Corner Rings | KEVOS® Mathematics

Q: Does (21.10) need any hypothesis on R or e?

None at all. R is any ring with identity and e any idempotent. The proof is two explicit inverse manipulations, so there is no chain condition, no semiperfectness and no fullness in play. This is unusual: most comparisons of radicals across subrings require substantial hypotheses.

Q: Is rad(eRe) an ideal of R?

No. It is an ideal of eRe and a subset of radR, but e(radR)e is not closed under multiplication by arbitrary elements of R unless e is central. The correct statement is the one in (21.10): it is the contraction of radR, not a sub-ideal in the ambient sense.

Q: Why does fullness make the ideal correspondence surjective?

Because the identity R(eBe)R = (ReR)B(ReR) collapses to RBR = B exactly when ReR = R. Conversely, if the map is onto then R itself is in the image, and its preimage must be eRe, forcing R(eRe)R = R, i.e. ReR = R. So fullness is equivalent to surjectivity.

Q: What is the relationship to Morita theory?

eR is always a finitely generated projective right R-module with End_R(eR) ≅ eRe and trace ideal ReR. Fullness therefore says eR is a progenerator, which is precisely the hypothesis under which the Morita theorems give an equivalence between the module categories of R and eRe. The ideal correspondence in (21.11)(2) is what that equivalence does to two-sided ideals.

Executive Summary

The corner ring $e R e$ is the natural home of everything happening at an idempotent: it is the endomorphism ring of $e R$ , and it is where questions about how $e R$ decomposes get answered. This page supplies the two facts that make it usable — how its radical relates to $rad R$ , and how its ideals relate to those of $R$ .

Both answers are as clean as one could hope. The radical is $rad (e R e) = e (rad R) e$ , with no hypotheses at all, so taking corners commutes with killing the radical. Ideals of $e R e$ inject into ideals of $R$ in a way that respects products, and the injection becomes a bijection exactly in the presence of the fullness condition $R e R = R$ — the same condition that makes $R$ and $e R e$ Morita equivalent.

$e (rad R) e$

rad (e R e)

NoneHypotheses needed

$R e R = R$ Fullness

7Properties inherited by

e R e

Overview

Let $R$ be a ring with identity and $e = e^{2} \in R$ . By the Peirce decomposition, $e R e = {r \in R : e r = r = r e}$ is closed under addition and multiplication and has $e$ as its identity element. It is not a unital subring of $R$ , and there is no ring homomorphism $R \to e R e$ ; nevertheless almost every structural invariant of $R$ has a well-behaved image in $e R e$ .

rad (e R e) = (rad R) \cap e R e = e (rad R) e, \frac{e R e}{rad (e R e)} ≅ \bar{e} \bar{R} \bar{e},

(21.10)

Here $\bar{R} = R / rad R$ and $\bar{e}$ is the image of $e$ . Corners commute with passage to the semiprimitive quotient.

The second half of the section compares ideal lattices. Extension $𝔄 \mapsto R 𝔄$ carries left ideals of $e R e$ into left ideals of $R$ , and contraction $𝔅 \mapsto 𝔅 \cap e R e$ undoes it. For two-sided ideals the extension is $𝔄 \mapsto R 𝔄 R$ , the contraction is $𝔅 \mapsto e 𝔅 e$ , and the composite is again the identity. Fullness makes the correspondence onto, and then $R$ and $e R e$ are indistinguishable at the level of ideals — a shadow of the Morita equivalence developed for module categories.

Learning Objectives

Prove the three inclusions that establish $rad (e R e) = (rad R) \cap e R e = e (rad R) e$ .
Deduce $e R e / rad (e R e) ≅ \bar{e} \bar{R} \bar{e}$ and explain why the map is well defined.
Prove $(R 𝔄) \cap e R e = 𝔄$ for a left ideal $𝔄$ of $e R e$ .
Prove $e (R 𝔄 R) e = 𝔄$ for an ideal $𝔄$ of $e R e$ , and that extension respects products.
Show the ideal correspondence is onto when $e$ is full, and locate $rad R$ under it.
Prove that semiprimitivity, semisimplicity, simplicity, primeness, semiprimeness and the chain conditions pass from $R$ to $e R e$ .

Definitions

Definition(21.11)(2)Full idempotent

An idempotent $e$ of $R$ is full if $R e R = R$ , i.e. the two-sided ideal generated by $e$ is all of $R$ . Equivalently, $1$ is a finite sum $\sum_{i} a_{i} e b_{i}$ with $a_{i}, b_{i} \in R$ . Every idempotent of a simple ring other than $0$ is full; $e_{11}$ is full in $M_{n} (k)$ ; $e_{11}$ is not full in the ring of upper triangular $2 \times 2$ matrices.

$e R e$: The corner ring at $e$ : a ring with identity $e$ , equal to ${r \in R : e r = r = r e}$ .
$R 𝔄$: For $𝔄$ a left ideal of $e R e$ , the set of finite sums $\sum r_{i} a_{i}$ with $r_{i} \in R$ , $a_{i} \in 𝔄$ — a left ideal of $R$ .
$R 𝔄 R$: For $𝔄$ an ideal of $e R e$ , the two-sided ideal of $R$ it generates.
$e 𝔅 e$: The contraction to $e R e$ of an ideal $𝔅$ of $R$ ; it is an ideal of $e R e$ , and equals $𝔅 \cap e R e$ .
Semiprimary ring: A ring $R$ with $rad R$ nilpotent and $R / rad R$ semisimple.

Throughout e is a fixed idempotent of R, and J denotes rad R. All statements below hold for arbitrary rings with identity unless a hypothesis is stated explicitly.

Core Concepts

Why the radical restricts so cleanly

The Jacobson radical has an element-wise characterisation: $y \in rad R$ if and only if $1 - x y$ is left-invertible for every $x \in R$ . The corner has its own version of that test with $1$ replaced by $e$ and $x$ ranging over $e R e$ . Comparing them is a matter of manufacturing an inverse in one ring from an inverse in the other, and both directions succeed because $e$ acts as a two-sided identity on everything in sight.

That is genuinely special. There is no analogous statement for, say, the prime radical restricted to an arbitrary subring, and none for the radical of a quotient. What makes the corner work is that $e R e$ is not a subring in the ordinary sense but the image of the idempotent projection $r \mapsto e r e$ , which is surjective and unital onto $e R e$ .

Extension and contraction of ideals

Two operations move ideals between $R$ and $e R e$ . Contraction sends $𝔅 \subseteq R$ to $e 𝔅 e$ ; extension sends $𝔄 \subseteq e R e$ to $R 𝔄 R$ . The composite contraction-after-extension is the identity on ideals of $e R e$ , so extension is injective and contraction is onto. In the other order the composite is generally not the identity: $R (e 𝔅 e) R \subseteq 𝔅$ can be strict, and equality for all $𝔅$ is exactly the fullness of $e$ .

𝔄 ⟼ R 𝔄 R, 𝔅 ⟼ e 𝔅 e, e (R 𝔄 R) e = 𝔄, R (e 𝔅 e) R \subseteq 𝔅 .

(21.11)

The Galois-style pair between the ideal lattices. The last inclusion is an equality for all $𝔅$ precisely when $R e R = R$ .

Fullness and Morita equivalence

The module $e R$ is always finitely generated projective, and $R e R$ is precisely its trace ideal. Fullness therefore says $e R$ is a progenerator, which is the hypothesis of the Morita theorems: $R$ and $e R e ≅ {End}_{R} (e R)$ then have equivalent module categories, and every Morita-invariant property transfers in both directions. The ideal correspondence of $(21.11)$ is the ideal-lattice shadow of that equivalence.

Key Results

Theorem(21.10)The radical of a corner ring

Let $R$ be a ring with identity, $e \in R$ an idempotent and $J = rad R$ . Then

rad (e R e) = J \cap (e R e) = e J e .

Moreover, writing $\bar{R} = R / J$ and $\bar{e}$ for the image of $e$ , there is a ring isomorphism $e R e / rad (e R e) ≅ \bar{e} \bar{R} \bar{e}$ .

Proof

It suffices to prove three inclusions: $rad (e R e) \subseteq J$ , then $J \cap e R e \subseteq e J e$ , then $e J e \subseteq rad (e R e)$ . Since $e J e \subseteq J \cap e R e$ trivially, the three together give equality throughout.

**Step 1: $rad (e R e) \subseteq J$ .** Let $r \in rad (e R e)$ ; note $e r = r = r e$ . Fix $y \in R$ ; we show $1 - y r$ is left-invertible in $R$ , which by the element test for the radical gives $r \in J$ . Since $e y e \in e R e$ and $rad (e R e)$ is an ideal of $e R e$ , the element $(e y e) r$ lies in $rad (e R e)$ , so $e - (e y e) r$ is left-invertible in $e R e$ : there is $b \in e R e$ with $b (e - (e y e) r) = e$ . Using $e r = r$ we have $(e y e) r = e y r$ , and using $b e = b$ this reads $b - b y r = e$ , that is $b (1 - y r) = e$ . Multiply on the left by $y r$ :

y r b (1 - y r) = y r e = y r,

since $r e = r$ . Adding $1 - y r$ to both sides gives $(1 + y r b) (1 - y r) = 1$ . So $1 - y r$ is left-invertible for every $y \in R$ , whence $r \in J$ .

**Step 2: $J \cap e R e \subseteq e J e$ .** If $r \in J$ and $r \in e R e$ then $r = e r e \in e J e$ because $r \in J$ .

**Step 3: $e J e \subseteq rad (e R e)$ .** Let $r \in e J e$ , so $r \in J$ and $e r = r = r e$ . Fix $y \in e R e$ ; we show $e - y r$ is left-invertible in $e R e$ . Since $y r \in J$ , there is $x \in R$ with $x (1 - y r) = 1$ . Multiplying on the left by $e$ and on the right by $e$ , and using $(1 - y r) e = e - y r e = e - y r$ together with $e - y r \in e R e$ :

e = e x (1 - y r) e = e x (e - y r) = (e x e) (e - y r),

so $e x e \in e R e$ is a left inverse of $e - y r$ in $e R e$ . Hence $r \in rad (e R e)$ .

The quotient. The assignment $e r e \mapsto \bar{e} \bar{r} \bar{e}$ is a well-defined surjective ring homomorphism $e R e \to \bar{e} \bar{R} \bar{e}$ (it is the restriction of $R \to \bar{R}$ ). Its kernel consists of the $e r e$ with $\bar{e} \bar{r} \bar{e} = 0$ , i.e. $e r e \in J \cap e R e = e J e$ . The first part identifies $e J e$ with $rad (e R e)$ , giving the stated isomorphism.

Theorem(21.11)Ideals of a corner ring

Let $e$ be an idempotent of a ring $R$ .

If $𝔄$ is a left ideal of $e R e$ , then $(R 𝔄) \cap e R e = 𝔄$ . Hence $𝔄 \mapsto R 𝔄$ is an injective, inclusion-preserving map from the left ideals of $e R e$ to the left ideals of $R$ .
If $𝔄$ is a two-sided ideal of $e R e$ , then $e (R 𝔄 R) e = 𝔄$ . Hence $𝔄 \mapsto R 𝔄 R$ is an injective, inclusion-preserving map from the ideals of $e R e$ to the ideals of $R$ ; it satisfies $(R 𝔄 R) (R 𝔄' R) = R (𝔄 𝔄') R$ ; and it is surjective when $e$ is full.

Proof

(1). Put $𝔄_{0} = (R 𝔄) \cap e R e$ , which certainly contains $𝔄$ . Every element of $𝔄_{0}$ is fixed by left multiplication by $e$ , so $𝔄_{0} = e 𝔄_{0} \subseteq e (R 𝔄)$ . Since $𝔄 \subseteq e R e$ we have $𝔄 = e 𝔄$ , hence $e (R 𝔄) = e R (e 𝔄) = (e R e) 𝔄 \subseteq 𝔄$ , the last step because $𝔄$ is a left ideal of $e R e$ . Therefore $𝔄_{0} \subseteq 𝔄$ and the two are equal. Injectivity follows: $R 𝔄 = R 𝔄'$ implies $𝔄 = (R 𝔄) \cap e R e = (R 𝔄') \cap e R e = 𝔄'$ .

(2), the retraction. Using $𝔄 = e 𝔄 e$ and the fact that $𝔄$ is an ideal of $e R e$ ,

e (R 𝔄 R) e = e R (e 𝔄 e) R e = (e R e) 𝔄 (e R e) \subseteq 𝔄,

and the reverse inclusion holds because $𝔄 = e 𝔄 e \subseteq e (R 𝔄 R) e$ . Injectivity follows as before.

(2), multiplicativity. For ideals $𝔄, 𝔄'$ of $e R e$ , using $𝔄 = 𝔄 e$ and $𝔄' = e 𝔄'$ ,

(R 𝔄 R) (R 𝔄' R) = R 𝔄 R 𝔄' R = R 𝔄 (e R e) 𝔄' R = R (𝔄 𝔄') R,

where $𝔄 (e R e) 𝔄' = 𝔄 𝔄'$ because $𝔄 (e R e) \subseteq 𝔄$ and $e \in e R e$ .

**(2), surjectivity for full $e$ .** Assume $R e R = R$ and let $𝔅$ be any ideal of $R$ . Take $𝔄 = e 𝔅 e$ , an ideal of $e R e$ . Since $R 𝔅 R = 𝔅$ ,

R (e 𝔅 e) R = R e (R 𝔅 R) e R = (R e R) 𝔅 (R e R) = R 𝔅 R = 𝔅,

so $𝔅$ is in the image.

Remark(21.12)Where the radical sits under the correspondence

If $e$ is full, the bijection of $(21.11) (2)$ matches $rad (e R e)$ with $rad R$ : contraction sends $rad R$ to $e (rad R) e = rad (e R e)$ by $(21.10)$ , and contraction is inverse to extension. In particular $R$ is semiprimitive if and only if $e R e$ is, for a full idempotent $e$ .

Corollary(21.13)Properties inherited by the corner

Let $e \neq 0$ be an idempotent of $R$ . If $R$ is semiprimitive (Jacobson semisimple), semisimple, simple, prime, semiprime, left noetherian, or left artinian, then $e R e$ has the same property.

Proof

Semiprimitive. By $(21.10)$ , $rad (e R e) = e (rad R) e = 0$ .

Chain conditions. By $(21.11) (1)$ , extension is an injective inclusion-preserving map from left ideals of $e R e$ to left ideals of $R$ with a retraction, so it is a strictly monotone embedding of posets. An infinite strictly ascending (resp. descending) chain in $e R e$ would produce one in $R$ ; hence $R$ left noetherian (resp. left artinian) forces the same for $e R e$ .

Simple. Let $0 \neq 𝔄$ be an ideal of $e R e$ . Then $R 𝔄 R$ is a nonzero ideal of $R$ , so $R 𝔄 R = R$ , and contracting gives $𝔄 = e (R 𝔄 R) e = e R e$ . As $e \neq 0$ , $e R e \neq 0$ .

Prime and semiprime. If $𝔄 𝔄' = 0$ for ideals of $e R e$ , then $(R 𝔄 R) (R 𝔄' R) = R (𝔄 𝔄') R = 0$ by the multiplicativity in $(21.11) (2)$ ; primeness of $R$ makes one factor zero, and injectivity of extension makes the corresponding $𝔄$ or $𝔄'$ zero. Taking $𝔄' = 𝔄$ gives the semiprime case.

Semisimple. A ring is semisimple exactly when it is left artinian with zero radical. Both properties have just been transferred, so $e R e$ is semisimple.

Remark(21.13a)The transfer is one-way

None of the implications in $(21.13)$ reverses without extra hypotheses. In the ring $R$ of upper triangular $2 \times 2$ matrices over a field, $e = e_{11}$ gives $e R e ≅ k$ , which is simple, semisimple and semiprimitive, while $R$ is none of those. Fullness is exactly what repairs this: for full $e$ the two rings are Morita equivalent and every property in the list transfers both ways.

Proof Techniques and Method

How these proofs work, and which move to reuse.

Move 1

Transport an inverse across the corner

Given $x (1 - y r) = 1$ in $R$ , sandwich by $e$ : $e = e x (1 - y r) e = (e x e) (e - y r)$ . Given $b (e - y r) = e$ in $e R e$ , expand to $(1 + y r b) (1 - y r) = 1$ . These two computations are the whole of $(21.10)$ .

Move 2

Absorb the idempotent

Every ideal-theoretic identity in $(21.11)$ is proved by rewriting $𝔄$ as $e 𝔄 e$ and letting the surrounding $e R$ and $R e$ collapse into $e R e$ , which the ideal absorbs.

Move 3

Retraction beats bijection

To prove injectivity of an extension map you rarely need a bijection — exhibiting a one-sided retraction suffices, and it also transfers chain conditions.

Move 1 is worth internalising because it is the standard way of comparing radicals across a non-unital inclusion. The asymmetry in it is instructive: going down into the corner you sandwich, going up out of it you expand a geometric-series-like identity. Neither direction is formal; both use $e r = r = r e$ at a precise point.

Note also what is not used. No chain condition appears anywhere in $(21.10)$ or $(21.11)$ ; the only finiteness in sight is the finiteness of the sums defining $R 𝔄 R$ .

Worked Example

A full idempotent: a matrix unit in a matrix ring

Let $k$ be any ring, $R = M_{n} (k)$ and $e = e_{11}$ . For $r = (r_{i j})$ one computes $e_{11} r e_{11} = r_{11} e_{11}$ , so $e R e = k e_{11} ≅ k$ . The idempotent is full: $e_{i 1} e_{11} e_{1 j} = e_{i j}$ , so $R e_{11} R$ contains every matrix unit and hence equals $R$ .

Now read off both theorems. From $(21.10)$ , $rad (k) ≅ e rad (M_{n} (k)) e$ , which is consistent with — and, given the correspondence, essentially equivalent to — the standard identity below.

rad (M_{n} (k)) = M_{n} (rad k) .

(E.1)

Checked against $(21.10)$ : contracting the right-hand side to the corner at $e_{11}$ returns $rad (k)$ .

From $(21.11) (2)$ with $e$ full, the ideals of $k$ correspond bijectively and multiplicatively to the ideals of $M_{n} (k)$ , by $𝔞 \mapsto M_{n} (𝔞)$ — the classical correspondence. In particular $M_{n} (k)$ is simple exactly when $k$ is.

A non-full idempotent: upper triangular matrices

Let $k$ be a field and $R = {(\begin{matrix} a & b \\ 0 & c \end{matrix})}$ , with $e = e_{11}$ . Then $e R e = k e_{11} ≅ k$ and $rad R = k e_{12}$ , so

e (rad R) e = e_{11} (k e_{12}) e_{11} = 0 = rad (k) = rad (e R e),

(E.2)

Theorem $(21.10)$ verified directly: the corner is semiprimitive even though $R$ is not.

Here $R e_{11} R = k e_{11} + k e_{12} \neq R$ , so $e$ is not full. Consistently, the ideal map is not onto: $R$ has the ideals $0$ , $k e_{12}$ , $k e_{11} + k e_{12}$ , $k e_{12} + k e_{22}$ and $R$ — five in all — while $e R e ≅ k$ has only two, and extension sends them to $0$ and $k e_{11} + k e_{12}$ .

A corner that is not a division ring

Take $R = M_{3} (k [[x]])$ and $e = e_{11} + e_{22}$ . Then $e R e ≅ M_{2} (k [[x]])$ , whose radical is $M_{2} (x k [[x]])$ . Contracting $rad R = M_{3} (x k [[x]])$ to the corner returns exactly that, again confirming $(21.10)$ . Here $R$ is left noetherian and semiprime but not semiprimitive, and $e R e$ inherits precisely those properties and no more.

Comparison and Classification

What transfers between $R$ and $e R e$ , and in which direction
Property	$R \Rightarrow e R e$	$e R e \Rightarrow R$	Reference
Semiprimitive	always	only if $e$ full	(21.10), (21.12)
Semisimple	always	only if $e$ full	(21.13)
Simple	always	only if $e$ full	(21.13)
Prime / semiprime	always	only if $e$ full	(21.13)
Left noetherian	always	only if $e$ full	(21.13)
Left artinian	always	only if $e$ full	(21.13)
Commutative	always (a corner sits inside $R$ )	no	$e_{11} M_{n} (k) e_{11} ≅ k$
Indecomposable as a ring	no	no	a corner of a connected path algebra can be $k \times k$

Behaviour of the two extension maps
	Injective	Inclusion-preserving	Respects products	Surjective
Left ideals, $𝔄 \mapsto R 𝔄$	yes	yes	not applicable	no
Ideals, $𝔄 \mapsto R 𝔄 R$	yes	yes	yes	partial
Ideals with $e$ full	yes	yes	yes	yes

Behaviour of the two extension maps

The entry part records that surjectivity of the ideal map is equivalent to fullness, not automatic. The left-ideal map is essentially never onto: for $e = e_{11}$ in $M_{2} (k)$ the corner has two left ideals and $M_{2} (k)$ has infinitely many when $k$ is infinite.

Relationship Map

e

idempotent

e R

finitely generated projective

e R e ≅ {End}_{R} (e R)

R e R = R

e R

a progenerator

R

and

e R e

Morita equivalent

All idempotents $e \neq 0$

rad (e R e) = e (rad R) e

; ideals of

e R e

inject into ideals of

R

Full idempotents, $R e R = R$ ideal lattices correspond bijectively;

R

and

e R e

Morita equivalent

Idempotents with $e R ≅ R_{R}$

e R e ≅ {End}_{R} (R_{R}) ≅ R

; forces

e = 1

when

R

is Dedekind-finite

$e = 1$ the corner is the whole ring

Reading outward: every statement true at an inner band is true at all bands containing it, but the converse fails at each step. The outermost band is where $(21.10)$ and $(21.11)$ live, which is why they carry no hypotheses.

Applications and Industry Use

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

Representation theory

Hecke algebras as corners

For a finite group $G$ with subgroup $H$ and $e = \frac{1}{| H |} \sum_{h \in H} h$ (when $| H |$ is invertible), $e k G e$ is the Hecke algebra of the pair. Its representation theory is that of the $H$ -invariant part of $k G$ -modules, extracted by exactly this construction.

Morita reduction

Basic algebras

Every finite-dimensional algebra is Morita equivalent to a basic one obtained as $e A e$ for a suitable idempotent $e$ collecting one primitive idempotent from each isomorphism class. Computations are done in the smaller corner and transported back.

Operator algebras

Hereditary subalgebras

For a projection $p$ in a $C^{*}$ -algebra $A$ , the corner $p A p$ is a hereditary subalgebra; full projections give Morita–Rieffel equivalence, and $K$ -theory is computed in whichever corner is convenient.

Computational algebra

Shrinking the problem

Working in $e A e$ rather than $A$ reduces the dimension quadratically in favourable cases. Since ideal lattices and radicals correspond for full idempotents, the reduction is lossless for the questions that matter.

The common pattern is compression: replace a large ring by a smaller one carrying the same categorical information, do the work there, and lift the answer back through the correspondence.

Design Considerations

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

Choose a full idempotent when you can. Fullness is what makes the reduction to $e R e$ lossless. If your idempotent is not full you are studying a genuinely smaller object and must say what is lost.
Check what you actually need. Many arguments need only $(21.10)$ , which is hypothesis-free. Do not invoke Morita equivalence when a radical computation suffices.
Left or right. $(21.11) (1)$ is stated for left ideals; the mirror statement for right ideals holds by symmetry, but $(21.13)$ 's chain conditions are genuinely one-sided and must be tracked.
Do not expect a ring map. There is no homomorphism $R \to e R e$ ; the projection $r \mapsto e r e$ is only additive. Any construction requiring functoriality in the ordinary sense will fail here and must be phrased through modules instead.
Watch the identity. In software, $e R e$ must be presented as a ring in its own right with identity $e$ , not as a subset of $R$ ; otherwise unit tests and inversion routines will use the wrong identity.

Standards and Notation

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

Corner ring

e R e

(universal); Peirce corner and compression also used

Radical

rad R

in this collection;

J (R)

elsewhere

Fullness

R e R = R

; also phrased as *

e

generates

R

as an ideal* or *

e R

is a generator*

Ideal extension

R 𝔄 R

; written

𝔄^{e}

in commutative-algebra style

Contraction

e 𝔅 e

; equals

𝔅 \cap e R e

MarkupPresentation MathML per ISO/IEC 40314; symbols per ISO 80000-2

SoftwareGAP RadicalOfAlgebra; Magma JacobsonRadical; Sage A.radical()

Computational Notes

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

For a finite-dimensional algebra $A$ over a field with $\dim A = n$ and an idempotent $e$ with $\dim e A e = m$ :

Constructing structure constants for $e A e$ costs $O (n m^{2})$ field operations: multiply basis elements in $A$ and compress by $x \mapsto e x e$ .
Radical computation is cubic in the dimension, so replacing $A$ by $e A e$ reduces the dominant cost from $O (n^{3})$ to $O (m^{3})$ . For a basic algebra of a group algebra the saving can be an order of magnitude.
$(21.10)$ makes the reduction sound: the radical computed in the corner is exactly the compression of the radical of $A$ , so no separate verification is needed.
Testing fullness means testing whether the ideal generated by $e$ is everything — a spanning computation on the products $a e b$ over basis elements $a, b$ , costing $O (n^{3})$ once.

Failure Modes and Common Mistakes

Do not assume $e (rad R) e = rad (R) \cap e R e$ needs $e$ to be central — it does not; but do not conclude that $rad (e R e)$ is an ideal of $R$ , which it generally is not.
Do not assume nonzero idempotents in a simple ring are all conjugate; fullness is automatic there, conjugacy is not.
Do not transfer local to a corner: a corner of a local ring need not be local, and a local corner does not make $R$ local.
Do not forget the hypothesis $e \neq 0$ in $(21.13)$ ; the zero corner satisfies no ring axioms of interest.

Best Practices

State whether the idempotent is full before quoting any corner-ring theorem; it is the single hypothesis that decides which direction results run.
Verify a computed $rad (e R e)$ against $e (rad R) e$ — the check is cheap and catches sign and side errors.
When transferring chain conditions, name the side; $(21.13)$ transfers left-handed hypotheses to left-handed conclusions.
Prefer the corner over the whole ring for computation, and record the correspondence used to lift results back.

Quick Reference

Corner

e R e = {r : e r = r = r e}

, identity

e

Radical

rad (e R e) = e (rad R) e = (rad R) \cap e R e

Quotient

e R e / rad (e R e) ≅ \bar{e} \bar{R} \bar{e}

Left ideals

(R 𝔄) \cap e R e = 𝔄

; extension injective

Ideals

e (R 𝔄 R) e = 𝔄

; extension multiplicative

Full

R e R = R

; then the ideal map is a bijection

Inheritedsemiprimitive, semisimple, simple, prime, semiprime, left noetherian, left artinian

Not inheritedcommutativity, locality, being a division ring

Reference numbers in Lam, §21
Reference	Statement
(21.10)	$rad (e R e) = e (rad R) e$ and the quotient description
(21.11)(1)	$(R 𝔄) \cap e R e = 𝔄$ for left ideals $𝔄$ of $e R e$
(21.11)(2)	$e (R 𝔄 R) e = 𝔄$ ; multiplicative; onto when $e$ is full
(21.12)	For full $e$ , $rad R$ corresponds to $rad (e R e)$
(21.13)	Seven properties inherited by $e R e$ from $R$

Frequently Asked Questions

Does

(21.10)

need any hypothesis on

R

e

None at all. $R$ is any ring with identity and $e$ any idempotent. The proof is two explicit inverse manipulations, so there is no chain condition, no semiperfectness and no fullness in play. This is unusual: most comparisons of radicals across subrings require substantial hypotheses.

rad (e R e)

an ideal of

R

No. It is an ideal of $e R e$ and a subset of $rad R$ , but $e (rad R) e$ is not closed under multiplication by arbitrary elements of $R$ unless $e$ is central. The correct statement is the one in $(21.10)$ : it is the contraction of $rad R$ , not a sub-ideal in the ambient sense.

Why does fullness make the ideal correspondence surjective?

Because the identity $R (e 𝔅 e) R = (R e R) 𝔅 (R e R)$ collapses to $R 𝔅 R = 𝔅$ exactly when $R e R = R$ . Conversely, if the map is onto then $R$ itself is in the image, and its preimage must be $e R e$ , forcing $R (e R e) R = R$ , i.e. $R e R = R$ . So fullness is equivalent to surjectivity.

What is the relationship to Morita theory?

$e R$ is always a finitely generated projective right $R$ -module with ${End}_{R} (e R) ≅ e R e$ and trace ideal $R e R$ . Fullness therefore says $e R$ is a progenerator, which is precisely the hypothesis under which the Morita theorems give an equivalence between the module categories of $R$ and $e R e$ . The ideal correspondence in $(21.11) (2)$ is what that equivalence does to two-sided ideals.

Can

e R e

be nicer than

R

Yes, and often dramatically so. For upper triangular matrices over a field the corner at $e_{11}$ is the field itself. That is exactly why $(21.13)$ runs only one way: passing to a corner can destroy pathology as easily as it preserves good behaviour.

Does

(21.13)

include left perfect or semiperfect?

Lam's list is the seven properties stated. Semiperfectness does transfer to corners as well, but that requires the idempotent-lifting machinery of the surrounding sections rather than $(21.11)$ alone, so it is best cited from the semiperfect ring theory rather than from this corollary.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §21, results (21.10)–(21.13) (pp. 323–326).
T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics 189, Springer-Verlag, 1999, §18 (Morita theory).
K. Morita, “Duality for modules and its applications to the theory of rings with minimum condition”, Science Reports of the Tokyo Kyoiku Daigaku, Section A, 6 (1958), 83–142.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §§21–22.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, §§1.1 and 4.1.

AI Suggested Questions

Prove that $e R$ is a progenerator if and only if $R e R = R$ , and deduce the Morita equivalence between $R$ and $e R e$ .
Give an example of a non-full idempotent for which the ideal lattices of $R$ and $e R e$ nevertheless have the same size.
Does the corner of a semiperfect ring have to be semiperfect? Sketch the argument or a counterexample.
Compute $rad (e R e)$ for $R$ a group algebra in characteristic $p$ and $e$ the idempotent attached to a Sylow subgroup.
Show that the centre of $e R e$ need not be the corner of the centre of $R$ , and identify when they agree.
How do the trace ideal $R e R$ and the annihilator of $e R$ interact for a non-full idempotent?
What is the analogue of $(21.10)$ for the prime radical or the Levitzki radical, and does it need hypotheses?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why the radical restricts so cleanly

Extension and contraction of ideals

Fullness and Morita equivalence

06Key Results

07Proof Techniques and Method

Transport an inverse across the corner

Absorb the idempotent

Retraction beats bijection

08Worked Example

A full idempotent: a matrix unit in a matrix ring

A non-full idempotent: upper triangular matrices

A corner that is not a division ring

09Comparison and Classification

10Relationship Map

11Applications and Industry Use

Hecke algebras as corners

Basic algebras

Hereditary subalgebras

Shrinking the problem

12Design Considerations

13Standards and Notation

14Computational Notes

15Failure Modes and Common Mistakes

16Best Practices

17Quick Reference

18Frequently Asked Questions

19Related KEVOS Topics

20References

21AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

Comparison and Classification

Relationship Map

Applications and Industry Use

Design Considerations

Standards and Notation

Computational Notes

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions