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✓ 7 results · all verified · 0 also independently AI-judged

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Modules, Submodules, Quotient Modules and the Isomorphism Theorems

1 · Prerequisites

2 · Summary

Rings and ideals provide the scalars and quotient constructions used here, while vector spaces supply the motivating special case over a field. Earlier group results establish the pattern for kernels, quotient maps, factorisations, and first isomorphism theorems.

This page defines left and right modules, submodules, generated and free modules, simple modules, annihilators, torsion, quotient modules, and module homomorphisms. It proves the submodule criterion, kernel and image laws, quotient-module laws, the quotient universal property, and the first isomorphism theorem for modules.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-03Open item page →

Unital left and right modules over a ring; unqualified module means left module

Definition

Let $R$ be a ring. A left $R$ -module is an abelian group $(M, +, 0_{M})$ with a scalar action $R \times M \to M$ , $(r, m) \mapsto r m$ , satisfying

$r (m + n) = r m + r n, (r + s) m = r m + s m, (r s) m = r (s m), 1_{R} m = m .$

A right $R$ -module has an action $M \times R \to M$ , $(m, r) \mapsto m r$ , with the analogous right-handed axioms. Unless “right” is stated, module means a unital left module.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

In a module, $0_{R} m = 0_{M}$ , $r 0_{M} = 0_{M}$ , $(- r) m = - (r m)$ and $r (- m) = - (r m)$

Statement

For every left $R$ -module $M$ , scalar $r \in R$ , and element $m \in M$ ,

$0_{R} m = 0_{M}, r 0_{M} = 0_{M}, (- r) m = - (r m), r (- m) = - (r m) .$

Facts & Assumptions

Given: A left $R$ -module $M$ , $r \in R$ , and $m \in M$ .

[L1]

The module action distributes over both addition operations, and $(M, +, 0_{M})$ is an abelian group (Unital left and right modules over a ring; unqualified module means left module).

[L2]

The additive structure $(R, +, 0_{R})$ is an abelian group, so $r + (- r) = 0_{R}$ (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L3]

Cancellation holds in every group, hence in the additive groups of $R$ and $M$ (Cancellation in a group: $g x = g y$ or $x g = y g$ forces $x = y$ ; equivalently left and right translation by $g$ are bijections of $G$ , so $g x = h$ and $x g = h$ each have exactly one solution).

Proof

technique · direct

1.1

Since $0_{R} = 0_{R} + 0_{R}$ , distributivity gives $0_{R} m = 0_{R} m + 0_{R} m$ ; cancellation yields $0_{R} m = 0_{M}$ .

L1L3given

1.2

Since $0_{M} = 0_{M} + 0_{M}$ , distributivity gives $r 0_{M} = r 0_{M} + r 0_{M}$ ; cancellation yields $r 0_{M} = 0_{M}$ .

L1L3given

2.1

From $(r + (- r)) m = 0_{R} m = 0_{M}$ , distributivity and step 1.1 give $r m + (- r) m = 0_{M}$ , so $(- r) m = - (r m)$ .

step 1.1L1L2L3given

3.1

From $r (m + (- m)) = r 0_{M} = 0_{M}$ , distributivity and step 1.2 give $r m + r (- m) = 0_{M}$ , so $r (- m) = - (r m)$ .

step 1.2L1L3given∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-03Open item page →

Submodule of a module

Definition

Let $M$ be a left $R$ -module. A subset $N \subseteq M$ is a submodule when it is a subgroup of the additive group of $M$ and is closed under scalars:

$r \in R, n \in N ⟹ r n \in N .$

The operations on $N$ are the restrictions of those of $M$ . Write $N \leq M$ when the ring and module are understood.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

The one-step submodule criterion; intersections and sums of submodules are submodules

Statement

Let $M$ be a left $R$ -module. A nonempty subset $S \subseteq M$ is a submodule if and only if

$r u + v \in S (r \in R, u, v \in S) .$

Consequently, the intersection of every nonempty family of submodules is a submodule, and, for submodules $A, B \leq M$ ,

$A + B : = {a + b : a \in A, b \in B}$

is a submodule of $M$ .

Facts & Assumptions

Given: A left $R$ -module $M$ .

[L1]

The module axioms include $1_{R} m = m$ and distributivity of scalar multiplication over both additions (Unital left and right modules over a ring; unqualified module means left module).

[L2]

In a module, $(- r) m = - (r m)$ ; taking $r = 1_{R}$ and using [L1] gives $(- 1_{R}) m = - m$ (In a module, $0_{R} m = 0_{M}$ , $r 0_{M} = 0_{M}$ , $(- r) m = - (r m)$ and $r (- m) = - (r m)$ ).

[L3]

A nonempty subset of a group is a subgroup exactly when it is closed under $x y^{- 1}$ (One-step subgroup test: a nonempty $H \subseteq G$ is a subgroup iff $g h^{- 1} \in H$ for all $g, h \in H$ ; the identity and the inverses of $H$ are then those of $G$ ).

[L4]

A submodule is an additive subgroup closed under scalar multiplication (Submodule of a module).

Proof

technique · direct

1.1

If $S$ is a submodule, then $u \in S$ implies $r u \in S$ , and then $r u + v \in S$ for $r \in R$ and $u, v \in S$ .

L4given

1.2

Conversely, suppose the displayed closure condition holds and choose $s \in S$ . With $r = - 1_{R}$ and both elements equal to $s$ , it gives $(- 1_{R}) s + s = 0_{M} \in S$ .

L1L2given

1.3

If $u, v \in S$ , the same condition with scalar $- 1_{R}$ , first element $v$ , and second element $u$ gives $(- 1_{R}) v + u = u - v \in S$ .

L1L2given

2.1

The additive subgroup test applies by steps 1.2--1.3; scalar closure follows from the displayed condition with $v = 0_{M}$ . Thus $S$ is a submodule.

step 1.2step 1.3L3L4given

3.1

For a nonempty family $(N_{i})$ of submodules, $0_{M}$ lies in every $N_{i}$ ; and if $u, v$ lie in their intersection, then $r u + v$ lies in every $N_{i}$ . The criterion proves $⋂_{i} N_{i}$ is a submodule.

step 2.1L4given

4.1

For $x = a + b$ and $y = a^{'} + b^{'}$ in $A + B$ , distributivity gives $r x + y = (r a + a^{'}) + (r b + b^{'}) \in A + B$ ; moreover $0_{M} = 0_{M} + 0_{M} \in A + B$ . The criterion proves $A + B$ is a submodule.

step 2.1L1L4given∎

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-03Open item page →

Generated submodule, cyclic and finitely generated modules, module basis and free module

Definition

Let $M$ be a left $R$ -module and $S \subseteq M$ . The submodule generated by $S$ is

$⟨ S ⟩_{R} : = ⋂ {N \leq M : S \subseteq N} .$

The family is nonempty because $M \leq M$ , and its intersection is a submodule by The one-step submodule criterion; intersections and sums of submodules are submodules. Thus $⟨ S ⟩_{R}$ is the smallest submodule of $M$ containing $S$ , just as The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups defines a generated subgroup.

The module $M$ is cyclic if $M = ⟨ m ⟩_{R}$ for some $m \in M$ , and finitely generated if $M = ⟨ S ⟩_{R}$ for some finite $S$ . A subset $B \subseteq M$ is a basis if every element of $M$ has a unique expression as a finite $R$ -linear combination of elements of $B$ . A module possessing a basis is a free $R$ -module.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-03Open item page →

Simple module: a nonzero module with no proper nonzero submodule

Definition

A left $R$ -module $M$ is simple if $M \neq 0$ and its only submodules are $0$ and $M$ . Equivalently, $M$ has no proper nonzero submodule.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-03Open item page →

Annihilators, torsion elements and the torsion subset of a module

Definition

For a left $R$ -module $M$ and $m \in M$ , the annihilator of $m$ is

${Ann}_{R} (m) : = {r \in R : r m = 0_{M}},$

and the annihilator of $M$ is

${Ann}_{R} (M) : = {r \in R : r m = 0_{M} for every m \in M} .$

If $R$ is an integral domain, an element $m \in M$ is a torsion element when $r m = 0_{M}$ for some nonzero $r \in R$ . The set of all torsion elements is denoted $Tor (M)$ ; $M$ is torsion-free when $Tor (M) = {0_{M}}$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-03Open item page →

Quotient module $M / N$ with scalar multiplication on additive cosets

Definition

Let $N \leq M$ be a submodule of a left $R$ -module. Since the additive group of $M$ is abelian, $N$ is normal, so the additive quotient group $M / N$ consists of the cosets $m + N$ . Its proposed scalar action is

$r (m + N) : = r m + N .$

The well-definedness and module laws are established in The quotient action is well defined and makes $M / N$ a module ↗. The resulting module is the quotient module $M / N$ .

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-03Open item page →

Module homomorphism and isomorphism, kernel, image and cokernel

Definition

For left $R$ -modules $M, N$ , a function $f : M \to N$ is an $R$ -module homomorphism if

$f (m + m^{'}) = f (m) + f (m^{'}) and f (r m) = r f (m)$

for all $m, m^{'} \in M$ and $r \in R$ . It is a module isomorphism if it is a bijective module homomorphism.

Its kernel and image are

$\ker f : = {m \in M : f (m) = 0_{N}}, im f : = {f (m) : m \in M} .$

Once Kernels and images of module homomorphisms are submodules, and injectivity is equivalent to trivial kernel ↗ establishes that the image is a submodule, the cokernel of $f$ is the quotient module $coker f : = N / im f$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

Kernels and images of module homomorphisms are submodules, and injectivity is equivalent to trivial kernel

Statement

For an $R$ -module homomorphism $f : M \to N$ , both $\ker f$ and $im f$ are submodules. Moreover,

$f is injective ⟺ \ker f = {0_{M}} .$

Facts & Assumptions

Given: A homomorphism $f : M \to N$ of left $R$ -modules.

[L1]

A module homomorphism preserves addition and scalar multiplication, and the displayed definitions give its kernel and image (Module homomorphism and isomorphism, kernel, image and cokernel).

[L2]

A nonempty subset of a module is a submodule exactly when it is closed under $r u + v$ (The one-step submodule criterion; intersections and sums of submodules are submodules).

[L3]

The additive underlying function of a module homomorphism is a group homomorphism and therefore sends $0_{M}$ to $0_{N}$ ; moreover $r 0_{N} = 0_{N}$ (Module homomorphism and isomorphism, kernel, image and cokernel, A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed, In a module, $0_{R} m = 0_{M}$ , $r 0_{M} = 0_{M}$ , $(- r) m = - (r m)$ and $r (- m) = - (r m)$ ).

[L4]

A group homomorphism is injective exactly when its kernel is trivial (A group homomorphism is injective if and only if its kernel is trivial).

Proof

technique · direct

1.1

The kernel contains $0_{M}$ because $f (0_{M}) = 0_{N}$ . If $u, v \in \ker f$ and $r \in R$ , then $f (r u + v) = r f (u) + f (v) = 0_{N}$ , so $r u + v \in \ker f$ .

L1L3given

1.2

The image contains $0_{N} = f (0_{M})$ . If $f (u), f (v) \in im f$ and $r \in R$ , then $r f (u) + f (v) = f (r u + v)$ lies in the image.

L1L3given

1.3

The additive underlying function has the same kernel described in [L1], so the group-homomorphism theorem gives $f$ injective exactly when $\ker f = {0_{M}}$ .

L1L4given

2.1

The submodule criterion applied to steps 1.1 and 1.2 proves that $\ker f$ and $im f$ are submodules.

step 1.1step 1.2L2

3.1

Together, steps 1.3 and 2.1 prove both assertions.

step 1.3step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

The quotient action is well defined and makes $M / N$ a module

Statement

Let $N \leq M$ be a submodule of a left $R$ -module. The rule

$r (m + N) : = r m + N$

is independent of the representative $m + N$ and, together with the additive quotient group, makes $M / N$ a left $R$ -module.

Facts & Assumptions

Given: A left $R$ -module $M$ and a submodule $N \leq M$ .

[L1]

Additive cosets satisfy $m + N = m^{'} + N$ exactly when $m^{'} - m \in N$ ( $x \in a H$ iff $a^{- 1} x \in H$ , and $a H = b H$ iff $a^{- 1} b \in H$ ).

[L2]

A submodule is closed under scalar multiplication and is an additive subgroup (Submodule of a module).

[L3]

The additive cosets form a quotient group with the inherited addition (For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$ ).

[L4]

The module axioms give distributivity, associativity of scalar action, and $1_{R} m = m$ (Unital left and right modules over a ring; unqualified module means left module).

[L5]

Scalar multiplication preserves additive negatives (In a module, $0_{R} m = 0_{M}$ , $r 0_{M} = 0_{M}$ , $(- r) m = - (r m)$ and $r (- m) = - (r m)$ ).

Proof

technique · direct

1.1

If $m + N = m^{'} + N$ , then $m^{'} - m \in N$ by [L1]. Thus $r (m^{'} - m) = r m^{'} - r m \in N$ by [L2, L4, L5], and [L1] gives $r m + N = r m^{'} + N$ ; the proposed scalar action is well defined.

L1L2L4L5given

1.2

The quotient addition is an abelian group operation: it is a quotient-group operation by [L3], and $(m + N) + (m^{'} + N) = (m + m^{'}) + N = (m^{'} + m) + N = (m^{'} + N) + (m + N)$ .

L3L4given

1.3

For cosets, the module identities in $M$ give $r ((m + N) + (m^{'} + N)) = r (m + m^{'}) + N = (r m + r m^{'}) + N = r (m + N) + r (m^{'} + N)$ , $(r + s) (m + N) = r (m + N) + s (m + N)$ , $(r s) (m + N) = r (s (m + N))$ , and $1_{R} (m + N) = m + N$ .

L4given

2.1

Steps 1.1--1.3 verify a well-defined scalar action on an abelian group satisfying all module axioms; hence $M / N$ is a left $R$ -module.

step 1.1step 1.2step 1.3L4∎

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

The canonical map $M \to M / N$ is a surjective module homomorphism with kernel $N$ ; thus every submodule is a kernel

Statement

For a submodule $N \leq M$ , the canonical map

$π : M ⟶ M / N, π (m) : = m + N,$

is a surjective $R$ -module homomorphism and has kernel $N$ . Hence every submodule is the kernel of a module homomorphism.

Facts & Assumptions

Given: A left $R$ -module $M$ and a submodule $N \leq M$ .

[L1]

The quotient action is a well-defined module action (The quotient action is well defined and makes $M / N$ a module).

[L2]

The canonical projection of the underlying additive groups is a surjective group homomorphism (The canonical projection $π : G \to G / N$ , $π (g) = g N$ , is a surjective group homomorphism).

[L3]

A module homomorphism preserves addition and scalar multiplication, and its kernel is the inverse image of zero (Module homomorphism and isomorphism, kernel, image and cokernel).

[L4]

The additive group of a submodule is closed under inverses, and the coset criterion consequently gives $m + N = N$ if and only if $m \in N$ (Submodule of a module, $x \in a H$ iff $a^{- 1} x \in H$ , and $a H = b H$ iff $a^{- 1} b \in H$ ).

Proof

technique · direct

1.1

The underlying additive map $π$ is a homomorphism and is surjective.

L2given

1.2

The quotient action gives $π (r m) = r m + N = r (m + N) = r π (m)$ .

L1given

1.3

Since $π (m) = 0_{M / N} = N$ exactly when $m \in N$ , its kernel is $N$ .

L3L4given

2.1

Steps 1.1--1.2 show that $π$ is a surjective module homomorphism.

step 1.1step 1.2L3

3.1

Together with step 1.3, this proves the claim and shows that every submodule is a kernel.

step 1.3step 2.1∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

A module homomorphism vanishing on $N$ factors uniquely through $M / N$

Statement

Let $f : M \to P$ be a module homomorphism and let $N \leq M$ satisfy $N \subseteq \ker f$ . There is a unique module homomorphism

$\overset{ˉ}{f} : M / N ⟶ P$

such that $\overset{ˉ}{f} (m + N) = f (m)$ , equivalently $f = \overset{ˉ}{f} \circ π$ .

Facts & Assumptions

Given: A module homomorphism $f : M \to P$ and a submodule $N \leq M$ with $N \subseteq \ker f$ .

[L1]

The canonical map $π : M \to M / N$ is a surjective module homomorphism with kernel $N$ (The canonical map $M \to M / N$ is a surjective module homomorphism with kernel $N$ ; thus every submodule is a kernel).

[L2]

A group homomorphism that kills a normal subgroup factors uniquely through the group quotient (A homomorphism that kills a normal subgroup factors uniquely through the quotient group).

[L3]

A module homomorphism is additive and scalar-preserving, and its kernel is the preimage of zero (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · constructive

1.1

By [L1], the canonical map $π : M \to M / N$ is an additive-group quotient map with kernel $N$ . Viewing the modules as additive groups, [L3] makes $f$ a group homomorphism and the hypothesis says it kills $N$ . By [L2], define the unique additive homomorphism $\overset{ˉ}{f}$ by $\overset{ˉ}{f} (m + N) = f (m)$ , with $f = \overset{ˉ}{f} \circ π$ .

L1L2L3givenconstruct

2.1

For every coset, $\overset{ˉ}{f} (r (m + N)) = \overset{ˉ}{f} (r m + N) = f (r m) = r f (m) = r \overset{ˉ}{f} (m + N)$ , so $\overset{ˉ}{f}$ is scalar-preserving.

step 1.1L3given

3.1

Thus $\overset{ˉ}{f}$ is a module homomorphism with the required factorisation.

step 1.1step 2.1L3

4.1

Any module-homomorphism factor is in particular an additive-group factor, so the uniqueness in step 1.1 proves its uniqueness as a module homomorphism.

step 1.1L2L3discharge-construct: final∎

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

First isomorphism theorem for modules: $M / \ker f ≅ im f$

Statement

For every module homomorphism $f : M \to N$ , there is a module isomorphism

$M / \ker f ≅ im f,$

given by $m + \ker f \mapsto f (m)$ .

Facts & Assumptions

Given: A module homomorphism $f : M \to N$ .

[L1]

Its kernel and image are submodules, and a module homomorphism is injective exactly when its kernel is trivial (Kernels and images of module homomorphisms are submodules, and injectivity is equivalent to trivial kernel).

[L2]

A homomorphism vanishing on a submodule factors uniquely through the quotient module (A module homomorphism vanishing on $N$ factors uniquely through $M / N$ ).

[L3]

Module isomorphisms are precisely bijective module homomorphisms; kernel and image have their displayed definitions (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct

1.1

Since $\ker f$ is a submodule and $f$ vanishes on it, [L2] gives a module homomorphism $\overset{ˉ}{f} : M / \ker f \to N$ with $\overset{ˉ}{f} (m + \ker f) = f (m)$ .

L1L2L3given

2.1

Every value of $\overset{ˉ}{f}$ lies in $im f$ , and every $f (m)$ is the value of $\overset{ˉ}{f}$ at $m + \ker f$ ; hence its corestriction $\overset{ˉ}{f} : M / \ker f \to im f$ is a surjective module homomorphism.

step 1.1L1L3given

2.2

The corestriction has trivial kernel: $\overset{ˉ}{f} (m + \ker f) = 0_{N}$ means $f (m) = 0_{N}$ , hence $m \in \ker f$ and $m + \ker f = 0_{M / \ker f}$ .

step 1.1L1L3given

3.1

The corestriction is injective by [L1], so it is bijective.

step 2.1step 2.2L1

4.1

By [L3], this bijective module homomorphism is the claimed module isomorphism.

step 3.1L3∎

5 · Examples, counterexamples and false statements

None yet.

Sources

McGerty, Algebra II: Rings and Modules, Section 3

Modules, Submodules, Quotient Modules and the Isomorphism Theorems

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Unital left and right modules over a ring; unqualified module means left module

Definition

In a module, 0Rm=0M0R​m=0M​, r0M=0Mr0M​=0M​, (−r)m=−(rm)(−r)m=−(rm) and r(−m)=−(rm)r(−m)=−(rm)

Statement

Facts & Assumptions

Proof

Submodule of a module

Definition

The one-step submodule criterion; intersections and sums of submodules are submodules

Statement

Facts & Assumptions

Proof

Generated submodule, cyclic and finitely generated modules, module basis and free module

Definition

Simple module: a nonzero module with no proper nonzero submodule

Definition

Annihilators, torsion elements and the torsion subset of a module

Definition

Quotient module M/NM/N with scalar multiplication on additive cosets

Definition

Module homomorphism and isomorphism, kernel, image and cokernel

Definition

Kernels and images of module homomorphisms are submodules, and injectivity is equivalent to trivial kernel

Statement

Facts & Assumptions

Proof

The quotient action is well defined and makes M/NM/N a module

Statement

Facts & Assumptions

Proof

The canonical map M→M/NM→M/N is a surjective module homomorphism with kernel NN; thus every submodule is a kernel

Statement

Facts & Assumptions

Proof

A module homomorphism vanishing on NN factors uniquely through M/NM/N

Statement

Facts & Assumptions

Proof

First isomorphism theorem for modules: M/ker⁡f≅im⁡fM/kerf≅imf

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources

In a module, $0_{R} m = 0_{M}$ , $r 0_{M} = 0_{M}$ , $(- r) m = - (r m)$ and $r (- m) = - (r m)$

Quotient module $M / N$ with scalar multiplication on additive cosets

The quotient action is well defined and makes $M / N$ a module

The canonical map $M \to M / N$ is a surjective module homomorphism with kernel $N$ ; thus every submodule is a kernel

A module homomorphism vanishing on $N$ factors uniquely through $M / N$

First isomorphism theorem for modules: $M / \ker f ≅ im f$