Idempotents in Module-Finite Algebras over Complete Local Rings

Q: Why does (21.35) require k to be semilocal as well as complete?

Because the proof passes through (21.34)(2), which needs k/I to be a finite product of fields so that R/IR is artinian. That is exactly semilocality together with I = rad k. Without it one still gets completeness of R and lifting of idempotents, but not the upgrade from *no nontrivial idempotents* to *local*, which is what Krull–Schmidt–Azumaya consumes.

Q: Is Z a legitimate base here?

No. rad Z = 0, so the only ideal for which Z is complete in the relevant sense is the zero ideal, and Z is not semilocal. This is not a technicality: Krull–Schmidt uniqueness genuinely fails for ZG-lattices, which is the historical reason the subject completes at each prime.

Q: Does (21.34) say that R is a complete local ring?

Part (1) says only that R is complete with respect to IR; R can have many idempotents and be far from local. Part (2) adds the hypothesis that R has no nontrivial idempotents, and only then concludes locality. Confusing the two is the most common misreading of the proposition.

Q: How do these results relate to semiperfect rings?

Under the hypotheses of (21.35), R/rad R is semisimple and idempotents lift modulo the radical, so R is semiperfect. That is the abstract form of what the completeness hypothesis buys, and it is why the Krull–Schmidt statement resembles the one for finite-dimensional algebras.

Q: Where exactly is the noetherian hypothesis used?

In two places. In (21.33) it licenses Krull's intersection theorem, hence the Hausdorff condition for a finitely generated module. In (21.35) it makes M a noetherian k-module, which gives the ACC needed for existence of a decomposition, and makes End_k(M_i) finitely generated so the endomorphism ring stays in the class.

Executive Summary

Completeness lifts idempotents, but complete noncommutative rings are not easy to produce by hand. This page gives the standard factory: take a commutative noetherian ring $k$ that is complete for an ideal $I$ , and any $k$ -algebra $R$ that is finitely generated as a $k$ -module. Then $R$ is automatically $I R$ -adically complete, so idempotents of $R / I R$ lift.

The payoff is $(21.35)$ . If in addition $k$ is semilocal and $I = rad k$ , then every finitely generated right $R$ -module has a Krull–Schmidt decomposition that is unique up to permutation and isomorphism. Uniqueness genuinely fails over noetherian local rings — Swan's example — so completeness is doing real work, and this is why integral representation theory is done over complete discrete valuation rings.

$I R$ The complete ideal

LocalEndomorphism rings of indecomposables

UniqueKrull–Schmidt decomposition

(21.35)The theorem

Overview

Throughout, $k$ is a commutative ring, $R$ is a $k$ -algebra — so there is a ring map $k \to Z (R)$ — and $R$ is module-finite, meaning finitely generated as a $k$ -module. The examples to hold in mind are $ℤ_{p} G$ for a finite group $G$ , $M_{n} (ℤ_{p})$ , an order in a semisimple $ℚ_{p}$ -algebra, and $k [[x]] [G]$ .

Three statements are proved in sequence, each feeding the next.

Completeness descends to modules

(21.33)

: a finitely generated module over a complete noetherian

k

is itself

I

-adically complete. Krull's intersection theorem plus Nakayama give the Hausdorff half; a coordinate computation gives convergence.

Completeness transfers to the algebra

(21.34) (1)

: applying the lemma to

M = R

and noting

(I R)^{n} = I^{n} R

shows

R

I R

-adically complete, so idempotents of

R / I R

lift.

Indecomposable becomes local

(21.34) (2)

: with

k

semilocal and

I = rad k

, a module-finite

R \neq 0

with no nontrivial idempotents is a local ring.

Krull–Schmidt is restored

(21.35)

: endomorphism rings of indecomposable finitely generated modules are themselves module-finite algebras with no nontrivial idempotents, hence local; Krull–Schmidt–Azumaya then applies.

Learning Objectives

Prove $(21.33)$ : finitely generated modules over a complete noetherian ring are complete.
Identify $(I R)^{n}$ with $I^{n} R$ and conclude that module-finite algebras inherit completeness.
Prove that a module-finite algebra over a complete semilocal base with only trivial idempotents is local.
Assemble the proof of $(21.35)$ from existence of decompositions under ACC and Krull–Schmidt–Azumaya.
Split $ℤ_{5} [C_{4}]$ into four factors by lifting idempotents, and check the lift modulo $25$ .
Explain, with Swan's example in mind, why locality of the base is not enough.

Definitions

$k$ -algebra: A ring $R$ with a homomorphism $k \to Z (R)$ of $k$ into the centre of $R$ ; then every $R$ -submodule of a module is a $k$ -submodule.
Module-finite: $R$ is generated as a $k$ -module by finitely many elements. This is much stronger than being a finitely generated $k$ -algebra.
$I R$: The ideal of $R$ generated by the image of $I \subseteq k$ . Because $k$ acts centrally, $(I R)^{n} = I^{n} R$ .
Semilocal: $R / rad R$ is semisimple. For commutative $k$ this means finitely many maximal ideals, and $k / rad k$ is then a finite product of fields.
Order: A module-finite algebra over a complete discrete valuation ring $𝒪$ that is $𝒪$ -free of finite rank and spans a semisimple algebra over the fraction field.

Definition—Complete base pair

By a complete base pair $(k, I)$ we mean a commutative noetherian ring $k$ together with an ideal $I \subseteq k$ such that $k$ is $I$ -adically complete. When we additionally require $k$ semilocal and $I = rad k$ we say the pair is semilocal complete. The archetypes are $(ℤ_{p}, p ℤ_{p})$ and $(k [[x]], (x))$ , both of which are complete discrete valuation rings, and finite products of such.

Completeness of $k$ already forces $I \subseteq rad k$ , so a complete base pair is automatically a pair in which $1 + I$ consists of units.

Core Concepts

Why noetherian is needed

Completeness of $k$ does not obviously pass to a finitely generated module $M$ , because $⋂_{n} I^{n} M$ need not vanish for formal reasons. The Krull intersection theorem supplies $I \cdot N = N$ for $N = ⋂_{n} I^{n} M$ , and this is where the noetherian hypothesis enters — via the Artin–Rees lemma. Nakayama then finishes the job, using $I \subseteq rad k$ and finite generation of $N$ .

Coordinates convert module convergence to ring convergence

Given generators $m_{1}, \dots, m_{r}$ of $M$ , a Cauchy sequence in $M$ can be written in coordinates as a family of Cauchy sequences in $k$ . The point is that $I^{n} M = \sum_{j} I^{n} m_{j}$ , so each successive difference $a_{n + 1} - a_{n}$ has coefficients in $I^{n}$ . Completeness of $k$ produces limits coefficient by coefficient, and reassembling them gives the limit in $M$ . Coordinates are not canonical, but the limit is, by the Hausdorff condition.

The extended ideal and its powers

(I R)^{n} = I^{n} R, hence \lim_{⟵} R / (I R)^{n} = \lim_{⟵} R / I^{n} R .

(C.1)

Valid because $k$ maps into $Z (R)$ , so scalars can be collected to the left. This identity is what lets a module-theoretic statement be read as a ring-theoretic one.

So the assertion * $R$ is complete as a $k$ -module for $I$ * and the assertion * $R$ is complete as a ring for $I R$ * are literally the same assertion, and $(21.31)$ applies to give lifting of idempotents modulo $I R$ .

From no idempotents to local

The last conceptual step is a reduction to the artinian case. With $k$ semilocal and $I = rad k$ , the ring $k / I$ is a finite product of fields; a module-finite algebra over it has finite length, hence is left and right artinian. A nonzero artinian ring whose only idempotents are $0$ and $1$ is local. Since $I R \subseteq rad R$ , locality of $R / I R$ lifts to locality of $R$ .

Key Results

Lemma(21.33)Completeness passes to finitely generated modules

Let $k$ be a commutative noetherian ring which is $I$ -adically complete with respect to an ideal $I \subseteq k$ , and let $M$ be a finitely generated $k$ -module. Then $M$ is $I$ -adically complete: the natural map $ι_{M} : M \to \lim_{⟵} M / I^{n} M$ is an isomorphism.

Proof

Injectivity. The kernel of $ι_{M}$ is $N = ⋂_{n \geq 1} I^{n} M$ . Since $k$ is noetherian and $M$ finitely generated, Krull's intersection theorem gives $I \cdot N = N$ . As $k$ is $I$ -adically complete we have $I \subseteq rad k$ , and $N$ is finitely generated because $M$ is noetherian; Nakayama's Lemma therefore forces $N = 0$ .

Surjectivity. Fix generators $m_{1}, \dots, m_{r}$ of $M$ and take an element of the inverse limit, represented by $a_{1}, a_{2}, \dots \in M$ with $a_{n + 1} - a_{n} \in I^{n} M$ . Since $I^{n} M = \sum_{j} I^{n} m_{j}$ , we may write

a_{n + 1} - a_{n} = \sum_{j = 1}^{r} β_{n j} m_{j}, β_{n j} \in I^{n} .

(21.33a)

Write $a_{1} = \sum_{j} α_{1 j} m_{j}$ with $α_{1 j} \in k$ , and set $α_{n j} = α_{1 j} + β_{1 j} + \dots + β_{n - 1, j}$ , so that $a_{n} = \sum_{j} α_{n j} m_{j}$ . Because $β_{n - 1, j} \in I^{n - 1}$ , the sequence $(α_{n j})_{n}$ satisfies $α_{n + 1, j} \equiv α_{n j} (\mod I^{n})$ and is therefore Cauchy in $k$ . Completeness of $k$ supplies $a_{j} \in k$ with $a_{j} \equiv α_{n j} (\mod I^{n})$ for every $n$ .

Put $a = \sum_{j} a_{j} m_{j} \in M$ . Then $a - a_{n} = \sum_{j} (a_{j} - α_{n j}) m_{j} \in I^{n} M$ for every $n$ , so $ι_{M} (a) = (a_{1}, a_{2}, \dots)$ as required.

Proposition(21.34)Module-finite algebras over a complete base

Let $k$ be a commutative noetherian ring which is $I$ -adically complete for an ideal $I \subseteq k$ , and let $R$ be a $k$ -algebra that is finitely generated as a $k$ -module. Then:

$R$ is $I R$ -adically complete, and idempotents of $R / I R$ can be lifted to $R$ .
Suppose in addition that $k$ is semilocal and $I = rad k$ . If $R \neq 0$ has no idempotents other than $0$ and $1$ , then $R$ is a local ring.

Proof

(1) Apply $(21.33)$ to the finitely generated $k$ -module $M = R$ : the map $R \to \lim_{⟵} R / I^{n} R$ is an isomorphism. Since $k$ maps into $Z (R)$ we have $(I R)^{n} = I^{n} R$ , so this says exactly that $R$ is $I R$ -adically complete as a ring. Lifting of idempotents is then $(21.31)$ .

(2) With $k$ semilocal and $I = rad k$ , the quotient $k / I$ is a commutative semisimple ring, that is, a finite product of fields. The ring $R / I R$ is a finitely generated module over $k / I$ , hence of finite length over $k / I$ ; every one-sided ideal of $R / I R$ is a $k / I$ -submodule, so $R / I R$ is left and right artinian.

Assume $R$ has no idempotents but $0$ and $1$ . By (1) every idempotent of $R / I R$ lifts to $R$ , so $R / I R$ likewise has no nontrivial idempotents; and $R / I R \neq 0$ because $I R \subseteq rad R$ is proper. A nonzero one-sided artinian ring with only trivial idempotents is local $(19.19)$ , so $R / I R$ is local. Finally $I R \subseteq rad R$ — a standard consequence of $R$ being module-finite over $k$ with $I \subseteq rad k$ — so $rad (R / I R) = (rad R) / I R$ and $R / rad R ≅ (R / I R) / rad (R / I R)$ is a division ring. Hence $R$ is local.

Theorem(21.35)Krull–Schmidt over a complete semilocal base

Let $k$ be a commutative noetherian semilocal ring which is $I$ -adically complete for $I = rad k$ , and let $R$ be a $k$ -algebra that is finitely generated as a $k$ -module. Then every finitely generated right $R$ -module $M$ admits a decomposition

M = M_{1} \oplus M_{2} \oplus \dots \oplus M_{r}

(21.35a)

with each $M_{i}$ an indecomposable $R$ -submodule.

Moreover $r$ is uniquely determined by $M$ , and the sequence of isomorphism types $M_{1}, \dots, M_{r}$ is uniquely determined up to a permutation.

Proof

Existence. $M$ is finitely generated over $R$ and $R$ is finitely generated over $k$ , so $M$ is a finitely generated module over the noetherian ring $k$ and hence noetherian as a $k$ -module. Every $R$ -submodule is a $k$ -submodule, so the $R$ -submodules of $M$ satisfy the ascending chain condition, and a module with ACC decomposes as a finite direct sum of indecomposable submodules $(19.20)$ .

Local endomorphism rings. Set $E_{i} = {End}_{R} (M_{i})$ . Indecomposability of $M_{i}$ says precisely that $E_{i}$ has no idempotents other than $0$ and $1$ . Now $M_{i}$ is a finitely generated module over the noetherian ring $k$ , so ${End}_{k} (M_{i})$ is a finitely generated $k$ -module; $E_{i}$ is a $k$ -submodule of it, hence also finitely generated over $k$ . Thus $E_{i}$ is a nonzero module-finite $k$ -algebra with only trivial idempotents, and $(21.34) (2)$ makes it a local ring.

Uniqueness. Each $M_{i}$ therefore has local endomorphism ring, i.e. is strongly indecomposable, and the Krull–Schmidt–Azumaya theorem $(19.21)$ delivers uniqueness of $r$ and of the isomorphism types up to permutation.

Counterexample—Completeness cannot be dropped

Over a Dedekind domain $𝒪$ with nontrivial class group, a non-principal ideal $𝔄$ satisfies $𝔄 \oplus 𝔅 ≅ 𝒪 \oplus 𝔄 𝔅$ . Taking $𝒪 = ℤ [\sqrt{- 5}]$ and $𝔄 = (2, 1 + \sqrt{- 5})$ , whose square is the principal ideal $(2)$ , gives $𝔄 \oplus 𝔄 ≅ 𝒪 \oplus 𝒪$ with all four summands indecomposable and $𝔄 not ≅ 𝒪$ .

That base is not local. Swan's example goes further: there is a commutative noetherian local domain $A$ and finitely generated $A$ -modules with $M_{1} \oplus M_{2} ≅ (M_{1} \cap M_{2}) \oplus B$ , all four indecomposable over $A$ and $B not ≅ M_{i}$ . So locality of the base does not restore uniqueness — completeness does.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Krull plus Nakayama for the Hausdorff condition

To show $⋂_{n} I^{n} M = 0$ , never argue directly. Show the intersection $N$ satisfies $I N = N$ by Artin–Rees, then kill it with Nakayama using $I \subseteq rad k$ and finite generation.

Move 2

Push structure through the centre

Because $k$ lands in $Z (R)$ , scalar ideals extend cleanly: $(I R)^{n} = I^{n} R$ . Any statement proved for $R$ as a $k$ -module is then a statement about the $I R$ -adic ring structure.

Move 3

Upgrade indecomposable to strongly indecomposable

Apply the structure theory to ${End}_{R} (M_{i})$ rather than to $M_{i}$ . Over a module-finite complete setting the endomorphism ring is again in the class, so hypotheses that were assumed about $R$ become available about $E_{i}$ .

Move 3 is the reusable idea and the reason $(21.34) (2)$ is stated for an arbitrary module-finite algebra rather than for $R$ alone: the theorem is applied not to $R$ but to the endomorphism rings its modules produce. The class of module-finite algebras over a fixed complete base is closed under the operation that the proof needs.

Worked Example

Splitting $ℤ_{5} [C_{4}]$ into four factors

Let $k = ℤ_{5}$ , $I = 5 ℤ_{5}$ , and $R = ℤ_{5} [C_{4}] = ℤ_{5} [g] / (g^{4} - 1)$ , free of rank $4$ over $k$ , so $(21.34)$ applies. Modulo $5$ , the field $𝔽_{5}$ contains a primitive fourth root of unity, namely $2$ , so $g^{4} - 1 = (g - 1) (g - 2) (g - 4) (g - 3)$ splits and

R / 5 R = 𝔽_{5} [C_{4}] ≅ 𝔽_{5} \times 𝔽_{5} \times 𝔽_{5} \times 𝔽_{5},

(E.1)

Maschke applies: $| C_{4} | = 4$ is invertible in $𝔽_{5}$ .

The four primitive orthogonal idempotents downstairs are ${\bar{e}}_{j} = \frac{1}{4} \sum_{m = 0}^{3} 2^{- j m} g^{m}$ . By $(21.34) (1)$ they lift to $R$ , and the lifts are the character idempotents $e_{j} = \frac{1}{4} \sum_{m} ω^{- j m} g^{m}$ , where $ω \in ℤ_{5}$ is the Teichmüller lift of $2$ : the unique fourth root of unity with $ω \equiv 2 (\mod 5)$ .

The lift to precision $25$

Solve $ω^{2} = - 1$ with $ω \equiv 2$ : writing $ω = 2 + 5 t$ gives $4 + 20 t \equiv 24 (\mod 25)$ , so $4 t \equiv 4 (\mod 5)$ and $t \equiv 1$ . Hence $ω \equiv 7 (\mod 25)$ , and indeed $7^{2} = 49 \equiv - 1 (\mod 25)$ . With $4^{- 1} \equiv 19 (\mod 25)$ and $ω^{- 1} = ω^{3} = - ω \equiv 18$ , $ω^{- 2} \equiv 24$ , $ω^{- 3} \equiv 7$ :

e_{1} \equiv 19 + 17 g + 6 g^{2} + 8 g^{3} (\mod 25), e_{1} \equiv 4 + 2 g + g^{2} + 3 g^{3} (\mod 5) .

(E.2)

Squaring in $(ℤ / 25) [C_{4}]$ confirms $e_{1}^{2} \equiv e_{1} (\mod 25)$ , and the four lifts $e_{0} \equiv 19 (1 + g + g^{2} + g^{3})$ , $e_{1}$ , $e_{2} \equiv 19 (1 - g + g^{2} - g^{3})$ , $e_{3}$ are pairwise orthogonal with $e_{0} + e_{1} + e_{2} + e_{3} = 1$ . Consequently

ℤ_{5} [C_{4}] ≅ ℤ_{5} \times ℤ_{5} \times ℤ_{5} \times ℤ_{5} .

(E.3)

A case where nothing splits

Take $R = ℤ_{p} [C_{p}] = ℤ_{p} [g] / (g^{p} - 1)$ . Here $R / p R = 𝔽_{p} [g] / ((g - 1)^{p})$ is local, so $R$ has no nontrivial idempotents; by $(21.34) (2)$ , $R$ is a local ring, with maximal ideal generated by $p$ and $g - 1$ . Its finitely generated lattices decompose uniquely by $(21.35)$ : classically there are exactly three indecomposable $ℤ_{p} [C_{p}]$ -lattices, namely $ℤ_{p}$ , $ℤ_{p} [ζ_{p}]$ and $ℤ_{p} [C_{p}]$ itself.

Comparison and Classification

What survives over which base
	Idempotents lift mod rad	End of indec. is local	Krull–Schmidt unique	Semiperfect
$k$ a field, $R$ finite-dimensional	yes	yes	yes	yes
$k$ complete noetherian semilocal, $R$ module-finite	yes	yes	yes	yes
$k$ noetherian local, not complete	no	no	no	no
$k$ a Dedekind domain with class number $> 1$	no	no	no	no
$k = ℤ$ , $R = ℤ G$	no	no	no	no

What survives over which base

The first two rows are the good cases and they are good for the same reason: the base is complete, trivially so for a field where the radical is zero. Rows three to five all fail, and they fail at the same point — an indecomposable module can have a non-local endomorphism ring, so nothing pins the decomposition down.

Hypotheses used by each result
Result	Needs $k$ noetherian	Needs $k$ complete	Needs $k$ semilocal, $I = rad k$
(21.33) modules are complete	yes	yes	no
(21.34)(1) $R$ complete, idempotents lift	yes	yes	no
(21.34)(2) trivial idempotents $\Rightarrow$ local	yes	yes	yes
(21.35) Krull–Schmidt uniqueness	yes	yes	yes

Relationship Map

$(k, I)$ complete noetherian, $R$ module-finite — the standing hypotheses
- gives immediately
  - $M$ is $I$ -adically complete for $M$ f.g. over $k$ $(21.33)$
  - $R$ is $I R$ -adically complete $(21.34) (1)$
  - $I R \subseteq rad R$ , so $1 + I R \subseteq U (R)$
- with $k$ semilocal, $I = rad k$ , gives
  - $R / rad R$ semisimple, so $R$ is semilocal
  - $R$ is semiperfect
  - indecomposable f.g. modules are strongly indecomposable
  - Krull–Schmidt uniqueness $(21.35)$
- fails without completeness
  - Swan's local noetherian domain
  - Dedekind domains of class number $> 1$

The chain of implications is short but each link needs its own hypothesis, which is why the theorem is stated with four separate conditions on $k$ rather than one.

Applications and Industry Use

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

Integral representation theory

Lattices over $𝒪 G$

For $G$ finite and $𝒪$ the completion of a ring of algebraic integers at a prime, $𝒪 G$ is module-finite over a complete base, so $𝒪 G$ -lattices decompose uniquely. Over $ℤ G$ they do not — which is precisely why the local-global method starts by completing.

Modular representation theory

Blocks and defect groups

Block theory is developed over a $p$ -modular system with $𝒪$ complete. Completeness makes the block idempotents of $k G$ lift to $𝒪 G$ and makes the Krull–Schmidt theorem available for the modules that carry defect group and vertex theory.

Orders and arithmetic

Genus and local-global

Two $ℤ G$ -lattices lie in the same genus when their completions agree at every prime. The theory is usable because at each prime the completed problem has unique decompositions; the genus records exactly what is lost on returning to $ℤ$ .

Computation

Working $p$ -adically

Computer algebra systems decompose modules over orders by reducing mod $p$ , splitting there, and lifting. The correctness of the lift is $(21.34) (1)$ ; the assertion that the answer is well posed is $(21.35)$ .

The honest description is that this is the theorem that makes $p$ -adic methods legitimate. Everything gained by completing would be worthless if the resulting decompositions were not canonical.

Design Considerations

Design considerations here means the choices made when modelling a problem with these algebraic structures.

Which base ring should carry the problem?

A fieldThe cleanest case: finite-dimensional algebras are semiprimary, so everything on this page holds trivially. Choose this whenever the arithmetic of the coefficients is not the point.

$ℤ$ or a ring of integersRetains arithmetic information but loses Krull–Schmidt uniqueness. Use it only when the global object is genuinely the object of study, and expect genus-type invariants rather than isomorphism invariants.

A complete DVRThe standard compromise. Retains characteristic-

p

and characteristic-

0

information simultaneously, and

(21.35)

guarantees unique decompositions. This is why

p

-modular systems are set up this way.

A complete semilocal ringNeeded when several primes must be handled at once, for instance a finite product of complete DVRs. All results here are stated at this generality precisely to allow it.

Module-finite, not finitely generated as an algebra. $k [x]$ is a finitely generated $k$ -algebra and none of this applies to it; $M_{n} (k)$ is module-finite and all of it does.
Left or right modules. The statements are symmetric because $k$ is central; choose the side that matches your representation convention and stay on it.
Which ideal. For $(21.34) (1)$ any ideal for which $k$ is complete will do; for $(21.34) (2)$ and $(21.35)$ the ideal must be $rad k$ , since the argument needs $k / I$ semisimple.

Failure Modes and Common Mistakes

Do not conclude that $R$ is a complete local ring from $(21.34) (1)$ ; completeness of $R$ for $I R$ says nothing about $R$ having a unique maximal one-sided ideal unless (2) applies.
Do not assume the number of factors in $R / I R$ equals the number in $R$ before checking liftability; the equality is the content of $(21.34) (1)$ , not a triviality.
Do not read $(21.35)$ as a classification: it says decompositions are unique, not that indecomposables are known. For $ℤ_{p} [C_{p^{2}}]$ the list of indecomposable lattices is already infinite in general.
Do not extend $(21.35)$ to arbitrary, non-finitely-generated modules; the ACC argument that gives existence of a decomposition breaks immediately.

Best Practices

State the base pair $(k, I)$ explicitly, including noetherian, semilocal and complete, before invoking any result of this page.
Verify module-finiteness by exhibiting generators, not by citing finite generation as an algebra.
When computing, lift idempotents to an explicit precision and record that precision alongside the answer.
Check the number of block factors against $R / rad R$ first: the semisimple quotient is where the count is visible.
Before claiming uniqueness of a decomposition over $ℤ G$ , complete at each relevant prime and work with the genus instead.

Quick Reference

Standing hypotheses

k

commutative noetherian,

I

-adically complete;

R

module-finite over

k

(21.33)f.g.

k

-modules are

I

-adically complete

Key identity

(I R)^{n} = I^{n} R

(21.34)(1)

R

I R

-adically complete; idempotents of

R / I R

lift

(21.34)(2)

k

semilocal,

I = rad k

: trivial idempotents

\Rightarrow

R

local

(21.35)f.g. right

R

-modules have unique Krull–Schmidt decompositions

Engine

{End}_{R} (M_{i})

is again module-finite over

k

FailureSwan: local noetherian base is not enough

Worked bases
$k$	$I$	Complete?	$(21.35)$ available?
$ℤ_{p}$	$p ℤ_{p}$	yes	yes
$k [[x]]$	$(x)$	yes	yes
A field	$0$	yes	yes
$ℤ_{(p)}$	$p ℤ_{(p)}$	no	no
$ℤ$	$rad ℤ = 0$	no	no
$ℤ [\sqrt{- 5}]$	$0$	no	no

Frequently Asked Questions

Why does

(21.35)

require

k

to be semilocal as well as complete?

Because the proof passes through $(21.34) (2)$ , which needs $k / I$ to be a finite product of fields so that $R / I R$ is artinian. That is exactly semilocality together with $I = rad k$ . Without it one still gets completeness of $R$ and lifting of idempotents, but not the upgrade from no nontrivial idempotents to local, which is what Krull–Schmidt–Azumaya consumes.

ℤ

a legitimate base here?

No. $rad ℤ = 0$ , so the only ideal for which $ℤ$ is complete in the relevant sense is the zero ideal, and $ℤ$ is not semilocal. This is not a technicality: Krull–Schmidt uniqueness genuinely fails for $ℤ G$ -lattices, which is the historical reason the subject completes at each prime.

Does

(21.34)

say that

R

is a complete local ring?

Part (1) says only that $R$ is complete with respect to $I R$ ; $R$ can have many idempotents and be far from local. Part (2) adds the hypothesis that $R$ has no nontrivial idempotents, and only then concludes locality. Confusing the two is the most common misreading of the proposition.

How do these results relate to semiperfect rings?

Under the hypotheses of $(21.35)$ , $R / rad R$ is semisimple and idempotents lift modulo the radical, so $R$ is semiperfect. That is the abstract form of what the completeness hypothesis buys, and it is why the Krull–Schmidt statement resembles the one for finite-dimensional algebras.

Where exactly is the noetherian hypothesis used?

In two places. In $(21.33)$ it licenses Krull's intersection theorem, hence the Hausdorff condition for a finitely generated module. In $(21.35)$ it makes $M$ a noetherian $k$ -module, which gives the ACC needed for existence of a decomposition, and makes ${End}_{k} (M_{i})$ finitely generated so the endomorphism ring stays in the class.

Does uniqueness extend to infinitely generated modules?

Not from this theorem. Existence of a decomposition into indecomposables relies on the ascending chain condition, and uniqueness in the infinite setting requires the Krull–Schmidt–Azumaya statement for arbitrary direct sums of modules with local endomorphism rings — a different theorem with different hypotheses.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §21, results (21.33)–(21.35), with (19.19)–(19.21).
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981, §§6, 30–33 (orders, lattices, complete local base rings).
I. Reiner, Maximal Orders, Academic Press, 1975, Chapters 5–6.
H. Matsumura, Commutative Ring Theory, Cambridge University Press, 1986, Chapter 8 (Artin–Rees, Krull intersection, completion).
R. G. Swan, “Projective modules over group rings and maximal orders”, Annals of Mathematics 76 (1962), 55–61.

AI Suggested Questions

Write out the Artin–Rees lemma and derive Krull's intersection theorem in the form used in the proof of (21.33).
Give an example of a non-noetherian commutative ring and a finitely generated module for which the intersection of the powers of an ideal is nonzero.
Classify the indecomposable $ℤ_{p} [C_{p}]$ -lattices and verify that there are exactly three.
Work out Swan's example in detail and identify precisely which step of the proof of (21.35) fails for it.
How does the genus of a $ℤ G$ -lattice encode the discrepancy between global and local decompositions?
Compute the block idempotents of $ℤ_{p} [S_{3}]$ for $p = 2$ and $p = 3$ and compare with the characteristic-zero case.
For which finite groups $G$ and primes $p$ is $ℤ_{p} G$ a local ring?

Idempotents in Complete Algebras

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Why noetherian is needed

Coordinates convert module convergence to ring convergence

The extended ideal and its powers

From no idempotents to local

Key Results

Proof Techniques and Method

Krull plus Nakayama for the Hausdorff condition

Push structure through the centre

Upgrade indecomposable to strongly indecomposable

Worked Example

Splitting $ℤ_{5} [C_{4}]$ into four factors

The lift to precision $25$

A case where nothing splits

Comparison and Classification

Relationship Map

Applications and Industry Use

Lattices over $𝒪 G$

Blocks and defect groups

Genus and local-global

Working $p$ -adically

Design Considerations

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why noetherian is needed

Coordinates convert module convergence to ring convergence

The extended ideal and its powers

From no idempotents to local

06Key Results

07Proof Techniques and Method

Krull plus Nakayama for the Hausdorff condition

Push structure through the centre

Upgrade indecomposable to strongly indecomposable

08Worked Example

Splitting ℤ5[C4] into four factors

The lift to precision 25

A case where nothing splits

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Lattices over 𝒪G

Blocks and defect groups

Genus and local-global

Working p-adically

12Design Considerations

13Failure Modes and Common Mistakes

14Best Practices

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

Worked Example

Splitting $ℤ_{5} [C_{4}]$ into four factors

The lift to precision $25$

Comparison and Classification

Relationship Map

Applications and Industry Use

Lattices over $𝒪 G$

Working $p$ -adically

Design Considerations

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions