Small (Superfluous) Submodules

Q: Is rad M always small in M?

No. It is small when M is finitely generated, and when R is right perfect it is small for every M. In general rad M is only the *sum* of the small submodules, and that sum can be all of M: for Q over Z, and for the Prüfer group, rad M = M.

Q: Does smallness depend on which side the module is on?

The definition is side-neutral in form — replace *right* by *left* everywhere and nothing changes. But the criteria that produce small submodules are not: MJ ⊆_s M for all right modules M requires J to be **right** T-nilpotent, and right perfect rings need not be left perfect. So the notion is symmetric; its supply is not.

Q: What is the relationship between small submodules and projective covers?

A projective cover of M is an epimorphism : P M from a projective module with ⊆_s P. Smallness of the kernel is exactly the minimality condition: it says no proper submodule of P already maps onto M, which is what makes the cover unique up to isomorphism.

Q: If S is small in M and M is small in a bigger module, is S small there?

Yes, and more simply than that: (24.2)(4) needs only S ⊆_s M' ⊆ M with no hypothesis on M' inside M. Transitivity is free in this direction. The direction that fails is going down: S ⊆_s M and S ⊆ M' ⊆ M do not force S ⊆_s M' without further information.

Q: Can a module be small in itself?

Only if it is zero. Taking N = 0 in the definition gives S + 0 = S, so S ⊆_s S forces 0 = S. This is why a projective cover of a nonzero module never has kernel equal to P, and why rad P P for nonzero projective P is a theorem worth proving.

Executive Summary

Bass's homological description of perfect and semiperfect rings rests on one elementary notion. A submodule $S \subseteq M$ is small (or superfluous) if it contributes nothing to any generating set: whenever $S + N = M$ for a submodule $N$ , already $N = M$ . Everything in §24 — the module radical, projective covers, and the two characterisation theorems — is built from this one definition.

Smallness is what survives when finite generation is removed. Nakayama's Lemma says $M J$ is small in a finitely generated $M$ for $J \subseteq rad R$ ; right T-nilpotence of $J$ buys the same conclusion for every module, and that is precisely the extra strength a right perfect ring supplies.

$S + N = M \Rightarrow N = M$ The definition

$0$ Only small submodule of a semisimple module

$M J$ Small when

J

is right T-nilpotent

(24.1)Lam's numbering

Overview

Throughout §24 modules are right $R$ -modules over a ring $R$ with identity, and $J$ denotes $rad R$ . Smallness is a relative notion: a module is never small in itself except when it is zero, and a submodule that is small in a submodule of $M$ is small in $M$ , but not conversely in any useful sense — smallness is inherited downwards and outwards, never upwards from a quotient.

S \subseteq_{s} M : ⟺ \forall N \subseteq M : S + N = M ⟹ N = M

(24.1)

The subscript $s$ notation is Lam's; Anderson-Fuller write $S ≪ M$ for the same relation.

The picture to carry is complement-free: $S$ is small exactly when it has no proper complement, in the weak sense that no proper submodule of $M$ can be enlarged to all of $M$ by adjoining $S$ . Direct summands are therefore the antithesis of small submodules, and the two notions collide productively in the uniqueness proof for projective covers.

The two consumers of this definition are the Radical of a Module, which turns out to be the sum of all small submodules, and Projective Covers, which are the epimorphisms with small kernel.

Learning Objectives

State $(24.1)$ and restate it as the absence of a proper submodule complementing $S$ .
Prove that a nonzero direct summand is never small, and identify the small submodules of a semisimple module.
Apply both halves of $(24.2) (2)$ : $M J \subseteq_{s} M$ for $M$ finitely generated, and for arbitrary $M$ when $J$ is right T-nilpotent.
Use the closure properties of $(24.2) (3)$ – $(24.2) (5)$ fluently, including the modular-law argument behind transitivity.
Explain why every maximal submodule contains every small submodule.
Compute the small submodules of $ℤ / 12 ℤ$ , of $ℚ$ , and of the Prüfer group $ℤ (p^{\infty})$ over $ℤ$ .

Definitions

Definition(24.1)Small (superfluous) submodule

Let $M$ be a right module over a ring $R$ . A submodule $S \subseteq M$ is small, or superfluous, in $M$ if for every submodule $N \subseteq M$ the equality $S + N = M$ forces $N = M$ . We write $S \subseteq_{s} M$ .

$S \subseteq_{s} M$: Lam's notation for * $S$ is small in $M$ *. Anderson–Fuller and Wisbauer write $S ≪ M$ ; Kasch says superfluous. The three are identical.
Right T-nilpotent: A subset $J \subseteq R$ is right T-nilpotent if for every sequence $a_{1}, a_{2}, \dots \in J$ there exists $n$ with $a_{n} a_{n - 1} \dots a_{1} = 0$ . Note the order: the sequence is applied on the left.
$rad M$: The intersection of all maximal submodules of $M$ , and $rad M = M$ when $M$ has none. For $M = R_{R}$ this is the Jacobson radical.
Direct summand: A submodule $S$ with $M = S \oplus N$ for some submodule $N$ ; equivalently the image of an idempotent in $End (M)$ .
Projective cover: An epimorphism $θ : P ↠ M$ with $P$ projective and $\ker θ \subseteq_{s} P$ .

Modules are unital right modules and rings have an identity. The empty sum convention gives $0 \subseteq_{s} M$ for every $M$ , and $M \subseteq_{s} M$ only when $M = 0$ .

Core Concepts

Smallness as a generation test

If $M = \sum_{i} x_{i} R$ and $S \subseteq_{s} M$ , then any subset of the $x_{i}$ whose $R$ -span together with $S$ is all of $M$ already spans $M$ . In other words, elements of a small submodule are never needed as generators. This is the working meaning of the definition and the reason smallness appears wherever minimal generating sets do.

S \subseteq_{s} M

S

contributes no generators

S \subseteq

every maximal submodule

S \subseteq rad M

The last arrow is not reversible in general: $rad M$ is the sum of all small submodules, and an infinite sum of small submodules can fail to be small. It is reversible when $M$ is finitely generated.

Two sources of small submodules

In practice small submodules are produced in exactly two ways, and both are Nakayama arguments. The first is classical and needs finite generation; the second removes that hypothesis at the cost of a strong condition on the ideal.

M finitely generated, J \subseteq rad R ⟹ M J \subseteq_{s} M

(24.2)(2a)

J \subseteq rad R right T-nilpotent ⟹ M J \subseteq_{s} M for every M_{R}

(24.2)(2b)

The second implication is the module-theoretic content of right perfectness; see the criterion (23.16).

Why summands are the enemy

If $M = S \oplus N$ with $S \neq 0$ then $S + N = M$ while $N \neq M$ , so $S$ is not small. Over a semisimple ring every submodule is a summand, so no nonzero submodule of any module is small — which is why $J$ -semisimple rings admit projective covers only for modules that are already projective.

Key Results

Proposition(24.2)(1)Summands are never small

Let $M$ be a right $R$ -module and $S \subseteq M$ a direct summand with $S \neq 0$ . Then $S$ is not small in $M$ . Consequently, if $M$ is a semisimple module then $0$ is its only small submodule.

Proof

Write $M = S \oplus N$ . Then $S + N = M$ , but $N \neq M$ because $S \cap N = 0$ and $S \neq 0$ . So the defining implication fails. If $M$ is semisimple, every submodule is a direct summand, so a small submodule must be zero.

Proposition(24.2)(2)Nakayama criteria for smallness

Let $M$ be a right $R$ -module and let $J \subseteq rad R$ be a right ideal of $R$ . If either (a) $M$ is finitely generated, or (b) $J$ is right T-nilpotent, then $M J \subseteq_{s} M$ .

Proof

Suppose $M J + N = M$ with $N \subseteq M$ a submodule, and put $\bar{M} = M / N$ . Applying the quotient map gives $\bar{M} J = \bar{M}$ .

Case (a). If $M$ is finitely generated then so is $\bar{M}$ , and Nakayama's Lemma $(4.22)$ — valid for any right ideal inside $rad R$ — gives $\bar{M} = 0$ , i.e. $N = M$ .

Case (b). If $J$ is right T-nilpotent, then by the criterion $(23.16)$ the equality $\bar{M} J = \bar{M}$ forces $\bar{M} = 0$ for any right module, finitely generated or not. Again $N = M$ .

In both cases every submodule $N$ with $M J + N = M$ equals $M$ , which is the definition of $M J \subseteq_{s} M$ .

Lemma(24.2)(3)Downward closure and finite sums

Let $M$ be a right $R$ -module. (i) If $S \subseteq_{s} M$ and $S' \subseteq S$ then $S' \subseteq_{s} M$ . (ii) If $S_{1}, \dots, S_{n} \subseteq_{s} M$ then $S_{1} + \dots + S_{n} \subseteq_{s} M$ . The finiteness in (ii) cannot be dropped.

Proof

(i) If $S' + N = M$ then $S + N \supseteq S' + N = M$ , so $S + N = M$ and hence $N = M$ .

(ii) It suffices to treat $n = 2$ and induct. Suppose $(S_{1} + S_{2}) + N = M$ . Read this as $S_{1} + (S_{2} + N) = M$ ; smallness of $S_{1}$ gives $S_{2} + N = M$ , and smallness of $S_{2}$ then gives $N = M$ .

Lemma(24.2)(4)Transitivity along a submodule

Let $S \subseteq M' \subseteq M$ be right $R$ -modules. If $S \subseteq_{s} M'$ then $S \subseteq_{s} M$ .

Proof

Let $N \subseteq M$ satisfy $S + N = M$ . Intersect with $M'$ and use the modular law, legitimate because $S \subseteq M'$ :

M' = M' \cap (S + N) = S + (M' \cap N) .

(P.1)

Since $S \subseteq_{s} M'$ , this forces $M' \cap N = M'$ , that is $M' \subseteq N$ . In particular $S \subseteq N$ , so $M = S + N = N$ .

Corollary(24.2)(5)Finite direct sums

If $S_{i} \subseteq_{s} M_{i}$ for $1 \leq i \leq n$ , then $⨁_{i = 1}^{n} S_{i} \subseteq_{s} ⨁_{i = 1}^{n} M_{i}$ .

Proof

Each $S_{i}$ is small in $M_{i}$ , and $M_{i}$ is a submodule of $M = ⨁_{j} M_{j}$ , so $S_{i} \subseteq_{s} M$ by $(24.2) (4)$ . A finite sum of small submodules of $M$ is small by $(24.2) (3)$ , and that sum is $⨁_{i} S_{i}$ .

Proposition(24.2)(6)Maximal submodules absorb small ones

If $N$ is a maximal submodule of $M$ and $S \subseteq_{s} M$ , then $S \subseteq N$ . Hence every small submodule of $M$ lies in $rad M$ .

Proof

If $S not \subseteq N$ then $S + N$ is a submodule strictly containing $N$ , so $S + N = M$ by maximality. Smallness gives $N = M$ , contradicting properness of a maximal submodule. Intersecting over all maximal $N$ gives $S \subseteq rad M$ .

Proposition—Homomorphic images of small submodules

Let $f : M \to N$ be a homomorphism of right $R$ -modules and $S \subseteq_{s} M$ . Then $f (S) \subseteq_{s} N$ . (This standard complement to $(24.2)$ is Anderson–Fuller $(5.18)$ ; Lam uses the special case where $f$ is the inclusion of a submodule.)

Proof

Let $L \subseteq N$ with $f (S) + L = N$ . Given $x \in M$ , write $f (x) = f (s) + ℓ$ with $s \in S$ , $ℓ \in L$ ; then $f (x - s) = ℓ \in L$ , so $x - s \in f^{- 1} (L)$ and $x \in S + f^{- 1} (L)$ . Hence $S + f^{- 1} (L) = M$ , and smallness of $S$ gives $f^{- 1} (L) = M$ , i.e. $f (M) \subseteq L$ . Then $f (S) \subseteq L$ as well, so $N = f (S) + L = L$ .

Proof Techniques and Method

How these proofs work, and which move to reuse.

The arguments above are short because they all reduce to one of three manoeuvres.

Move 1

Quotient and apply Nakayama

To prove $S \subseteq_{s} M$ , assume $S + N = M$ and pass to $M / N$ . The hypothesis becomes a statement about a module killed by its own radical action, and Nakayama — in its finitely generated or its T-nilpotent form — finishes.

Move 2

Re-bracket the sum

Smallness of a sum is proved by reading $S_{1} + S_{2} + N = M$ as $S_{1} + (S_{2} + N) = M$ and peeling one summand at a time. The induction is on the number of summands, which is why infinitude breaks it.

Move 3

Cut down by the modular law

To move smallness from $M'$ to $M$ , intersect the equation $S + N = M$ with $M'$ . Because $S \subseteq M'$ , modularity yields $M' = S + (M' \cap N)$ and the hypothesis applies inside $M'$ .

Move 1 is the one that recurs throughout §24: the proof that $rad P ⊊ P$ for projective $P$ , the construction of projective covers over semiperfect rings, and the perfect-ring characterisation all run it. Move 3 is what makes smallness usable inside a direct sum decomposition, where one constantly slides between a summand and the whole.

Worked Example

A finite example: $ℤ / 12 ℤ$

Take $R = M = ℤ / 12 ℤ$ as a module over $ℤ$ (equivalently over itself). Its submodules are the cyclic groups generated by $0, 6, 4, 3, 2, 1$ , and its maximal submodules are $(2)$ and $(3)$ , of index $2$ and $3$ . Hence

rad (ℤ / 12 ℤ) = (2) \cap (3) = (6) = {0, 6} .

(E.1)

Since $M$ is finitely generated, the small submodules are exactly the submodules of $rad M$ , namely $0$ and $(6)$ . Check $(6)$ directly: $(6) + (2) = (2)$ , $(6) + (3) = (3)$ , $(6) + (4) = (2)$ , and $(6) + (6) = (6)$ — no proper submodule is enlarged to $M$ . By contrast $(4)$ is not small, because $(4) + (3) = M$ while $(3) \neq M$ ; consistently, $M ≅ ℤ / 4 \oplus ℤ / 3$ makes $(4) ≅ ℤ / 3$ a nonzero direct summand.

An infinite example: $ℚ$ over $ℤ$

Claim: $ℤ \subseteq_{s} ℚ$ as $ℤ$ -modules. Suppose $ℤ + N = ℚ$ for a subgroup $N \subseteq ℚ$ . If $N = 0$ then $ℚ = ℤ$ , false; so pick $0 \neq p / q \in N$ and note $a : = p \in N \cap ℤ$ is nonzero. For $r \geq 1$ write

\frac{1}{r a} = r' + b, r' \in ℤ, b \in N,

(E.2)

and multiply by $a$ : $\frac{1}{r} = a r' + a b$ . Both terms lie in $N$ — the first because $a \in N$ , the second because $N$ is a $ℤ$ -module — so $1 / r \in N$ for all $r$ , whence $N = ℚ$ . Multiplying by any nonzero rational is a $ℤ$ -automorphism of $ℚ$ , so every cyclic subgroup $\frac{a}{b} ℤ$ is small in $ℚ$ .

The Prüfer group

Let $M = ℤ (p^{\infty}) = ℤ [1 / p] / ℤ$ . Its proper submodules are the finite cyclic groups $C_{p^{k}}$ , and they form a chain $C_{p} \subset C_{p^{2}} \subset \dots$ whose union is $M$ . If $S + N = M$ with $S, N$ proper, then both are members of the chain, so one contains the other and $S + N$ is proper — a contradiction. Hence *every proper submodule of $ℤ (p^{\infty})$ is small*, $M$ has no maximal submodule, and $rad M = M$ .

Process and Workflow

Is $S$ small in $M$ ?

$S$ is a nonzero direct summandNo, immediately, by

(24.2) (1)

. This is the fastest disqualifier and covers every nonzero submodule of a semisimple module.

$M$ is finitely generatedThen

S \subseteq_{s} M

iff

S \subseteq rad M

. Compute

rad M

once and read off all small submodules at a stroke.

$S = M J$ with $J \subseteq rad R$ Yes if

M

is finitely generated, or if

J

is right T-nilpotent — that is, if

R

is right perfect and

J = rad R

$M$ has no maximal submoduleThen

rad M = M

and smallness must be checked by hand against the lattice of submodules;

ℚ

and

ℤ (p^{\infty})

are the model cases.

For a general module the practical route is: test the finitely generated criterion first, then look for a T-nilpotent ideal, and only then attack the submodule lattice directly.

Comparison and Classification

Small submodules of familiar modules
Module $M$ over $R$	$rad M$	Small submodules	Maximal submodules exist?
$ℤ / 12 ℤ$ over $ℤ$	$(6)$	$0$ and $(6)$	yes: $(2)$ , $(3)$
$ℤ$ over $ℤ$	$0$	$0$ only	yes: $(p)$ for each prime
$ℚ$ over $ℤ$	$ℚ$	every cyclic (indeed every finitely generated) submodule	no
$ℤ (p^{\infty})$ over $ℤ$	$ℤ (p^{\infty})$	every proper submodule	no
Any $M$ over a semisimple $R$	$0$	$0$ only	yes, in abundance
$M$ finitely generated, any $R$	$M rad R$	the submodules of $rad M$	yes
Any $M$ over a right perfect $R$	$M rad R$	the submodules of $M rad R$	yes when $M \neq 0$

The last two rows are the point of the section: finite generation and right perfectness are two different ways to force $rad M$ itself to be small, and the second works uniformly across all modules.

Relationship Map

$S \subseteq_{s} M$ — small submodule
- is implied by
  - $S \subseteq S'$ with $S' \subseteq_{s} M$
  - $S \subseteq_{s} M'$ for some $M' \subseteq M$
  - $S = M J$ , $M$ finitely generated, $J \subseteq rad R$
  - $S = M J$ , $J$ right T-nilpotent
- implies
  - $S \subseteq N$ for every maximal submodule $N$
  - $S \subseteq rad M$
  - $f (S) \subseteq_{s} N$ for every $f : M \to N$
  - $S$ is not a nonzero direct summand
- is preserved by
  - finite sums
  - finite direct sums
  - passage to a larger ambient module
- is NOT preserved by
  - infinite sums
  - infinite direct sums in general
  - passage to a quotient of $M$ in general

R

semisimple

R

semiperfect

R

right perfect

M J \subseteq_{s} M

for all

M

Reading the chain right to left: uniform smallness of $M rad R$ is one of the defining features of right perfect rings, semiperfect rings supply it only for finitely generated modules, and semisimple rings trivialise it by making $rad R = 0$ .

Applications and Industry Use

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

Smallness is infrastructure rather than a headline object; its value is that it makes minimality well-behaved in settings where no finiteness is available.

Minimal projective resolutions. Over a semiperfect ring, requiring each syzygy map to have small kernel produces a resolution with no redundant summands; the Betti numbers read off from it are genuine invariants. This underlies the computation of $Ext$ over group algebras and quiver algebras.
Quiver and path-algebra software. Systems that compute with finite-dimensional algebras (GAP's QPA, Magma, Sage) build projective covers of modules over basic algebras; smallness of the kernel is the correctness condition being enforced.
Representation theory of finite groups. For $k G$ with $char k = p$ dividing $| G |$ , the radical series and the socle series are both governed by which submodules are superfluous, and Brauer theory manipulates them constantly.
Lifting and approximation arguments. Statements of the form a solution mod a small submodule lifts to a solution are the module-theoretic version of Hensel-style lifting used in idempotent lifting and in block theory.

The honest summary: outside algebra this notion is not used directly. It is used constantly inside the algorithms that decompose modules, and those algorithms are what serve coding theory, symbolic computation and computational representation theory.

Failure Modes and Common Mistakes

Do not assume smallness passes to quotients: the image of a small submodule under an epimorphism is small in the image, but a submodule of $M / K$ can fail to be small even when its preimage is not.
Do not confuse $rad M$ (intersection of maximal submodules of the module $M$ ) with $rad R$ (a two-sided ideal). They agree only for $M = R_{R}$ .
Do not use the convention $rad M = M$ for a module with no maximal submodule as evidence that $M$ is small in itself — it is not, unless $M = 0$ .
Do not apply Nakayama's Lemma without checking that the right ideal really lies inside $rad R$ ; for $J not \subseteq rad R$ the conclusion fails at once.

Best Practices

State which of the two Nakayama criteria you are invoking; the finitely generated one and the T-nilpotent one have different scopes and mixing them silently is the commonest error in this area.
When proving smallness, always name the test submodule $N$ explicitly and quotient by it — the argument is then two lines rather than a search.
Check candidate small submodules against the summand test first: it is instantaneous and rules out most non-examples.
In a direct sum decomposition, move smallness outward with $(24.2) (4)$ before combining, not after; combining first invites an illegitimate infinite sum.
Record whether your ambient module is finitely generated. That single fact decides whether small and contained in the radical are the same condition.

Quick Reference

Definition

S \subseteq_{s} M iff (S + N = M \Rightarrow N = M)

Alternative notation

S ≪ M

; the word superfluous is a synonym

Never smallA nonzero direct summand

Always small

0

, and any submodule of a small submodule

ClosureFinite sums, finite direct sums, homomorphic images, larger ambient modules

Fails forInfinite sums

Nakayama form

M

finitely generated,

J \subseteq rad R

\Rightarrow M J \subseteq_{s} M

Perfect form

J

right T-nilpotent

\Rightarrow M J \subseteq_{s} M

for all

M

Link to radical

rad M

is the sum of all small submodules

The statements of (24.2) at a glance
Item	Statement	Hypotheses
(24.2)(1)	A nonzero direct summand is not small	none
(24.2)(2)	$M J \subseteq_{s} M$	$J \subseteq rad R$ a right ideal; $M$ finitely generated or $J$ right T-nilpotent
(24.2)(3)	Submodules of small are small; finite sums of small are small	finitely many summands
(24.2)(4)	$S \subseteq_{s} M' \subseteq M \Rightarrow S \subseteq_{s} M$	none
(24.2)(5)	$⨁_{i = 1}^{n} S_{i} \subseteq_{s} ⨁_{i = 1}^{n} M_{i}$	finitely many summands
(24.2)(6)	Every maximal submodule contains every small submodule	none

Frequently Asked Questions

Why is the definition phrased with sums rather than with intersections?

Because smallness is about generation. The dual notion — a submodule $E \subseteq M$ with $E \cap N = 0 \Rightarrow N = 0$ — is essentiality, and it governs injective hulls exactly as smallness governs projective covers. The two theories are formally dual, but not equally well behaved: injective hulls always exist, projective covers usually do not.

rad M

always small in

M

No. It is small when $M$ is finitely generated, and when $R$ is right perfect it is small for every $M$ . In general $rad M$ is only the sum of the small submodules, and that sum can be all of $M$ : for $ℚ$ over $ℤ$ , and for the Prüfer group, $rad M = M$ .

Does smallness depend on which side the module is on?

The definition is side-neutral in form — replace right by left everywhere and nothing changes. But the criteria that produce small submodules are not: $M J \subseteq_{s} M$ for all right modules $M$ requires $J$ to be right T-nilpotent, and right perfect rings need not be left perfect. So the notion is symmetric; its supply is not.

What is the relationship between small submodules and projective covers?

A projective cover of $M$ is an epimorphism $θ : P ↠ M$ from a projective module with $\ker θ \subseteq_{s} P$ . Smallness of the kernel is exactly the minimality condition: it says no proper submodule of $P$ already maps onto $M$ , which is what makes the cover unique up to isomorphism.

S

is small in

M

and

M

is small in a bigger module, is

S

small there?

Yes, and more simply than that: $(24.2) (4)$ needs only $S \subseteq_{s} M' \subseteq M$ with no hypothesis on $M'$ inside $M$ . Transitivity is free in this direction. The direction that fails is going down: $S \subseteq_{s} M$ and $S \subseteq M' \subseteq M$ do not force $S \subseteq_{s} M'$ without further information.

Can a module be small in itself?

Only if it is zero. Taking $N = 0$ in the definition gives $S + 0 = S$ , so $S \subseteq_{s} S$ forces $0 = S$ . This is why a projective cover of a nonzero module never has kernel equal to $P$ , and why $rad P ⊊ P$ for nonzero projective $P$ is a theorem worth proving.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §24, (24.1)–(24.2) (pp. 358–359).
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §5 and §9 (superfluous submodules and projective covers).
H. Bass, “Finitistic dimension and a homological generalization of semi-primary rings”, Transactions of the American Mathematical Society 95 (1960), 466–488.
F. Kasch, Modules and Rings, London Mathematical Society Monographs 17, Academic Press, 1982, Chapter 5.
R. Wisbauer, Foundations of Module and Ring Theory, Gordon and Breach, 1991, §19 and §21.

AI Suggested Questions

Give a proof that the sum of all small submodules of $M$ equals $rad M$ , and identify exactly where finite generation is used.
Construct a module $M$ and small submodules $S_{1}, S_{2}, \dots$ whose infinite direct sum is not small in the corresponding infinite direct sum.
Dualise: state the definition of an essential submodule and compare the existence theory of injective hulls with that of projective covers.
Which rings have the property that $0$ is the only small submodule of every module, and how is that class characterised?
Show that for a finitely generated module $M$ , a submodule is small if and only if it lies in $rad M$ .
Explain why right T-nilpotence of $rad R$ , rather than nilpotence, is the correct hypothesis for uniform smallness of $M rad R$ .
How do small submodules behave under Morita equivalence, and is superfluity a categorical notion?

Small Submodules

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Smallness as a generation test

Two sources of small submodules

Why summands are the enemy

Key Results

Proof Techniques and Method

Quotient and apply Nakayama

Re-bracket the sum

Cut down by the modular law

Worked Example

A finite example: $ℤ / 12 ℤ$

An infinite example: $ℚ$ over $ℤ$

The Prüfer group

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Smallness as a generation test

Two sources of small submodules

Why summands are the enemy

06Key Results

07Proof Techniques and Method

Quotient and apply Nakayama

Re-bracket the sum

Cut down by the modular law

08Worked Example

A finite example: ℤ/12ℤ

An infinite example: ℚ over ℤ

The Prüfer group

09Process and Workflow

10Comparison and Classification

11Relationship Map

12Applications and Industry Use

13Failure Modes and Common Mistakes

14Best Practices

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

A finite example: $ℤ / 12 ℤ$

An infinite example: $ℚ$ over $ℤ$

Process and Workflow

Comparison and Classification

Relationship Map

Applications and Industry Use

Failure Modes and Common Mistakes

Best Practices

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions