---
title: Introduction to Modules
module: Module Theory
moduleNumber: 8
lessonNumber: 1
order: 801
summary: >
  A module is an abelian group on which a ring acts, generalizing both vector
  spaces (when the ring is a field) and abelian groups (when the ring is the
  integers). Submodules, homomorphisms, quotients, and the isomorphism theorems
  carry over from groups, and an F[x]-module is the same datum as a vector space
  with a chosen linear operator — the correspondence behind the canonical forms.
topics: [Module Theory]
draft: false
sources:
  - book: Dummit & Foote
    ref: "Ch. 10 — Introduction to Module Theory; §10.1 Basic Definitions and Examples"
  - book: Dummit & Foote
    ref: "§10.2 Quotient Modules and Module Homomorphisms"
---

A [group action](/abstract-algebra/groups-and-symmetry/homomorphisms-and-group-actions)
lets a group $G$ permute the elements of a set. A **module** is the same idea one
level up: a [ring](/abstract-algebra/ring-theory/rings-definitions-and-examples)
$R$ acts on an abelian group $M$, and the action is required to respect both the
addition of $M$ and the addition and multiplication of $R$. Two familiar theories
are special cases. When $R$ is a field the axioms are exactly those of a vector
space; when $R = \mathbb{Z}$ they are exactly those of an abelian group. Module
theory studies both at once: a single structure theorem for
modules over a [principal ideal domain](/abstract-algebra/factorization-and-polynomials/euclidean-domains-pids-ufds)
reproduces the classification of finite abelian groups and the Jordan and
rational canonical forms of a matrix from one argument.[^df-intro]

## The definition

> **Definition (Left $R$-module).** Let $R$ be a ring. A **left $R$-module** is an
> abelian group $(M, +)$ together with an action $R \times M \to M$, written
> $(r, m) \mapsto rm$, satisfying, for all $r, s \in R$ and $m, n \in M$,
>
> - $(r + s)m = rm + sm$,
> - $(rs)m = r(sm)$,
> - $r(m + n) = rm + rn$.
>
> If $R$ has an identity $1$, we further require $1m = m$ for all $m$; such a module
> is called **unital**. Throughout, rings have $1$ and modules are unital.

The word "left" records that ring elements are written on the left of module
elements. A **right module** puts them on the right, and axiom $(rs)m = r(sm)$
becomes $m(rs) = (mr)s$; when $R$ is commutative the two notions coincide and we
drop the adjective. The action is a ring analogue of a group action: writing
$R^{\times}$ for the units of $R$, the map $m \mapsto rm$ for a unit $r$ is a
bijection of $M$, and the axioms make $R^{\times}$ act on the set $M$ by additive
automorphisms.

Two boundary cases:

- **$R$ a field $F$.** The axioms are verbatim the vector-space axioms. Modules over
  a field _are_ vector spaces over that field.
- **$R = \mathbb{Z}$.** For any abelian group $A$ the rule $na = a + \cdots + a$ ($n$
  times, with $0a = 0$ and $(-n)a = -(na)$) is the only unital action of $\mathbb{Z}$
  on $A$. So **$\mathbb{Z}$-modules are the same as abelian groups**, and there is
  nothing to choose: the group structure forces the action.

$$
% caption: A module over a general ring $R$ specializes to a vector space when $R$
% is a field, to an abelian group when $R$ is the integers, and to a vector space
% equipped with a linear operator when $R$ is a polynomial ring.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  box/.style={draw, minimum width=30mm, minimum height=11mm, align=center, font=\footnotesize}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[box, draw=acc, text=acc, thick] (mod) at (0,2.1) {left $R$-module $M$};
  \node[box] (vs)  at (-4.6,0) {vector space};
  \node[box] (ab)  at (0,0)    {abelian group};
  \node[box] (op)  at (4.6,0)  {space $+$ operator};
  \draw[->, acc, thick] (mod) -- (vs) node[midway, above, sloped, font=\scriptsize, text=black] {$R = F$};
  \draw[->, acc, thick] (mod) -- (ab) node[midway, right, font=\scriptsize, text=black] {$R = \mathbf{Z}$};
  \draw[->, acc, thick] (mod) -- (op) node[midway, above, sloped, font=\scriptsize, text=black] {$R = F[x]$};
\end{tikzpicture}
$$

### Submodules and the submodule criterion

> **Definition (Submodule).** A **submodule** of an $R$-module $M$ is a subgroup
> $N \le M$ closed under the action: $rn \in N$ for all $r \in R$, $n \in N$.

Submodules are the subobjects of the theory. Over a field they are subspaces; over
$\mathbb{Z}$ they are subgroups. Every $M$ has the two trivial submodules $0$ and
$M$. Testing the definition directly means checking closure under subtraction and
under the action separately; a single condition does both.

> **Proposition (Submodule criterion).** A subset $N \subseteq M$ is a submodule if
> and only if
>
> - $N \ne \varnothing$, and
> - $x + ry \in N$ for all $r \in R$ and all $x, y \in N$.

> **Proof.** Taking $r = -1$ recovers closure under subtraction, so $N$ is a
> subgroup; taking $x = 0$ recovers closure under the action. Conversely a
> submodule contains $x + ry$ for any $x, y \in N$, $r \in R$, since $ry \in N$
> and $N$ is closed under addition.[^df-sub] $\square$
> **Worked example.** In the $\mathbb{Z}$-module $\mathbb{Z}^2$, let
> $N = \{(a, b) : a - 2b \equiv 0 \pmod 5\}$. Check the criterion:
> $(0,0) \in N$, so $N \ne \varnothing$. For $(a_1, b_1), (a_2, b_2) \in N$ and
> $r \in \mathbb{Z}$,
>
> $$
> (a_1 + r a_2) - 2(b_1 + r b_2) = (a_1 - 2b_1) + r(a_2 - 2b_2) \equiv 0 + r \cdot 0 = 0 \pmod 5,
> $$
>
> so $(a_1, b_1) + r(a_2, b_2) \in N$ and $N$ is a submodule. It is the kernel of
> the homomorphism $\mathbb{Z}^2 \to \mathbb{Z}/5\mathbb{Z}$,
> $(a, b) \mapsto a - 2b \bmod 5$. By contrast $N' = \{(a, b) : a - 2b \equiv 1
> \pmod 5\}$ fails immediately: $(0,0) \notin N'$, and $N'$ is not closed under
> addition, since the sum of two elements has $a - 2b \equiv 2$.

## First examples

- **$R$ over itself.** Any ring $R$ is a left module over itself, the action being
  ring multiplication. Its submodules are precisely the **left ideals** of $R$. A
  cyclic submodule $Ra$ is a principal (left) ideal, so a
  [principal ideal domain](/abstract-algebra/factorization-and-polynomials/euclidean-domains-pids-ufds)
  is exactly a commutative domain in which every submodule of $R$ is cyclic.
- **Free module of rank $n$.** For $n \in \mathbb{Z}^{+}$, the set
  $R^n = \{(a_1, \dots, a_n) : a_i \in R\}$ with componentwise addition and action
  $r(a_1, \dots, a_n) = (ra_1, \dots, ra_n)$ is the **free module of rank $n$**. It
  is the ring analogue of coordinate space $F^n$.
- **Abelian groups as $\mathbb{Z}$-modules.** Any abelian group $A$ carries its
  unique $\mathbb{Z}$-action. If $A$ has an element $x$ of finite order $n$, then
  $nx = 0$ with $n \ne 0$: unlike a vector space, a module can have a nonzero element
  killed by a nonzero scalar. These are **torsion** elements; they distinguish module theory from linear
  algebra.
- **Killing an ideal.** If a two-sided ideal $I$ annihilates $M$ (that is $am = 0$
  for all $a \in I$, $m \in M$), then $M$ becomes an $R/I$-module via
  $(r + I)m = rm$. When $I$ is a maximal ideal and $IM = 0$, the module $M$ is a
  vector space over the field $R/I$. For an elementary abelian $p$-group $A$ with
  $px = 0$, this makes $A$ a vector space over $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$.

$$
% caption: The integer action on an abelian group is repeated addition: the scalar
% $n$ sends a generator $a$ to its $n$-fold sum, so a cyclic group of order six is
% the $\mathbf{Z}$-module generated by one element with $6a = 0$.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \foreach \k/\lab/\ang in {0/{0}/90, 1/{a}/30, 2/{2a}/-30, 3/{3a}/-90, 4/{4a}/-150, 5/{5a}/150}{
    \node[draw, circle, minimum size=8mm, inner sep=0pt, font=\scriptsize] (n\k) at (\ang:2) {$\lab$};
  }
  \draw[->, acc, thick] (n0) to[bend left=18] (n1);
  \draw[->, acc, thick] (n1) to[bend left=18] (n2);
  \draw[->, acc, thick] (n2) to[bend left=18] (n3);
  \draw[->, acc, thick] (n3) to[bend left=18] (n4);
  \draw[->, acc, thick] (n4) to[bend left=18] (n5);
  \draw[->, acc, thick] (n5) to[bend left=18] (n0);
  \node[font=\scriptsize, text=acc] at (0,0) {add $a$};
\end{tikzpicture}
$$

### Vector spaces as $F[x]$-modules

This example makes a vector space into a module over a
[polynomial ring](/abstract-algebra/factorization-and-polynomials/polynomial-rings-over-fields).
Let $F$ be a field, $V$ a vector space over $F$, and $T : V \to V$ a linear
transformation. Define an action of $F[x]$ on $V$ by letting $x$ act as $T$ and
extending: for $p(x) = a_n x^n + \cdots + a_1 x + a_0$,

$$
p(x)\,v \;=\; a_n T^n(v) + \cdots + a_1 T(v) + a_0 v,
$$

where $T^n$ is the $n$-fold composite and $T^0 = I$. The module axioms hold, and the
constant polynomials act as the original scalar multiplication, so this action
_extends_ the $F$-module structure to an $F[x]$-module structure. The choice of $T$
is the only freedom, so a single vector space $V$ carries many different
$F[x]$-module structures — one for each linear operator on it.

> **Theorem (The $F[x]$-module dictionary).** There is a bijection between
> $F[x]$-modules and pairs $(V, T)$ where $V$ is a vector space over $F$ and
> $T : V \to V$ is a linear transformation, with $x$ acting as $T$. Under this
> correspondence, the $F[x]$-submodules of $V$ are the **$T$-stable**
> subspaces and only those — the subspaces $U$ with $T(U) \subseteq U$.

> **Proof.** Every $F[x]$-module is an $F$-vector space (restrict scalars to
> constants), and the action of $x$ is a linear map $T$; the axioms force the
> action of every polynomial. A submodule must be an $F$-subspace closed under
> $x$, hence under $T$, and any $T$-stable subspace is closed under all powers of
> $T$ and all polynomials in $T$.[^df-fx] $\square$
Because $F[x]$ is a principal ideal domain, its module structure is tightly
constrained, and that constraint becomes detailed information about $T$: its
[rational canonical form](/abstract-algebra/modules-over-pids/rational-canonical-form)
and [Jordan canonical form](/abstract-algebra/modules-over-pids/jordan-canonical-form).

$$
% caption: A linear operator $T$ on a vector space is the same datum as an
% $F[x]$-module structure on that space: the indeterminate $x$ acts as $T$, and
% $T$-stable subspaces correspond exactly to submodules.
\begin{tikzpicture}[>=stealth, font=\footnotesize,
  box/.style={draw, minimum width=34mm, minimum height=13mm, align=center, font=\footnotesize}]
  \definecolor{acc}{HTML}{4A6FA5}
  \node[box] (pair) at (0,0) {space $V$ with operator $T$};
  \node[box, draw=acc, text=acc, thick] (mod) at (6.4,0) {$F[x]$-module $V$};
  \draw[->, acc, thick] (pair.10) -- (mod.170) node[midway, above, font=\scriptsize, text=black] {$x$ acts as $T$};
  \draw[->, black, thick] (mod.190) -- (pair.-10) node[midway, below, font=\scriptsize] {restrict to constants};
\end{tikzpicture}
$$

> **Worked example.** Let $V = F^2$ and let $T$ be the shift $T(a, b) = (b, 0)$,
> so $T e_2 = e_1$ and $T e_1 = 0$. Find all $F[x]$-submodules of $V$, i.e. all
> $T$-stable subspaces.
>
> The candidates are $0$, $V$, and the one-dimensional subspaces
> $W = \operatorname{Span}(a e_1 + b e_2)$. If $b \ne 0$, then
> $T(a e_1 + b e_2) = b e_1$, and $b e_1 \in W$ forces $b e_1 = \lambda(a e_1 + b e_2)$
> for some $\lambda \in F$; comparing $e_2$-coordinates gives $\lambda b = 0$, so
> $\lambda = 0$ and $b e_1 = 0$, a contradiction. Hence $b = 0$ and the only
> $T$-stable line is $\operatorname{Span}(e_1)$. The full submodule lattice is
>
> $$
> 0 \;\subset\; \operatorname{Span}(e_1) \;\subset\; V.
> $$
>
> As an $F[x]$-module, $V$ is cyclic with generator $e_2$: applying $1$ and $x$
> gives $e_2$ and $e_1$, which span. Since $x^2 \cdot e_2 = T^2 e_2 = 0$, the
> annihilator of $V$ is $(x^2)$ and $V \cong F[x]/(x^2)$ — a torsion module whose
> proper nonzero submodule $\operatorname{Span}(e_1)$ corresponds to the ideal
> $(x)/(x^2)$.

## Homomorphisms

> **Definition (Module homomorphism).** An **$R$-module homomorphism** is a map
> $\varphi : M \to N$ of $R$-modules with $\varphi(x + y) = \varphi(x) + \varphi(y)$
> and $\varphi(rx) = r\varphi(x)$ for all $x, y \in M$, $r \in R$. It is an
> **isomorphism** if it is bijective, written $M \cong N$. The **kernel** is
> $\ker \varphi = \{m : \varphi(m) = 0\}$ and the **image** is $\varphi(M)$.

Equivalently, a single condition $\varphi(rx + y) = r\varphi(x) + \varphi(y)$
captures both requirements. Every module homomorphism is in particular a
homomorphism of the underlying abelian groups, but not conversely: the group map
$x \mapsto 2x$ on $\mathbb{Z}$ is a $\mathbb{Z}$-module homomorphism, while over
$R = F[x]$ the map $f(x) \mapsto f(x^2)$ is a ring homomorphism that fails to be
$F[x]$-linear. The names specialize as expected.

- Over a field, module homomorphisms are **linear transformations**.
- Over $\mathbb{Z}$, they are **abelian-group homomorphisms** — the action condition
  is automatic, since scalar multiplication is repeated addition.

Kernels and images are submodules, by the submodule criterion applied to $\varphi$.
The set $\operatorname{Hom}_R(M, N)$ of all homomorphisms $M \to N$ is itself an
abelian group under pointwise addition, and an $R$-module when $R$ is commutative.
The endomorphisms $\operatorname{End}_R(M) = \operatorname{Hom}_R(M, M)$ form a ring
under composition — the **endomorphism ring** of $M$.

$$
% caption: A module homomorphism collapses each coset of its kernel to a single
% point of the image; the kernel is the fiber over zero, and every fiber is a
% translate of it.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  % domain M as a stack of coset-fibers; the middle fiber is the kernel
  \draw[black] (-0.3,-2.2) rectangle (2.3,2.2);
  \node[anchor=south, font=\footnotesize] at (1.0,2.25) {$M$};
  \fill[acc!14] (-0.3,-0.5) rectangle (2.3,0.5);
  \foreach \y in {1.5, 0.5, -0.5, -1.5}
    \draw[black] (-0.3,\y) -- (2.3,\y);
  \node[font=\scriptsize, text=acc] at (1.0,0) {kernel};
  % codomain
  \draw[black] (6.2,-2.2) rectangle (7.4,2.2);
  \node[anchor=south, font=\footnotesize] at (6.8,2.25) {$N$};
  \foreach \yy in {1.5, 0.75, -0.75, -1.5}
    \fill[black] (6.8,\yy) circle (2pt);
  \fill[acc] (6.8,0) circle (2.6pt);
  \node[font=\scriptsize, text=acc, anchor=west] at (6.93,0) {$0$};
  % one arrow per fiber, leaving from the fiber's center
  \draw[->, black] (2.4,1.85)  -- (6.7,1.5);
  \draw[->, black] (2.4,1.0)   -- (6.7,0.75);
  \draw[->, acc, thick] (2.4,0) -- (6.7,0);
  \draw[->, black] (2.4,-1.0)  -- (6.7,-0.75);
  \draw[->, black] (2.4,-1.85) -- (6.7,-1.5);
  \node[font=\scriptsize] at (4.5,2.05) {$f$};
\end{tikzpicture}
$$

## Quotient modules

Every submodule is the kernel of a projection, so quotients always exist. This is
simpler than the group case: a module is an abelian group, so every submodule is a
normal subgroup, and there is no obstruction to forming the quotient.

> **Proposition (Quotient module).** Let $N \le M$ be a submodule. The quotient
> abelian group $M/N$ is an $R$-module under $r(x + N) = (rx) + N$, and the
> projection $\pi : M \to M/N$, $\pi(x) = x + N$, is a surjective $R$-module
> homomorphism with $\ker \pi = N$.

> **Proof.** The action is well defined because $N$ is closed under scalar
> multiplication: if $x + N = y + N$ then $x - y \in N$, so $r(x - y) \in N$ and
> $rx + N = ry + N$. The module axioms for $M/N$ follow from those of $M$ applied
> to representatives, and $\pi$ is $R$-linear by construction with kernel $N$.
> $\square$

The elements of $M/N$ are the **cosets** $x + N$, translates of the submodule $N$
partitioning $M$.

$$
% caption: A submodule $N$ tiles $M$ into parallel cosets; the quotient $M/N$ has
% one point for each tile, and scalars act by acting on any representative.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \foreach \i in {0,1,2,3}{
    \fill[acc!10] (0, \i*0.85) rectangle (5.2, \i*0.85+0.6);
    \draw[black] (0, \i*0.85) rectangle (5.2, \i*0.85+0.6);
  }
  \node[anchor=west, font=\scriptsize, text=acc] at (5.35,0.3) {$N$};
  \node[anchor=west, font=\scriptsize] at (5.35,1.15) {$x + N$};
  \node[anchor=west, font=\scriptsize] at (5.35,2.0) {$2x + N$};
  \node[anchor=west, font=\scriptsize] at (5.35,2.85) {$3x + N$};
  \node[anchor=east, font=\footnotesize] at (-0.15,1.7) {$M$};
\end{tikzpicture}
$$

> **Worked example.** Compute $M/N$ for $M = \mathbb{Z}^2$ and
> $N = \{(2s, 3t) : s, t \in \mathbb{Z}\}$, the submodule generated by $(2, 0)$
> and $(0, 3)$.
>
> Two pairs $(a, b)$ and $(a', b')$ lie in the same coset exactly when
> $a \equiv a' \pmod 2$ and $b \equiv b' \pmod 3$, so the map
>
> $$
> \varphi : \mathbb{Z}^2 \to \mathbb{Z}/2 \oplus \mathbb{Z}/3,
> \qquad
> \varphi(a, b) = (a \bmod 2,\; b \bmod 3)
> $$
>
> is a surjective $\mathbb{Z}$-module homomorphism with kernel $N$, and
> $M/N \cong \mathbb{Z}/2 \oplus \mathbb{Z}/3$ — six cosets. The quotient is
> cyclic: the coset $(1,1) + N$ has order $6$, since $n \cdot (1,1) \in N$
> requires $2 \mid n$ and $3 \mid n$, i.e. $6 \mid n$. By the Chinese remainder
> theorem $M/N \cong \mathbb{Z}/6$.

## The isomorphism theorems

Because a module is an abelian group with extra structure, the four isomorphism
theorems carry over verbatim from [groups](/abstract-algebra/subgroups-and-quotients/isomorphism-theorems);
one only checks that the group isomorphisms are $R$-linear.[^df-iso]

> **Theorem (Isomorphism theorems for modules).** Let $R$ be a ring and $M$ an
> $R$-module.
>
> - **First.** For a homomorphism $\varphi : M \to N$, $\ker \varphi$ is a submodule
>   and $M / \ker \varphi \cong \varphi(M)$.
> - **Second.** For submodules $A, B \le M$, $(A + B)/B \cong A/(A \cap B)$.
> - **Third.** For submodules $A \le B \le M$, $(M/A)/(B/A) \cong M/B$.
> - **Fourth (Lattice).** For a submodule $N \le M$, the submodules of $M/N$
>   correspond bijectively to the submodules of $M$ containing $N$, via
>   $A \mapsto A/N$, and this correspondence preserves sums and intersections.

The first theorem matters most: every homomorphic image of $M$ is a
quotient of $M$, so understanding $M$ up to isomorphism reduces to understanding its
submodules and the quotients by them. The sum $A + B = \{a + b : a \in A, b \in B\}$
appearing in the second theorem is the smallest submodule containing both $A$ and
$B$.

> **Worked example.** Let $M = \mathbb{Z}^n$ and let
> $\varphi : \mathbb{Z}^n \to \mathbb{Z}/d\mathbb{Z}$ send
> $(a_1, \dots, a_n) \mapsto a_1 \bmod d$. This is a $\mathbb{Z}$-module
> homomorphism, surjective, with kernel $\{(a_1, \dots, a_n) : d \mid a_1\}$. The
> first isomorphism theorem gives
>
> $$
> \mathbb{Z}^n \big/ \{(a_1, \dots, a_n) : d \mid a_1\} \;\cong\; \mathbb{Z}/d\mathbb{Z}.
> $$
>
> More generally, for an ideal $I \le R$ and the free module $R^n$, one has
> $R^n / I R^n \cong (R/I)^n$: the map reducing every coordinate mod $I$ is
> surjective with kernel $IR^n$.

Reductions of this shape, specialized to $R = \mathbb{Z}$, package statements
about finite abelian groups; the
[structure theorem over a PID](/abstract-algebra/modules-over-pids/structure-theorem-over-pids)
turns them into a complete classification.

## Fields versus principal ideal domains

The definition splits the subject cleanly. Over a field, the added ring structure
degenerates and one recovers linear algebra: every module is
[free](/abstract-algebra/module-theory/vector-spaces-and-linear-maps) and dimension
is a complete invariant. Over $\mathbb{Z}$ or $F[x]$ — more generally over a
principal ideal domain — torsion appears, modules need not have a basis, and the
finer invariants of the structure theorem are needed.

| Ring $R$ | Modules are | Torsion? | Basis always exists? |
| --- | --- | --- | --- |
| field $F$ | vector spaces | no | yes |
| $\mathbb{Z}$ | abelian groups | yes (finite order) | no ($\mathbb{Z}/n\mathbb{Z}$) |
| $F[x]$ | space $+$ operator | yes | no |
| general $R$ | $R$-modules | possible | rarely |

[^df-intro]: **Dummit & Foote**, _Abstract Algebra_, 3rd ed., Ch. 10 introduction and §10.1 — modules as the representation objects of a ring, with vector spaces (over a field) and abelian groups (over $\mathbb{Z}$) as the two guiding special cases.
[^df-sub]: **Dummit & Foote**, §10.1, Proposition 1 — the submodule criterion: $N$ nonempty and $x + ry \in N$ for all $r \in R$, $x, y \in N$.
[^df-fx]: **Dummit & Foote**, §10.1 — the $F[x]$-module associated to a vector space $V$ and a linear transformation $T$, with $F[x]$-submodules equal to the $T$-stable subspaces; the source of the canonical-form theory of §12.2–12.3.
[^df-iso]: **Dummit & Foote**, §10.2, Proposition 3 and Theorem 4 — the quotient module $M/N$ and the four isomorphism theorems for modules, deduced from the corresponding theorems for abelian groups.