Can I use these theorems to compute blocks of ZG?

Not directly, since Z is not complete. The standard route is to complete at each prime dividing the group order, apply (22.11) to relate Z_pG to F_pG, decompose there, and then reassemble the global information; the reassembly is genuine work and not a formal consequence of §22.

Lifting Central Idempotents

Q: Why does passing to R/I^2 instead of R/I fix the problem?

The proof of (22.8) improves the inclusion eRf ⊆ I^n to eRf ⊆ I^n+1 by rewriting eI ⊆ Ie + I^2, which consumes one factor of I per step. Starting from I^2 leaves a factor to spare; starting from I leaves none, and the conclusion is then false.

Q: What exactly does completeness give in (22.11)?

Two things. It puts I inside radk, which is what Dade's Lemma needs, and it makes idempotents of R/IR liftable by successive approximation even though IR is not nilpotent. The example of ZC_2 reduced modulo 3 shows the conclusion genuinely fails without it.

Q: Does the block correspondence preserve more than the number of blocks?

Yes, it is induced by a ring surjection, so each block c_iR of R maps onto the matching block c_iR with kernel the intersection of that block with the ideal being killed. Invariants defined by the quotient — the simple modules of a block, for instance — therefore match up automatically.

Executive Summary

Reducing modulo the Jacobson radical is the standard simplification in ring theory, and idempotents survive it: if $rad R$ is nil, every idempotent of $R / rad R$ lifts. Centrality does not survive. Upper triangular $n \times n$ matrices over a division ring form an indecomposable ring whose radical quotient is a product of $n$ division rings with $2^{n}$ central idempotents — so all but two of those idempotents lift to non-central ones.

This page collects the two positive results. The first replaces the quotient: for a nilpotent ideal $I$ , central idempotents of $R$ correspond bijectively to those of $R / I^{2}$ , so a right artinian ring has the same blocks as $R / (rad R)^{2}$ . The second changes the direction of the hypothesis: over a complete noetherian local coefficient ring, central idempotents of a module-finite algebra correspond to those of its reduction. The second is why blocks of $ℤ_{p} G$ and of $𝔽_{p} G$ are the same.

$2^{n}$ vs $2$ Central idempotents of

T_{n} (k) / rad

T_{n} (k)

$R / I^{2}$ The quotient that works

CompleteThe coefficient hypothesis

$(22.10)$ Dade's Lemma

Overview

Two facts about idempotents pull in opposite directions. Idempotents lift modulo any nil ideal, so the idempotent theory of $R$ and $R / rad R$ agree whenever the radical is nil. But being central is a condition involving all of $R$ , and a quotient forgets most of $R$ ; there is no reason for it to be detectable downstairs, and in general it is not.

e central in R iff e R f = f R e = 0, f = 1 - e,

(21.5)

Centrality is the vanishing of the two off-diagonal Peirce corners — a condition on products, not on a single element.

Both positive results have the same shape: they find a hypothesis making the corners $e R f$ and $f R e$ small enough that vanishing downstairs forces vanishing upstairs. Lemma $(22.8)$ does it with an intersection of powers, Dade's Lemma $(22.10)$ does it with Nakayama's Lemma. Everything else is bookkeeping about lifting and uniqueness.

The practical upshot for right artinian rings is a normalisation. With $J = rad R$ , the blocks of $R$ and of $R / J^{2}$ coincide, and $rad (R / J^{2}) = J / J^{2}$ has square zero. Block theory for right artinian rings therefore reduces to the case of radical square zero — which is far from trivial, but is a genuine reduction.

Learning Objectives

Prove that the ring of upper triangular matrices over a division ring is indecomposable.
Explain why that ring shows central idempotents do not lift modulo the radical.
Prove $(22.8)$ : with $⋂_{n} I^{n} = 0$ , an idempotent is central in $R$ if and only if it is central in $R / I^{2}$ .
Deduce the bijection $(22.9)$ of central idempotents and of blocks between $R$ and $R / I^{2}$ for nilpotent $I$ .
Prove Dade's Lemma $(22.10)$ from the Peirce decomposition and Nakayama's Lemma.
Apply $(22.11)$ to relate the blocks of a group ring over a complete local ring and over its residue field.

Definitions

Lifts: An element $y$ of $R / I$ lifts to $x \in R$ if $x + I = y$ . For idempotents one additionally requires $x^{2} = x$ .
$⋂_{n \geq 1} I^{n} = 0$: The separatedness condition on the $I$ -adic filtration; automatic when $I$ is nilpotent, and valid for the maximal ideal of a noetherian local domain by the Krull intersection theorem.
Module-finite algebra: A $k$ -algebra $R$ that is finitely generated as a $k$ -module, not merely as a $k$ -algebra. Group rings $k G$ with $G$ finite are the standard example.
$I$ -adically complete: The natural map from $k$ to the inverse limit of the rings $k / I^{n}$ is an isomorphism. Then $I \subseteq rad k$ , since $1 - x$ has the convergent inverse $1 + x + x^{2} + \dots$ for $x \in I$ .
$T_{n} (k)$: The ring of upper triangular $n \times n$ matrices over a ring $k$ ; its radical is the strictly upper triangular part, nilpotent of index $n$ .

In $(22.10)$ and $(22.11)$ the letter $k$ denotes a commutative coefficient ring, not necessarily a field.

Core Concepts

The obstruction, in one example

Proposition(22.8a)Triangular matrix rings are indecomposable

Let $k$ be a division ring and $R = T_{n} (k)$ the ring of upper triangular $n \times n$ matrices over $k$ . Then the only central idempotents of $R$ are $0$ and $1$ , so $R$ is indecomposable.

Proof

Let $c = (c_{p q}) \in R$ be central. Comparing entries in $c E_{i i} = E_{i i} c$ gives $c_{p i} = 0$ for $p \neq i$ , for every $i$ , so $c$ is diagonal. Comparing entries in $c E_{i, i + 1} = E_{i, i + 1} c$ — legitimate, since $E_{i, i + 1} \in R$ — gives $c_{i i} = c_{i + 1, i + 1}$ . Hence $c = λ I_{n}$ for a single $λ \in k$ ; commuting with the diagonal matrix $μ E_{11}$ for all $μ \in k$ forces $λ \in Z (k)$ .

If in addition $c^{2} = c$ then $λ^{2} = λ$ in the field $Z (k)$ , so $λ \in {0, 1}$ and $c \in {0, I_{n}}$ .

Now $rad R$ is the strictly upper triangular part, which is nilpotent, and

R / rad R ≅ {\underset{⏟}{k \times \dots \times k}}_{n},

(22.8b)

a product with $2^{n}$ central idempotents, while $R$ itself has only $2$ .

Idempotents do lift here, because $rad R$ is nilpotent; what fails is that the lifts are not central. This single example is the reason §22 does not simply reduce block theory to the semisimple case, and it is the source of the genuine difficulty in the structure theory of right artinian rings.

Why the square of the ideal is the right quotient

Suppose $\bar{e}$ is central in $R / I^{2}$ ; this says $e R f \subseteq I^{2}$ with $f = 1 - e$ . The extra factor of $I$ is exactly what allows a bootstrapping argument: it feeds a copy of $I$ into an inductive step that improves $e R f \subseteq I^{n}$ to $e R f \subseteq I^{n + 1}$ . Starting from $e R f \subseteq I$ alone there is nothing to bootstrap with, and indeed the conclusion is false — the triangular example has $e R f \subseteq rad R$ for many non-central idempotents $e$ .

\bar{e}

central mod

I^{2}

e R f \subseteq I^{2}

e R f \subseteq I^{n}

for all

n

e R f = 0

The Nakayama route

Dade's Lemma replaces the filtration argument by a finiteness argument. The Peirce decomposition exhibits $e R f$ as a direct summand of $R$ as a $k$ -module. If $R$ is finitely generated over $k$ , so is $e R f$ ; if $e R f \subseteq I R$ then $e R f = I \cdot e R f$ because the Peirce decomposition is a decomposition of $k$ -modules; and Nakayama's Lemma over $k$ finishes the job. No filtration, no nilpotency — only module-finiteness and $I \subseteq rad k$ .

Key Results

Lemma(22.8)Detecting centrality modulo the square

Let $I$ be an ideal of a ring $R$ with $⋂_{n \geq 1} I^{n} = 0$ , and let $e \in R$ be an idempotent. Then $e$ is central in $R$ if and only if its image $\bar{e}$ is central in $\bar{R} = R / I^{2}$ .

Proof

The "only if" direction is trivial. Assume $\bar{e}$ is central in $R / I^{2}$ and put $f = 1 - e$ , so that $e R f \subseteq I^{2}$ and $f R e \subseteq I^{2}$ .

A preliminary inclusion. For $x \in I$ write $e x = e x e + e x f$ . Here $e x \in I$ , so $e x e \in I e$ ; and $e x f \in e R f \subseteq I^{2}$ . Hence $e I \subseteq I e + I^{2}$ .

Induction. We claim $e R f \subseteq I^{n}$ for every $n \geq 2$ . The case $n = 2$ is the hypothesis. Assume it for some $n \geq 2$ . For $r \in R$ , idempotency gives $e r f = e (e r f) f$ , so

e R f \subseteq e I^{n} f = (e I) I^{n - 1} f \subseteq (I e + I^{2}) I^{n - 1} f \subseteq I (e I^{n - 1} f) + I^{n + 1} \subseteq I \cdot I^{n} + I^{n + 1} = I^{n + 1},

(22.8c)

using $e I^{n - 1} f \subseteq e R f \subseteq I^{n}$ at the last step.

Therefore $e R f \subseteq ⋂_{n} I^{n} = 0$ . The same argument with $e$ and $f$ interchanged — note $\bar{f} = 1 - \bar{e}$ is central too — gives $f R e = 0$ . By $(21.5)$ , $e$ is central in $R$ .

Theorem(22.9)Central idempotents modulo the square of a nilpotent ideal

Let $R$ be a ring and let $I \subseteq rad R$ be a nilpotent ideal. Then reduction modulo $I^{2}$ gives a bijection

{central idempotents of R} ⟶ {central idempotents of \bar{R} = R / I^{2}}, e \mapsto \bar{e} .

(22.9a)

Moreover $e$ is centrally primitive in $R$ if and only if $\bar{e}$ is centrally primitive in $\bar{R}$ . In particular $R$ is indecomposable if and only if $\bar{R}$ is, and $R$ and $\bar{R}$ have the same blocks.

Proof

Surjectivity. Let $x$ be a central idempotent of $\bar{R}$ . Since $I^{2}$ is nilpotent, hence nil, $x$ lifts to an idempotent $e \in R$ . Nilpotency of $I$ gives $⋂_{n} I^{n} = 0$ , so $(22.8)$ makes $e$ central in $R$ .

Injectivity. Suppose $e, e'$ are central idempotents of $R$ with $\bar{e} = \bar{e}'$ , and put $u = e - e'$ . As $e$ and $e'$ commute, $u^{3} = e^{3} - 3 e^{2} e' + 3 e e'^{2} - e'^{3} = e - e' = u$ , so $u (1 - u^{2}) = 0$ . But $u \in I^{2} \subseteq rad R$ , hence $u^{2} \in rad R$ and $1 - u^{2}$ is a unit. Therefore $u = 0$ .

Central primitivity. If $e = e_{1} + e_{2}$ with $e_{1}, e_{2}$ nonzero orthogonal central idempotents of $R$ , then $\bar{e} = {\bar{e}}_{1} + {\bar{e}}_{2}$ is a corresponding decomposition in $\bar{R}$ , and ${\bar{e}}_{i} \neq 0$ because $I^{2} \subseteq rad R$ contains no nonzero idempotent. Conversely, let $\bar{e} = x_{1} + x_{2}$ with $x_{1}, x_{2}$ nonzero orthogonal central idempotents of $\bar{R}$ , and lift them to central idempotents $e_{1}, e_{2}$ of $R$ . Then $e_{1} e_{2}$ is an idempotent mapping to $0$ , so $e_{1} e_{2} \in I^{2}$ and therefore $e_{1} e_{2} = 0$ . Now $e_{1} + e_{2}$ is a central idempotent lifting $\bar{e}$ , so injectivity forces $e = e_{1} + e_{2}$ , a nontrivial decomposition. Hence central primitivity is preserved in both directions.

Corollary(22.9b)Reduction to radical square zero

Let $R$ be a right artinian ring and $J = rad R$ , which is nilpotent. Then the blocks of $R$ correspond bijectively to the blocks of $R / J^{2}$ , and $rad (R / J^{2}) = J / J^{2}$ has square zero. Block theory for right artinian rings therefore reduces to the case of a ring whose radical squares to zero.

Lemma(22.10)Dade's Lemma

Let $k$ be a commutative ring, $I \subseteq rad k$ an ideal, and $R$ a $k$ -algebra which is finitely generated as a $k$ -module. Then an idempotent $e \in R$ is central in $R$ if and only if its image $\bar{e}$ is central in $\bar{R} = R / I R$ .

Proof

Only the "if" direction needs argument. Let $f = 1 - e$ . Centrality of $\bar{e}$ gives $\bar{e} \bar{R} \bar{f} = 0$ , that is $e R f \subseteq I R$ .

The Peirce decomposition $R = e R e \oplus e R f \oplus f R e \oplus f R f$ is a decomposition of $k$ -modules, since $k$ acts centrally. Hence $I R$ decomposes accordingly, and intersecting with the summand $e R f$ gives $I R \cap e R f = I \cdot e R f$ . Combined with the inclusion above, $e R f = I \cdot e R f$ .

As a $k$ -module direct summand of the finitely generated $k$ -module $R$ , the module $e R f$ is finitely generated. Since $I \subseteq rad k$ , Nakayama's Lemma gives $e R f = 0$ . The symmetric argument gives $f R e = 0$ , so $e$ is central by $(21.5)$ .

Theorem(22.11)Complete coefficient rings

Let $k$ be a commutative noetherian ring which is $I$ -adically complete with respect to an ideal $I \subseteq k$ , and let $R$ be a $k$ -algebra finitely generated as a $k$ -module. Then $e \mapsto \bar{e}$ is a bijection between the central idempotents of $R$ and those of $\bar{R} = R / I R$ ; moreover $e$ is centrally primitive if and only if $\bar{e}$ is. In particular $R$ is indecomposable if and only if $\bar{R}$ is, and the blocks of $R$ correspond bijectively to those of $\bar{R}$ .

Proof

Completeness gives $I \subseteq rad k$ , and for a module-finite algebra $(rad k) R \subseteq rad R$ ; hence $I R \subseteq rad R$ and $I R$ contains no nonzero idempotent of $R$ . Under the stated hypotheses every idempotent of $R / I R$ lifts to an idempotent of $R$ , and Dade's Lemma $(22.10)$ upgrades a central idempotent downstairs to a central idempotent upstairs. Injectivity and the statement about central primitivity are proved exactly as in $(22.9)$ , using $I R \subseteq rad R$ in place of $I^{2} \subseteq rad R$ .

Note also that $R$ is then noetherian on both sides, being module-finite over a commutative noetherian ring, so $(22.2)$ gives both $R$ and $\bar{R}$ a unique block decomposition and the bijection is a bijection of blocks.

Corollary(22.11a)Blocks in integral and modular representation theory

Let $(k, 𝔪)$ be a commutative noetherian local ring that is $𝔪$ -adically complete, with residue field $\bar{k} = k / 𝔪$ , and let $G$ be a finite group. Then the blocks of $k G$ correspond bijectively to the blocks of $\bar{k} G$ . Taking $k = ℤ_{p}$ , the blocks of $ℤ_{p} G$ are the blocks of $𝔽_{p} G$ — the fact that lets integral and modular block theory be conducted simultaneously.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Improve an inclusion, then intersect

Prove $X \subseteq I^{n}$ for all $n$ by induction, then use separatedness $⋂ I^{n} = 0$ to conclude $X = 0$ . The starting point must be $I^{2}$ , since the induction consumes one factor of $I$ per step.

Move 2

Nakayama on a Peirce corner

A Peirce corner is a $k$ -module direct summand, so it inherits finite generation. Any statement of the form 'corner is contained in $I R$ ' becomes 'corner equals $I$ times itself', and Nakayama kills it.

Move 3

Uniqueness by a unit

If $e, e'$ are commuting idempotents with $u = e - e'$ then $u^{3} = u$ . When $u$ lies in the radical, $1 - u^{2}$ is a unit and $u = 0$ . This one-line argument replaces every ad hoc uniqueness computation.

Move 3 is worth keeping in reserve: it shows that a central idempotent has at most one lift through any ideal contained in the radical, whether or not that ideal is nil. Existence is the hard half; uniqueness is free.

Worked Example

Completeness is not a technicality

Compare two group rings of the cyclic group $C_{2} = {1, s}$ , reduced modulo $3$ .

Over $ℤ$ , the correspondence fails

Let $R = ℤ C_{2}$ , a commutative ring. If $e = a + b s$ satisfies $e^{2} = e$ then comparing coefficients gives $a^{2} + b^{2} = a$ and $2 a b = b$ . The second equation forces $b (2 a - 1) = 0$ , and $2 a - 1 \neq 0$ in $ℤ$ , so $b = 0$ ; then $a^{2} = a$ gives $a \in {0, 1}$ . Hence $ℤ C_{2}$ has only the trivial idempotents and is indecomposable.

ℤ C_{2} / 3 ℤ C_{2} ≅ 𝔽_{3} C_{2} ≅ 𝔽_{3} \times 𝔽_{3},

(E.1)

since $2$ is invertible modulo $3$ and $\frac{1}{2} (1 \pm s)$ are orthogonal idempotents.

So an indecomposable ring has a decomposable reduction: the conclusion of $(22.11)$ fails. The hypothesis that fails is completeness — $ℤ$ is noetherian and $ℤ C_{2}$ is module-finite, but $ℤ$ is not $3$ -adically complete, and $3 ℤ$ is not contained in $rad ℤ = 0$ .

Over $ℤ_{3}$ , it holds

Now take $k = ℤ_{3}$ , the ring of $3$ -adic integers: noetherian, local with maximal ideal $3 ℤ_{3}$ , and $3$ -adically complete. Here $2$ is a unit, so $\frac{1}{2} (1 + s)$ and $\frac{1}{2} (1 - s)$ lie in $ℤ_{3} C_{2}$ and

ℤ_{3} C_{2} ≅ ℤ_{3} \times ℤ_{3}, 𝔽_{3} C_{2} ≅ 𝔽_{3} \times 𝔽_{3} .

(E.2)

Two blocks upstairs, two blocks downstairs, in bijection — as $(22.11)$ predicts.

A modular example where the reduction is indecomposable

Take $k = ℤ_{3}$ again and $G = S_{3}$ . The reduction $𝔽_{3} S_{3}$ has centre spanned by $1$ , the transposition class sum $T$ , and the $3$ -cycle class sum $A$ ; setting $x = A + 1$ one finds $x^{2} = x T = T^{2} = 0$ , so the centre is local and $𝔽_{3} S_{3}$ is indecomposable. By $(22.11)$ , $ℤ_{3} S_{3}$ is therefore indecomposable as well: the integral group ring has a single block.

Process and Workflow

You want to transfer central idempotents from a quotient of $R$ back to $R$ . Which theorem applies?

The quotient is $R / I^{2}$ with $I$ a nilpotent idealUse

(22.9)

. You get a bijection of central idempotents, of centrally primitive idempotents and of blocks. For right artinian

R

, take

I = rad R

$R$ is module-finite over a complete noetherian $k$ , quotient $R / I R$ Use

(22.11)

. Dade's Lemma supplies centrality, completeness supplies the lift. This is the group-ring situation.

The quotient is $R / rad R$ Nothing applies. Central idempotents genuinely fail to lift;

T_{n} (k)

is the standard counterexample. Compute the blocks of

R

by linkage instead.

You only need centrality tested, not liftedUse

(22.8)

(22.10)

directly: both are statements that centrality can be checked in the quotient, independently of any lifting.

Locate a suitable idealA nilpotent ideal

I

R

, or an ideal

I

of a complete commutative base

k

over which

R

is module-finite.

ReducePass to

R / I^{2}

or to

R / I R

, whichever the hypotheses allow. Never to

R / rad R

Find the central idempotents downstairsThe quotient is smaller and usually more tractable; for a group ring it is an algebra over a field.

LiftLift each idempotent — Newton iteration, or successive approximation in the complete case — and invoke

(22.8)

(22.10)

to conclude centrality.

Match up the blocksCentrally primitive idempotents correspond, so the blocks correspond, and each block of

R

reduces onto the matching block downstairs.

Comparison and Classification

What survives which reduction
	Idempotents lift	Central idempotents lift	Blocks correspond
$R \to R / I$ , $I$ nil	yes	no	no
$R \to R / rad R$ , $R$ artinian	yes	no	no
$R \to R / I^{2}$ , $I$ nilpotent	yes	yes	yes
$R \to R / I R$ , $k$ complete noetherian, $R$ module-finite	yes	yes	yes
$R \to R / I R$ , $k$ not complete	partial	no	no

What survives which reduction

The two centrality criteria side by side
	$(22.8)$	Dade's Lemma $(22.10)$
Hypothesis on the ideal	$⋂_{n} I^{n} = 0$ in $R$	$I \subseteq rad k$ in a commutative base
Hypothesis on the ring	none	$R$ finitely generated as a $k$ -module
Quotient used	$R / I^{2}$	$R / I R$
Engine of the proof	induction along the $I$ -adic filtration	Nakayama's Lemma on a Peirce corner
Typical application	right artinian rings, $I = rad R$	group rings over complete local rings

Relationship Map

ℤ_{p} G

𝔽_{p} G

blocks in bijectionBrauer theory

The chain above is the practical payoff. A $p$ -modular system consists of a complete discrete valuation ring $k$ of characteristic $0$ with residue field of characteristic $p$ ; $(22.11)$ says the middle term of the system — the group ring over $k$ — has the same blocks as the modular group algebra. Ordinary characters and modular representations can then be discussed inside one block simultaneously, which is the entire technical basis of Brauer's theory.

Lifting problems for idempotents
- Plain idempotents
  - lift modulo any nil ideal
  - lift modulo $I$ when $R$ is $I$ -adically complete
  - may fail modulo a non-nil ideal, for example $ℤ \to ℤ / 6$ is fine but $ℤ [x] \to$ generic quotients are not
- Central idempotents
  - fail modulo $rad R$ , even when it is nilpotent
  - lift modulo $I^{2}$ for $I$ nilpotent
  - lift modulo $I R$ over a complete base
- Orthogonal families
  - lift together whenever single idempotents do
  - preserve central primitivity in both directions

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

$p$ -modular systems

Blocks of $𝒪 G$ for $𝒪$ a complete discrete valuation ring coincide with blocks of $k G$ for its residue field. Brauer characters, decomposition matrices and defect groups are all defined relative to this identification.

Integral representation theory

Orders and genus theory

For an order in a semisimple algebra over a $p$ -adic field, indecomposability of the order is decided by its reduction. Completing at each prime is standard practice precisely because $(22.11)$ makes the reduction faithful.

Computer algebra

Computing blocks over the integers

Block decompositions of $ℤ G$ -orders are computed prime by prime: complete at $p$ , reduce to characteristic $p$ , decompose there, lift. Without the correspondence one would have to work with integral coefficients throughout.

Structure theory

The radical-square-zero reduction

Since blocks of a right artinian ring match those of $R / (rad R)^{2}$ , classification efforts concentrate on algebras with radical square zero — a class where quiver methods are effective.

Computational Notes

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

Idempotent lifting through a nilpotent ideal is done by the Newton step $e \mapsto 3 e^{2} - 2 e^{3}$ , which squares the error each round; $O (\log m)$ iterations suffice when $I^{m} = 0$ .
In the $I$ -adic setting the same iteration converges $I$ -adically, so lifting from $𝔽_{p} G$ to $ℤ_{p} G$ is performed to any requested $p$ -adic precision at linear cost per digit doubling.
Centrality of a lifted idempotent never has to be verified directly: $(22.8)$ and $(22.10)$ guarantee it. Verifying $e R f = 0$ by brute force costs $O (n^{3})$ in $\dim_{k} R = n$ , which is worth avoiding.
Blocks of $R$ can be computed on $R / (rad R)^{2}$ , whose dimension is often far smaller than that of $R$ ; the block idempotents are then lifted back.

Failure Modes and Common Mistakes

Do not assume completeness comes for free from being local: $ℤ_{(p)}$ is noetherian local but not complete, and the correspondence for $ℤ_{(p)} C_{2}$ has to be checked separately.
Do not confuse the two roles of $I$ : in $(22.9)$ it is an ideal of the noncommutative ring $R$ ; in $(22.10)$ and $(22.11)$ it is an ideal of the commutative coefficient ring $k$ .
Do not expect a lifted central idempotent to be centrally primitive just because you lifted a primitive one carelessly — check orthogonality of the lifted family, which $(22.9)$ supplies automatically only for lifts obtained by the theorem itself.

Quick Reference

Failurecentral idempotents do not lift modulo

rad R

;

T_{n} (k)

is indecomposable with

R / rad R ≅ k^{n}

$(22.8)$

⋂_{n} I^{n} = 0

e

central in

R

iff

\bar{e}

central in

R / I^{2}

$(22.9)$

I \subseteq rad R

nilpotent: central idempotents of

R

\leftrightarrow

those of

R / I^{2}

Consequenceright artinian

R

: blocks of

R

= blocks of

R / (rad R)^{2}

$(22.10)$ Dade

I \subseteq rad k

R

module-finite over

k

: centrality is detected in

R / I R

$(22.11)$

k

noetherian and

I

-adically complete,

R

module-finite: blocks of

R

= blocks of

R / I R

Group ringsblocks of

ℤ_{p} G

= blocks of

𝔽_{p} G

Uniquenessa central idempotent has at most one lift through an ideal inside

rad R

Hypotheses that cannot be dropped
Hypothesis	Result	What breaks without it
Quotient by $I^{2}$ , not $I$	$(22.8)$ , $(22.9)$	$T_{n} (k)$ : indecomposable with a decomposable radical quotient
$I$ nilpotent	$(22.9)$	no lifting of idempotents, and $⋂ I^{n}$ may be nonzero
$R$ module-finite over $k$	$(22.10)$	Nakayama's Lemma does not apply to the Peirce corner
$k$ complete	$(22.11)$	$ℤ C_{2}$ is indecomposable but $𝔽_{3} C_{2}$ is not
$k$ noetherian	$(22.11)$	idempotent lifting from the reduction can fail

Frequently Asked Questions

Idempotents lift modulo a nilpotent ideal, so why do central idempotents cause trouble?

Because centrality is not a property of a single element but of its products with the whole ring. Lifting produces an idempotent with the right image; nothing forces its Peirce corners $e R f$ and $f R e$ to vanish. The upper triangular matrix ring shows the failure concretely: all $2^{n}$ central idempotents of the radical quotient lift, but only two of the lifts are central.

Why does passing to

R / I^{2}

instead of

R / I

fix the problem?

The proof of $(22.8)$ improves the inclusion $e R f \subseteq I^{n}$ to $e R f \subseteq I^{n + 1}$ by rewriting $e I \subseteq I e + I^{2}$ , which consumes one factor of $I$ per step. Starting from $I^{2}$ leaves a factor to spare; starting from $I$ leaves none, and the conclusion is then false.

Is the hypothesis

I \subseteq rad R

(22.9)

doing any work?

It is automatic, since every nilpotent ideal is nil and every nil ideal lies in the Jacobson radical. It is stated because the proof uses it twice explicitly: to know that $I^{2}$ contains no nonzero idempotent, and to invert $1 - u^{2}$ in the uniqueness argument.

What exactly does completeness give in

(22.11)

Two things. It puts $I$ inside $rad k$ , which is what Dade's Lemma needs, and it makes idempotents of $R / I R$ liftable by successive approximation even though $I R$ is not nilpotent. The example of $ℤ C_{2}$ reduced modulo $3$ shows the conclusion genuinely fails without it.

Does the block correspondence preserve more than the number of blocks?

Yes, it is induced by a ring surjection, so each block $c_{i} R$ of $R$ maps onto the matching block ${\bar{c}}_{i} \bar{R}$ with kernel the intersection of that block with the ideal being killed. Invariants defined by the quotient — the simple modules of a block, for instance — therefore match up automatically.

Can I use these theorems to compute blocks of

ℤ G

Not directly, since $ℤ$ is not complete. The standard route is to complete at each prime dividing the group order, apply $(22.11)$ to relate $ℤ_{p} G$ to $𝔽_{p} G$ , decompose there, and then reassemble the global information; the reassembly is genuine work and not a formal consequence of §22.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §22 (pp. 342–344), results (22.8) through (22.11).
T. Y. Lam, A First Course in Noncommutative Rings, §21, for idempotent lifting modulo nil ideals and modulo complete filtrations, and §5 for the containment of the radical of the coefficient ring.
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley-Interscience, 1981, on $p$ -modular systems, orders and block correspondences.
H. Matsumura, Commutative Ring Theory, Cambridge Studies in Advanced Mathematics 8, Cambridge University Press, 1986, for Nakayama's Lemma and adic completion.
J. L. Alperin, Local Representation Theory, Cambridge Studies in Advanced Mathematics 11, Cambridge University Press, 1986, for the use of complete discrete valuation rings in block theory.

AI Suggested Questions

Prove that a nil ideal contains no nonzero idempotent, and locate that fact in the proof of $(22.9)$ .
Give an example of a ring $R$ and a nilpotent ideal $I$ where $R / I$ and $R / I^{2}$ have different numbers of blocks.
How does the Newton iteration for lifting idempotents behave $p$ -adically, and what precision is needed to separate two block idempotents?
State the standard idempotent lifting theorem for $I$ -adically complete rings and identify where noetherianness is used.
What are the blocks of $ℤ_{2} S_{3}$ , and how do they compare with the blocks of $𝔽_{2} S_{3}$ ?
Explain why the study of right artinian rings with radical square zero is a genuine reduction rather than a special case.
Where does Dade's Lemma appear in the theory of orders over complete discrete valuation rings?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

The obstruction, in one example

Why the square of the ideal is the right quotient

The Nakayama route

06Key Results

07Proof Techniques and Method

Improve an inclusion, then intersect

Nakayama on a Peirce corner

Uniqueness by a unit

08Worked Example

Completeness is not a technicality

Over ℤ, the correspondence fails

Over ℤ3, it holds

A modular example where the reduction is indecomposable

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

p-modular systems

Orders and genus theory

Computing blocks over the integers

The radical-square-zero reduction

13Computational Notes

14Failure Modes and Common Mistakes

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

Over $ℤ$ , the correspondence fails

Over $ℤ_{3}$ , it holds

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

$p$ -modular systems

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions