When Are eR and fR Isomorphic?

Q: Why is the relation written e ≅ f rather than e = f up to something?

Because it is literally an isomorphism of modules: eR ≅ fR. The elements e and f are usually different, and the right ideals they generate are usually different subsets of R. What agrees is the abstract module structure, and the notation is chosen to keep that in view.

Q: If ab = e, why does ba have to be checked separately?

Because ba is always idempotent when e is — (ba)^2 = b(ab)a = bea = ba if a = ea and b = be — but it need not equal the f you started with. In a ring that is not Dedekind-finite, ab = 1 with ba a proper idempotent is exactly this phenomenon. The single equation gives a split surjection fR eR, not an isomorphism.

Q: Is isomorphism of idempotents an equivalence relation?

Yes. Reflexivity uses a = b = e; symmetry swaps a and b; transitivity composes the module isomorphisms, or at the element level, if a_1b_1 = e, b_1a_1 = f, a_2b_2 = f, b_2a_2 = g, then (a_1a_2)(b_2b_1) = a_1fb_1 = a_1b_1a_1b_1 = e and symmetrically (b_2b_1)(a_1a_2) = g.

Q: How does this relate to Morita theory?

The isomorphism Hom_R(eR, fR) ≅ fRe and its consequence End_R(eR) ≅ eRe are the first steps of the Morita correspondence. When e is a full idempotent, meaning ReR = R, the functor M Me is an equivalence between right R-modules and right eRe-modules, and the invariance of ReR under e ≅ f shows fullness is a property of the isomorphism class.

Q: Can two isomorphic idempotents be orthogonal?

Yes, and that is the useful case. In M_n(k) the diagonal matrix units E_11, …, E_nn are pairwise orthogonal and pairwise isomorphic; a family of r pairwise orthogonal pairwise isomorphic idempotents summing to e makes eRe an r × r matrix ring over e_1Re_1.

Executive Summary

An idempotent $e$ of a ring $R$ cuts out a projective right module $e R$ , and $R = e R \oplus (1 - e) R$ . The natural question — when do two idempotents cut out the same module up to isomorphism? — has an answer that is purely multiplicative and requires no module theory to state.

$e R ≅ f R$ iff there exist $a, b \in R$ with $a b = e$ and $b a = f$ . The condition is symmetric under reversing the multiplication, so it simultaneously characterises $R e ≅ R f$ . Lam records this as $(21.20)$ and writes $e ≅ f$ for the resulting equivalence relation on idempotents.

4Equivalent conditions

$a b = e$ The test

SymmetricLeft vs right

$e R e ≅ f R f$ Consequence

Overview

Let $R$ be a ring with identity and $e = e^{2} \in R$ . The Peirce decomposition $R = e R \oplus (1 - e) R$ exhibits $e R$ as a direct summand of $R_{R}$ , hence as a finitely generated projective right module. Every finitely generated projective right module arises this way, from an idempotent in some matrix ring $M_{n} (R)$ . Classifying projectives therefore reduces to classifying idempotents up to the relation studied here.

e R ≅ f R ⟺ \exists a, b \in R : a b = e, b a = f

(21.20)

The factorisation criterion. No reference to modules survives on the right-hand side.

The criterion looks slight, but it is the reason so much idempotent theory is side-neutral. Primitivity, locality and the structure of the corner ring $e R e$ are all invariants of the isomorphism class of $e$ , and each of them is therefore computable from either side.

Three relations on idempotents must be kept apart: equality of the right ideals $e R = f R$ , conjugacy $f = u^{- 1} e u$ , and isomorphism $e ≅ f$ . They are strictly decreasing in strength, and the last gap — isomorphic but not conjugate — is exactly the failure of Dedekind-finiteness in disguise.

Learning Objectives

State $(21.20)$ with all four equivalent conditions and their side conventions.
Prove $(21.20)$ by representing module maps $e R \to f R$ as left multiplications by elements of $f R e$ .
Deduce that $e ≅ f$ implies $e R e ≅ f R f$ as rings and $R e R = R f R$ as ideals.
Show that conjugate idempotents are isomorphic, and exhibit a ring where the converse fails.
Recognise the relation as Murray–von Neumann equivalence when $R$ is an operator algebra.
Compute the isomorphism classes of idempotents in $M_{n} (k)$ and in a triangular matrix ring.

Definitions

Definition(21.20)Isomorphic idempotents

Idempotents $e, f$ in a ring $R$ are isomorphic, written $e ≅ f$ , if $e R ≅ f R$ as right $R$ -modules. By the Proposition below this does not depend on the choice of side, and it is an equivalence relation on the set of idempotents of $R$ .

$e R$: The principal right ideal generated by $e$ ; equal to ${r \in R : e r = r}$ when $e$ is idempotent.
$e R f$: The additive group ${e r f : r \in R}$ , an $(e R e, f R f)$ -bimodule. It is where the homomorphisms $f R \to e R$ live.
$U (R)$: The group of units of $R$ .
Orthogonal: Idempotents $e, f$ with $e f = f e = 0$ ; then $e + f$ is again idempotent.
Primitive idempotent: A nonzero idempotent $e$ that is not the sum of two nonzero orthogonal idempotents; equivalently $e R$ is indecomposable, equivalently $e R e$ has no nontrivial idempotents.

All rings have an identity, all modules are unital, and homomorphisms of right modules are written on the left of their arguments, so that composition matches the ring multiplication in the corner rings.

Core Concepts

Homomorphisms are elements

The whole proof rests on one identification, established on the page Idempotents and Direct Decompositions of Modules: for an idempotent $e$ and any right $R$ -module $M$ ,

{Hom}_{R} (e R, M) ≅ M e, θ ⟼ θ (e),

(21.6)

an isomorphism of additive groups. A homomorphism out of $e R$ is determined by where it sends the generator $e$ , and the possible images are exactly the elements $m$ with $m e = m$ . Specialising $M = f R$ gives ${Hom}_{R} (e R, f R) ≅ f R e$ , and specialising $M = e R$ gives the ring isomorphism ${End}_{R} (e R) ≅ e R e$ .

Map

e R \to f R

Its value

θ (e)

Element of

f R e

So an isomorphism $e R \to f R$ is the same data as an element $b \in f R e$ that is invertible in the appropriate sense; the composite of $θ$ and $θ^{- 1}$ turns that vague phrase into the two equations $a b = e$ , $b a = f$ .

Why the relation is side-neutral

Condition (3) of $(21.20)$ — the existence of $a, b$ with $a b = e$ and $b a = f$ — mentions no module and no side. Passing to the opposite ring $R^{op}$ interchanges $a b$ and $b a$ , hence interchanges the roles of $e$ and $f$ while turning right modules into left modules. That single observation converts the right-handed statement into the left-handed one at no cost.

Isomorphism versus conjugacy

A unit $u$ conjugates $e$ to $u^{- 1} e u$ , and conjugation is an automorphism of $R$ , so it preserves everything in sight — including the complementary idempotent, which is carried to $u^{- 1} (1 - e) u$ . Isomorphism makes no promise about complements. The obstruction is precisely that: $e$ and $f$ are conjugate iff $e ≅ f$ and $1 - e ≅ 1 - f$ .

Key Results

Proposition(21.20)Isomorphism criterion for idempotents

Let $R$ be a ring with identity and let $e, f \in R$ be idempotents. The following statements are equivalent.

(1) $e R ≅ f R$ as right $R$ -modules.
**(1 $'$ )** $R e ≅ R f$ as left $R$ -modules.
(2) There exist $a \in e R f$ and $b \in f R e$ with $a b = e$ and $b a = f$ .
(3) There exist $a, b \in R$ with $a b = e$ and $b a = f$ .

When they hold, $e$ and $f$ are called isomorphic idempotents, written $e ≅ f$ .

Proof

**(1) $\Rightarrow$ (2).** Let $θ : e R \to f R$ be an isomorphism of right $R$ -modules with inverse $θ^{- 1}$ . Put $b = θ (e)$ and $a = θ^{- 1} (f)$ . Then $b \in f R$ , and $b = θ (e \cdot e) = θ (e) e = b e$ , so $b \in f R e$ ; symmetrically $a \in e R$ and $a = θ^{- 1} (f \cdot f) = θ^{- 1} (f) f = a f$ , so $a \in e R f$ . Since $e R$ is generated by $e$ and $θ$ is right $R$ -linear, $θ (e x) = b x$ for all $x \in R$ ; likewise $θ^{- 1} (f y) = a y$ . Now

e = θ^{- 1} (θ (e)) = θ^{- 1} (b) = θ^{- 1} (f b) = a b,

using $b = f b$ , and symmetrically $f = θ (θ^{- 1} (f)) = θ (a) = θ (e a) = b a$ .

**(2) $\Rightarrow$ (3)** is trivial.

**(3) $\Rightarrow$ (1).** Suppose $a b = e$ and $b a = f$ . Then

b e = b (a b) = (b a) b = f b \in f R, a f = a (b a) = (a b) a = e a \in e R .

Define $θ : e R \to f R$ by $θ (x) = b x$ and $θ' : f R \to e R$ by $θ' (y) = a y$ . These land where claimed: for $x = e r$ we get $b x = (b e) r \in f R$ , and for $y = f r$ we get $a y = (a f) r \in e R$ . Both maps are visibly right $R$ -linear. Finally, for all $r \in R$ ,

θ' θ (e r) = a b e r = e \cdot e r = e r, θ θ' (f r) = b a f r = f \cdot f r = f r,

using idempotency of $e$ and $f$ . Hence $θ$ is an isomorphism.

**(3) $\Leftrightarrow$ (1 $'$ ).** Given $a, b$ as in (3), define $ψ : R e \to R f$ by $ψ (x) = x a$ and $ψ' : R f \to R e$ by $ψ' (y) = y b$ ; the computations $e a = a f \in R f$ and $f b = b e \in R e$ show these are well defined, and $ψ' ψ (r e) = r e a b = r e$ , $ψ ψ' (r f) = r f b a = r f$ . Conversely, the argument for (1) $\Rightarrow$ (2) run in $R^{op}$ produces $a, b$ from an isomorphism $R e ≅ R f$ . $■$

Corollary—Corner rings are invariants

If $e ≅ f$ in $R$ , then $e R e ≅ f R f$ as rings, and $R e R = R f R$ as two-sided ideals.

Proof

Choose $a \in e R f$ , $b \in f R e$ with $a b = e$ , $b a = f$ . Define $Φ : e R e \to f R f$ by $Φ (x) = b x a$ ; it lands in $f R f$ because $b \in f R e$ and $a \in e R f$ . It is additive, and for $x, y \in e R e$ , using $x e = x$ ,

Φ (x) Φ (y) = (b x a) (b y a) = b x (a b) y a = b x e y a = b (x y) a = Φ (x y),

while $Φ (e) = b e a = b a = f$ is the identity of $f R f$ . The map $Ψ (y) = a y b$ is a two-sided inverse: $Ψ Φ (x) = a (b x a) b = (a b) x (a b) = e x e = x$ , and symmetrically. For the ideals, $f = b a = b (a b) a = b e a \in R e R$ , so $R f R \subseteq R e R$ , and the reverse inclusion follows by exchanging the roles of $e$ and $f$ . $■$

Corollary—Transport of idempotent properties

Let $e ≅ f$ be isomorphic idempotents of $R$ . Then $e$ is primitive iff $f$ is primitive; $e$ is a local idempotent (that is, $e R e$ is a local ring) iff $f$ is; and $e$ is right irreducible (that is, $e R$ is a minimal right ideal) iff $f$ is. The first two follow from $e R e ≅ f R f$ , the third from $e R ≅ f R$ .

Proposition—Conjugate implies isomorphic

Let $e$ be an idempotent of $R$ and $u \in U (R)$ . Then $u^{- 1} e u$ is an idempotent and $e ≅ u^{- 1} e u$ .

Proof

Set $f = u^{- 1} e u$ , so $f^{2} = u^{- 1} e u u^{- 1} e u = u^{- 1} e^{2} u = f$ . Take $a = e u$ and $b = u^{- 1} e$ . Then $a b = e u u^{- 1} e = e^{2} = e$ and $b a = u^{- 1} e \cdot e u = u^{- 1} e u = f$ . Condition (3) of $(21.20)$ holds. $■$

Remark—The commutative case is empty

If $R$ is commutative and $e ≅ f$ , then $f = b a = a b = e$ . So the relation collapses to equality, and the theory of isomorphic idempotents has content only for noncommutative $R$ .

Counterexample—Isomorphic but not conjugate

Let $R$ be any ring that is not Dedekind-finite: there are $a, b \in R$ with $a b = 1$ and $e : = b a \neq 1$ . Then $e^{2} = b (a b) a = b a = e$ , and $(21.20) (3)$ applied to this very pair gives $1 ≅ e$ . But $u^{- 1} \cdot 1 \cdot u = 1$ for every unit $u$ , so $1$ is conjugate only to itself; hence $1$ and $e$ are isomorphic and not conjugate. Concretely, take $V$ a vector space over a field $k$ with basis $v_{1}, v_{2}, \dots$ , put $R = {End}_{k} (V)$ , and let $b$ be the shift $v_{i} \mapsto v_{i + 1}$ and $a$ the map $v_{1} \mapsto 0$ , $v_{i + 1} \mapsto v_{i}$ . Then $a b = 1$ while $e = b a$ is the projection onto the span of $v_{2}, v_{3}, \dots$ .

Here $1 - e$ is a rank-one projection and $1 - 1 = 0$ , so indeed $1 - e$ and $1 - 1$ are not isomorphic: the missing hypothesis is exactly the one that separates conjugacy from isomorphism.

Proposition—Orthogonal isomorphic idempotents give matrix units

Let $e_{1}, \dots, e_{r}$ be pairwise orthogonal idempotents of $R$ that are pairwise isomorphic, and put $e = e_{1} + \dots + e_{r}$ . Then $e R e ≅ M_{r} (e_{1} R e_{1})$ as rings.

Proof

For each $i$ choose $a_{i} \in e_{1} R e_{i}$ and $b_{i} \in e_{i} R e_{1}$ with $a_{i} b_{i} = e_{1}$ and $b_{i} a_{i} = e_{i}$ , possible by $(21.20)$ after the normalisation above; take $a_{1} = b_{1} = e_{1}$ . Put $E_{i j} = b_{i} a_{j} \in e_{i} R e_{j}$ . Because $a_{j} \in e_{1} R e_{j}$ and $b_{k} \in e_{k} R e_{1}$ , the product $a_{j} b_{k}$ contains the factor $e_{j} e_{k}$ , which vanishes unless $j = k$ ; hence $a_{j} b_{k} = δ_{j k} e_{1}$ and

E_{i j} E_{k ℓ} = b_{i} (a_{j} b_{k}) a_{ℓ} = δ_{j k} b_{i} e_{1} a_{ℓ} = δ_{j k} E_{i ℓ} .

Also $\sum_{i} E_{i i} = \sum_{i} b_{i} a_{i} = \sum_{i} e_{i} = e$ , the identity of $e R e$ . So ${E_{i j}}$ is a full set of $r \times r$ matrix units in $e R e$ , and the standard recognition theorem gives $e R e ≅ M_{r} (S)$ with $S = E_{11} (e R e) E_{11} = e_{1} R e_{1}$ . $■$

Proof Techniques and Method

How these proofs work, and which move to reuse.

Three moves carry every argument on this page.

Move 1

Replace maps by elements

${Hom}_{R} (e R, f R) ≅ f R e$ turns a question about modules into a question about products in $R$ . Composition of maps becomes multiplication, in the order dictated by writing maps on the left.

Move 2

Normalise into corners

Given any witnesses $a b = e$ , $b a = f$ , replace them by $e a f$ and $f b e$ . Idempotency absorbs the extra factors, so nothing is lost and everything now sits in the right bimodule.

Move 3

Read the condition in $R^{op}$

A criterion phrased only in products is automatically two-sided. Whenever a definition can be reduced to such a form, the left–right symmetry theorem is free.

Move 2 is the reason condition (2) can be stated at all. It is also what makes the matrix-unit construction work: without normalisation the products $a_{j} b_{k}$ would not visibly vanish for $j \neq k$ .

Worked Example

Matrix units in $M_{2} (k)$

Let $k$ be a field and $R = M_{2} (k)$ , with matrix units $E_{11}, E_{12}, E_{21}, E_{22}$ . Put $e = E_{11}$ , $f = E_{22}$ , $a = E_{12}$ , $b = E_{21}$ . Then $a \in e R f$ , $b \in f R e$ , and

a b = E_{12} E_{21} = E_{11} = e, b a = E_{21} E_{12} = E_{22} = f,

(E.1)

so $e ≅ f$ . Concretely $e R$ is the set of matrices with second row zero and $f R$ the set with first row zero; both are two-dimensional over $k$ and isomorphic to the unique simple right $R$ -module. The corner rings are $e R e = k E_{11}$ and $f R f = k E_{22}$ , both isomorphic to $k$ , as the Corollary predicts. Here $e$ and $f$ are also conjugate, by the permutation matrix $u = E_{12} + E_{21}$ .

Equal principal ideals, unequal idempotents

Still in $M_{2} (k)$ , set $e' = e + E_{12} = (\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix})$ . One checks $(e')^{2} = e'$ , and $e' = e + e E_{12} (1 - e)$ , so $e' R = e R$ — the two idempotents generate the same right ideal. Setting $n = E_{12}$ , we have $n^{2} = 0$ , $u = 1 + n$ is a unit with $u^{- 1} = 1 - n$ , and

u^{- 1} e u = (1 - n) e (1 + n) = e + e n - n e = e + n = e',

(E.2)

Using $n e = E_{12} E_{11} = 0$ and $e n = E_{11} E_{12} = E_{12} = n$ .

So equality of principal right ideals is strictly stronger than conjugacy, which is strictly stronger than isomorphism — and in $M_{2} (k)$ the last two happen to coincide, because idempotent matrices of equal rank are similar.

Certifying non-isomorphism in a triangular ring

Let $R = {(\begin{matrix} a & b \\ 0 & c \end{matrix}) : a, b, c \in k}$ and take $e = E_{11}$ , $f = E_{22}$ , which are idempotents of $R$ . Then $e R f = k E_{12}$ is nonzero, but

f R e = E_{22} R E_{11} = (0),

(E.3)

because every element of $R$ has zero in the $(2, 1)$ position. Condition (2) of $(21.20)$ cannot be met, so $e R$ and $f R$ are not isomorphic. This is confirmed by dimension: $\dim_{k} e R = 2$ while $\dim_{k} f R = 1$ .

Process and Workflow

Compute the two cornersForm

e R f

and

f R e

. If either is zero,

e R

and

f R

cannot be isomorphic and you are done.

Look for a factorisationSearch for

a \in e R f

and

b \in f R e

with

a b = e

. This is a linear-algebra problem once

R

is finite-dimensional over a field.

Check the second equationVerify

b a = f

. It does not follow from

a b = e

; without it you have only a split epimorphism.

Record the consequencesTransport primitivity, locality, and the structure of the corner ring across the isomorphism.

Decide about conjugacy separatelyIf you need

f = u^{- 1} e u

, also test whether

1 - e ≅ 1 - f

You have $a \in e R f$ , $b \in f R e$ with $a b = e$ . What can you conclude?

Also $b a = f$

e ≅ f

: the maps

x \mapsto b x

and

y \mapsto a y

are mutually inverse.

$b a \neq f$

g : = b a

is an idempotent with

g \leq f

in the usual partial order, and

e ≅ g

. So

e R

embeds as a direct summand of

f R

Nothing else is knownOnly that

f R ↠ e R

splits. In a Dedekind-finite ring with suitable finiteness this can still be upgraded; in general it cannot.

Comparison and Classification

Three relations on idempotents, from strongest to weakest
Relation	Definition	Equivalent form	What it forces
$e R = f R$	Equality of principal right ideals	$e f = f$ and $f e = e$ ; also $R (1 - e) = R (1 - f)$	$f = e + e r (1 - e)$ for some $r$ , hence conjugacy by $1 + e r (1 - e)$
Conjugacy	$f = u^{- 1} e u$ for some $u \in U (R)$	$e ≅ f$ and $1 - e ≅ 1 - f$	Every ring-theoretic property of $e$ transfers, complements included
Isomorphism $e ≅ f$	$e R ≅ f R$ as right modules	$a b = e$ , $b a = f$ for some $a, b \in R$	$R e ≅ R f$ , $e R e ≅ f R f$ , $R e R = R f R$

How the relation behaves in specific rings
	$e ≅ f$ forces $e = f$	$1 ≅ e$ with $e \neq 1$ occurs	Numerical invariant classifies
$R$ commutative	yes	no	trivial
$M_{n} (k)$ , $k$ a field	no	no	yes: rank
$R$ Dedekind-finite	no	no	no
${End}_{k} (V)$ , $\dim_{k} V$ infinite	no	yes	no

How the relation behaves in specific rings

The third column is a warning: over $M_{n} (k)$ the rank makes the classification finite and easy, and that convenience does not survive to general rings, where $e R ≅ f R$ is a genuine module isomorphism problem.

Relationship Map

The implications below are always valid. None of them reverses in a general ring.

e = f

e R = f R

f = u^{- 1} e u

e ≅ f

$e ≅ f$ — isomorphism of idempotents
- implies
  - $R e ≅ R f$ (left version)
  - $e R e ≅ f R f$ as rings
  - $R e R = R f R$ as ideals
  - $e$ primitive $\Leftrightarrow$ $f$ primitive
  - $e$ local $\Leftrightarrow$ $f$ local
- does not imply
  - $e = f$
  - $e$ and $f$ conjugate
  - $1 - e ≅ 1 - f$
  - $e f = f e$
- is implied by
  - $e R = f R$
  - conjugacy by a unit
  - $e$ and $f$ orthogonal with a common matrix-unit family

All rings

e ≅ f

defined; side-neutral; implies

e R e ≅ f R f

Dedekind-finite

1 ≅ e

forces

e = 1

Semiperfectisomorphism classes of primitive idempotents index the simple modules

Semisimple

e ≅ f

iff

e R

and

f R

have the same composition factors

Applications and Industry Use

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

Operator algebras

Murray–von Neumann equivalence

In a $C^{*}$ -algebra or von Neumann algebra, projections $p, q$ are equivalent when $p = v^{*} v$ and $q = v v^{*}$ for a partial isometry $v$ . That is exactly condition (3) with $a = v^{*}$ , $b = v$ . The comparison theory of projections, and hence the type classification of factors, is built on this relation.

K-theory

Building $K_{0}$

$K_{0} (R)$ is assembled from idempotents in the matrix rings $M_{n} (R)$ modulo isomorphism and stabilisation. The criterion of $(21.20)$ is the equivalence relation at the bottom of that construction.

Representation theory

Counting the simples

In a decomposition $1 = e_{1} + \dots + e_{n}$ into orthogonal primitive idempotents of a semiperfect ring, the isomorphism classes among the $e_{i}$ correspond to the isomorphism classes of simple modules. Discarding duplicates produces the basic ring.

Computer algebra

Deduplicating idempotents

Wedderburn decomposition routines in GAP, Magma and Sage produce a list of primitive idempotents; the list must then be reduced modulo isomorphism to obtain the distinct simple modules and the block structure.

The honest summary is that this criterion is infrastructure. It is rarely the endpoint of a calculation; it is what lets a calculation about modules be carried out with elements, which is what a machine — and a proof — can actually manipulate.

Standards and Notation

Standards here covers notation, symbol and markup standards, and reference implementations, rather than material or design codes.

Lam's notation

e ≅ f

for isomorphic idempotents, matching the module isomorphism

e R ≅ f R

Common variant

e \sim f

, standard in operator algebras (Murray–von Neumann) and in much of the

K

-theory literature.

Conflict to watchSome ring-theory texts use

e \sim f

for conjugate. Always say which relation you mean on first use.

Corner notation

e R e

is universal;

e R f

for the bimodule is standard, and matches the Peirce block notation.

Matrix units

E_{i j}

e_{i j}

, with

e_{i j} e_{k ℓ} = δ_{j k} e_{i ℓ}

Computational Notes

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

Let $A$ be a finite-dimensional algebra over a field $k$ , given by structure constants with $\dim_{k} A = n$ , and let $e, f \in A$ be idempotents.

$e A f$ and $f A e$ are computed as images of linear maps; each costs $O (n^{3})$ field operations.
Deciding $e A ≅ f A$ is a module isomorphism test. Over a finite field the Meataxe supplies a practical algorithm; for modules given by matrices the standard isomorphism test runs in time polynomial in the dimension and the field size.
Searching directly for $a \in e A f$ , $b \in f A e$ with $a b = e$ is a system of quadratic equations in $\dim_{k} (e A f) + \dim_{k} (f A e)$ unknowns and is not the recommended route.
A cheap necessary condition: $\dim_{k} e A = \dim_{k} f A$ . A cheaper sufficient obstruction: $e A f = 0$ or $f A e = 0$ .
Over a semisimple $A$ , comparing composition-factor multiplicities of $e A$ and $f A$ decides the question outright.

Failure Modes and Common Mistakes

Do not read $e ≅ f$ as an equality of subsets. The right ideals $e R$ and $f R$ are usually different subsets of $R$ that happen to be abstractly isomorphic.
Do not assume isomorphic idempotents commute, or that their sum is idempotent. Orthogonality is an extra hypothesis, and it is what the matrix-unit construction needs.
Do not forget that ${Hom}_{R} (e R, f R) ≅ f R e$ has the letters in that order; getting it backwards inverts every composition in the proof.
Isomorphic corner rings do not detect isomorphic idempotents — the triangular example has $e R e ≅ f R f ≅ k$ with $e R$ and $f R$ of different dimensions.

Quick Reference

Criterion

e ≅ f

iff

\exists a, b \in R

with

a b = e

b a = f

Normalised formMay take

a \in e R f

b \in f R e

; replace

a, b

e a f, f b e

Hom formula

{Hom}_{R} (e R, f R) ≅ f R e

;

{End}_{R} (e R) ≅ e R e

Consequences

R e ≅ R f

e R e ≅ f R f

R e R = R f R

Strictly stronger

e R = f R

\Rightarrow

conjugate

\Rightarrow

isomorphic.

Commutative rings

e ≅ f

iff

e = f

Tests at a glance
Question	Answer	Reference
Is $e R ≅ f R$ ?	Find $a \in e R f$ , $b \in f R e$ with $a b = e$ , $b a = f$	(21.20)
Is the relation side-neutral?	Yes — condition (3) survives passage to $R^{op}$	(21.20)(1 $'$ )
Does $e ≅ f$ give $e R e ≅ f R f$ ?	Yes, via $x \mapsto b x a$	Corollary above
Does $e ≅ f$ give conjugacy?	No — need $1 - e ≅ 1 - f$ as well	Counterexample above
When is $1 ≅ e$ with $e \neq 1$ ?	Exactly when $R$ is not Dedekind-finite	(21.26)

Frequently Asked Questions

Why is the relation written

e ≅ f

rather than

e = f

up to something?

Because it is literally an isomorphism of modules: $e R ≅ f R$ . The elements $e$ and $f$ are usually different, and the right ideals they generate are usually different subsets of $R$ . What agrees is the abstract module structure, and the notation is chosen to keep that in view.

a b = e

, why does

b a

have to be checked separately?

Because $b a$ is always idempotent when $e$ is — $(b a)^{2} = b (a b) a = b e a = b a$ if $a = e a$ and $b = b e$ — but it need not equal the $f$ you started with. In a ring that is not Dedekind-finite, $a b = 1$ with $b a$ a proper idempotent is exactly this phenomenon. The single equation gives a split surjection $f R ↠ e R$ , not an isomorphism.

Does

e ≅ f

say anything about

1 - e

and

1 - f

Nothing. That is precisely the gap between isomorphism and conjugacy: $e$ and $f$ are conjugate by a unit if and only if $e ≅ f$ and $1 - e ≅ 1 - f$ . In ${End}_{k} (V)$ with $V$ infinite-dimensional, $1$ is isomorphic to a projection $e$ whose complement has rank one, while the complement of $1$ is zero.

Is isomorphism of idempotents an equivalence relation?

Yes. Reflexivity uses $a = b = e$ ; symmetry swaps $a$ and $b$ ; transitivity composes the module isomorphisms, or at the element level, if $a_{1} b_{1} = e$ , $b_{1} a_{1} = f$ , $a_{2} b_{2} = f$ , $b_{2} a_{2} = g$ , then $(a_{1} a_{2}) (b_{2} b_{1}) = a_{1} f b_{1} = a_{1} b_{1} a_{1} b_{1} = e$ and symmetrically $(b_{2} b_{1}) (a_{1} a_{2}) = g$ .

Why does the criterion make the theory left-right symmetric when so much of ring theory is not?

Because condition (3) is a statement about products only. Left and right enter a ring-theoretic definition through the choice of module side; a condition expressed purely by equations between products is invariant under $R \mapsto R^{op}$ , so the theorem transfers automatically. Contrast right irreducibility of an idempotent, which genuinely depends on the side.

How does this relate to Morita theory?

The isomorphism ${Hom}_{R} (e R, f R) ≅ f R e$ and its consequence ${End}_{R} (e R) ≅ e R e$ are the first steps of the Morita correspondence. When $e$ is a full idempotent, meaning $R e R = R$ , the functor $M \mapsto M e$ is an equivalence between right $R$ -modules and right $e R e$ -modules, and the invariance of $R e R$ under $e ≅ f$ shows fullness is a property of the isomorphism class.

Can two isomorphic idempotents be orthogonal?

Yes, and that is the useful case. In $M_{n} (k)$ the diagonal matrix units $E_{11}, \dots, E_{n n}$ are pairwise orthogonal and pairwise isomorphic; a family of $r$ pairwise orthogonal pairwise isomorphic idempotents summing to $e$ makes $e R e$ an $r \times r$ matrix ring over $e_{1} R e_{1}$ .

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §21, especially (21.6), (21.7) and (21.20) (pp. 319–326).
T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics 189, Springer-Verlag, 1999, §18 (Morita theory) for the role of full idempotents.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §§7 and 21.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapter III.
K. R. Goodearl, von Neumann Regular Rings, 2nd edition, Krieger, 1991, Chapter 1, for comparison of idempotents and the equivalence relation in the regular setting.

AI Suggested Questions

Prove that $e$ and $f$ are conjugate in $R$ if and only if $e ≅ f$ and $1 - e ≅ 1 - f$ .
Give an example of a ring in which two isomorphic idempotents have non-isomorphic complements but the ring is still Dedekind-finite, or prove none exists.
How does the isomorphism relation on idempotents descend to $R / I$ when $I \subseteq rad R$ ?
Work out the isomorphism classes of idempotents in the ring of upper triangular $3 \times 3$ matrices over a field.
Explain the precise relationship between $(21.20)$ and Murray–von Neumann equivalence of projections in a von Neumann algebra.
Show that a ring $R$ is Dedekind-finite if and only if $1$ is isomorphic to no idempotent other than $1$ .
Describe how $K_{0}$ of a ring is built from isomorphism classes of idempotents over matrix rings, and what stabilisation adds.
For which rings does $e ≅ f$ imply that $e$ and $f$ are conjugate?

Isomorphism of eR and fR

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Homomorphisms are elements

Why the relation is side-neutral

Isomorphism versus conjugacy

Key Results

Proof Techniques and Method

Replace maps by elements

Normalise into corners

Read the condition in $R^{op}$

Worked Example

Matrix units in $M_{2} (k)$

Equal principal ideals, unequal idempotents

Certifying non-isomorphism in a triangular ring

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Murray–von Neumann equivalence

Building $K_{0}$

Counting the simples

Deduplicating idempotents

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Homomorphisms are elements

Why the relation is side-neutral

Isomorphism versus conjugacy

06Key Results

07Proof Techniques and Method

Replace maps by elements

Normalise into corners

Read the condition in Rop

08Worked Example

Matrix units in M2(k)

Equal principal ideals, unequal idempotents

Certifying non-isomorphism in a triangular ring

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

Murray–von Neumann equivalence

Building K0

Counting the simples

Deduplicating idempotents

13Standards and Notation

14Computational Notes

15Failure Modes and Common Mistakes

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

Read the condition in $R^{op}$

Worked Example

Matrix units in $M_{2} (k)$

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Building $K_{0}$

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions