Lifting Idempotents Modulo an Ideal

Q: Why insist that I lie inside the Jacobson radical?

Two consequences are used constantly. First, 1 + I ⊆ U(R), which lets you invert 1 - in the orthogonalisation lemma. Second, I contains no nonzero idempotent, so a nonzero idempotent of R stays nonzero in the quotient. Without the first, the correction step is unavailable; without the second, a nontrivial decomposition upstairs could collapse to a trivial one downstairs.

Q: Does the converse of (21.22) really need liftability?

Yes. Its proof lifts a nontrivial decomposition of e back to R, and there is nothing to lift otherwise. The semilocal domain Z_(p) Z_(q) shows a ring where 1 is primitive but its image in F_p × F_q is not, and this is possible precisely because idempotents do not lift there.

Q: Why can only countably many idempotents be lifted at once?

The proof is an induction that at stage n orthogonalises against the single idempotent = e_1 + … + e_n. That partial sum has to exist as an element of R, which requires finitely many terms at each stage, and the induction then exhausts a countable index set. Nothing in the argument produces a limit for an uncountable family.

Q: Is liftability inherited by quotients or matrix rings?

Matrix rings behave well: if idempotents lift modulo I in R, they lift modulo M_n(I) in M_n(R), since M_n(R)/M_n(I) ≅ M_n(R/I) and the constructive proofs used in practice — nil or complete — pass to matrix rings. For quotients, the safest statement is the transitivity one: if idempotents lift modulo I and modulo J/I in R/I with I ⊆ J, they lift modulo J.

Executive Summary

Quotienting a ring can create idempotents out of nothing. In $ℤ / 6 ℤ$ the class of $3$ is idempotent, yet $ℤ$ itself has only $0$ and $1$ . Lifting asks the reverse question: given an idempotent $x \in R / I$ , is there an idempotent $e \in R$ with $\bar{e} = x$ ?

For a general ideal the answer is no. But when $I \subseteq rad R$ , a great deal survives the quotient even without liftability: isomorphism classes of idempotents are faithfully detected in $R / I$ , and primitivity in $R / I$ forces primitivity in $R$ . When lifting is available, primitivity becomes a two-way street and entire countable orthogonal families can be pulled back.

$rad R$ Where

I

must live

0Only idempotent in the radical

CountableFamilies liftable at once

$1 + rad R$ Source of the units used

Overview

Let $I$ be an ideal of a ring $R$ and write $\bar{R} = R / I$ with $r \mapsto \bar{r}$ the quotient map. An idempotent $x \in \bar{R}$ lifts if $x = \bar{e}$ for some $e = e^{2} \in R$ . Liftability is a property of the pair $(R, I)$ , not of $x$ alone.

The reason lifting matters is that idempotents encode direct decompositions: a decomposition $1 = e_{1} + \dots + e_{n}$ into orthogonal idempotents is the same thing as a direct sum decomposition $R_{R} = e_{1} R \oplus \dots \oplus e_{n} R$ . Quotients are usually simpler — $R / rad R$ is semisimple when $R$ is semilocal — so one wants to compute the decomposition downstairs and transport it upstairs.

Decomposition of

\bar{1}

R / I

Lift each idempotentOrthogonalise the liftsDecomposition of

1

R

Two hypotheses do the work. The condition $I \subseteq rad R$ makes the quotient map faithful on isomorphism classes of idempotents; liftability makes it surjective on idempotents. Rings for which both hold with $I = rad R$ , and for which $R / rad R$ is semisimple, are precisely the semiperfect rings.

Learning Objectives

State precisely what it means for idempotents to lift modulo an ideal, and exhibit a failure.
Prove $(21.23)$ : the radical contains no nonzero idempotent.
Prove $(21.21)$ : for $I \subseteq rad R$ , $e ≅ f$ in $R$ if and only if $\bar{e} ≅ \bar{f}$ in $R / I$ .
Prove $(21.22)$ : primitivity of $\bar{e}$ implies primitivity of $e$ , with converse under liftability.
Carry out the orthogonalisation $(21.24)$ and extend it to countable families $(21.25)$ .
Use $(21.26)$ to produce infinitely many orthogonal idempotents in a non Dedekind-finite ring and deduce $(21.27)$ .

Definitions

Definition—Liftable idempotent

Let $I$ be an ideal of $R$ . An idempotent $x \in R / I$ **can be lifted to $R$ ** if there exists an idempotent $e \in R$ whose image under $R \to R / I$ equals $x$ . We say *idempotents lift modulo $I$ * when every idempotent of $R / I$ can be lifted.

$rad R$: The Jacobson radical, the intersection of the maximal left ideals; equivalently the largest left ideal $U$ with $1 + U \subseteq U (R)$ . Also written $J (R)$ .
$\bar{r}$: The image of $r \in R$ in the quotient $\bar{R} = R / I$ .
$e ≅ f$: Isomorphic idempotents: $e R ≅ f R$ as right modules, equivalently $a b = e$ and $b a = f$ for some $a, b \in R$ .
Nontrivial decomposition: A writing $e = α + β$ with $α, β$ nonzero orthogonal idempotents.
Semilocal ring: A ring $R$ with $R / rad R$ semisimple.

Rings have an identity; ideal means two-sided ideal unless stated otherwise; modules are unital right modules unless stated otherwise.

Core Concepts

The radical sees no idempotents

Everything on this page rests on one small observation, which is also the reason $I \subseteq rad R$ is the right hypothesis: an idempotent lying in the radical must be zero. It follows that the quotient map is injective on the property of being nonzero: a nonzero idempotent of $R$ has nonzero image in $R / I$ whenever $I \subseteq rad R$ .

Faithful, but not surjective

For $I \subseteq rad R$ the quotient map is faithful on idempotents in a strong sense: $\bar{e} ≅ \bar{f}$ already forces $e ≅ f$ . What it is not, in general, is surjective onto the idempotents of $\bar{R}$ . Those two properties are logically independent, and the theory keeps them apart.

What each hypothesis provides
	$\bar{e} ≅ \bar{f} \Rightarrow e ≅ f$	$\bar{e}$ primitive $\Rightarrow e$ primitive	$e$ primitive $\Rightarrow \bar{e}$ primitive	Countable families lift
$I$ arbitrary	no	no	no	no
$I \subseteq rad R$	yes	yes	no	no
$I \subseteq rad R$ and idempotents lift	yes	yes	yes	yes

What each hypothesis provides

Nearly orthogonal is good enough

If $α, β$ are idempotents of $R$ whose images are orthogonal, then $α β$ and $β α$ lie in $I \subseteq rad R$ , so $1 - β α$ is a unit. Conjugating $β$ by that unit kills one of the two products; multiplying by $1 - α$ on the left kills the other. Two moves, and the pair is genuinely orthogonal without changing anything modulo $I$ .

Key Results

Lemma(21.23)No idempotents in the radical

Let $R$ be a ring and $a \in rad R$ with $a^{2} = a$ . Then $a = 0$ .

Proof

Since $a \in rad R$ , the element $1 - a$ is a unit. But $(1 - a) a = a - a^{2} = 0$ , and multiplying on the left by $(1 - a)^{- 1}$ gives $a = 0$ . $■$

Proposition(21.21)Isomorphism is detected in the quotient

Let $I \subseteq rad R$ be an ideal of $R$ and let $e, f \in R$ be idempotents. Then $e ≅ f$ in $R$ if and only if $\bar{e} ≅ \bar{f}$ in $\bar{R} = R / I$ . In particular, if $\bar{e} = \bar{f}$ then $e ≅ f$ .

Proof

Necessity. If $a b = e$ and $b a = f$ in $R$ , reduce modulo $I$ : $\bar{a} \bar{b} = \bar{e}$ and $\bar{b} \bar{a} = \bar{f}$ , so $\bar{e} ≅ \bar{f}$ .

Sufficiency. Note first that $e R$ is a finitely generated projective right $R$ -module with $e R \cdot I = e I$ and $e R / e I ≅ \bar{e} \bar{R}$ as $\bar{R}$ -modules. Assume $\bar{e} \bar{R} ≅ \bar{f} \bar{R}$ , that is, $e R / e I ≅ f R / f I$ . Because $e R$ is projective and $f R \to f R / f I$ is surjective, the composite $e R \to e R / e I \tilde{⟶} f R / f I$ lifts to an $R$ -homomorphism $θ : e R \to f R$ .

Surjectivity. By construction $θ (e R) + f I = f R$ . Since $f R$ is finitely generated and $I \subseteq rad R$ , Nakayama's Lemma gives $θ (e R) = f R$ .

Injectivity. As $f R$ is projective, the surjection $θ$ splits: $e R = \ker θ \oplus C$ with $C ≅ f R$ . Reducing modulo $I$ , the induced map $e R / e I \to f R / f I$ is the given isomorphism, so the image of $\ker θ$ in $e R / e I$ is zero, i.e. $\ker θ \subseteq e R \cdot I = \ker θ \cdot I \oplus C \cdot I$ . Projecting along the decomposition gives $\ker θ \subseteq \ker θ \cdot I$ , hence $\ker θ = \ker θ \cdot I$ . Now $\ker θ$ is a direct summand of the cyclic module $e R$ , hence finitely generated, so Nakayama's Lemma forces $\ker θ = 0$ .

Therefore $θ$ is an isomorphism $e R ≅ f R$ , that is, $e ≅ f$ . $■$

Proposition—Two lifts are conjugate

Let $I \subseteq rad R$ be an ideal and let $e, f \in R$ be idempotents with $e - f \in I$ . Then $f = u^{- 1} e u$ for the unit $u = e f + (1 - e) (1 - f)$ .

Proof

Modulo $I$ we have $\bar{e} = \bar{f}$ , so $\bar{u} = {\bar{e}}^{2} + (1 - \bar{e})^{2} = \bar{e} + 1 - \bar{e} = 1$ . Hence $u \in 1 + I \subseteq 1 + rad R \subseteq U (R)$ . Next,

e u = e (e f + (1 - e) (1 - f)) = e f + (e - e^{2}) (1 - f) = e f,

u f = (e f + (1 - e) (1 - f)) f = e f^{2} + (1 - e) (f - f^{2}) = e f .

So $e u = u f$ , and multiplying on the left by $u^{- 1}$ gives $u^{- 1} e u = f$ . $■$

Lemma(21.24)Orthogonalising a lifted idempotent

Let $I \subseteq rad R$ be an ideal and let $α, β \in R$ be idempotents whose images in $R / I$ are orthogonal, that is $α β \in I$ and $β α \in I$ . Then there is an idempotent $β' \in R$ with $α β' = β' α = 0$ and $β' \equiv β (\mod I)$ .

Proof

Since $β α \in I \subseteq rad R$ , the element $u = 1 - β α$ is a unit. Put $β_{0} = u^{- 1} β u$ , an idempotent congruent to $β$ modulo $I$ because $u \equiv 1$ . Using $β^{2} = β$ and $α^{2} = α$ ,

β u α = β (1 - β α) α = β α - β^{2} α^{2} = β α - β α = 0,

so $β_{0} α = u^{- 1} (β u α) = 0$ . The other product $α β_{0}$ need not vanish, so set $β' = (1 - α) β_{0}$ . Then:

$α β' = α (1 - α) β_{0} = (α - α^{2}) β_{0} = 0$ .
$β' α = (1 - α) β_{0} α = 0$ .
$(β')^{2} = (1 - α) β_{0} (1 - α) β_{0} = (1 - α) (β_{0} - β_{0} α) β_{0} = (1 - α) β_{0}^{2} = β'$ .
$β' = β_{0} - α β_{0} \equiv β - α β \equiv β (\mod I)$ , since $α β \in I$ .

So $β'$ is an idempotent orthogonal to $α$ and congruent to $β$ . $■$

Proposition(21.22)Primitivity across the quotient

Let $I \subseteq rad R$ be an ideal of $R$ , $\bar{R} = R / I$ , and let $e \in R$ be an idempotent. If $\bar{e}$ is primitive in $\bar{R}$ , then $e$ is primitive in $R$ . The converse holds under the additional hypothesis that idempotents of $\bar{R}$ can be lifted to $R$ .

Proof

Forward. Suppose $e = α + β$ with $α, β$ nonzero orthogonal idempotents of $R$ . If $\bar{α} = 0$ then $α \in I \subseteq rad R$ , so $α = 0$ by $(21.23)$ , a contradiction; likewise $\bar{β} \neq 0$ . Hence $\bar{e} = \bar{α} + \bar{β}$ is a nontrivial decomposition into orthogonal idempotents, contradicting primitivity of $\bar{e}$ .

Converse. Assume idempotents lift, and suppose $\bar{e} = x + y$ with $x, y$ nonzero orthogonal idempotents of $\bar{R}$ . Lift $x$ to an idempotent $α \in R$ and $y$ to an idempotent $β \in R$ . Since $x y = y x = 0$ , we have $α β, β α \in I$ , so $(21.24)$ supplies an idempotent $β'$ orthogonal to $α$ with $\bar{β'} = y$ .

Put $e' = α + β'$ . This is an idempotent (a sum of orthogonal idempotents), and it is not primitive because $α$ and $β'$ are nonzero — their images $x$ and $y$ are nonzero. Moreover $\bar{e'} = x + y = \bar{e}$ , so $e' ≅ e$ by $(21.21)$ . Primitivity is an invariant of the isomorphism class of an idempotent, so $e$ is not primitive either. $■$

Proposition(21.25)Lifting countable orthogonal families

Let $I \subseteq rad R$ be an ideal such that idempotents of $\bar{R} = R / I$ can be lifted to $R$ . Then for any countable (possibly finite) set ${x_{1}, x_{2}, \dots}$ of pairwise orthogonal idempotents of $\bar{R}$ there is a set ${e_{1}, e_{2}, \dots}$ of pairwise orthogonal idempotents of $R$ with $\bar{e_{i}} = x_{i}$ for every $i$ .

Proof

Induct. Suppose pairwise orthogonal idempotents $e_{1}, \dots, e_{n}$ with $\bar{e_{i}} = x_{i}$ have been found; it suffices to produce $e_{n + 1}$ . Put $α = e_{1} + \dots + e_{n}$ , an idempotent with $\bar{α} = x_{1} + \dots + x_{n}$ , and let $β$ be any idempotent of $R$ lifting $x_{n + 1}$ .

Because $x_{n + 1}$ is orthogonal to each $x_{i}$ with $i \leq n$ , the images $\bar{α}$ and $\bar{β}$ are orthogonal. By $(21.24)$ there is an idempotent $e_{n + 1} : = β'$ orthogonal to $α$ with $\bar{e_{n + 1}} = x_{n + 1}$ . Finally, $e_{i} = α e_{i} = e_{i} α$ for $i \leq n$ , so $e_{n + 1} e_{i} = e_{n + 1} α e_{i} = 0$ and $e_{i} e_{n + 1} = e_{i} α e_{n + 1} = 0$ : the new idempotent is orthogonal to each of the old ones. $■$

Example(21.26)Jacobson's matrix units

Let $R$ be a ring that is not Dedekind-finite: there exist $a, b \in R$ with $a b = 1$ and $e : = b a \neq 1$ . Then $e^{2} = b (a b) a = b a = e$ , and for $i, j \geq 0$ the elements

e_{i j} = b^{i} (1 - e) a^{j}

form a set of matrix units: $e_{i j} e_{k ℓ} = δ_{j k} e_{i ℓ}$ , and every $e_{i j}$ is nonzero. In particular ${e_{i i} : i \geq 0}$ is an infinite family of nonzero pairwise orthogonal idempotents, and $⨁_{i \geq 0} e_{i i} R$ is an infinite direct sum of nonzero right ideals inside $R$ .

Proof

From $a b = 1$ we get $a^{i} b^{i} = 1$ for all $i \geq 0$ . Also $a (1 - e) = a - a (b a) = a - (a b) a = 0$ and $(1 - e) b = b - (b a) b = b - b (a b) = 0$ ; hence $a^{m} (1 - e) = 0$ and $(1 - e) b^{m} = 0$ for every $m \geq 1$ .

Now compute $e_{i j} e_{k ℓ} = b^{i} (1 - e) a^{j} b^{k} (1 - e) a^{ℓ}$ . If $j = k$ then $a^{j} b^{k} = 1$ and, since $1 - e$ is idempotent, the product is $b^{i} (1 - e) a^{ℓ} = e_{i ℓ}$ . If $j > k$ then $a^{j} b^{k} = a^{j - k}$ and the middle factor $(1 - e) a^{j - k} (1 - e)$ vanishes because $a^{j - k} (1 - e) = 0$ . If $k > j$ then $a^{j} b^{k} = b^{k - j}$ and the middle factor vanishes because $(1 - e) b^{k - j} = 0$ .

Nonvanishing: if $b^{i} (1 - e) a^{j} = 0$ , multiply on the left by $a^{i}$ and on the right by $b^{j}$ to get $1 \cdot (1 - e) \cdot 1 = 0$ , i.e. $e = 1$ , contrary to hypothesis. $■$

Corollary(21.27)A finiteness criterion for Dedekind-finiteness

Let $S$ be a ring such that $R : = S / rad S$ contains no infinite direct sum of nonzero right ideals — for instance, $R$ right noetherian. Then $S$ is Dedekind-finite.

Proof

By $(21.26)$ , a ring that is not Dedekind-finite contains an infinite direct sum of nonzero right ideals. The hypothesis therefore forces $R$ to be Dedekind-finite.

Now let $a b = 1$ in $S$ . Passing to $R$ gives $\bar{a} \bar{b} = 1$ , hence $\bar{b} \bar{a} = 1$ , so $b a - 1 \in rad S$ and therefore $b a \in 1 + rad S \subseteq U (S)$ . Choose $u \in S$ with $b a u = 1$ . Multiplying on the left by $a$ gives $a = a (b a u) = (a b) (a u) = a u$ . Substituting back, $1 = b a u = b (a u) = b a$ . Hence $S$ is Dedekind-finite. $■$

Proof Techniques and Method

How these proofs work, and which move to reuse.

The arguments above use a small, highly reusable toolkit.

Move 1

Conjugate by $1 - β α$

An error term lying in the radical makes $1 - (error)$ a unit. Conjugating by it is a change of coordinates that does not move anything modulo $I$ but does kill one unwanted product.

Move 2

Truncate with $1 - α$

Left multiplication by the complementary idempotent annihilates the remaining product. Idempotency of $β_{0}$ combined with $β_{0} α = 0$ keeps the result idempotent.

Move 3

Nakayama twice

Surjectivity of a lifted map comes from $θ (e R) + f I = f R$ ; injectivity comes from applying Nakayama to $\ker θ$ , which is finitely generated because it is a summand.

Move 1 followed by Move 2 is the standard orthogonalisation. It appears again whenever a family of idempotents has to be replaced by an orthogonal family with the same image, and it is the reason $(21.25)$ can be proved by a bare induction with no extra hypotheses.

Worked Example

A semilocal ring where idempotents do not lift

Let $p \neq q$ be primes and let $R = ℤ_{(p)} \cap ℤ_{(q)}$ , that is, the subring of $ℚ$ consisting of fractions $m / n$ with $n$ coprime to $p q$ . This is a semilocal principal ideal domain with exactly two maximal ideals, $p R$ and $q R$ , so

rad R = p R \cap q R = p q R, R / rad R ≅ ℤ / p q ℤ ≅ 𝔽_{p} \times 𝔽_{q},

(E.1)

the middle isomorphism because every integer coprime to $p q$ is already invertible modulo $p q$ . Now $𝔽_{p} \times 𝔽_{q}$ has four idempotents, including $(1, 0)$ and $(0, 1)$ . But $R$ is an integral domain, so $x^{2} = x$ forces $x (x - 1) = 0$ and hence $x \in {0, 1}$ .

Orthogonalisation in a nilpotent extension

Let $k$ be a field and $R = {(\begin{matrix} a & b \\ 0 & c \end{matrix}) : a, b, c \in k}$ , with $I = rad R = k E_{12}$ , which satisfies $I^{2} = 0$ ; here $R / I ≅ k \times k$ . Suppose the two orthogonal idempotents of $R / I$ have been lifted carelessly, to $α = E_{11}$ and $β = E_{22} + E_{12}$ .

Both are idempotent: $β^{2} = E_{22} + E_{12} E_{22} = E_{22} + E_{12} = β$ , and $\bar{β} = {\bar{E}}_{22}$ is orthogonal to $\bar{α} = {\bar{E}}_{11}$ . But in $R$ the lifts are not orthogonal:

α β = E_{11} (E_{22} + E_{12}) = E_{12} \neq 0, β α = (E_{22} + E_{12}) E_{11} = 0 .

(E.2)

Apply $(21.24)$ . Here $β α = 0$ , so $u = 1 - β α = 1$ and $β_{0} = β$ . The second move gives

β' = (1 - α) β_{0} = E_{22} (E_{22} + E_{12}) = E_{22},

(E.3)

which is idempotent, orthogonal to $α = E_{11}$ on both sides, and congruent to $β$ modulo $I$ . The recipe has done exactly what it promises.

Process and Workflow

Reduce modulo the radicalCompute

\bar{R} = R / rad R

. If

R

is semilocal this is a finite product of matrix rings over division rings, where idempotents are transparent.

Decompose $\bar{1}$ Write

\bar{1} = x_{1} + \dots + x_{n}

as a sum of orthogonal primitive idempotents in

\bar{R}

Lift each $x_{i}$ separatelyUse nilness of the radical, or completeness, to obtain idempotents

β_{i} \in R

with

\bar{β_{i}} = x_{i}

. Ignore orthogonality at this stage.

Orthogonalise inductivelyApply (21.24) with

α = e_{1} + \dots + e_{i - 1}

to replace

β_{i}

by an idempotent orthogonal to all its predecessors.

Check the sumThe idempotent

e_{1} + \dots + e_{n}

is congruent to

1

modulo the radical, hence equals

1

by (21.23) applied to its complement.

Read off the decomposition

R_{R} = e_{1} R \oplus \dots \oplus e_{n} R

, with each

e_{i} R

indecomposable by (21.22).

You have an idempotent $x \in R / I$ and want a lift. What is known about $I$ ?

$I$ is nilLift exists and can be written down explicitly from a binomial expansion; see Lifting Idempotents Modulo Nil Ideals.

$R$ is $I$ -adically completeLift exists by successive approximation through the tower

R / I^{n}

; see Idempotents in I-Adically Complete Rings.

Only $I \subseteq rad R$ No lift is guaranteed. You may still use (21.21) and the forward half of (21.22) to transfer information downwards.

NothingExpect failure. Even

ℤ

modulo

6 ℤ

defeats you.

Comparison and Classification

When do idempotents lift modulo $I$ ?
Hypothesis on $I$	Idempotents lift?	Reason or reference
$I$ nilpotent	Yes	Special case of the nil result (21.28)
$I$ nil	Yes	(21.28): a binomial-expansion construction
$R$ is $I$ -adically complete	Yes	(21.31): lift step by step through $R / I^{n}$
$I \subseteq rad R$ , no further hypothesis	Not in general	$ℤ_{(p)} \cap ℤ_{(q)}$ modulo its radical
$I$ arbitrary	Not in general	$ℤ$ modulo $6 ℤ$
$R$ semiperfect, $I = rad R$	Yes	Part of the definition of semiperfect

The table splits into two ideas. Nilness and completeness are constructive: they give an algorithm producing the lift. Semiperfectness is definitional: it packages liftability with semisimplicity of the quotient because that combination is what makes the theory of projective covers work.

Relationship Map

The following implications hold for an ideal $I \subseteq rad R$ .

$I \subseteq rad R$ — the standing hypothesis
- always gives
  - only idempotent in $I$ is $0$
  - $\bar{e} ≅ \bar{f} \Rightarrow e ≅ f$
  - $\bar{e}$ primitive $\Rightarrow e$ primitive
  - nearly orthogonal pairs can be orthogonalised
- gives, if idempotents lift
  - $e$ primitive $\Rightarrow \bar{e}$ primitive
  - countable orthogonal families lift
  - decompositions of $1$ lift
- never gives by itself
  - existence of a lift for a given idempotent
  - conjugacy of two lifts of the same idempotent

All ringslifting may fail entirely

$I \subseteq rad R$ faithfulness results hold

Idempotents lift mod $rad R$ primitivity is detected both ways

Semiperfectlifting plus semisimple quotient

Semiprimary / one-sided artinianradical nil, so lifting is automatic

Applications and Industry Use

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

Structure theory

Semiperfect rings

A ring is semiperfect exactly when it is semilocal and idempotents lift modulo the radical. Everything about projective covers, basic rings and block decompositions is downstream of the results on this page.

Modular representation theory

Blocks of group algebras

For a $p$ -modular system, the block idempotents of a group algebra over a complete discrete valuation ring are obtained by lifting the central idempotents of the residue algebra. Orthogonality of the lifted family is exactly $(21.25)$ .

Integral representations

Orders over complete local rings

Decomposing a lattice over an order reduces to decomposing its reduction, then lifting the idempotent decomposition of the endomorphism ring.

Computer algebra

Wedderburn decomposition

Algorithms that split a finite-dimensional algebra compute idempotents in the semisimple quotient and lift them through the nilpotent radical, orthogonalising as they go.

Honestly stated, this material is internal machinery. Its value is that it converts a hard computation in $R$ into an easy computation in $R / rad R$ plus a mechanical transport step — which is precisely what a computer algebra system needs.

Computational Notes

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

For a finite-dimensional algebra $A$ over a field $k$ given by structure constants:

$rad A$ is nilpotent, so lifting is always possible and the workflow above always terminates.
The dominant cost is computing $rad A$ and splitting $A / rad A$ , not the lifting itself.
Orthogonalisation costs one inversion of $1 - β α$ per step; since $β α$ is nilpotent, the inverse is a finite geometric series and no general linear solve is needed.
The number of steps in $(21.25)$ equals the number of summands, so the total work is linear in the number of primitive idempotents.
GAP, Magma and Sage all expose primitive idempotent computation for finite-dimensional algebras; the returned family is orthogonal, which is the output of exactly this procedure.

Failure Modes and Common Mistakes

The forward direction of $(21.22)$ needs no liftability; the converse does. Quoting the proposition without its hypothesis is a genuine error, not a technicality.
$(21.25)$ is stated for countable families. Do not silently extend it to arbitrary index sets.
In $(21.24)$ the order of the two moves matters: conjugating first makes $β_{0} α = 0$ , and only then does multiplication by $1 - α$ preserve idempotency.
$(21.27)$ concerns $S / rad S$ , not $S$ . A ring may itself contain an infinite direct sum of right ideals while its semisimple quotient does not.

Historical Notes and Lessons Learned

1900sWedderburn's principal theoremFor a finite-dimensional algebra over a perfect field, the semisimple quotient lifts to a subalgebra. Lifting idempotents is the elementary shadow of this phenomenon.
1945Jacobson's radicalThe identification of $rad R$ as the largest ideal $U$ with $1 + U \subseteq U (R)$ is what makes the conjugation trick in (21.24) available.
1956Jacobson's matrix unitsThe construction (21.26) turning a one-sided inverse into an infinite orthogonal family becomes the standard route to Dedekind-finiteness criteria.
1960Bass introduces perfect and semiperfect ringsLiftability of idempotents modulo the radical is promoted from a technical lemma to part of a definition, alongside semisimplicity of the quotient.
1991Lam's textbook treatment§21 separates the two sufficient conditions — nilness and completeness — and isolates the orthogonalisation lemma as a reusable step.

The methodological lesson is the same one that shaped the radical itself: a property defined by an internal construction (here, an explicit formula for the lift) is powerful but narrow, whereas a property defined by what it enables (here, semiperfectness) travels further. The theory keeps both, and uses the constructive versions to verify the axiomatic one.

Quick Reference

Lifting

x \in R / I

lifts if

x = \bar{e}

for some

e = e^{2} \in R

Key hypothesis

I \subseteq rad R

; then

1 + I \subseteq U (R)

and

I

contains no nonzero idempotent.

Faithfulness

\bar{e} ≅ \bar{f} \Rightarrow e ≅ f

; in particular

\bar{e} = \bar{f} \Rightarrow e ≅ f

Orthogonalisation

β' = (1 - α) (1 - β α)^{- 1} β (1 - β α)

Primitivity

\bar{e}

primitive

\Rightarrow

e

primitive; converse needs liftability.

FamiliesCountably many pairwise orthogonal idempotents lift simultaneously.

The five results at a glance
Number	Statement	Hypotheses
(21.21)	$e ≅ f$ iff $\bar{e} ≅ \bar{f}$	$I \subseteq rad R$
(21.22)	Primitivity descends; ascends under lifting	$I \subseteq rad R$
(21.23)	Only idempotent in $rad R$ is $0$	none
(21.24)	Orthogonalise $β$ against $α$	$I \subseteq rad R$ , images orthogonal
(21.25)	Countable orthogonal families lift	$I \subseteq rad R$ , idempotents lift
(21.26)–(21.27)	Matrix units from $a b = 1 \neq b a$ ; Dedekind-finiteness criterion	$S / rad S$ has no infinite direct sum of right ideals

Frequently Asked Questions

Why insist that

I

lie inside the Jacobson radical?

Two consequences are used constantly. First, $1 + I \subseteq U (R)$ , which lets you invert $1 - β α$ in the orthogonalisation lemma. Second, $I$ contains no nonzero idempotent, so a nonzero idempotent of $R$ stays nonzero in the quotient. Without the first, the correction step is unavailable; without the second, a nontrivial decomposition upstairs could collapse to a trivial one downstairs.

If two idempotents have the same image, are they equal?

No, but they are conjugate — hence isomorphic — by the unit $u = e f + (1 - e) (1 - f)$ , which lies in $1 + I$ . For a concrete gap take the triangular ring over a field with $I = rad R = k E_{12}$ : the idempotents $E_{11}$ and $E_{11} + E_{12}$ have the same image, are conjugate by $1 + E_{12}$ , and are not equal.

Does the converse of

(21.22)

really need liftability?

Yes. Its proof lifts a nontrivial decomposition of $\bar{e}$ back to $R$ , and there is nothing to lift otherwise. The semilocal domain $ℤ_{(p)} \cap ℤ_{(q)}$ shows a ring where $1$ is primitive but its image in $𝔽_{p} \times 𝔽_{q}$ is not, and this is possible precisely because idempotents do not lift there.

Why can only countably many idempotents be lifted at once?

The proof is an induction that at stage $n$ orthogonalises against the single idempotent $α = e_{1} + \dots + e_{n}$ . That partial sum has to exist as an element of $R$ , which requires finitely many terms at each stage, and the induction then exhausts a countable index set. Nothing in the argument produces a limit for an uncountable family.

What does

(21.26)

have to do with lifting?

Directly, nothing: it is placed here because it manufactures a large orthogonal family from a single failure of Dedekind-finiteness, which is the same currency $(21.25)$ trades in. Indirectly it shows why the finiteness hypotheses in this area are natural: a ring with a one-sided inverse that is not two-sided is forced to be very large.

Is liftability inherited by quotients or matrix rings?

Matrix rings behave well: if idempotents lift modulo $I$ in $R$ , they lift modulo $M_{n} (I)$ in $M_{n} (R)$ , since $M_{n} (R) / M_{n} (I) ≅ M_{n} (R / I)$ and the constructive proofs used in practice — nil or complete — pass to matrix rings. For quotients, the safest statement is the transitivity one: if idempotents lift modulo $I$ and modulo $J / I$ in $R / I$ with $I \subseteq J$ , they lift modulo $J$ .

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §21, (21.21)–(21.27) (pp. 326–329).
H. Bass, “Finitistic dimension and a homological generalization of semi-primary rings”, Transactions of the American Mathematical Society 95 (1960), 466–488.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §27.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964.
C. W. Curtis and I. Reiner, Methods of Representation Theory, Volume I, Wiley, 1981, for idempotent lifting in the modular representation theory of finite groups.

AI Suggested Questions

Give a ring $R$ and ideal $I \subseteq rad R$ where idempotents do not lift but $R / I$ is not semisimple.
Prove that idempotents lift modulo $I$ in $R$ if and only if they lift modulo $M_{n} (I)$ in $M_{n} (R)$ .
Show that two commuting idempotents congruent modulo an ideal inside the radical must be equal.
Extend $(21.25)$ to uncountable families under a chain condition, or explain the obstruction.
Work out the block idempotents of $ℤ_{p} [S_{3}]$ by lifting from $𝔽_{p} [S_{3}]$ for $p = 2$ and $p = 3$ .
How does liftability of idempotents relate to the existence of projective covers?
Prove Bergman's example of a ring $R$ with $R / R e R ≅ R \times R$ whose nontrivial idempotents do not lift.
Is Dedekind-finiteness inherited by matrix rings over a Dedekind-finite ring?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

The radical sees no idempotents

Faithful, but not surjective

Nearly orthogonal is good enough

06Key Results

07Proof Techniques and Method

Conjugate by 1−βα

Truncate with 1−α

Nakayama twice

08Worked Example

A semilocal ring where idempotents do not lift

Orthogonalisation in a nilpotent extension

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Semiperfect rings

Blocks of group algebras

Orders over complete local rings

Wedderburn decomposition

13Computational Notes

14Failure Modes and Common Mistakes

15Historical Notes and Lessons Learned

16Quick Reference

17Frequently Asked Questions

18Related KEVOS Topics

19References

20AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Conjugate by $1 - β α$

Truncate with $1 - α$

Worked Example

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