄) ≅ R$ is not local.

Relationship Map

(19.16)

Fitting

(19.16 d)

automorphism or nilpotent

(19.17)

local

End

(19.21)

uniqueness

(19.22)

Krull–Schmidt

$(19.17)$ feeds $(19.19)$ by taking $M = R_{R}$ and using Hopkins–Levitzki to supply finite length.
$(19.20)$ is logically independent of $(19.17)$ : it needs one chain condition and no locality, and it is what supplies the decomposition that $(19.21)$ then pins down.
$(19.23)$ is $(19.22)$ plus Hopkins–Levitzki, and it is the form used in modular representation theory, where $R = k G$ is finite-dimensional.
Dickson's theorem $(19.30)$ — that the dimension of a principal indecomposable $k G$ -module is divisible by $| H |$ for $H$ a Sylow $p$ -subgroup — depends on $(19.23)$ together with the freeness of projectives over the local ring $k H$ .
The Noether–Deuring theorem $(19.25)$ , that $M^{K} ≅ N^{K}$ over an extension field implies $M ≅ N$ , uses $(19.23)$ to cancel: $t \cdot M ≅ t \cdot N$ forces $M ≅ N$ .

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

Principal indecomposables

Decomposing $k G$ as a right module over itself into indecomposables is well posed only because of $(19.23)$ . The summands are the principal indecomposable modules, and their dimensions carry the divisibility information in Dickson's theorem.

Linear algebra

Jordan form

For $M$ a finite-dimensional $k [x]$ -module, Fitting's decomposition is the generalised eigenspace splitting, and Krull–Schmidt is the uniqueness of the Jordan blocks. The abstract theorem is the reason the block sizes are an invariant.

Computer algebra

Meataxe and friends

Algorithms that decompose modules over finite-dimensional algebras search $End (M)$ for idempotents; termination and correctness rest on finite length, and the canonical form of the answer rests on $(19.22)$ .

Integral representation theory

Lattices and genera

Krull–Schmidt fails for lattices over orders in general but holds after completion, where endomorphism rings become local. The systematic exploitation of that difference is the genus theory of lattices.

Coding theory

Canonical generator matrices

Codes over finite chain rings decompose uniquely into indecomposable submodules because the ambient module has finite length, which is what makes standard generator matrix forms canonical.

Homological algebra

Auslander–Reiten theory

The Krull–Remak–Schmidt property of an additive category — every object a finite sum of objects with local endomorphism rings — is the standing hypothesis under which AR quivers are defined.

Computational Notes

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

For $M$ a module of dimension $d$ over a finite-dimensional algebra, $End (M)$ is the solution space of the linear system asserting commutation with each algebra generator — an $O (d^{6})$ nullspace computation in the naive formulation, and much less when the generators are sparse.
The Fitting decomposition of a single endomorphism $f$ is computed by taking $n = d$ and forming $\ker (f^{d})$ and $im (f^{d})$ ; repeated squaring computes $f^{d}$ in $O (\log d)$ matrix multiplications, so the whole split costs $O (d^{3} \log d)$ .
Deciding indecomposability of $M$ amounts to searching $End (M)$ for a nontrivial idempotent. Over a finite field the Meataxe does this in expected polynomial time; over $ℚ$ one works modulo a prime and lifts.
Once $End (M)$ is known to be local, $(19.17)$ gives $𝔪^{n} = 0$ for free, which bounds the number of Loewy layers and hence the depth of any recursion over the radical filtration.
There is no algorithm for the general case: without finite length, both the Fitting split and the decomposition into indecomposables can fail to exist, and no finite computation certifies their absence.

Failure Modes and Common Mistakes

Do not assume $im (f^{n})$ and $\ker (f^{n})$ are complementary for the first $n$ at which either chain stabilises; you need both to have stabilised.
Do not conclude from $(19.19)$ that a local ring is artinian. The corollary is a criterion within the class of right artinian rings; $k [[x]]$ is local with no nontrivial idempotents and is not artinian.
Do not apply $(19.17)$ to an indecomposable module over an artinian ring without checking finite generation — an infinitely generated module over an artinian ring need not have finite length.
Do not read $(19.21)$ as requiring both families to be strongly indecomposable: one family strongly indecomposable and the other merely indecomposable suffices, and that asymmetry is what makes the theorem usable.

Historical Notes and Lessons Learned

1909WedderburnThe first uniqueness statement for direct decompositions, in the setting of finite groups and finite-dimensional algebras.
1911–13Remak and SchmidtRemak proves uniqueness of the decomposition of a finite group into directly indecomposable factors; Schmidt extends the argument to groups with both chain conditions.
1925KrullKrull formulates the theorem for groups with operators, the setting from which the module-theoretic version descends.
1930sFittingFitting isolates the decomposition $M = \ker (f^{n}) \oplus im (f^{n})$ and the resulting dichotomy for endomorphisms, giving the argument its modern, purely module-theoretic form.
1950AzumayaAzumaya replaces the chain conditions by the hypothesis that the summands have local endomorphism rings, obtaining the general uniqueness theorem and covering infinite decompositions.
1956–62Atiyah and GabrielAtiyah extends the theorem to sheaves; Gabriel formulates it for abstract additive categories, where Krull–Remak–Schmidt category becomes standard terminology.

The lesson is the same as for the Jacobson radical. Fitting's chain conditions are a property of the module; Azumaya's locality is a property of a ring attached to the module. Relocating the hypothesis to the endomorphism ring is what freed the theorem from finiteness and let it move to sheaves and categories.

Quick Reference

Fitting (19.16)

M

finite length

\Rightarrow

M = \ker (f^{n}) \oplus im (f^{n})

for large

n

Dichotomy

M

indecomposable finite length

\Rightarrow

each

f

is an automorphism or nilpotent

(19.17)

End (M_{R})

local,

𝔪 =

nilpotent endomorphisms,

𝔪^{n} = 0

(19.19)

R \neq 0

right artinian is local iff no nontrivial idempotents

(19.20)ACC or DCC

\Rightarrow

finite direct sum of indecomposables

(19.21)Uniqueness needs the

M_{i}

strongly indecomposable, the

N_{j}

only indecomposable

(19.23)Applies to f.g. modules over right artinian rings

Failure caseDedekind domain of class number

2

: existence yes, uniqueness no

Numbered results and what they need
Result	Hypothesis	Conclusion
(19.16)	$M$ of finite length, $f$ any endomorphism	Fitting decomposition
(19.17)	$M$ indecomposable of length $n$	$End$ local, $𝔪^{n} = 0$
(19.19)	$R \neq 0$ right artinian	local iff no nontrivial idempotents
(19.20)	$M$ with ACC or DCC	finite sum of indecomposables
(19.21)	two decompositions, $M_{i}$ strongly indec.	$r = s$ and matching up to reindexing
(19.22)	$M$ of finite length	existence and uniqueness
(19.23)	$M$ f.g. over right artinian $R$	existence and uniqueness

Frequently Asked Questions

Why must one wait for both chains to stabilise?

The proof compares $f^{n}$ with $f^{2 n}$ twice: directness of the sum uses $\ker (f^{2 n}) = \ker (f^{n})$ , and surjectivity uses $im (f^{2 n}) = im (f^{n})$ . Either alone gives only one half. In practice one takes $n$ to be the composition length, which certainly works for both.

Is the nilpotency bound

𝔪^{n} = 0

sharp?

Yes. For $M = ℤ / p^{n} ℤ$ the composition length is $n$ , $End (M) ≅ ℤ / p^{n} ℤ$ , and $𝔪^{n - 1} = (p^{n - 1}) \neq 0$ . For other modules the true index is often much smaller — for the three-dimensional module of Lam's $(19.15)$ it is $2$ against a bound of $3$ .

Does

(19.17)

have a converse?

Not as stated: a module can have a local endomorphism ring without finite length, as $ℤ (p^{\infty})$ over $ℤ$ shows, and $ℚ$ over $ℤ$ likewise. What is true is that if $End (M_{R})$ is local then $M$ is indecomposable, which is the trivial direction, and the nilpotency of the radical genuinely requires finite length.

Where exactly does uniqueness fail without local endomorphism rings?

At the step $(19.21 a)$ . One has $1_{M_{1}} = \sum_{j} (α_{1} β_{j}) |_{M_{1}}$ , and without locality none of the summands need be invertible — the identity can be a sum of non-units. Over a Dedekind domain with class number $2$ that is exactly what happens, and $𝔄 \oplus 𝔄 ≅ R \oplus R$ with $𝔄 ncong R$ results.

Why is

(19.20)

stated with ACC or DCC rather than both?

Because the contradiction produced by an infinite bad decomposition violates both at once: the bad summands descend strictly and the discarded complements accumulate strictly upward. So a single chain condition suffices to rule the situation out. Uniqueness, by contrast, needs more than a chain condition — it needs locality.

How does this relate to the Jordan canonical form?

Take $R = k [x]$ acting on a finite-dimensional space $V$ via a linear operator. Fitting's decomposition for $f = x - λ$ is the splitting into the generalised $λ$ -eigenspace and its complement; iterating gives the primary decomposition, and $(19.22)$ says the resulting multiset of Jordan blocks is an invariant of the operator.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §19 (pp. 293–310), especially (19.16)–(19.23).
G. Azumaya, “Corrections and supplementaries to my paper concerning Krull–Remak–Schmidt's theorem”, Nagoya Mathematical Journal 1 (1950), 117–124.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §12.
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981, §6 and §30.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964.
D. J. Benson, Representations and Cohomology, Volume I, Cambridge Studies in Advanced Mathematics 30, Cambridge University Press, 1991, Chapter 1.

AI Suggested Questions

Prove that $End (M_{R})$ is semiprimary for any module of finite length, without assuming indecomposability.
Give a module with ACC whose endomorphism ring is local but whose radical is not nilpotent.
How does Azumaya's theorem extend Krull-Schmidt to infinite direct sums, and what replaces the counting argument?
Work out the Fitting decomposition for a nilpotent operator on a finite-dimensional vector space and relate it to Jordan blocks.
For which orders in a semisimple algebra does the Krull-Schmidt property hold for lattices?
What is the complexity of finding a nontrivial idempotent in the endomorphism ring of a module over a finite-dimensional algebra?
State and prove the cancellation property that follows from Krull-Schmidt-Azumaya, and give a counterexample when it fails.

01Executive Summary

02Overview

03Learning Objectives

04Core Concepts

Why both chain conditions appear

The two conclusions of (19.17) are independent

Surjective endomorphisms of finite length modules

Existence versus uniqueness

05Key Results

06Proof Techniques and Method

Stabilise both chains at once

Length counts steps

Surjective plus finite length equals bijective

Faithful action transfers bounds

Good versus bad, then contradict

Split 1 and pick a unit

07Worked Example

A: the Fitting decomposition of multiplication by 2 on ℤ/12ℤ

B: the bound 𝔪n=0 is sharp

08Process and Workflow

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Principal indecomposables

Jordan form

Meataxe and friends

Lattices and genera

Canonical generator matrices

Auslander–Reiten theory

12Computational Notes

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

Core Concepts

Key Results

Proof Techniques and Method

Split $1$ and pick a unit

Worked Example

A: the Fitting decomposition of multiplication by 2 on $ℤ / 12 ℤ$

B: the bound $𝔪^{n} = 0$ is sharp

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Computational Notes

Failure Modes and Common Mistakes

Historical Notes and Lessons Learned

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions