The Radical and Quotients of Semilocal Rings

Q: Does (20.7) need I to be finitely generated, or R noetherian?

No. I is an arbitrary two-sided ideal and no chain condition on R is used beyond what semilocality already provides — and semilocality itself imposes no chain condition on R, only on R/rad R.

Q: How does this relate to lifting idempotents?

It does not, directly. (20.7) controls the radical of a quotient; idempotent lifting is an independent property that upgrades semilocal to semiperfect. A semilocal ring need not lift idempotents modulo its radical.

Q: Can R/I be semisimple while R is only semilocal?

Yes — take I = rad R. More generally R/I is semisimple exactly when I rad R, since by (20.7) the radical of R/I is the image of rad R, which vanishes precisely then.

Q: Why is the corollary about GL_n placed here rather than with Bass' Theorem?

Because it needs both halves: (20.7) to know R/I is semilocal and to identify units modulo the radical, and Bass' Theorem to produce an actual unit in a coset. It is stated as an exercise in §20 for that reason.

Executive Summary

Radicals do not usually survive quotients. For any surjection $R ↠ R / I$ the image of $rad R$ is contained in $rad (R / I)$ , but the containment is frequently strict: $rad ℤ = 0$ while $rad (ℤ / 4) \neq 0$ . The semilocal hypothesis removes the discrepancy entirely.

Lam's $(20.7)$ says: if $R$ is semilocal then for every ideal $I \subseteq R$ ,

rad (R / I) = (rad R + I) / I, and R / I is semilocal.

(20.7)

The radical is computed once, at the top, and then simply pushed down.

$\subseteq$ General containment

$=$ Semilocal case

ClosedQuotients stay semilocal

$(20.7)$ Lam

Overview

Two facts about radicals and surjections are true for all rings, and they bracket the problem.

For any surjective ring homomorphism $ϕ : R \to S$ one has $ϕ (rad R) \subseteq rad S$ . Reason: if $y \in rad R$ and $s \in S$ , write $s = ϕ (x)$ ; then $1 - s ϕ (y) = ϕ (1 - x y)$ is the image of a unit, hence a unit.
If $I \subseteq rad R$ then equality holds: $rad (R / I) = (rad R) / I$ . This is Lam $(4.6)$ and needs no hypothesis on $R$ beyond $I$ sitting inside the radical.

Between these lies the interesting case: an ideal $I$ that is not inside $rad R$ . Then killing $I$ can create new radical — quotienting can manufacture non-units out of units. $(20.7)$ says a semilocal ring has no room for this to happen, because its radical quotient is semisimple and semisimple rings have no proper quotients with nonzero radical.

Learning Objectives

Prove $ϕ (rad R) \subseteq rad S$ for any surjection $ϕ$ .
State $(4.6)$ and identify exactly which hypothesis it needs.
Prove $(20.7)$ in full: equality of radicals and semilocality of the quotient.
Exhibit a non-semilocal ring where the containment is strict.
Deduce that the simple modules of $R / I$ are a subfamily of those of $R$ .
Compute $rad (R / I)$ for $R = ℤ / 12$ and for an upper triangular matrix ring.

Definitions

$rad R$: The Jacobson radical of R: the intersection of the maximal left ideals; equivalently the largest left ideal U with 1 + U inside U(R).
$\bar{R}$: Shorthand for the quotient R/I with the natural surjection r maps to r-bar; used throughout the proof of (20.7).
Semilocal: R/rad R is semisimple.
Semiprimitive: rad R = 0. A semisimple ring is semiprimitive and left artinian.

$I$ always denotes a two-sided ideal. The statement of $(20.7)$ is false for one-sided ideals, for which $R / I$ is not a ring at all.

Core Concepts

Why the radical can grow under a quotient

$rad R$ is an intersection of maximal left ideals. Passing to $R / I$ keeps only those maximal left ideals that contain $I$ , so the intersection is taken over a smaller family and may therefore be larger. Concretely, $ℤ$ has one maximal ideal per prime and their intersection is $0$ ; $ℤ / 4$ retains only $2 ℤ / 4 ℤ$ , whose intersection with itself is itself.

maximal left ideals of

R

those containing

I

rad (R / I)

Why semilocality closes the gap

Write $J = rad R$ . The composite $R \to R / I \to R / (I + J)$ factors through the semisimple ring $R / J$ , so $R / (I + J)$ is a quotient of a semisimple ring and hence semisimple. Semisimple rings have zero radical. That single observation forces the radical of $R / I$ to be no bigger than the image of $J$ , because a further quotient by that image is already radical-free.

R / I / \frac{J + I}{I} ≅ R / (I + J) = \frac{R / J}{(I + J) / J}

(20.7a)

The key isomorphism: the double quotient is visibly a quotient of the semisimple ring $R / J$ .

What survives, module-theoretically

Since $R / (I + J)$ is a quotient of $\prod_{i = 1}^{r} M_{n_{i}} (D_{i})$ , it is the product over a subset of the indices: quotients of semisimple rings are obtained by deleting Wedderburn factors. Hence the simple $R / I$ -modules are exactly those simple $R$ -modules killed by $I$ , and there are at most $r$ of them. Quotienting a semilocal ring never creates a new simple module.

Key Results

Lemma(4.6)Quotient by an ideal inside the radical

Let $R$ be any ring and $I \subseteq rad R$ a two-sided ideal. Then $rad (R / I) = (rad R) / I$ . In particular $rad (R / rad R) = 0$ .

Proposition(20.7)Radical of a quotient of a semilocal ring

Let $R$ be a semilocal ring and let $I \subseteq R$ be any two-sided ideal. Then

rad (R / I) = (rad R + I) / I,

and $R / I$ is again semilocal.

Proof

Write $J = rad R$ , let $\bar{R} = R / I$ and let $\bar{J} = (J + I) / I$ denote the image of $J$ .

**Step 1: $\bar{J} \subseteq rad \bar{R}$ .** For $y \in J$ and any $\bar{x} \in \bar{R}$ , the element $1 - \bar{x} \bar{y}$ is the image of the unit $1 - x y$ , hence a unit. By the unit characterisation of the radical, $\bar{y} \in rad \bar{R}$ .

Step 2: the double quotient is semisimple. Since $\bar{J} \subseteq rad \bar{R}$ , Lemma $(4.6)$ applies to the ring $\bar{R}$ and its ideal $\bar{J}$ :

rad (\bar{R} / \bar{J}) = (rad \bar{R}) / \bar{J} .

On the other hand $\bar{R} / \bar{J} ≅ R / (I + J)$ , which is a quotient of $R / J$ . By hypothesis $R / J$ is semisimple, and every quotient ring of a semisimple ring is semisimple; hence $\bar{R} / \bar{J}$ is semisimple and therefore has zero radical.

Step 3: conclude. Combining the two displays, $(rad \bar{R}) / \bar{J} = 0$ , that is $rad \bar{R} = \bar{J} = (J + I) / I$ . Finally

\bar{R} / rad \bar{R} = \bar{R} / \bar{J} ≅ R / (I + J)

was just shown to be semisimple, so $\bar{R} = R / I$ is semilocal. $□$

CorollarySemilocality is inherited by quotients

Every homomorphic image of a semilocal ring is semilocal. Consequently the class of semilocal rings is closed under quotients, and combined with $(20.4)$ it is closed under $R \mapsto M_{n} (R)$ and finite products.

CorollarySimple modules do not multiply

Let $R$ be semilocal with $R / rad R ≅ \prod_{i = 1}^{r} M_{n_{i}} (D_{i})$ . For any ideal $I$ , the simple left $R / I$ -modules are precisely the simple left $R$ -modules annihilated by $I$ , and $R / I$ has at most $r$ of them up to isomorphism.

CorollaryReduction of the general linear group

If $R$ is semilocal and $I \subseteq R$ is any ideal, the natural map ${GL}_{n} (R) \to {GL}_{n} (R / I)$ is surjective for every $n \geq 1$ .

Sketch. For $n = 1$ : if $\bar{u}$ is a unit of $R / I$ then $R u + I = R$ , so Bass' Theorem $(20.9)$ produces a unit in the coset $u + I$ . For general $n$ , apply the case $n = 1$ to the semilocal ring $M_{n} (R)$ and its ideal $M_{n} (I)$ , using $U (M_{n} (R)) = {GL}_{n} (R)$ and $M_{n} (R) / M_{n} (I) ≅ M_{n} (R / I)$ .

Proof Techniques and Method

How this proof works, and which move to reuse.

The proof of $(20.7)$ is a three-line argument dressed as a two-page one. The reusable content is the following pattern.

1. Push the radical downThe image of

rad R

always lands inside

rad

of the target. This gives you a candidate ideal

\bar{J}

for free.

2. Apply $(4.6)$ to the targetBecause

\bar{J}

sits inside

rad \bar{R}

, the radical of

\bar{R} / \bar{J}

is exactly

rad \bar{R} / \bar{J}

— no hypotheses needed.

3. Show the double quotient has zero radicalRewrite

\bar{R} / \bar{J}

as a quotient of

R / rad R

. Semilocality makes this semisimple, hence semiprimitive.

4. Read off equality

rad \bar{R} / \bar{J} = 0

forces

rad \bar{R} = \bar{J}

, and the same rewriting shows

\bar{R}

is semilocal.

Worked Example

The same target ring, two different sources

Take the target $ℤ / 4$ , whose radical is $2 ℤ / 4 ℤ$ , and reach it in two ways.

Semilocal source versus non-semilocal source
Source $R$	$rad R$	Ideal $I$	Image $(rad R + I) / I$	$rad (R / I)$
$ℤ$ (not semilocal)	$0$	$4 ℤ$	$0$	$2 ℤ / 4 ℤ$ — strictly larger
$ℤ / 12$ (semilocal)	$6 ℤ / 12 ℤ$	$4 ℤ / 12 ℤ$	$2 ℤ / 4 ℤ$	$2 ℤ / 4 ℤ$ — equal

Checking row two. $ℤ / 12 ≅ ℤ / 4 \times ℤ / 3$ is a finite product of local rings, so it is semilocal with $rad = (6)$ . The ideal $I = (4)$ satisfies $I not \subseteq rad$ , so $(4.6)$ does not apply and $(20.7)$ is doing real work. In $ℤ / 12$ the ideal $(6) + (4) = (2)$ , whose image in $ℤ / 12 / (4) ≅ ℤ / 4$ is $(2)$ — exactly $rad (ℤ / 4)$ . The quotient $ℤ / 4 / (2) ≅ 𝔽_{2}$ is semisimple, confirming that $ℤ / 4$ is semilocal.

A noncommutative check

Let $R = T_{3} (k)$ be the upper triangular $3 \times 3$ matrices over a field $k$ . It is finite-dimensional, hence semilocal, with

rad R = {(\begin{matrix} 0 & * & * \\ 0 & 0 & * \\ 0 & 0 & 0 \end{matrix})}, R / rad R ≅ k \times k \times k .

(20.7b)

The strictly upper triangular matrices form a nilpotent ideal with semisimple quotient.

Let $I = k e_{13}$ , the span of the corner matrix unit; it is a two-sided ideal, since $e_{11} e_{13} = e_{13}$ , $e_{13} e_{33} = e_{13}$ , and all other products with matrix units either reproduce $e_{13}$ or vanish. Since $I \subseteq rad R$ , $(4.6)$ already gives $rad (R / I) = (rad R) / I$ , which is spanned by the images of $e_{12}$ and $e_{23}$ and has square zero because $e_{12} e_{23} = e_{13} \in I$ . Then $R / I$ is semilocal with $(R / I) / rad (R / I) ≅ k \times k \times k$ : the number of simple modules is unchanged, as the corollary above permits.

Process and Workflow

In practice one meets a quotient and wants its radical. The decision below records when $(20.7)$ may be invoked.

You need $rad (R / I)$ . Which tool applies?

$I \subseteq rad R$ Use

(4.6)

rad (R / I) = (rad R) / I

. No hypothesis on

R

is required.

$R$ semilocal, $I$ arbitraryUse

(20.7)

rad (R / I) = (rad R + I) / I

, and

R / I

is semilocal.

NeitherYou only get the containment

(rad R + I) / I \subseteq rad (R / I)

. Compute the radical of

R / I

directly — for instance by locating its maximal left ideals or by finding a nilpotent ideal with semiprimitive quotient.

Comparison and Classification

Behaviour of the radical under standard constructions
	Radical computed by image or entrywise	Needs a hypothesis on $R$
$R \to R / I$ , $I \subseteq rad R$	yes	no
$R \to R / I$ , $I$ arbitrary, $R$ semilocal	yes	yes
$R \to R / I$ , $I$ arbitrary, $R$ general	no	—
$R \mapsto M_{n} (R)$	yes	no
$R \mapsto R_{1} \times \dots \times R_{m}$ (finite)	yes	no
$R \mapsto R [x]$	no	—
Subring $S \subseteq R$	no	—

Behaviour of the radical under standard constructions

$rad M_{n} (R) = M_{n} (rad R)$ holds for every ring. For polynomial rings, $rad (R [x])$ is a nil ideal extended by $x$ and is not the image of $rad R$ ; Amitsur's theorem describes it.

Relationship Map

$R$ semilocal — $R / rad R$ semisimple
- $(20.7)$ — $rad (R / I) = (rad R + I) / I$
  - Every quotient of $R$ is semilocal
  - Simple $R / I$ -modules form a subfamily of the simple $R$ -modules
  - ${GL}_{n} (R) ↠ {GL}_{n} (R / I)$ (via Bass' Theorem)
- $(20.8)$ – $(20.9)$ — Dedekind finiteness and stable range one
  - Cancellation of modules
  - Invariant basis number

The two branches are independent consequences of the same hypothesis; $(20.7)$ is about ideals, the other branch about units. They meet in the surjectivity of ${GL}_{n}$ , which needs both.

Applications and Industry Use

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

Algebraic K-theory. Surjectivity of ${GL}_{n} (R) \to {GL}_{n} (R / I)$ for semilocal $R$ makes the relative exact sequence of $K_{1}$ behave, and it is the reason the $K$ -theory of semilocal rings is computable in terms of units.
Modular representation theory. For a complete discrete valuation ring $𝒪$ with residue field $k$ , reduction $𝒪 G \to k G$ is a quotient of semilocal rings; $(20.7)$ says the radical reduces correctly, which is what allows blocks and their defect groups to be tracked from characteristic zero to characteristic $p$ .
Computer algebra. Once the radical of a finite-dimensional algebra is computed, the radical of any quotient by an ideal is obtained by linear algebra alone — image plus ideal — instead of rerunning a radical algorithm.
Coding over rings. For a finite chain ring $R$ and an ideal $I$ , code constructions descend from $R$ to $R / I$ with the radical filtration intact, which is what makes the tower $R \supseteq rad R \supseteq (rad R)^{2} \supseteq \dots$ usable as a chain of nested codes.

Computational Notes

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

For a finite-dimensional algebra $A$ over a field, the radical is computed by the Dickson–Friedl–Rónyai style algorithms: in characteristic zero, $rad A$ is the kernel of the trace form on the regular representation, obtained by one $n \times n$ linear solve in $O (n^{ω})$ field operations; in positive characteristic a sequence of refined trace forms is needed.

Given $rad A$ and an ideal $I$ , $(20.7)$ gives $rad (A / I)$ as the image of $rad A + I$ — a subspace sum and a projection, both cheap.
Certifying semilocality is trivial for finite-dimensional algebras and undecidable in general; there is no finitary certificate for an arbitrary presented ring.
GAP's RadicalOfAlgebra, Magma's JacobsonRadical and Sage's radical for finite-dimensional algebras all return a basis of $rad A$ ; none exposes an operation that computes the radical of a quotient without recomputation, so applying $(20.7)$ by hand is a genuine saving.

Failure Modes and Common Mistakes

Quick Reference

Universal factFor any surjection,

(rad R + I) / I \subseteq rad (R / I)

$(4.6)$

I \subseteq rad R \Rightarrow rad (R / I) = (rad R) / I

, any ring.

$(20.7)$

R

semilocal,

I

any ideal

\Rightarrow

equality, and

R / I

semilocal.

Failure

ℤ \to ℤ / 4

0 ⊊ 2 ℤ / 4 ℤ

Consequence

{GL}_{n} (R) \to {GL}_{n} (R / I)

is onto for semilocal

R

Checklist before applying the formula
Question	If yes	If no
Is $I$ two-sided?	Proceed.	$R / I$ is not a ring; the question is malformed.
Is $I \subseteq rad R$ ?	Use $(4.6)$ ; no further hypothesis.	Continue.
Is $R / rad R$ semisimple?	Use $(20.7)$ .	Only the containment is available.
Is $R / I$ artinian?	Radical = largest nilpotent ideal.	Compute maximal left ideals directly.

Frequently Asked Questions

Why does the radical of a quotient tend to be bigger than the image of the radical?

The radical is an intersection of maximal left ideals, and the quotient sees only those maximal left ideals containing $I$ . Intersecting over a smaller family gives a larger ideal. Semilocality prevents this because the surviving family is already forced to cut out exactly the image of $rad R$ .

Does

(20.7)

need

I

to be finitely generated, or

R

noetherian?

No. $I$ is an arbitrary two-sided ideal and no chain condition on $R$ is used beyond what semilocality already provides — and semilocality itself imposes no chain condition on $R$ , only on $R / rad R$ .

Is the converse true: if

rad (R / I) = (rad R + I) / I

for all

I

, must

R

be semilocal?

No. Any semiprimitive ring all of whose proper quotients are semiprimitive would satisfy the formula trivially, and there are such rings that are not semilocal. The condition in $(20.7)$ is sufficient, not necessary; it is the usable hypothesis, not a characterisation.

How does this relate to lifting idempotents?

It does not, directly. $(20.7)$ controls the radical of a quotient; idempotent lifting is an independent property that upgrades semilocal to semiperfect. A semilocal ring need not lift idempotents modulo its radical.

Can

R / I

be semisimple while

R

is only semilocal?

Yes — take $I = rad R$ . More generally $R / I$ is semisimple exactly when $I \supseteq rad R$ , since by $(20.7)$ the radical of $R / I$ is the image of $rad R$ , which vanishes precisely then.

Why is the corollary about

{GL}_{n}

placed here rather than with Bass' Theorem?

Because it needs both halves: $(20.7)$ to know $R / I$ is semilocal and to identify units modulo the radical, and Bass' Theorem to produce an actual unit in a coset. It is stated as an exercise in §20 for that reason.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §20, especially (20.7), pp. 312–313.
T. Y. Lam, A First Course in Noncommutative Rings, §4, (4.6) and Exercise 4.21, for the radical of a quotient and lifting units.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapter I.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §15.
L. Rónyai, “Computing the structure of finite algebras”, Journal of Symbolic Computation 9 (1990), 355–373.

AI Suggested Questions

Prove that every quotient ring of a semisimple ring is semisimple, and identify the quotient as a product over a subset of the Wedderburn factors.
Give a ring $R$ and ideal $I$ with $(rad R + I) / I$ strictly smaller than $rad (R / I)$ where $R$ is semiprimitive but not semilocal.
Describe $rad (R [x])$ via Amitsur's theorem and explain why it is not the image of $rad R$ .
Work out the full proof that ${GL}_{n} (R) \to {GL}_{n} (R / I)$ is surjective for semilocal $R$ .
For $𝒪$ a complete DVR and $G$ finite, compare $rad (𝒪 G)$ with $rad (k G)$ under reduction.
Is the property "the radical commutes with all quotients" closed under matrix rings and finite products?
How expensive is it to recompute the Jacobson radical of a finite-dimensional algebra versus using (20.7)?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why the radical can grow under a quotient

Why semilocality closes the gap

What survives, module-theoretically

06Key Results

07Proof Techniques and Method

08Worked Example

The same target ring, two different sources

A noncommutative check

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

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

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Computational Notes

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions