The Radical of a Module

Q: Is rad M ever equal to ann(M) or to rad R?

No — the types differ. rad M is a submodule of M, ann(M) is an ideal of R, and rad R is an ideal of R. The only coincidence is that for M = R_R the module radical is the ideal rad R, because submodules of R_R are right ideals.

Q: Does rad behave well for quotients?

Only downwards. If N ⊆ rad M then rad(M/N) = (rad M)/N, because the maximal submodules of M/N are exactly the images of those of M. For a general N the radical of the quotient can be much larger: radZ = 0 but rad(Z/4Z) = (2)/(4).

Q: Does rad P = PJ hold for flat modules too?

Not in general. Q is flat over Z and QradZ = 0, while radQ = Q. The formula uses that P is a summand of a free module, which flatness does not supply. Over a right perfect ring, however, flat and projective coincide and the formula returns.

Q: How does the radical relate to minimal generating sets?

For finitely generated M over a semilocal ring, M/rad M = M/MJ is a semisimple module and lifting any of its generating sets gives a generating set of M by Nakayama. Minimal generating sets of M then correspond to minimal generating sets of M/MJ, which is where the invariance of the number of generators over a local ring comes from.

Executive Summary

The Jacobson radical of a ring has a module-theoretic twin. For a right $R$ -module $M$ , $rad M$ is the intersection of the maximal submodules of $M$ , with the convention $rad M = M$ when there are none. Two facts make it useful: it is exactly the sum of the small submodules of $M$ , and it always contains $M rad R$ , with equality whenever $R$ is semilocal.

The section's main theorem is Bass's observation that a nonzero projective module $P$ satisfies $rad P = P rad R ⊊ P$ — with no finiteness hypothesis whatever. That properness is the engine of every existence and uniqueness statement about projective covers.

$⋂ N_{\max}$ Definition

$\sum S_{small}$ Equivalent description

$P rad R ⊊ P$ Projective case (24.7)

$rad R$ The case

M = R_{R}

Overview

Write $J = rad R$ throughout. The definition of $rad M$ imitates the ring case verbatim, and for the right regular module $M = R_{R}$ the two notions literally coincide: the maximal submodules of $R_{R}$ are the maximal right ideals. Beyond that the module version is strictly wilder, because a module need not have any maximal submodule at all.

rad M = ⋂ {N \subseteq M : N a maximal submodule}, rad M : = M if there are none.

(24.3)

The convention is forced: an empty intersection of submodules of $M$ is $M$ .

Existence of maximal submodules is a Zorn's Lemma argument that needs a finiteness input. If $M \neq 0$ is finitely generated, a chain of proper submodules has proper union — a generating set cannot be swallowed at any finite stage — so maximal submodules exist and $rad M \neq M$ . Drop finite generation and this collapses: the $p$ -primary component of $ℚ / ℤ$ has no maximal submodule, and neither does $ℚ$ over $ℤ$ .

The payoff is $(24.7)$ : for projective modules the radical is computed by a single formula, $rad P = P J$ , and is never everything. Since a module admitting a projective cover inherits $rad M \neq M$ , this is also the first obstruction to the existence of covers.

Learning Objectives

State $(24.3)$ including the convention for modules with no maximal submodule.
Prove $(24.4) (1)$ : $rad M$ is the sum of all small submodules of $M$ .
Prove $M J \subseteq rad M$ and identify when equality holds.
Compute $rad$ of a submodule, of a direct sum, and of a free module using $(24.6)$ .
Reproduce the matrix-invertibility argument showing $rad P ⊊ P$ for nonzero projective $P$ .
Use $rad M = M$ as a certificate that $M$ is neither projective nor possessed of a projective cover.

Definitions

Definition(24.3)Radical of a module

For a right $R$ -module $M$ , $rad M$ denotes the intersection of all maximal submodules of $M$ . If $M$ possesses no maximal submodule, set $rad M = M$ . For $M = R_{R}$ this is the Jacobson radical $rad R$ .

$rad M$: The module radical. Also written $Rad (M)$ or $J (M)$ in the literature; ambiguity with the ring radical arises only when $M$ carries a ring structure of its own.
Maximal submodule: A proper submodule $N \subseteq M$ with $M / N$ simple.
Semilocal ring: $R / rad R$ is semisimple. Every left or right artinian ring, every semiperfect ring and every local ring is semilocal.
$M J$: The submodule of $M$ generated by all $m j$ with $m \in M$ , $j \in J$ , for a right ideal $J$ of $R$ .
$soc M$: The socle: the sum of all simple submodules. It is the dual invariant to the radical, obtained by replacing maximal with minimal.

Modules are unital right $R$ -modules; $J$ always abbreviates $rad R$ . Simple modules are nonzero by convention.

Core Concepts

Two descriptions, one object

The radical has a top-down description — intersect the maximal submodules — and a bottom-up one — add up the small submodules. The top-down version is the definition and is what one intersects with in proofs; the bottom-up version is what makes the radical computable and what connects it to the material on Small Submodules.

rad M = ⋂_{N maximal} N = \sum_{S \subseteq_{s} M} S

(24.4)(1)

Both descriptions are valid even when $M$ has no maximal submodule: then both sides equal $M$ .

The two-line reason the descriptions agree: a small submodule lies in every maximal submodule, so the sum is contained in the intersection; conversely, for $m \in rad M$ the cyclic module $m R$ is small, because a proper $N$ with $N + m R = M$ would produce a maximal submodule of the nonzero cyclic module $M / N$ , hence a maximal submodule of $M$ missing $m$ .

The ring acts through its own radical

Every simple quotient $M / N$ is killed by $J$ , so $M J \subseteq N$ for each maximal $N$ , hence $M J \subseteq rad M$ . Whether this inclusion is an equality is a question about $R$ rather than about $M$ : it is an equality for every $M$ as soon as $R$ is semilocal, since then $M / M J$ is a module over the semisimple ring $R / J$ and therefore has zero radical.

Why projectives are rigid

The radical commutes with arbitrary direct sums, so on a free module it is computed coordinatewise: $rad F = F J$ . Splitting off a projective summand transports the formula to $rad P = P J$ . What is not formal is that $P J \neq P$ ; for finitely generated $P$ that is Nakayama, but the general case needs the matrix computation of $(24.8)$ , and that computation is where $rad M_{n} (R) = M_{n} (rad R)$ earns its keep.

Key Results

Proposition(24.4)Radical as a sum of small submodules

Let $M$ be a right $R$ -module and $J = rad R$ . Then:

$rad M$ is the sum of all small submodules of $M$ ;
$M J \subseteq rad M$ , with equality for every $M$ if $R$ is semilocal.

Proof

(1). Let $T$ be the sum of all $S \subseteq_{s} M$ . Every small submodule lies in every maximal submodule by $(24.2) (6)$ , so $T \subseteq rad M$ .

For the reverse inclusion it is enough to show $m R \subseteq_{s} M$ for each $m \in rad M$ . Suppose $N + m R = M$ for a submodule $N$ , and assume $m \notin N$ . Then $M / N$ is a nonzero cyclic module, so it has a maximal submodule $N' / N$ by Zorn's Lemma. Its preimage $N'$ is a maximal submodule of $M$ containing $N$ ; and $m \notin N'$ , since $m \in N'$ would give $M = N + m R \subseteq N'$ . This contradicts $m \in rad M \subseteq N'$ . Hence $m \in N$ , so $M = N + m R = N$ , proving $m R \subseteq_{s} M$ and therefore $rad M \subseteq T$ .

(2). For a maximal submodule $N$ , the simple module $M / N$ is annihilated by $J$ , so $M J \subseteq N$ ; intersecting gives $M J \subseteq rad M$ . Now let $R$ be semilocal. The quotient $M / M J$ is a module over the semisimple ring $R / J$ , hence a direct sum of simple $R / J$ -modules, so $rad (M / M J) = 0$ . Since $M J \subseteq rad M$ , maximal submodules of $M / M J$ correspond to maximal submodules of $M$ , giving $rad (M / M J) = (rad M) / M J$ . Therefore $rad M = M J$ .

Example(24.5)A module equal to its own radical

Let $R$ be a commutative domain with quotient field $K ⊋ R$ . Then $rad (K_{R}) = K$ ; in particular $K$ has no maximal $R$ -submodule. The proof shows every cyclic $R$ -submodule $\frac{a}{b} R \subseteq K$ is small, and $K$ is the sum of these.

Proof

Multiplication by $b / a$ is an $R$ -module automorphism of $K$ carrying $\frac{a}{b} R$ to $R$ , so it suffices to prove $R \subseteq_{s} K$ . Let $N \subseteq K$ be an $R$ -submodule with $R + N = K$ . If $N = 0$ then $K = R$ , excluded by hypothesis; so choose $0 \neq x \in N$ and clear denominators to get $0 \neq a \in N \cap R$ . Given $0 \neq r \in R$ , write $\frac{1}{r a} = r' + β$ with $r' \in R$ and $β \in N$ . Multiplying by $a$ gives $\frac{1}{r} = a r' + a β$ , and both summands lie in $N$ because $a \in N$ and $N$ is an $R$ -module. Hence $\frac{s}{r} \in N$ for all $s \in R$ , $r \neq 0$ , i.e. $N = K$ . So $R \subseteq_{s} K$ , and $(24.4) (1)$ gives $rad K = K$ .

Proposition(24.6)Functorial behaviour of the radical

Let $R$ be a ring and $J = rad R$ .

If $M' \subseteq M$ are right $R$ -modules then $rad M' \subseteq rad M$ ;
$rad (⨁_{i \in I} M_{i}) = ⨁_{i \in I} rad M_{i}$ for any index set $I$ ;
if $F$ is a free right $R$ -module then $rad F = F J$ .

Proof

(1). By $(24.4) (1)$ , $rad M'$ is the sum of the submodules small in $M'$ ; each of these is small in $M$ by $(24.2) (4)$ ; so by $(24.4) (1)$ again their sum lies in $rad M$ .

(2). Containment $\supseteq$ follows from (1) applied to each $M_{i} \subseteq ⨁_{j} M_{j}$ . For $\subseteq$ , let $m = (m_{i}) \in rad (⨁_{j} M_{j})$ and fix $i$ . If $N \subseteq M_{i}$ is a maximal submodule, then $N \oplus ⨁_{j \neq i} M_{j}$ is a maximal submodule of the direct sum, so $m$ lies in it, forcing $m_{i} \in N$ . Intersecting over all maximal $N \subseteq M_{i}$ gives $m_{i} \in rad M_{i}$ (and if $M_{i}$ has none, $rad M_{i} = M_{i}$ and there is nothing to prove).

(3). Write $F = ⨁_{i \in I} e_{i} R$ with each $e_{i} R ≅ R_{R}$ . By (2), $rad F = ⨁_{i} rad (e_{i} R) = ⨁_{i} e_{i} J = F J$ .

Theorem(24.7)The radical of a projective module

Let $P$ be a nonzero projective right $R$ -module and $J = rad R$ . Then $rad P = P J$ and $P J ⊊ P$ . In particular every nonzero projective module has a maximal submodule.

Proof

Choose a module $Q$ with $F : = P \oplus Q$ free. By $(24.6) (2)$ and $(24.6) (3)$ ,

rad P \oplus rad Q = rad F = F J = P J \oplus Q J,

(P.1)

and since $rad P, P J \subseteq P$ while $rad Q, Q J \subseteq Q$ , comparing the $P$ -components gives $rad P = P J$ .

It remains to prove $P J \neq P$ ; for finitely generated $P$ this is Nakayama, but no such hypothesis is available. Suppose $P J = P$ and pick $0 \neq p \in P$ . Write $F = ⨁_{i \in I} e_{i} R$ and $p = \sum_{i = 1}^{n} e_{i} r_{i}$ , indexing so that the support of $p$ is ${1, \dots, n}$ . Let $π : F \to P$ be the projection along $Q$ . Since $π (e_{i}) \in P = P J \subseteq F J = ⨁_{i} e_{i} J$ , we may write

π (e_{i}) = \sum_{j = 1}^{m} e_{j} a_{j i}, a_{j i} \in J, m \geq n,

(P.2)

Enlarge $m$ so that all supports occurring for $i = 1, \dots, n$ are covered.

Applying $π$ to $p$ , which fixes $p$ , and collecting coefficients:

p = π (p) = \sum_{i = 1}^{n} π (e_{i}) r_{i} = \sum_{j = 1}^{m} e_{j} (\sum_{i = 1}^{n} a_{j i} r_{i}) .

(24.8)

Comparing with $p = \sum_{i = 1}^{n} e_{i} r_{i}$ and using freeness of $F$ on the $e_{i}$ gives, for $j = 1, \dots, n$ , the system $\sum_{i = 1}^{n} (δ_{j i} - a_{j i}) r_{i} = 0$ . Its coefficient matrix lies in

I_{n} + M_{n} (rad R) = I_{n} + rad M_{n} (R) \subseteq U (M_{n} (R)),

(P.3)

Using $rad M_{n} (R) = M_{n} (rad R)$ and the maximality property of the Jacobson radical.

so the matrix is invertible and $r_{1} = \dots = r_{n} = 0$ , forcing $p = 0$ — a contradiction. Hence $P J ⊊ P$ . Since $rad P \neq P$ , $P$ has at least one maximal submodule.

Corollary—A non-projectivity test

If $M \neq 0$ satisfies $rad M = M$ , then $M$ is not projective. For example $ℚ$ is not a projective $ℤ$ -module, and the Prüfer group $ℤ (p^{\infty})$ is not projective over $ℤ$ .

Proof Techniques and Method

How these proofs work, and which move to reuse.

Reduce to cyclic submodulesTo show

rad M \subseteq \sum S

, it suffices to prove

m R

is small for each

m

in the radical. Cyclic modules are nonzero and finitely generated, so Zorn applies to their quotients.

Manufacture a maximal submoduleGiven a proper

N

, the quotient

M / N

is cyclic when

N + m R = M

; a maximal submodule upstairs is then pulled back. This is the standard way to contradict membership in the radical.

Split off and compare componentsFor projective

P

, embed in a free

F = P \oplus Q

, apply an identity valid on

F

, and read off the

P

-component. Radical, socle and torsion arguments all use this.

Finish with matrix invertibilityA finite linear system whose coefficient matrix lies in

I_{n} + M_{n} (J)

has only the trivial solution, because

I_{n} + rad M_{n} (R)

consists of units.

Step 4 is the genuinely new move in this section. It replaces Nakayama's Lemma — which needs finite generation of the module — by finite generation of a single element's support, which is automatic in a direct sum. That substitution is the whole reason $(24.7)$ holds without finiteness assumptions.

Worked Example

A mixed direct sum over $ℤ$

Let $M = ℤ / 12 ℤ \oplus ℤ \oplus ℚ$ as a $ℤ$ -module. Apply $(24.6) (2)$ componentwise. The maximal submodules of $ℤ / 12$ are $(2)$ and $(3)$ , so its radical is $(6)$ ; the maximal submodules of $ℤ$ are the $p ℤ$ , whose intersection is $0$ ; and $rad (ℚ) = ℚ$ by $(24.5)$ with $R = ℤ$ , $K = ℚ$ . Hence

rad M = (6) / 12 ℤ \oplus 0 \oplus ℚ .

(E.1)

Note $M J = 0$ here, since $J = rad ℤ = 0$ , so the inclusion $M J \subseteq rad M$ is very strict. $ℤ$ is not semilocal, so $(24.4) (2)$ promises nothing more.

A local ring: $R = ℤ_{(p)}$

Let $R = ℤ_{(p)}$ , the localisation of $ℤ$ at a prime $p$ . This is a local ring with $J = p R$ and residue field $R / J ≅ 𝔽_{p}$ , hence semilocal, so $rad M = M J$ for every $R$ -module $M$ .

For $M = R$ : $rad R = p R$ , and $R / p R ≅ 𝔽_{p}$ is the unique simple module.
For $M = R / p^{3} R$ : $rad M = p M = (p) / (p^{3})$ , a module of length $2$ ; the radical series is $M ⊋ p M ⊋ p^{2} M ⊋ 0$ .
For $M = ℚ$ : every element of $ℚ$ is $p$ times another, so $ℚ p = ℚ$ and $rad (ℚ) = ℚ J = ℚ$ — matching $(24.5)$ , since $ℤ_{(p)}$ is a domain with quotient field $ℚ ⊋ ℤ_{(p)}$ .

Consistency check on a free module

Take $F = R^{(I)}$ free over $R = ℤ_{(p)}$ with $I$ infinite. Then $rad F = F J = p F$ , and $F / p F ≅ 𝔽_{p}^{(I)}$ is nonzero, confirming $rad F ⊊ F$ without any appeal to finite generation. The maximal submodules of $F$ are the preimages of the hyperplanes of the $𝔽_{p}$ -vector space $F / p F$ .

Comparison and Classification

Which properties of $rad M$ hold in which setting
	$M$ finitely generated	$M$ projective, nonzero	$R$ semilocal	$R$ right perfect	General $M$ , general $R$
$rad M \neq M$	yes	yes	partial	yes	no
$rad M = M J$	partial	yes	yes	yes	no
$rad M \subseteq_{s} M$	yes	partial	partial	yes	no
$M$ has a maximal submodule	yes	yes	partial	yes	no
$rad$ commutes with $⨁$	yes	yes	yes	yes	yes

Which properties of $rad M$ hold in which setting

The last row is the only unconditional entry, and it is the one that makes the projective case tractable. Note the second column: projectivity substitutes for finite generation in every row but the third, where smallness of the radical still needs an extra hypothesis on $R$ .

Ring radical versus module radical
Feature	$rad R$ (ring)	$rad M$ (module)
Definition	Intersection of maximal right ideals	Intersection of maximal submodules
Can equal the whole object?	Only if $R = 0$	Yes: $ℚ$ over $ℤ$
Left–right symmetry	Symmetric	Not applicable; fixed side
Element test	$1 - x y$ left-invertible for all $x$	No element test in general
Behaviour under $⨁$	$rad (R \times S) = rad R \times rad S$	$rad ⨁ M_{i} = ⨁ rad M_{i}$

Relationship Map

All modules $M_{R}$

M J \subseteq rad M

, and

rad M

is the sum of the small submodules

$R$ semilocal

rad M = M J

for every

M

$R$ semiperfect…and every finitely generated

M

has a projective cover

$R$ right perfect…and

rad M = M J \subseteq_{s} M

for every

M

, so every

M

has a projective cover

$R$ right artinian…and

J

is nilpotent, so the radical series terminates in finitely many steps

P

projective,

P \neq 0

rad P = P J

rad P ⊊ P

P

has a maximal submodule

Reading the chain backwards gives the contrapositive used in practice: a nonzero module equal to its own radical is not projective, and cannot admit a projective cover either, since a cover induces a bijection between the maximal submodules of $P$ and those of $M$ .

Computational Notes

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

For a module over a finite-dimensional algebra the radical is entirely mechanical, and for anything else it is generally not computable at all.

If $A$ is a finite-dimensional algebra over a field $k$ , then $A$ is artinian, hence semilocal, so $rad M = M rad A$ for every $A$ -module $M$ . Computing $rad M$ reduces to computing $rad A$ once and then forming a matrix image — $O (d^{2} n)$ field operations for $\dim_{k} M = d$ and $n$ algebra generators after the radical is known.
Computing $rad A$ itself is a nullspace computation for the trace form in characteristic $0$ , and the Friedl–Rónyai algorithm in characteristic $p$ ; both are polynomial time in $\dim_{k} A$ .
The radical series $M \supseteq rad M \supseteq {rad}^{2} M \supseteq \dots$ terminates in at most the nilpotency index of $rad A$ steps, giving the Loewy length; the Meataxe uses it to split modules into composition factors.
GAP's RadicalOfAlgebra and the QPA package's radical routines, Magma's JacobsonRadical, and Sage's radical() all implement this pattern; none of them accept infinite-dimensional input.
For modules over $ℤ$ or a general Noetherian ring, $rad M$ is not finitely presentable from a presentation of $M$ in any uniform way — $rad (ℚ) = ℚ$ shows the answer need not even be a proper submodule.

Failure Modes and Common Mistakes

Do not read $rad M$ as an annihilator: it is a submodule of $M$ , whereas $rad R$ is an ideal of $R$ . The two live in different places even when the notation looks the same.
Do not assume $rad (M / N) = (rad M) / N$ for arbitrary $N$ ; this needs $N \subseteq rad M$ , exactly as in the ring case.
Do not conclude that a module with a maximal submodule has small radical — $rad M \neq M$ and $rad M \subseteq_{s} M$ are different statements outside the finitely generated case.
Do not apply $(24.7)$ to $P = 0$ : the conclusion $rad P ⊊ P$ fails there for the trivial reason that both sides are $0$ .

Best Practices

Decide first whether $R$ is semilocal. If it is, use $rad M = M J$ and compute with the ring; if not, work with maximal submodules directly.
For direct sums, always compute the radical componentwise — it is the one unconditional rule available.
When you need $rad P ⊊ P$ , cite $(24.7)$ rather than Nakayama unless $P$ is known to be finitely generated.
Use $rad M = M$ as a fast certificate of non-projectivity before attempting any resolution argument.
Say module radical or ring radical explicitly in any write-up where $M$ carries an algebra structure; the notation $rad$ alone is genuinely ambiguous there.

Quick Reference

Definition

rad M = ⋂ {N : N maximal in M}

Empty caseNo maximal submodule

\Rightarrow rad M = M

Bottom-up form

rad M = \sum {S : S \subseteq_{s} M}

Ring action

M J \subseteq rad M

; equality if

R

is semilocal

Submodules

M' \subseteq M \Rightarrow rad M' \subseteq rad M

Direct sums

rad ⨁_{i} M_{i} = ⨁_{i} rad M_{i}

Free modules

rad F = F J

Projective modules

P \neq 0

projective

\Rightarrow rad P = P J ⊊ P

Certificate

rad M = M \neq 0 \Rightarrow M

not projective

Radicals of standard modules
Ring	Module	$rad M$
$ℤ$	$ℤ / n ℤ$	$m ℤ / n ℤ$ , where $m$ is the product of the distinct primes dividing $n$
$ℤ$	$ℤ$	$0$
$ℤ$	$ℚ$	$ℚ$
$ℤ_{(p)}$	$R / p^{k} R$	$p R / p^{k} R$
Any $R$	Free $F$	$F rad R$
Any $R$	Projective $P \neq 0$	$P rad R ⊊ P$
Semisimple $R$	Any $M$	$0$

Frequently Asked Questions

Why define

rad M = M

when there are no maximal submodules?

Because the intersection of an empty family of submodules of $M$ is $M$ , so the convention is the only consistent one. It also keeps $(24.4) (1)$ true in that case: a module with no maximal submodule is the sum of its small submodules, as $ℚ$ over $ℤ$ illustrates.

rad M

ever equal to

ann (M)

or to

rad R

No — the types differ. $rad M$ is a submodule of $M$ , $ann (M)$ is an ideal of $R$ , and $rad R$ is an ideal of $R$ . The only coincidence is that for $M = R_{R}$ the module radical is the ideal $rad R$ , because submodules of $R_{R}$ are right ideals.

Does

rad

behave well for quotients?

Only downwards. If $N \subseteq rad M$ then $rad (M / N) = (rad M) / N$ , because the maximal submodules of $M / N$ are exactly the images of those of $M$ . For a general $N$ the radical of the quotient can be much larger: $rad ℤ = 0$ but $rad (ℤ / 4 ℤ) = (2) / (4)$ .

What is the role of

rad M_{n} (R) = M_{n} (rad R)

in the proof of

(24.7)

It converts an infinite problem into a finite one. The element $p$ has finite support, so the obstruction is a single $n \times n$ linear system whose matrix is $I_{n}$ minus a matrix over $J$ . Because $M_{n} (J) = rad M_{n} (R)$ , that matrix is a unit of $M_{n} (R)$ and the system has only the zero solution.

Does

rad P = P J

hold for flat modules too?

Not in general. $ℚ$ is flat over $ℤ$ and $ℚ rad ℤ = 0$ , while $rad ℚ = ℚ$ . The formula uses that $P$ is a summand of a free module, which flatness does not supply. Over a right perfect ring, however, flat and projective coincide and the formula returns.

How does the radical relate to minimal generating sets?

For finitely generated $M$ over a semilocal ring, $M / rad M = M / M J$ is a semisimple module and lifting any of its generating sets gives a generating set of $M$ by Nakayama. Minimal generating sets of $M$ then correspond to minimal generating sets of $M / M J$ , which is where the invariance of the number of generators over a local ring comes from.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §24, (24.3)–(24.8) (pp. 359–361).
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §9 (the radical of a module).
H. Bass, “Finitistic dimension and a homological generalization of semi-primary rings”, Transactions of the American Mathematical Society 95 (1960), 466–488.
N. Jacobson, Structure of Rings, American Mathematical Society Colloquium Publications 37, revised edition, 1964, Chapter I.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 2.

AI Suggested Questions

Prove that $rad M \subseteq_{s} M$ whenever $M$ is finitely generated, and give a non-finitely-generated counterexample.
Show by example that $rad$ does not commute with infinite direct products of modules.
Work out the radical and socle series of the regular module over the group algebra $𝔽_{p} C_{p^{2}}$ .
Give a self-contained proof that $rad M_{n} (R) = M_{n} (rad R)$ and explain why it is a Morita-invariance statement.
Characterise the rings over which $rad M = M rad R$ holds for every module, and compare with the semilocal condition.
Dualise the theory: define the socle, and state the analogue of $(24.7)$ for injective modules.
Which nonzero modules over $ℤ$ satisfy $rad M = M$ , and how are they classified?

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Two descriptions, one object

The ring acts through its own radical

Why projectives are rigid

06Key Results

07Proof Techniques and Method

08Worked Example

A mixed direct sum over ℤ

A local ring: R=ℤ(p)

Consistency check on a free module

09Comparison and Classification

10Relationship Map

11Computational Notes

12Failure Modes and Common Mistakes

13Best Practices

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Proof Techniques and Method

Worked Example

A mixed direct sum over $ℤ$

A local ring: $R = ℤ_{(p)}$

Comparison and Classification

Relationship Map

Computational Notes

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions