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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 RR be a ring. A left RR-module is an abelian group (M,+,0M)(M,+,0_M) with a scalar action R×M→MR\times M\to M, (r,m)↦rm(r,m)\mapsto rm, satisfying

r(m+n)=rm+rn,(r+s)m=rm+sm,(rs)m=r(sm),1Rm=m.r(m+n)=rm+rn,\quad (r+s)m=rm+sm,\quad (rs)m=r(sm),\quad 1_Rm=m.

A right RR-module has an action M×R→MM\times R\to M, (m,r)↦mr(m,r)\mapsto mr, 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, 0Rm=0M0_Rm=0_M, r0M=0Mr0_M=0_M, (−r)m=−(rm)(-r)m=-(rm) and r(−m)=−(rm)r(-m)=-(rm)

Statement

For every left RR-module MM, scalar r∈Rr\in R, and element m∈Mm\in M,

0Rm=0M,r0M=0M,(−r)m=−(rm),r(−m)=−(rm).0_Rm=0_M,\qquad r0_M=0_M,\qquad (-r)m=-(rm),\qquad r(-m)=-(rm).

Facts & Assumptions

Given: A left RR-module MM, r∈Rr\in R, and m∈Mm\in M.

[L1]

The module action distributes over both addition operations, and (M,+,0M)(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,+,0R)(R,+,0_R) is an abelian group, so r+(−r)=0Rr+(-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 RR and MM (Cancellation in a group: gx=gygx = gy or xg=ygxg = yg forces x=yx = y; equivalently left and right translation by gg are bijections of GG, so gx=hgx = h and xg=hxg = h each have exactly one solution).

Proof

technique · direct
1.1

Since 0R=0R+0R0_R=0_R+0_R, distributivity gives 0Rm=0Rm+0Rm0_Rm=0_Rm+0_Rm; cancellation yields 0Rm=0M0_Rm=0_M.

L1L3given
1.2

Since 0M=0M+0M0_M=0_M+0_M, distributivity gives r0M=r0M+r0Mr0_M=r0_M+r0_M; cancellation yields r0M=0Mr0_M=0_M.

L1L3given
2.1

From (r+(−r))m=0Rm=0M(r+(-r))m=0_Rm=0_M, distributivity and step 1.1 give rm+(−r)m=0Mrm+(-r)m=0_M, so (−r)m=−(rm)(-r)m=-(rm).

step 1.1L1L2L3given
3.1

From r(m+(−m))=r0M=0Mr(m+(-m))=r0_M=0_M, distributivity and step 1.2 give rm+r(−m)=0Mrm+r(-m)=0_M, so r(−m)=−(rm)r(-m)=-(rm).

step 1.2L1L3given∎
DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-03Open item page →

Submodule of a module

Definition

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

r∈R, n∈N⟹rn∈N.r\in R,\ n\in N\quad\Longrightarrow\quad rn\in N.

The operations on NN are the restrictions of those of MM. Write N≤MN\le 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 MM be a left RR-module. A nonempty subset S⊆MS\subseteq M is a submodule if and only if

ru+v∈S(r∈R,u,v∈S).ru+v\in S\qquad(r\in R, u,v\in S).

Consequently, the intersection of every nonempty family of submodules is a submodule, and, for submodules A,B≤MA,B\le M,

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

is a submodule of MM.

Facts & Assumptions

Given: A left RR-module MM.

[L1]

The module axioms include 1Rm=m1_Rm=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=−(rm)(-r)m=-(rm); taking r=1Rr=1_R and using [L1] gives (−1R)m=−m(-1_R)m=-m (In a module, 0Rm=0M0_Rm=0_M, r0M=0Mr0_M=0_M, (−r)m=−(rm)(-r)m=-(rm) and r(−m)=−(rm)r(-m)=-(rm)).

[L3]

A nonempty subset of a group is a subgroup exactly when it is closed under xy−1xy^{-1} (One-step subgroup test: a nonempty H⊆GH \subseteq G is a subgroup iff gh−1∈Hgh^{-1} \in H for all g,h∈Hg, h \in H; the identity and the inverses of HH are then those of GG).

[L4]

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

Proof

technique · direct
1.1

If SS is a submodule, then u∈Su\in S implies ru∈Sru\in S, and then ru+v∈Sru+v\in S for r∈Rr\in R and u,v∈Su,v\in S.

L4given
1.2

Conversely, suppose the displayed closure condition holds and choose s∈Ss\in S. With r=−1Rr=-1_R and both elements equal to ss, it gives (−1R)s+s=0M∈S(-1_R)s+s=0_M\in S.

L1L2given
1.3

If u,v∈Su,v\in S, the same condition with scalar −1R-1_R, first element vv, and second element uu gives (−1R)v+u=u−v∈S(-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=0Mv=0_M. Thus SS is a submodule.

step 1.2step 1.3L3L4given
3.1

For a nonempty family (Ni)(N_i) of submodules, 0M0_M lies in every NiN_i; and if u,vu,v lie in their intersection, then ru+vru+v lies in every NiN_i. The criterion proves ⋂iNi\bigcap_iN_i is a submodule.

step 2.1L4given
4.1

For x=a+bx=a+b and y=a′+b′y=a'+b' in A+BA+B, distributivity gives rx+y=(ra+a′)+(rb+b′)∈A+Brx+y=(ra+a')+(rb+b')\in A+B; moreover 0M=0M+0M∈A+B0_M=0_M+0_M\in A+B. The criterion proves A+BA+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 MM be a left RR-module and S⊆MS\subseteq M. The submodule generated by SS is

⟨S⟩R:=⋂{N≤M:S⊆N}.\langle S\rangle_R:=\bigcap\{N\le M:S\subseteq N\}.

The family is nonempty because M≤MM\le M, and its intersection is a submodule by The one-step submodule criterion; intersections and sums of submodules are submodules. Thus ⟨S⟩R\langle S\rangle_R is the smallest submodule of MM containing SS, just as The subgroup ⟨S⟩\langle S \rangle generated by a subset, the cyclic subgroup ⟨g⟩\langle g \rangle, and cyclic groups defines a generated subgroup.

The module MM is cyclic if M=⟨m⟩RM=\langle m\rangle_R for some m∈Mm\in M, and finitely generated if M=⟨S⟩RM=\langle S\rangle_R for some finite SS. A subset B⊆MB\subseteq M is a basis if every element of MM has a unique expression as a finite RR-linear combination of elements of BB. A module possessing a basis is a free RR-module.

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

Simple module: a nonzero module with no proper nonzero submodule

Definition

A left RR-module MM is simple if M≠0M\ne0 and its only submodules are 00 and MM. Equivalently, MM 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 RR-module MM and m∈Mm\in M, the annihilator of mm is

Ann⁡R(m):={r∈R:rm=0M},\operatorname{Ann}_R(m):=\{r\in R:rm=0_M\},

and the annihilator of MM is

Ann⁡R(M):={r∈R:rm=0M for every m∈M}.\operatorname{Ann}_R(M):=\{r\in R:rm=0_M\text{ for every }m\in M\}.

If RR is an integral domain, an element m∈Mm\in M is a torsion element when rm=0Mrm=0_M for some nonzero r∈Rr\in R. The set of all torsion elements is denoted Tor⁡(M)\operatorname{Tor}(M); MM is torsion-free when Tor⁡(M)={0M}\operatorname{Tor}(M)=\{0_M\}.

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

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

Definition

Let N≤MN\le M be a submodule of a left RR-module. Since the additive group of MM is abelian, NN is normal, so the additive quotient group M/NM/N consists of the cosets m+Nm+N. Its proposed scalar action is

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

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

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

Module homomorphism and isomorphism, kernel, image and cokernel

Definition

For left RR-modules M,NM,N, a function f:M→Nf:M\to N is an RR-module homomorphism if

f(m+m′)=f(m)+f(m′)andf(rm)=rf(m)f(m+m')=f(m)+f(m')\quad\text{and}\quad f(rm)=rf(m)

for all m,m′∈Mm,m'\in M and r∈Rr\in R. It is a module isomorphism if it is a bijective module homomorphism.

Its kernel and image are

ker⁡f:={m∈M:f(m)=0N},im⁡f:={f(m):m∈M}.\ker f:=\{m\in M:f(m)=0_N\},\qquad\operatorname{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 ff is the quotient module coker⁡f:=N/im⁡f\operatorname{coker}f:=N/\operatorname{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 RR-module homomorphism f:M→Nf:M\to N, both ker⁡f\ker f and im⁡f\operatorname{im}f are submodules. Moreover,

f is injective⟺ker⁡f={0M}.f\text{ is injective}\quad\Longleftrightarrow\quad\ker f=\{0_M\}.

Facts & Assumptions

Given: A homomorphism f:M→Nf:M\to N of left RR-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 ru+vru+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 0M0_M to 0N0_N; moreover r0N=0Nr0_N=0_N (Module homomorphism and isomorphism, kernel, image and cokernel, A group homomorphism automatically satisfies f(e)=e′f(e) = e' and f(g−1)=f(g)−1f(g^{-1}) = f(g)^{-1}, and f(gn)=f(g)nf(g^{n}) = f(g)^{n} for every n∈Zn \in \mathbb{Z}; for monoid homomorphisms preservation of the identity must be assumed, In a module, 0Rm=0M0_Rm=0_M, r0M=0Mr0_M=0_M, (−r)m=−(rm)(-r)m=-(rm) and r(−m)=−(rm)r(-m)=-(rm)).

[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 0M0_M because f(0M)=0Nf(0_M)=0_N. If u,v∈ker⁡fu,v\in\ker f and r∈Rr\in R, then f(ru+v)=rf(u)+f(v)=0Nf(ru+v)=rf(u)+f(v)=0_N, so ru+v∈ker⁡fru+v\in\ker f.

L1L3given
1.2

The image contains 0N=f(0M)0_N=f(0_M). If f(u),f(v)∈im⁡ff(u),f(v)\in\operatorname{im}f and r∈Rr\in R, then rf(u)+f(v)=f(ru+v)rf(u)+f(v)=f(ru+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 ff injective exactly when ker⁡f={0M}\ker f=\{0_M\}.

L1L4given
2.1

The submodule criterion applied to steps 1.1 and 1.2 proves that ker⁡f\ker f and im⁡f\operatorname{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/NM/N a module

Statement

Let N≤MN\le M be a submodule of a left RR-module. The rule

r(m+N):=rm+Nr(m+N):=rm+N

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

Facts & Assumptions

Given: A left RR-module MM and a submodule N≤MN\le M.

[L1]

Additive cosets satisfy m+N=m′+Nm+N=m'+N exactly when m′−m∈Nm'-m\in N (x∈aHx\in aH iff a−1x∈Ha^{-1}x\in H, and aH=bHaH=bH iff a−1b∈Ha^{-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⊴GN\mathrel{\trianglelefteq}G, the cosets form a group with identity NN and inverse (gN)−1=g−1N(gN)^{-1}=g^{-1}N).

[L4]

The module axioms give distributivity, associativity of scalar action, and 1Rm=m1_Rm=m (Unital left and right modules over a ring; unqualified module means left module).

[L5]

Scalar multiplication preserves additive negatives (In a module, 0Rm=0M0_Rm=0_M, r0M=0Mr0_M=0_M, (−r)m=−(rm)(-r)m=-(rm) and r(−m)=−(rm)r(-m)=-(rm)).

Proof

technique · direct
1.1

If m+N=m′+Nm+N=m'+N, then m′−m∈Nm'-m\in N by [L1]. Thus r(m′−m)=rm′−rm∈Nr(m'-m)=rm'-rm\in N by [L2, L4, L5], and [L1] gives rm+N=rm′+Nrm+N=rm'+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)(m+N)+(m'+N)=(m+m')+N=(m'+m)+N=(m'+N)+(m+N).

L3L4given
1.3

For cosets, the module identities in MM give r((m+N)+(m′+N))=r(m+m′)+N=(rm+rm′)+N=r(m+N)+r(m′+N)r((m+N)+(m'+N))=r(m+m')+N=(rm+rm')+N=r(m+N)+r(m'+N), (r+s)(m+N)=r(m+N)+s(m+N)(r+s)(m+N)=r(m+N)+s(m+N), (rs)(m+N)=r(s(m+N))(rs)(m+N)=r(s(m+N)), and 1R(m+N)=m+N1_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/NM/N is a left RR-module.

step 1.1step 1.2step 1.3L4∎
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03Open item page →

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

Statement

For a submodule N≤MN\le M, the canonical map

π:M⟶M/N,π(m):=m+N,\pi:M\longrightarrow M/N,\qquad \pi(m):=m+N,

is a surjective RR-module homomorphism and has kernel NN. Hence every submodule is the kernel of a module homomorphism.

Facts & Assumptions

Given: A left RR-module MM and a submodule N≤MN\le M.

[L1]

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

[L2]

The canonical projection of the underlying additive groups is a surjective group homomorphism (The canonical projection π:G→G/N\pi:G\to G/N, π(g)=gN\pi(g)=gN, 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=Nm+N=N if and only if m∈Nm\in N (Submodule of a module, x∈aHx\in aH iff a−1x∈Ha^{-1}x\in H, and aH=bHaH=bH iff a−1b∈Ha^{-1}b\in H).

Proof

technique · direct
1.1

The underlying additive map π\pi is a homomorphism and is surjective.

L2given
1.2

The quotient action gives π(rm)=rm+N=r(m+N)=rπ(m)\pi(rm)=rm+N=r(m+N)=r\pi(m).

L1given
1.3

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

L3L4given
2.1

Steps 1.1--1.2 show that π\pi 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 NN factors uniquely through M/NM/N

Statement

Let f:M→Pf:M\to P be a module homomorphism and let N≤MN\le M satisfy N⊆ker⁡fN\subseteq\ker f. There is a unique module homomorphism

fˉ:M/N⟶P\bar f:M/N\longrightarrow P

such that fˉ(m+N)=f(m)\bar f(m+N)=f(m), equivalently f=fˉ∘πf=\bar f\circ\pi.

Facts & Assumptions

Given: A module homomorphism f:M→Pf:M\to P and a submodule N≤MN\le M with N⊆ker⁡fN\subseteq\ker f.

[L1]

The canonical map π:M→M/N\pi:M\to M/N is a surjective module homomorphism with kernel NN (The canonical map M→M/NM\to M/N is a surjective module homomorphism with kernel NN; 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→M/N\pi:M\to M/N is an additive-group quotient map with kernel NN. Viewing the modules as additive groups, [L3] makes ff a group homomorphism and the hypothesis says it kills NN. By [L2], define the unique additive homomorphism fˉ\bar f by fˉ(m+N)=f(m)\bar f(m+N)=f(m), with f=fˉ∘πf=\bar f\circ\pi.

L1L2L3givenconstruct
2.1

For every coset, fˉ(r(m+N))=fˉ(rm+N)=f(rm)=rf(m)=rfˉ(m+N)\bar f(r(m+N))=\bar f(rm+N)=f(rm)=rf(m)=r\bar f(m+N), so fˉ\bar f is scalar-preserving.

step 1.1L3given
3.1

Thus fˉ\bar 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⁡fM/\ker f\cong\operatorname{im}f

Statement

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

M/ker⁡f ≅ im⁡f,M/\ker f\ \cong\ \operatorname{im}f,

given by m+ker⁡f↦f(m)m+\ker f\mapsto f(m).

Facts & Assumptions

Given: A module homomorphism f:M→Nf: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 NN factors uniquely through M/NM/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\ker f is a submodule and ff vanishes on it, [L2] gives a module homomorphism fˉ:M/ker⁡f→N\bar f:M/\ker f\to N with fˉ(m+ker⁡f)=f(m)\bar f(m+\ker f)=f(m).

L1L2L3given
2.1

Every value of fˉ\bar f lies in im⁡f\operatorname{im}f, and every f(m)f(m) is the value of fˉ\bar f at m+ker⁡fm+\ker f; hence its corestriction fˉ:M/ker⁡f→im⁡f\bar f:M/\ker f\to\operatorname{im}f is a surjective module homomorphism.

step 1.1L1L3given
2.2

The corestriction has trivial kernel: fˉ(m+ker⁡f)=0N\bar f(m+\ker f)=0_N means f(m)=0Nf(m)=0_N, hence m∈ker⁡fm\in\ker f and m+ker⁡f=0M/ker⁡fm+\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

Standard references

Recommended treatments; not extraction sources.