Is (4.6) true for one-sided ideals A?

The statement does not typecheck: R/A is only a ring when A is two-sided. What is true is that the lattice argument works for the module R/A and shows that the radical of that module is (rad R)/A when A ⊆ rad R, where the radical of a module means the intersection of its maximal submodules.

The Radical of a Quotient Ring

Q: Does R/rad R determine R in any useful sense?

No, but it determines a lot: by (4.8) the two rings have the same simple left modules, and an element of R is invertible exactly when its image is. What is lost is everything nilpotent, and recovering R from the quotient requires extra hypotheses such as idempotent lifting, which is the subject of the semiperfect pages.

Q: Can R/A be semiprimitive for two different ideals inside the radical?

No. If I ⊆ rad R and R/I is semiprimitive then minimality gives rad R ⊆ I, so I = rad R. Uniqueness holds only under the containment hypothesis; above the radical there can be many semiprimitive quotients.

Executive Summary

The Jacobson radical does not commute with arbitrary quotients. It commutes with the quotients that matter. If $𝔄$ is a two-sided ideal of $R$ contained in $rad R$ , then $rad (R / 𝔄) = (rad R) / 𝔄$ . Lam records this as $(4.6)$ and suppresses the proof as immediate; the consequences are anything but.

Two of those consequences organise the rest of radical theory. Taking $𝔄 = rad R$ shows that every ring has a canonical semiprimitive quotient. Reading the statement in reverse gives the working mathematician's method for computing a radical: exhibit a nil ideal $J$ , show that $R / J$ has zero radical, and conclude that $rad R = J$ exactly.

(4.6)Lam's numbering

$𝔄 \subseteq rad R$ The one hypothesis

$0$

rad (R / rad R)

SmallestRank of rad R among ideals with semiprimitive quotient

Overview

Let $R$ be a ring with identity and $𝔄 ⊴ R$ a two-sided ideal. The correspondence theorem gives an inclusion-preserving bijection between the left ideals of $R / 𝔄$ and the left ideals of $R$ that contain $𝔄$ ; maximality is preserved in both directions. So the maximal left ideals of $R / 𝔄$ are precisely the images $𝔪 / 𝔄$ of the maximal left ideals $𝔪 \supseteq 𝔄$ of $R$ .

The radical of the quotient is therefore an intersection over a *sub*family of the maximal left ideals of $R$ — those that happen to contain $𝔄$ . Intersecting fewer things gives something bigger, which is why the radical can grow under a quotient. The hypothesis of $(4.6)$ removes the discrepancy by forcing the subfamily to be the whole family.

𝔄 \subseteq rad R ⟹ rad (R / 𝔄) = (rad R) / 𝔄

(4.6)

The hypothesis is on $𝔄$ alone; no chain condition, no commutativity, no finiteness.

The companion facts are that the radical only ever shrinks along a surjection, and that $rad R$ is the smallest ideal whose quotient is semiprimitive. Together with The Jacobson Radical: Definition and Equivalent Characterisations these three statements are the complete account of how $rad$ behaves under change of rings by quotients.

Learning Objectives

State $(4.6)$ with its hypothesis $𝔄 \subseteq rad R$ and know why the hypothesis cannot be dropped.
Prove $(4.6)$ from the correspondence between left ideals of $R$ and of $R / 𝔄$ .
Deduce that $rad (R / rad R) = 0$ for every ring $R$ .
Prove that $f (rad R) \subseteq rad S$ for a surjective homomorphism $f : R \to S$ , and find a non-surjective counterexample.
Show that $rad R$ is the smallest ideal $I$ with $R / I$ semiprimitive.
Compute $rad T_{3} (k)$ using $(4.6)$ rather than by intersecting maximal left ideals.

Definitions

$rad R$: The intersection of all maximal left ideals of $R$ ; equivalently the intersection of the annihilators of the simple left $R$ -modules, hence a two-sided ideal.
$𝔄 ⊴ R$: A two-sided ideal. Quotients are only rings when $𝔄$ is two-sided, so $(4.6)$ is stated for two-sided ideals even though the radical is defined by one-sided ones.
Semiprimitive: $rad R = 0$ . Also called Jacobson semisimple or J-semisimple; see Jacobson Semisimple (Semiprimitive) Rings.
Core of a left ideal: For a left ideal $𝔅$ , the largest two-sided ideal of $R$ contained in $𝔅$ , equal to $ann (R / 𝔅)$ .
Nil ideal: A one-sided or two-sided ideal all of whose elements are nilpotent. Every nil one-sided ideal lies in $rad R$ , which is what makes the verification method below work.

All rings have an identity and all modules are unital. Ideal means two-sided ideal unless the word left or right appears.

Core Concepts

The correspondence, drawn out

Write $π : R ↠ R / 𝔄$ for the quotient map. For a left ideal $𝔅 \subseteq R$ containing $𝔄$ , the image $π (𝔅) = 𝔅 / 𝔄$ is a left ideal of $R / 𝔄$ , and $π^{- 1}$ inverts this assignment. Proper corresponds to proper and maximal to maximal, because the lattice of left ideals above $𝔄$ is carried isomorphically onto the lattice of left ideals of $R / 𝔄$ .

Maximal left ideals

𝔪 \supseteq 𝔄

Maximal left ideals of

R / 𝔄

rad (R / 𝔄)

Consequently $rad (R / 𝔄) = (⋂ {𝔪 : 𝔪 maximal left, 𝔪 \supseteq 𝔄}) / 𝔄$ for every ideal $𝔄$ . This identity is the whole content of the correspondence theorem, and $(4.6)$ is the observation that the qualifier $𝔪 \supseteq 𝔄$ becomes vacuous once $𝔄 \subseteq rad R$ .

One inclusion is free

For an arbitrary ideal $𝔄$ the family of maximal left ideals containing $𝔄$ is smaller, so its intersection is larger: $rad (R / 𝔄) \supseteq (rad R + 𝔄) / 𝔄$ . The radical can only grow when you quotient, never shrink relative to the image. In particular a ring can acquire a radical it did not have: $ℤ$ is semiprimitive while $ℤ / 4 ℤ$ is not.

Key Results

Proposition(4.6)Radical of a quotient by an ideal inside the radical

Let $R$ be a ring and let $𝔄$ be a two-sided ideal of $R$ with $𝔄 \subseteq rad R$ . Then

rad (R / 𝔄) = (rad R) / 𝔄 .

Proof

Let $𝔪$ be a maximal left ideal of $R$ . Then $rad R \subseteq 𝔪$ by definition of the radical, and hence $𝔄 \subseteq 𝔪$ by hypothesis. So every maximal left ideal of $R$ contains $𝔄$ , and the correspondence theorem makes $𝔪 \mapsto 𝔪 / 𝔄$ a bijection from the maximal left ideals of $R$ onto the maximal left ideals of $R / 𝔄$ .

Intersecting, and using that $π (⋂_{i} 𝔪_{i}) = (⋂_{i} 𝔪_{i}) / 𝔄$ whenever each $𝔪_{i}$ contains $𝔄$ ,

rad (R / 𝔄) = ⋂_{𝔪} (𝔪 / 𝔄) = (⋂_{𝔪} 𝔪) / 𝔄 = (rad R) / 𝔄 .

The middle equality is the elementary fact that for subgroups containing $𝔄$ , taking images commutes with intersections. If $R = 𝔄$ both sides are the zero ring's radical, namely zero, so the degenerate case is covered.

Corollary(4.6a)Every ring has a semiprimitive quotient

For any ring $R$ , $rad (R / rad R) = 0$ ; that is, $R / rad R$ is semiprimitive.

Proof

Apply $(4.6)$ with $𝔄 = rad R$ , which certainly satisfies the hypothesis. The right-hand side is $(rad R) / (rad R) = 0$ .

Proposition—Radicals shrink along surjections

Let $f : R \to S$ be a surjective ring homomorphism. Then $f (rad R) \subseteq rad S$ . Equivalently, for every two-sided ideal $𝔄$ of $R$ , $(rad R + 𝔄) / 𝔄 \subseteq rad (R / 𝔄)$ .

Proof

Take $y \in rad R$ and $s \in S$ . By surjectivity write $s = f (x)$ . By the element characterisation of the radical there is $u \in R$ with $u (1 - x y) = 1$ . Applying $f$ gives $f (u) (1 - s f (y)) = 1$ , so $1 - s f (y)$ is left-invertible in $S$ for every $s \in S$ . Hence $f (y) \in rad S$ .

CorollaryEx. 4.11Minimality of the radical

Let $I$ be a two-sided ideal of $R$ such that $R / I$ is semiprimitive. Then $rad R \subseteq I$ . Consequently $rad R$ is the smallest ideal of $R$ with semiprimitive quotient, and it is the unique ideal $I \subseteq rad R$ with $R / I$ semiprimitive.

Proof

By the correspondence theorem the maximal left ideals of $R / I$ are the $𝔪 / I$ with $𝔪$ a maximal left ideal of $R$ containing $I$ . Semiprimitivity of $R / I$ says that these intersect in zero, i.e. $⋂ {𝔪 : 𝔪 \supseteq I} = I$ . Since $rad R$ is contained in every maximal left ideal, it is in particular contained in every $𝔪 \supseteq I$ , hence in their intersection $I$ .

For the last clause: if $I \subseteq rad R$ and $R / I$ is semiprimitive, then $rad R \subseteq I$ by the above, so $I = rad R$ .

Remark—Surjectivity is not decoration

For a non-surjective homomorphism the conclusion fails outright. The inclusion $k [[x]] ↪ k ((x))$ sends the radical $(x)$ into a field, whose radical is zero, so the image of the radical is not contained in the radical of the target. The reason is visible in the proof: without surjectivity one cannot realise an arbitrary $s \in S$ as $f (x)$ .

Proof Techniques and Method

The reusable moves behind (4.6) and its corollaries.

Move 1

Guess a nil ideal, then verify

Produce a candidate $J$ that is visibly nil or nilpotent, so $J \subseteq rad R$ . Then show $R / J$ is semiprimitive. By $(4.6)$ , $(rad R) / J = rad (R / J) = 0$ , so $rad R = J$ . No maximal left ideal is ever listed.

Move 2

Push the lattice, not the elements

Every statement here is proved by transporting the lattice of left ideals through $π$ . Once you know which maximal left ideals survive, the radical of the quotient is determined; no computation with individual elements is needed.

Move 3

Characterise by a universal property

Minimality turns $rad R$ into the kernel of the universal surjection onto a semiprimitive ring. Questions of the form *is $y$ in the radical?* become *does $y$ die in every semiprimitive quotient?*

Move 1 is the reason $(4.6)$ appears so early in Lam's development: it is the tool used to evaluate almost every explicit radical in the book, including the triangular matrix rings below and the group algebras of §6.

Worked Example

Upper triangular matrices in two lines

Let $k$ be a division ring and $R = T_{3} (k)$ the ring of upper triangular $3 \times 3$ matrices over $k$ . Let $J$ be the set of matrices in $R$ with zero diagonal.

J = (\begin{matrix} 0 & * & * \\ 0 & 0 & * \\ 0 & 0 & 0 \end{matrix}), J^{2} = (\begin{matrix} 0 & 0 & * \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix}), J^{3} = 0 .

(E.1)

$J$ is a two-sided ideal of $R$ and is nilpotent of index exactly $3$ .

Because $J$ is nilpotent it is nil, so $J \subseteq rad R$ . Reduction modulo $J$ kills the off-diagonal entries and leaves the diagonal:

R / J ≅ k \times k \times k,

(E.2)

a finite product of division rings, hence semisimple and in particular semiprimitive.

Now apply $(4.6)$ : since $J \subseteq rad R$ we get $(rad R) / J = rad (R / J) = 0$ , so $rad R = J$ . The three simple left $R$ -modules are the one-dimensional modules on which a matrix acts through its $i$ -th diagonal entry, $i = 1, 2, 3$ .

An arithmetic check

Take $R = ℤ / 72 ℤ$ . Its maximal ideals are $(2)$ and $(3)$ , so $rad R = (6)$ . Choose $𝔄 = (12)$ , which satisfies $𝔄 \subseteq (6) = rad R$ . Then $R / 𝔄 ≅ ℤ / 12 ℤ$ , whose radical is $(6)$ — precisely the image of $(6)$ under reduction. Both sides of $(4.6)$ equal the two-element ideal ${0, 6}$ of $ℤ / 12 ℤ$ .

Where the hypothesis bites

Let $R = k [x]$ with $k$ a field. There are infinitely many monic irreducible polynomials and a nonzero polynomial is divisible by only finitely many, so $rad R = 0$ . Take $𝔄 = (x^{2})$ , which is not contained in $rad R = 0$ . Then $R / 𝔄 = k [x] / (x^{2})$ is local with maximal ideal $(x) / (x^{2})$ , so

rad (k [x] / (x^{2})) = (x) / (x^{2}) \neq 0 = (rad k [x] + (x^{2})) / (x^{2}) .

(E.3)

Strict growth. The containment of the previous section is all one can say in general.

Process and Workflow

Find a candidateLook for an obvious nil or nilpotent ideal

J

: strictly triangular entries, an augmentation ideal, a maximal ideal of a local ring.

Get $J \subseteq rad R$ Nil one-sided ideals always lie in the radical, so nilpotency of

J

settles this step with no further work.

Identify $R / J$ Recognise the quotient — a product of division rings, a polynomial ring, a matrix ring over a field — and show its radical is zero.

Close the loop with (4.6)

(rad R) / J = rad (R / J) = 0

gives

rad R = J

on the nose.

You want $rad (R / 𝔄)$ . Where does $𝔄$ sit?

$𝔄 \subseteq rad R$ Use

(4.6)

: the answer is

(rad R) / 𝔄

, and nothing else needs checking.

$𝔄$ nilNil implies

𝔄 \subseteq rad R

, so you are in the first branch. This is the common case in practice.

NeitherOnly the containment

(rad R + 𝔄) / 𝔄 \subseteq rad (R / 𝔄)

is available. Compute the radical of

R / 𝔄

from scratch — it may be strictly larger.

Comparison and Classification

What happens to the radical under a quotient
Situation	$rad (R / 𝔄)$ versus $(rad R + 𝔄) / 𝔄$	Reference or witness
$𝔄 \subseteq rad R$	equal	(4.6)
$𝔄 = rad R$	equal, both zero	(4.6) with the radical itself
$𝔄$ nil	equal	nil ideals lie in the radical, then (4.6)
$𝔄$ arbitrary	containment only, can be strict	$ℤ ↠ ℤ / 4 ℤ$
$R / 𝔄$ semiprimitive	forces $𝔄 \supseteq rad R$	Exercise 4.11
non-surjective homomorphism	no containment at all	$k [[x]] ↪ k ((x))$

The pattern to memorise: radicals never shrink under quotients and never grow under the hypothesis of $(4.6)$ .

Relationship Map

𝔄 \subseteq rad R

rad (R / 𝔄) = (rad R) / 𝔄

rad (R / rad R) = 0

R / rad R

semiprimitive

Ideals $I ⊴ R$ — sorted by what the quotient's radical does
- $I \subseteq rad R$
  - $rad (R / I) = (rad R) / I$ by $(4.6)$
  - $R / I$ is semiprimitive precisely when $I = rad R$
- $I \supseteq rad R$
  - $R / I$ may or may not be semiprimitive
  - every $I$ with $R / I$ semiprimitive lands here
- incomparable with the radical
  - only the free containment is available
  - $R = ℤ$ , $I = (4)$ is the standard case

The tree also explains why the radical is the right invariant to quotient by: it is the unique ideal that is simultaneously small enough for $(4.6)$ to apply and large enough to kill the radical of the quotient.

Applications and Industry Use

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

Symbolic computation

The first step of every algebra decomposition

Structure-recognition routines for finite-dimensional algebras compute $rad A$ , form $A / rad A$ , and apply Wedderburn–Artin to the semiprimitive quotient. $(4.6)$ is the guarantee that the second step is not circular: the quotient really has no radical left.

Modular representation theory

Brauer's reduction

For $k G$ with $char k = p$ dividing $| G |$ , the simple modules of $k G$ are exactly those of $k G / rad (k G)$ , and the radical quotient is where character-theoretic invariants live.

Coding theory

Codes over finite chain rings

A finite chain ring $R$ has $rad R = (π)$ nilpotent and residue field $R / (π)$ . Codes are analysed through the tower $R / (π^{i})$ , and $(4.6)$ says each stage has radical $(π) / (π^{i})$ , giving the graded pieces used for weight computations.

Deformation and lifting

Complete local rings

Obstruction theory works over rings with nilpotent ideals inside the radical, and lifts a solution along $R / 𝔄^{i + 1} ↠ R / 𝔄^{i}$ . $(4.6)$ keeps the radical predictable at every level of the tower.

None of this is an application outside algebra in the civil-engineering sense. The honest description is that $(4.6)$ is the licence for a reduction step used everywhere that a noncommutative ring is decomposed by machine or by hand.

Computational Notes

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

For a finite-dimensional algebra $A$ given by structure constants, a system computes $rad A$ once and then works in $A / rad A$ ; forming the quotient is a change of basis plus a linear-algebra reduction, so it costs no more than the radical computation itself.
The verification method — check $J$ nilpotent, check $A / J$ semiprimitive — is often much cheaper than the general radical algorithm when $J$ is visible from a presentation, for example the strictly triangular part of a triangular algebra.
Nilpotency of a finitely generated ideal $J$ in a finite-dimensional algebra is decided by computing $J, J^{2}, J^{4}, \dots$ by repeated squaring, which needs about $\log_{2} \dim A$ ideal products rather than $\dim A$ of them.
GAP exposes RadicalOfAlgebra and the quotient via A / RadicalOfAlgebra(A); Magma and Sage offer equivalent constructors. All of them assume finite dimension over a field.

Failure Modes and Common Mistakes

Do not confuse $(4.6)$ with the false statement that $rad$ is a functor on all ring maps; it behaves only along surjections.
Do not assume $𝔄 \subseteq rad R$ from $𝔄$ being small or nilpotent-looking. Verify nilpotency or nilness; that is what licenses the containment.
Do not read minimality as uniqueness among all ideals: many ideals $I ⊋ rad R$ have semiprimitive quotients, for example every maximal ideal of a commutative ring.

Best Practices

State the hypothesis $𝔄 \subseteq rad R$ every time you invoke $(4.6)$ ; it is the single point where the argument can fail.
When claiming $rad R = J$ , present both halves: $J \subseteq rad R$ and $rad (R / J) = 0$ . Half an argument gives only a containment.
Prefer the nilness route to $J \subseteq rad R$ ; it is one line and needs no chain condition.
Record what the quotient forgets before you pass to it — nilpotents, non-split extensions, Loewy structure — so that a later lifting step is not a surprise.

Quick Reference

Statement (4.6)

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

Always true

rad (R / 𝔄) \supseteq (rad R + 𝔄) / 𝔄

Fixed point

rad (R / rad R) = 0

Minimality

R / I

semiprimitive

\Rightarrow I \supseteq rad R

Surjections

f

onto

\Rightarrow f (rad R) \subseteq rad S

Failure mode

rad ℤ = 0

but

rad (ℤ / 4 ℤ) \neq 0

Computation recipe

J

nil and

R / J

semiprimitive

\Rightarrow rad R = J

Notation

J (R)

is a common alternative to

rad R

Worked values
$R$	$𝔄$	$rad (R / 𝔄)$	(4.6) applies?
$T_{3} (k)$	$J^{2}$ (corner entry)	$J / J^{2}$	yes
$ℤ / 72 ℤ$	$(12)$	$(6) / (12)$	yes
$k [x]$	$(x^{2})$	$(x) / (x^{2})$	no — radical grows
$ℤ$	$(4)$	$(2) / (4)$	no — radical grows
$k [[x]]$	$(x^{n})$	$(x) / (x^{n})$	yes

Frequently Asked Questions

Why does Lam call the proof of

(4.6)

immediate?

Because once you notice that the hypothesis forces every maximal left ideal to contain $𝔄$ , the statement is the correspondence theorem applied to an intersection. The content is entirely in the observation, not in the manipulation.

(4.6)

true for one-sided ideals

𝔄

The statement does not typecheck: $R / 𝔄$ is only a ring when $𝔄$ is two-sided. What is true is that the lattice argument works for the module $R / 𝔄$ and shows that the radical of that module is $(rad R) / 𝔄$ when $𝔄 \subseteq rad R$ , where the radical of a module means the intersection of its maximal submodules.

Does

R / rad R

determine

R

in any useful sense?

No, but it determines a lot: by $(4.8)$ the two rings have the same simple left modules, and an element of $R$ is invertible exactly when its image is. What is lost is everything nilpotent, and recovering $R$ from the quotient requires extra hypotheses such as idempotent lifting, which is the subject of the semiperfect pages.

How is

(4.6)

used to compute a radical when I cannot see a nil ideal?

It is not, directly. In that situation you fall back on the element characterisation or, for a finite-dimensional algebra, on a trace-form or Friedl–Rónyai computation. $(4.6)$ then re-enters as a verification step: whatever $J$ the algorithm returns, confirm $R / J$ is semiprimitive.

Is there an analogue for the other radicals?

Yes, and it is generally cleaner. The lower and upper nilradicals satisfy ${Nil}_{*} (R / {Nil}_{*} R) = 0$ and ${Nil}^{*} (R / {Nil}^{*} R) = 0$ , and each is the smallest ideal with the corresponding quotient property. The formal pattern — a radical is idempotent as an operation on rings — is what abstract radical theory axiomatises.

Can

R / 𝔄

be semiprimitive for two different ideals inside the radical?

No. If $I \subseteq rad R$ and $R / I$ is semiprimitive then minimality gives $rad R \subseteq I$ , so $I = rad R$ . Uniqueness holds only under the containment hypothesis; above the radical there can be many semiprimitive quotients.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §4, (4.6) and Exercises 4.10–4.11 (pp. 54–55, 67–68).
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. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Show that for any ideal $𝔄$ of $R$ , $rad (R / 𝔄)$ is the intersection of the maximal left ideals of $R$ containing $𝔄$ , divided by $𝔄$ .
Give an example of a surjection $f : R \to S$ with $f (rad R)$ strictly smaller than $rad S$ .
Does $rad (R \times S) = rad R \times rad S$ hold for infinite products, and how does that interact with quotients?
Formulate and prove the module-theoretic version: for $N \subseteq rad M$ , is $rad (M / N) = (rad M) / N$ ?
How does $(4.6)$ interact with the ring homomorphism $R \to M_{n} (R)$ and the identity $rad M_{n} (R) = M_{n} (rad R)$ ?
Which of the abstract Kurosh–Amitsur radical axioms does the Jacobson radical satisfy, and which fail for the Wedderburn radical?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

The correspondence, drawn out

One inclusion is free

06Key Results

07Proof Techniques and Method

Guess a nil ideal, then verify

Push the lattice, not the elements

Characterise by a universal property

08Worked Example

Upper triangular matrices in two lines

An arithmetic check

Where the hypothesis bites

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

The first step of every algebra decomposition

Brauer's reduction

Codes over finite chain rings

Complete local rings

13Computational Notes

14Failure Modes and Common Mistakes

15Best Practices

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

Worked Example

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Computational Notes

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions