module_theory (13).pdf

A Study on Module Theory
M.Sc. MATHEMATICS
2021 − 2023

A STUDY ON MODULE THEORY
Dissertation submitted to the University of Kerala, in partial
fulfillment of the requirements for the award of the
Degree of Master of Science
In
Mathematics
By
SREELEKSHMI M S
Candidate code : 62021126014
Exam code : 62020402
DEPARTMENT OF MATHEMATICS
ST. GREGORIOS COLLEGE
KOTTARAKARA
2023

CERTIFICATE
This is to certify that the dissertation entitled A Study on Module Theory
is a bonafide record of the work carried out by Sreelekshmi M S under my
supervision and guidance in partial fulfillment of the requirements for the award
of Master of Science in Mathematics.
Dr. Jino Nainan Mrs. Beena G P
Assistant Professor Head of the Department
Department of Mathematics Department of Mathematics
St. Gregorios College St. Gregorios College
Kottarakara Kottarakara

ACKNOWLEDGEMENT
First and foremost we concede the surviving presence and flourishing refinement
of almighty god for concealed hand yet substantial supervision althrough the dis-
sertation. I would like to express sincere thanks to Dr. Jino Nainan, Assistant
Professor, Department of Mathematics, St. Gregorios College, Kottarakara for his
inspiring guidence and support to complete this dissertation. I also wish to express
my profound thanks to Mrs. Beena G P, Head of the department and all other
teachers of Mathematics Department for their constant help throughout the course
of this work. I extend our sincere thanks to librarian and other non-teaching staffs
for their co-operation and support. Above all I would like to express my sincere
gratitude and thanks to my family members, all my friends and well-wishers for
their valuable comments and suggestions and making this work a success.
Kottarakara Sreelekshmi M S
14 August 2023

ABSTRACT
The concept of module is a generalisation of that of a vector space. In a
vector space the scalars are elements of a field, while in a module we shall allow
the scalars to be elements of an arbitrary ring. So the concept of module represents
a significant generalisation of vector spaces. This project surveys on some of the
basic module structures and its properties.

Contents
Introduction 1
1 Preliminaries 2
2 Modules 6
2.1 Definitions and examples . . . . . . . . . . . . . . . . . . . . . 6
2.2 Quotient modules and module homomorphisms . . . . . . 13
2.3 Generation of modules, direct sums, and free modules . . 21
3 Modules with chain conditions 27
3.1 Artinian modules . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.2 Noetherian Modules . . . . . . . . . . . . . . . . . . . . . . . . 30
Conclusion 35
Bibliography 36

Introduction
It is well-known that modules is an extension of the concept of a vector space
over a field to the vector space over an arbitrary ring. Modules are representation
objects for ring, that is, they are by definition, algebraic objects on which rings act.
Modules also generalizes the notion of abelian groups. A module, like vector space,
is an additive abelian group, where the multiplication is defined between elements
of the ring and the elements of module, and this multiplication is associative and
distributive. Since, both ideals and quotient rings are modules, many arguments
about ideals or quotient rings can be combined into a single argument about
modules.
This dissertation is divided into three chapters, the first chapter consist of
some basic concepts which are required for the succeeding chapters. The second
chapter deals with introduction of modules and submodules, and it discusses about
the module homomorphisms, quotient modules, generation of modules and direct
sums. The third chapter presents the basic properties of an important class of
modules and rings, ”Artinian” and ”Noetherian”, which have some very special
properties.
1

Chapter 1
Preliminaries
It is well known that, a set is a well defined collection of objects. A binary
operation ∗ : G × G → G, that is, for all a, b ∈ G we have a ∗ b is in G. For
example, the addition and usual multiplication on the set of all integers Z are
binary operations on Z.
A group is an ordered pair (G, ∗) where G is a set and ∗ is a binary operation
on G satisfying the following axioms:
1. (a ∗ b) = a ∗ (b ∗ c), for all a, b, c ∈ G, that is, ∗ is associative.
2. there is an element e ∈ G, called an identity of G, such that for all a ∈ G,
we have a ∗ e = e ∗ a = a,
3. for each a ∈ G there is an element a−1
of G, called an inverse of a, such that
a ∗ a−1
= a−1
∗ a = e.
The group (G, ∗) is said to be an abelian or commutative group if a ∗ b = b ∗ a, for
all a, b ∈ G. A subset H of G is said to be a subgroup if it is non empty and H
itself is a group under the operation induced from G and we denote it by H ≤ G.
For example, the set of all real numbers R under addition is a commutative group
while R is not a group under usual multiplication. Also the set of all n × n
matrices under usual addition is a commutative group and the set of all invertible
operators under usual multiplication is a group. The set of all functions from R to
2

R is a commutative group under point wise addition and is a group under function
composition.
A ring R is a set with two binary operations, addition (denoted by a + b) and
multiplication (denoted by ab), such that for all a, b, c ∈ R
1. a + b = b + a
2. (a + b) + c = a + (b + c)
3. there is an additive identity 0. That is, there is an element 0 ∈ R such that
a + 0 = a for all a ∈ R.
4. there is element −a ∈ R such that a + (−a) = 0.
5. a(bc) = (ab)c.
6. a(b + c) = ab + ac and (b + c)a = ba + ca.
So, a ring is an abelian group under addition, also having an associative multipli-
cation that is left and right distributive over addition.
Note that multiplication need not be commutative in a ring structure. When
it is, we say that the ring is commutative. Also, a ring need not have an identity
under multiplication. A unity (or identity) in a ring is a non zero element that is
an identity under multiplication. A non zero element of a commutative ring with
unity need not have a multiplicative inverse. When it does, we say that its is a
unit of the ring. Thus, a is a unit if a−1
exists. For example, the set of integers
(Z), rational numbers (Q), real numbers (R) and complex numbers (C) with usual
addition and multiplication are rings. Let n ∈ N, the set of all n×n matrices over
R is a ring with respect to usual addition and multiplication of matrices.
A subset S of a ring R is a subring of R if S itself is a ring under the operations
of R. For example, the subsets 0 and R are subrings of any ring R and are called
trivial subrings of R. The subset Z(R) = {a ∈ R : ax + xa, for every x ∈ R}
is a subring of R, called the center of R. Any subring of Z(R) is called the
central subring of R. The subsets Z ⊂ Q ⊂ R are all subrings of C. A field is a
3

commutative ring with unity in which every non zero element is a unit. The rings
Q, R, C are examples of fields.
Let R be a ring, a subset I of R is called a left ideal of R if, I is a subgroup of
(R, +), that is, a, b ∈ I =⇒ a − b ∈ I and I is closed for arbitrary multiplication
on the left by elements in R, that is, a ∈ I and x ∈ R =⇒ ax ∈ I. A subset I of
R is called a right ideal of R if a, b ∈ I =⇒ a − b ∈ I and a ∈ I, x ∈ R =⇒ ax ∈ I.
For example, in any ring R the subsets {0} and R are both ideals. If R is a field
these are the only ideals. Let F be a field, the set of all polynomials over F,
denoted by F[x] = {
Pn
i=1 aixi
: ai ∈ F} is a ring under usual polynomial addition
and multiplication, called polynomial ring over F.
Let R and S be rings, a ring homomorphism is a map ϕ : R −→ S satisfying
1. ϕ(a + b) = ϕ(a + b), for all a, b ∈ R.
2. ϕ(ab) = ϕ(a) + ϕ(b), for all a, b ∈ R
The kernel of the ring homomorphism ϕ, denoted by kerϕ is the set of all elements
of R that maps to 0 in S. That is kerϕ = {x ∈ R : ϕ(x) = 0s}. An isomorphism
is a ring homomorphism which is both injective and surjective. For example,
consider two rings Z and 2Z. These are isomorphic as groups, since the function
Z −→ 2Z which sends n −→ 2n, is a group homomorphism which is one to one and
onto. However ϕ is not an isomorphism of rings (infact they are not isomorphic
as rings). Indeed, ϕ(1 × 1) = ϕ(1) = 2 while ϕ(1)ϕ(1) = 2 × 2 = 4 ̸= 2. Thus
ϕ(1×1) ̸= ϕ(1)ϕ(1). Let R[x] denote the ring of polynomials with real coefficients.
The mapping f(x) −→ f(1) is a ring homomorphism from R[X] onto R.
A vector space V over the field F is a set with two binary operations - addition
and scalar multiplication on V satisfying the following:
1. V is an abelian group under addition.
2. 1v = v for all v ∈ V .
3. a(u + v) = au + av, (a + b)u = au + bu and
4. a(b.v) = (a.b)v for all a, b ∈ F, v ∈ V.
4

A subset U of a vector space V is called a subspace of V if U itself is a vector space
(using the same addition and scalar multiplication on V ). A list (v1, v2 . . . vm) of
vectors in V is called linearly independent if the only choices of a1, a2, . . . am ∈ F
that makes a1v1 + a2v2 + . . . amvm = 0 is a1 = a2 = . . . am = 0.
A list of vectors in V is called linearly dependent if it is not linearly indepen-
dent. A basis of V is a list of vectors in V that is linearly independent and spans
V . A list (v1, v2, . . . vn) of vectors in V is a basis of V if and only if every v ∈ V can
be written uniquely in the form v = a1v1 +a2v2 +. . . anvn, where a1, a2, . . . an ∈ F.
For example, the space V = F[x] of polynomials in the variable x with coefficients
from the field F is in particular a vector space over F. The elements 1, x, x2
, . . .
are linearly independent by definition, that is, a polynomial is 0 if and only if all
its coefficients are 0. Since these elements spans V , by definition, they are basis
for V .
Let V and W be vector spaces over a field F, a linear map T from V into
W is a function which satisfies: T(u + v) = T(u) + T(v) and T(av) = aT(v), for
all a ∈ F and u, v ∈ V . The set of all linear maps from V into W is denoted
by L(V, W) and we can see that L(V, W) is a vector space over the field F under
usual addition and scalar multipication.
5

Chapter 2
Modules
2.1 Definitions and examples
In this section we give the definition, examples and basic results on modules.
Definition 2.1.1. Let R be a ring (not necessarily commutative nor with 1). A
left R-module or a left module over R is a set M together with
1. a binary operation + on M under which M is an abelian group,and
2. an action of R on M (that is,a map R × M → M) denoted by rm, for all
r ∈ R and for all m ∈ M which satisfies
(a) (r + s)m = rm + sm, for all r, s ∈ R, m ∈ M
(b) (rs)m = r(sm), for all r, s ∈ R, m ∈ M,and
(c) r(m + n) = rm + rn, for all r ∈ R, m, n ∈ M.
If the ring R has a 1 we impose the additional axiom:
(d) 1m = m, for all m ∈ M
The descriptor ”left” in the above definition indicates that the ring elements
appear on the left; ”right” R-modules can be defined analogously. If the ring R is
commutative and M is a left R-module we can make M into a right R-module by
defining mr = rm for m ∈ M and r ∈ R. If R is not commutative, axiom 2(b) in
general will not hold with this definition (so not every left R-module is also a right
6

R-module). Unless explicitly mentioned otherwise the term ”module” will always
mean ”left module.” Modules satisfying axiom 2(d) are called unital modules and
in this book all our modules will be unital (this is to avoid ”pathologies” such as
having rm = 0 for allr ∈ R and m ∈ M).
When R is a field F the axioms for an R-module are precisely the same as
those for a vector space over F, so that modules over a field F and vector spaces
over F are the same.
Before giving other examples of R-modules we record the obvious definition of
submodules.
Definition 2.1.2. Let R be a ring and let M be an R-module. An R-submodule
of M is a subgroup N of M which is closed under the action of ring elements, that
is, rn ∈ N, for all r ∈ R, n ∈ N.
Submodules of M are therefore just subsets of M which are themselves modules
under the restricted operations. In particular, if R = F is a field, submodules are
the same as subspaces. Every R-module M has the two submodules M and 0 (the
latter is called the trivial submodule).
Example 2.1.3. Let R be any ring. Then M = R is a left R -module, where the
action of a ring element on a module element is just the usual multiplication in the
ring R (similarly, R is a right module over itself). In particular, every field can be
considered as a (1-dimensional) vector space over itself. When R is considered as
a left module over itself in this fashion, the submodules of R are precisely the left
ideals of R (and if R is considered as a right R-module over itself, its submodules
are the right ideals). Thus if R is not commutative it has a left and right module
structure over itself and these structures may be different.
Example 2.1.4. Let R = Z, let A be any abelian group (finite or infinite) and
write the operation of A as +. Make A into Z-module as follows : for any n ∈ Z
7

and a ∈ A define
na =













a + a + · · · + a(n times) if n > 0
0 if n = 0
−a − a − · · · − a(-n times) if n < 0
(here 0 is the identity of the additive group A). This definition of an action of the
integers on A makes A into a Z-module, and the module axioms show that this
is the only possible action of Z on A making it a (unital) Z-module. Thus every
abelian group is a Z-module. Conversely, if M is any Z-module, a fortiori M is
an abelian group, so Z-modules are the same as abelian groups. Furthermore, it
is immediate from the definition that Z-submodules are the same as subgroups.
Example 2.1.5. Let L be a left ideal of R. Then,
1. L is an additive abelian group induced by that of R, and
2. rm ∈ L, for all m ∈ L and r ∈ R.
Moreover, (a), (b) in module definition are clearly satisfied for all m, n ∈ L and
r, s ∈ R. Hence every ideal L of R has the structure of left R-module in which
1. the addition is the one induced by addition in R, and
2. the scalar multiplication r · m of an element m ∈ L by an element r ∈ R is
the product rm of r and m in the ring R
Similarly, right ideal I of R has the structure of a right R-module in which ad-
dition and scalar multiplication are induced respectively from the addition and
multiplication in the ring R.
In particular, taking L = R, I = R, we see that the ring R, has in the above
manner, the structure of left as well as right R-module.
Example 2.1.6. If R is any ring, then, Rn
, the set of all n-tuples with components
in R is an R-module, with usual definitions of addition and scalar multiplication.
8

Example 2.1.7. Let F be a field, let x be an indeterminate and let R be the
polynomial ring F[x]. Let V be a vector space over F and let T be a linear
transformation from V to V . We have already seen that V is an F-module; the
linear map T will enable us to make V into an F[x]-module.
First, for the nonnegative integer n, define
T0 = I
.
.
.
Tn
= T ◦ T ◦ . . . ◦ T(n times)
where I is the identity map from V to V and ◦ denotes function composition
(which makes sense because the domain and codomain of T are the same). Also,
for any two linear transformations A, B from V to V and elements α, β ∈ F, let
αA + βB defined by
(αA + βB)(v) = α(A(v)) + β(B(v))
(that is addition and scalar multiplication of linear transformations are defined
pointwise). Then αA + βB is easily seen to be a linear transformation from
V to V , so that linearcombinations of linear transformations are again linear
transformations.
We now define the action of any polynomial in x on V . Let p(x) be the polynomial
p(x) = anxn
+ an−1xn−1
+ .... + a1 + a0, where a0, a1, .., an ∈ F. For each v ∈ V
define an action of the ring element p(x) on the module element v by
p(x)v = (anTn
+ an−1Tn−1 + . . . + a1T + a0)(v)
= anTn
(v) + an−1Tn−1(v) + . . . + a1T(v) + a0v
(that is p(x) acts by substituting the linear transformation T for x in p(x) and
applying the resulting linear transformation to v). Put another way, x acts on V
as the linear transformation T and we extend this to an action of all of F[x] on V
in a natural way. It is easy to check that this definition of an action of F[x] on V
satisfies all the module axioms and makes V into an F[x]-module.
9

Remark. We can easily verify that the polynomial ring R[x] over a ring R is an
R-module.
Proposition 2.1.8. Let R be a ring and let M be an R-module. A subset N of
M is a submodule of M if and only if
1. N ̸= ϕ and
2. x + ry ∈ N for all r ∈ R and for all x, y ∈ N
Proof. If N is a submodule, then 0 ∈ N so N = I = 0. Also N is closed under
addition and is sent to itself under the action of elements of R. Conversely, suppose
(1) and (2) hold. Let r = −1 and apply the subgroup criterion (in additive form)
to see that N is a subgroup of M. In particular, 0 ∈ N. Now let x = 0 and apply
hypothesis (2) to see that N is sent to itself under the action of R. This completes
the proof.
Proposition 2.1.9. Let M be an R-module. Then for all r ∈ R and for all
m ∈ M.We have
1. 0R = 0M
2. r.0M = 0M
3. (−r)m = −(rm) =r(−m)
4. (−r)(−m)=rm
Proof. Since 0R + 0R =0R, the definition of an R-module shows that 0Rm =
(0R + 0R)m =0Rm + 0Rm from which 0Rm =0M because M is a group. Next,
from 0M + 0M = 0M follows r0M = r(0M + 0M ) = r0M + r0M , whence r0M = 0M .
Thus (1) and (2) are proved. Again 0R = r + (−r) and therefore, by (1), 0M =
0Rm= (r + (−r))m = rm + (−r)m which yields (−r)m = −(rm). A similar
argument shows that r(−x) = −(rx). Finally, using (3), we obtain (−r)(−x) =
−((−r)x)=−(−rx)) = rx. This completes the proof.
10

Proposition 2.1.10. For an abelian group M, let EndZ(M) be the ring of all
(additive) endomorphisms of M. Let R be any ring. Then we have the following.
1. M is a left R-module ⇐⇒ there exists a homomorphism of rings
ψ : R → EndZ(M).
2. M is a right R-module ⇐⇒ there exits an anti-homomorphism of rings
ψ̀ : R → EndZ(M) that is ψ̀ preserves addition but reverses the multiplica-
tion.
3. M is R- unitary ⇐⇒ ψ(1R) = idM (resp. ψ̀(1R) = idM )
Proof. Let M be a left R-module with scalar multiplication,
R × M → M, (a, x) 7→ ax. Now define ψ : R → EndZ(M) by a 7→ ψ(a), where
ψ(a) : M → M is defined by ψ(a)(x) = ax, for all a ∈ R and x ∈ M.
Claim(i): ψ is a homomorphism of rings.
For let a, b ∈ R and x ∈ M. We have
ψ(a + b) = (a+b)x = ψ(a)(x) + ψ(b)(x) =[ψ(a) + ψ(b)](x).
Thus ψ(a + b) = ψ(a) + ψ(b). Similarly, we have for x ∈ M,
ψ(ab)(x) = (ab)(x) = a(bx) = a(ψ(b)(x))
= ψ(a)(ψ(b)(x)) = (ψ(a)oψ(b))(x).
Thus ψ(ab) = ψ(a)ψ(b) and hence ψ is a homomorphism of rings.
Conversely, suppose that ψ : R → EndZ(M) is a homomorphism of rings. Now
define the scalar multiplication by
R × M → (a, x) 7→ ax = (ψ(a))(x).
Claim(ii): This defines a left R-module structure on M.
For, let a, b ∈ R and x, y ∈ M. Since ψ(a) ∈ EndZ(M), we have
a(x + y) = (ψ(a))(x + y) = ψ(a)(x) + ψ(a)(y) = ax + ay.
Similarly, we have
(a + b)(x) = ψ(a + b)(x) = [ψ(a) + ψ(b)](x)
= ψ(a)(x) + ψ(b)(x) = ax + bx
11

and (ab)(x) = ψ(ab)(x) = (ψ(a)oψ(b))(x) = ψ(a)(ψ(b)(x)) = ψ(a)(bx) = a(bx).
Thus M is an R-module. Proof of (2) is similar to (1).
(3) Suppose M is R-unitary and ψ : R → EndZ(M) is the corresponding homo-
morphism of rings . We have ψ(1) : M → M, x 7→ ψ(1)(x) = 1 · x = x.
Hence ψ(1R) = idM . Conversely, suppose that ψ(1R) = idM where ψ : R →
EndZ(M) is a homomorphism of rings. Look at ths scalar multiplication deined
as above, R × M → M, (a, x) 7→ ax = (ψ(a))(x). We have 1 · x = (ψ(1))(x) =
idM (x) = x, as required.
Corollory 2.1.11. M is a left R-module implies that M is a right R0
-module
where R0
is the ring opposite to R.
Proof. We have a homomorphism of rings ψ : R → EndZ(M). Compose this
with the identity map id : R0
→ R which is an anti-isomorphism, to get an anti-
homomorphism R0
→ EndZ(M) which means M is a right R0
-module.
Conversely, suppose that we have an anti-homomorphism of rings ψ̀ : R0
→
EndZ(M). Compose this with the identity map id : R → R0
which is an anti-
isomorphism, to get a homomorphism R → EndZ(M). Therefore M is a left
R-module.
Definition 2.1.12. Let R be a commutative ring with identity. An R -algebra is
a ring A with identity together with a ring homomorphism f : R → A mapping 1R
to 1A such that the subring f(R) of A is contained in the center of A.
If A is an R-algebra then it is easy to check that A has a natural left and right
(unital) R-module structure defined by r · a = a · r = f(r)a where f(r)a is just
the multiplication in the ring A and this is the same as af(r) since by assumption
f(r) lies in the center of A. In general it is possible for an R-algebra A to have
other left (or right) R-module structures, but unless otherwise stated, this natural
module structure on an algebra will be assumed.
Definition 2.1.13. If A and B are two R-algebras, an R-algebra homomorphism
(or isomorphism) is a ring homomorphism (isomorphism, respectively) ϕ : A → B
mapping 1A to 1B such that ϕ(r · a) = r · ϕ(a) for all r ∈ R and a ∈ A.
12

Example 2.1.14. Let R be a commutative ring with 1.
1. Any ring with identity is a Z-algebra
2. For any ring A with identity, if R is a subring of the center of A containing
the identity of A then A is an R-algebra. In particular, a commutative ring
A containing 1 is an R-algebra for any subring R of A containing 1.
2.2 Quotient modules and module homomorphisms
This section contains the basics of quotient modules and module homomorphisms.
Definition 2.2.1. Let R be a ring and let M and N be R-modules.
1. A map ϕ : M → N is an R-module homomorphism if it respects the R-
module structures of M and N, that is,
(a) ϕ(x + y) = ϕ(x) + ϕ(y), for all x, y ∈ M and
(b) ϕ(rx) = rϕ(x), for all r ∈ R, x ∈ M.
2. An R-module homomorphism is an isomorphism (of R-modules) if it is both
injective and surjective. The modules M and N are said to be isomorphic,
denoted M ∼
= N, if there is some R-module isomorphism ϕ : M → N.
3. If ϕ : M → N is an R-module homomorphism, let kerϕ = {m ∈ M|ϕ(m) =
0} (the kernel of ϕ) and let ϕ(M) = {n ∈ N|n = ϕ(m)for some m ∈ M}
(the image of ϕ, as usual).
4. Let M and N be R-modules and define HomR(M, N) to be the set of all
R-module homomorphisms from M into N.
Example 2.2.2. If R is a ring and M = R is a module over itself, then R-module
homomorphisms (even from R to itself) need not be ring homomorphisms and ring
homomorphisms need not be R-module homomorphisms. For example, when
R = Z the Z-module homomorphism x 7→ 2x is not a ring homomorphism (1 does
not map to 1). When R = F[x] the ring homomorphism ϕ : f(x) 7→ f(x2
) is not
13

an F[x]-module homomorphism (if it were, we would have x2
= ϕ(x) = ϕ(x · 1) =
xϕ(1) = x).
Example 2.2.3. Let R be a ring, let n ∈ Z+
and let M = Rn
, for each
i ∈ {1, ..., n} the projection map πi : Rn
→ R by πi(x1, ..., xn) = xi is a surjective
R-module homomorphism with kernel equal to the submodule of n-tuples which
have a zero in position i.
Example 2.2.4. If R is a field, R-module homomorphisms are called linear trans-
formations.
Example 2.2.5. For the ring R = Z the action of ring elements (integers) on any
Z-module amounts to just adding and subtracting within the (additive) abelian
group structure of the module so that in this case condition (b) of a homomorphism
is implied by condition (a). For example, ϕ(2x) = ϕ(x + x) = ϕ(x) + ϕ(x) =
2ϕ(x), It follows that Z-module homomorphisms are the same as abelian group
homomorphisms.
Proposition 2.2.6. Let M, N and L be R-modules
1. A map ϕ : M → N is an R-module homomorphism if and only if
ϕ(rx + y) = rϕ(x) + ϕ(y) for all x, y ∈ Mand all r ∈ R.
2. Let ϕ, ψ be elements of HomR(M, N). Define ϕ + ψ by , for all m ∈ M,
(ϕ + ψ)(m) = ϕ(m) + ψ(m).
Then ϕ + ψ ∈ HomR(M, N) and with this operation HomR(M, N) is an
abelian group. If R is a commutative ring then for r ∈ R define rϕ by for
all m ∈ M,
(rϕ)(m) = r(ϕ(m))
Then rϕ ∈ HomR(M, N) and with this action of the commutative ring R
the abelian group HomR(M, N) is an R-module.
3. If ϕ ∈ HomR(L, M) and ψ ∈ HomR(M, N) then ψ ◦ ϕ ∈ HomR(L, N).
14

4. With addition as above and multiplication defined as function composition,
HomR(M, M) is a ring with 1. When R is commutative HomR(M, M) is
an R-algebra.
Proof. 1. Certainly ϕ(rx+y) = ϕ(x)+ϕ(y) if ϕ is an R-module homomorphism.
Conversely, if ϕ(rx + y) = ϕ(x) + ϕ(y), take r = 1 to see that ϕ is additive
and take y = 0 to see that ϕ commutes with the action of R on M (that is,
is homogeneous).
2. It is straightforward to check that all the abelian group and R-module axioms
hold with these definitions. We note that the commutativity of R is used
to show that rϕ satisfies the second axiom of an R-module homomorphism,
namely,
(r1ϕ)(r2m) = r1ϕ(r2m) (by the definition of r1ϕ)
= r1r2(ϕ(m) (Since ϕ is a homomorphism)
= r2r1ϕ(m) (Since R commutative)
= r2r1ϕ(m) ( by the definition of r1ϕ)
Verification of the axioms relies ultimately on the hypothesis that N is an
R-module. The domain M could in fact be any set - it does not have to be
an R-module nor an abelian group.
3. Let ϕ and ψ be as given and let r ∈ R, x, y, ∈ L. Then
(ψ ◦ ϕ)(rx + y) = ψ(ϕ(rx + y)
= ψ(rϕ(x + ϕ(y)) (by (1) applied to ϕ)
= rψ(ϕ(x)) + ψ(ϕ(y)) (by(1) applied to ψ)
= r(ψ ◦ ϕ)(x) + (ψ ◦ ϕ)(y)
so, by (1), ψ ◦ ϕ is an R-module homomorphism.
4. Note that since the domain and codomain of the elements of HomR(M, M)
are the same, function composition is defined. By (3), it is a binary op-
eration on HomR(M, M). As usual, function composition is associative.
15

The remaining ring axioms are straightforward to check. The identity func-
tion, I, (as usual, I(x) = x, for all x ∈ M) is seen to be the multiplica-
tive identity of HomR(M, M). If R is commutative, then (2) shows that
the ring HomR(M, M) is a left R-module and defining ϕr = rϕ for all
ϕ ∈ HomR(M, M) and r ∈ R makes HomR(M, M) into an R-algebra.
Definition 2.2.7. The ring HomR(M, M) is called the endomorphism ring of M
and will often be denoted by EndR(M), or just End(M) when the ring R is clear
from the context. Elements of End(M) are called endomorphisms.
When R is commutative there is a natural map from R into End(M) given
by r 7→ rI, where the latter endomorphism of M is just multiplication by r on
M. The image of R is contained in the center of End(M) so if R has an identity,
End(M) is an R-algebra. The ring homomorphism from R to EndR(M) may not
be injective since for some r we may have rm = 0 for all m ∈ M
(example, R = Z , M = Z/2Z, and r = 2). When R is a field, however, this map
is injective (in general, no unit is in the kernel of this map) and the copy of R in
EndR(M) is called the (subring of) scalar transformations. Next we prove that
every submodule N of an R-module M is ”normal” in the sense that we can always
form the quotient module M/N, and the natural projection π : M → M/N is an R-
module homomorphism with kernel N. The proof of this fact and, more generally,
the subsequent proofs of the isomorphism theorems for modules follow easily from
the corresponding facts for groups. The reason for this is because a module is
first of all an abelian group and so every submodule is automatically a normal
subgroup and any module homomorphism is, in particular, a homomorphism of
abelian groups. What remains to be proved in order to extend results on abelian
groups to corresponding results on modules is to check that the action of R is
compatible with these group quotients and homomorphisms.
Proposition 2.2.8. Let R be a ring, let M be an R-module and let N be a
submodule of M. The (additive, abelian) quotient group M/N can be made into
16

an R-module by defining an action of elements of R by
r(x + N) = (rx) + N, for all r ∈ R, x+N ∈ M/N.
The natural projection map π : M → M/N defined by π(x) = x + N is an R-
module homomorphism with kernel N.
Proof. Since M is an abelian group under + the quotient group M/N is defined
and is an abelian group. To see that the action of the ring element r on the coset
x+N is well defined, suppose x+N = y+N, that is x−y ∈ N. Since N is a (left)
R-submodule, r(x − y) ∈ N. Thus rx − ry ∈ N and rx + N = ry + N, as desired.
Now since the operations in M/N are ”compatible” with those of M, the axioms for
an R-module are easily checked in the same way as was done for quotient groups.
For example, axiom 2(b) holds as follows: for all r1, r2 ∈ R and x + N ∈ M/N, by
definition of the action of ring elements on elements of M/N
(r1r2)(x + N) = (r1r2x) + N
= r1(r2x + N)
= r1(r2(x + N))
The other axioms are similarly checked. The natural projection map π described
above is, in particular, the natural projection of the abelian group M onto the
abelian group M/N hence is a group homomorphism with kernel N. The kernel
of any module homomorphism is the same as its kernel when viewed as a homo-
morphism of the abelian group structures. It remains only to show π is a module
homomorphism, that is π(rm) = rπ(m). But
π(rm) = rm + N
= r(m + N)( by definition of the action of R on M/N)
= rπ(m).
This completes the proof.
17

Definition 2.2.9. Let A, B be submodules of the R-module M. The sum of A
and B is the set
A + B = {a + b ; a ∈ A, b ∈ B}.
The sum of two submodules A and B is a submodule and is the smallest submodule
which contains both A and B.
Theorem 2.2.10. 1. (The First Isomorphism Theorem for Modules) Let M, N
be R-modules and let ϕ : M → N be an R-module homomorphism. Then
kerϕ is a submodule of M and M/kerϕ ∼
= ϕ(M)
2. (The Second Isomorphism Theorem) Let A, B be submodules of the R-module
M . Then (A + B)/B ∼
= A/(A ∩ B).
3. (The Third Isomorphism Theorem) Let M be an R-module, and let A and
B be submodules of M with A ⊆ B. Then (M/A)/(B/A) ∼
= M/B.
Proof. 1. By definition of kerϕ , kerϕ = {m ∈ M; ϕ(m) = 0}. Since ϕ(0) =
0 , 0 ∈ kerϕ , kerϕ is a non empty subset of M. Now, let m, n be any two
arbitrary elements of kerϕ. Then we have ϕ(m) = 0 and ϕ(n) = 0.
ϕ(m − n) = ϕ(m) − ϕ(n) = 0. This implies m, n ∈ kerϕ, for every m,n ∈
kerϕ. Hence kerϕ is an additive subgroup of M. Again for r ∈ R and m ∈
M, we have ϕ(rm) = rϕ(m) = r.0 = 0. This implies rm ∈ kerϕ , for all r ∈
R. Therefore from (2.1) and (2.2), we have kerϕ is an R-submodule of M.
Since kerϕ is a submodule of M, M/kerϕ is an R-module defined by r(x +
kerϕ) = rx + kerϕ for all r ∈ R, x + kerϕ ∈ M/kerϕ. We need to show
that M/kerϕ ∼
= ϕ(M). For consider a map, ψ : M / ker ϕ → ϕ(M). We
will prove that ψ is an R - module homomorphism that is both injective and
18

surjective. To prove ψ is one to one, consider,
ψ(m + kerϕ) = ψ(n + kerϕ)
=⇒ ϕ(m) = ϕ(n)
=⇒ ϕ(m − n) = 0
=⇒ m − n ∈ kerϕ
=⇒ m + kerϕ = n + kerϕ.
To prove ψ is onto, consider for any ϕ(m) ∈ ϕ(M) we can find m + kerϕ ∈
M/kerϕ such that ψ (m + ker ϕ ) = ϕ(m).
To prove ψ is an R-module homomorphism, let m+kerϕ, n+kerϕ ∈ M/kerϕ,
ψ(m + kerϕ + n + kerϕ) = ψ(m + n + kerϕ)
= ϕ(m + n)
= ϕ(m) + ϕ(n)
= ψ(m + kerϕ) + ψ(n + kerϕ).
and
ψ(r(m + kerϕ)) = ψ(rm + kerϕ)
= ϕ(rm)
= rϕ(m)
= rψ(m + kerϕ)
It follows that, M/kerϕ ∼
= ϕ(M).
2. Let f : M → N be homomorphism of R-modules, let A be a submodule
of M and B be a submodule of N. Then f(A) and f−1
(B) are submoodules
of N and M respectively. In particular, Imf is a submodule of N while
kerf is a submodule of M. Finally f−1
(f(A)) = A + kerf = A + B
whence f(A + B) = f(A). But f(A + B) = (A + B)/B and therefore
f(A) + (A + B)/B. Thus by restricting f to A there is produced an epi-
morphism ϕ : A → (A + B)/B. First isomorphism theorem for modules
shows that we have an isomorphism A/kerϕ ∼
= (A + B)/B. However, the
19

elements of A which are mapped by ϕ into zero are just the elements of A
which belong to B. Accordingly kerϕ = A ∩ B and the proof is complete.
3. Since A and B are submodules of M with A ⊆ B, we have B/A is a
submodule of M/A. Define ϕ : M/A → M/B by ϕ(m + A) = m + B, for all
m ∈ M.
To prove ϕ is well defined,
Let m1 + A , m2 + A ∈ M/A. Suppose m1 + A = m2 + A
=⇒ m1 − m2 ∈ A
=⇒ m1 − m2 ∈ B
=⇒ m1 + B = m2 + B
=⇒ ϕ(m1 + A) = ϕ(m2 + A).
To prove ϕ is an R-module homomorphism,
ϕ((m1 + A) + (m2 + B)) = ϕ(m1 + m2 + A)
= ϕ(m1 + m2 + A)
= m1 + m2 + B
= m1 + B + m2 + B
= ϕ(m1 + A) + ϕ(m2 + A)
and ϕ(r(m + A)) = ϕ(rm + A)
= rm + B
= r(m + B)
= rϕ(m + A).
To prove ϕ is onto, for any m + B ∈ M/B we can find m + A ∈ M/A such
that ϕ(m + A) = m + B. Hence it follows that, (M/A)/(kerϕ) ∼
= M/B.
Now, kerϕ = {m + A ∈ M/A : ϕ(m + A) = B}
= {m + A ∈ M/A : m + B = B}
= {m + A ∈ M/A : m ∈ B}
= B/A.
20

Therefore, (M/A)/(B/A) ∼
= M/B.
2.3 Generation of modules, direct sums, and free modules
Let R be a ring with 1. As in the preceding sections the term ”module” will mean
”left module.” We first extend the notion of the sum of two submodules to sums of
any finite number of submodules and define the submodule generated by a subset.
Definition 2.3.1. Let M be an R-module and let N1, ..., Nn be submodules of M.
1. The sum of N1, ..., Nn is the set of all finite sums of elements from the sets
Ni : {a1 + a2 + + an : ai ∈ Ni for all i }. Denote this sum by N1 + . . . + Nn.
2. For any subset A of M let RA = {r1al + r2a2 + + rmam : r1, ..., rm ∈
R, ai, ..., am ∈ A, m ∈ Z+
} (where by convention RA = {0} if A = ϕ). lf
A is the finite set {a1, a2, ..., an} we shall write Ra1 + Ra2 + . . . + Ran for
RA. Call RA the submodule of M generated by A. If N is a submodule of
M (possibly N = M) and N = RA, for some subset A of M, we call A a
set of generators or generating set for N, and we say N is generated by A.
3. A submodule N of M (possibly N = M) is finitely generated if there is some
finite subset A of M such that N = RA, that is, if N is generated by some
finite subset.
4. A submodule N of M (possibly N = M) is cyclic if there exists an element
a ∈ M such that N = Ra, that is, if N is generated by one element:
N = Ra = {ra : r ∈ R}
Note that these definitions do not require that the ring R contain a 1, however
this condition ensures that A is contained in RA. It is easy to see using the
submodule criterion that for any subset A of M, RA is indeed a submodule of M
and is the smallest submodule of M which contains A (that is, any submodule of
21

M which contains A also contains RA). In particular, for submodules N1, ..., Nn
of M, N1 + . . . + Nn is just the submodule generated by the set N1 ∪ . . . ∪ Nn and
is the smallest submodule of M containing Ni, for all i. If N1, ..., Nn are generated
by sets A1, . . . , An respectively, then N1 + . . . + Nn is generated by A1 ∪ . . . ∪ An.
Note that cyclic modules are, a fortiori, finitely generated.
A submodule N of an R-module M may have many different generating sets (for
instance the set N itself always generates N). If N is finitely generated, then there
is a smallest nonnegative integer d such that N is generated by d elements (and
no fewer). Any generating set consisting of d elements will be called a minimal set
of generators for N (it is not unique in general). If N is not finitely generated, it
need not have a minimal generating set.
The process of generating submodules of an R-module M by taking subsets A of M
and forming all finite ”R-linear combinations” of elements of A will be our primary
way of producing submodules (this notion is perhaps familiar from vector space
theory where it is referred to as taking the span of A). The obstruction which made
the analogous process so difficult for groups in general was the noncommutativity
of groupoperations. For abelian groups, G, however, it was much simpler to
control the subgroup < A > generated by A, for a subset A of G. The situation
for R-modules is similar to that of abelian groups.
Example 2.3.2. Let R = Z and let M be any R-module, that is, any abelian
group. If a ∈ M, then Z is just the cyclic subgroup of M generated by a: < a >.
More generally, M is generated as a Z-module by a set A if and only if M is
generated as a group by A (that is, the action of ring elements in this instance
produces no elements that cannot already be obtained from A by addition and
subtraction). The definition of finitely generated for Z-modules is identical to that
for abelian groups.
Example 2.3.3. Let R be a ring with 1 and let M be the (left) R-module R itself.
Note that R is a finitely generated, in fact cyclic, R-module because R = R1 (that
is we can take A = {1} ). Recall that the submodules of R are precisely the left
ideals of R, so saying I is a cyclic R-submodule of the left R-module R is the
22

same as saying I is a principal ideal of R. (Also, saying I is a finitely generated
R-submodule of R is the same as saying I is a finitely generated ideal. When
R is a commutative ring we often write AR or aR for the submodule (ideal)
generated by A or a respectively, as we have been doing for Z when we wrote nZ.
In this situation AR = RA and aR = Ra (elementwise as well). Thus a Principal
Ideal Domain is a (commutative) integral domain R with identity in which every
R-submodule of R is cyclic.
Definition 2.3.4. Let M1, . . . , Mk be a collection of R-modules. The collection of
k-tuples (m1, m2, . . . , mk) where mi ∈ M; with addition and action of R defined
componentwise is called the direct product of M1, M2, . . . , Mk denoted
M1 ⊕ . . . ⊕ Mk.
Proposition 2.3.5. Le N1, N2, . . . , Nk be submodules of the R-module M. Then
the following are equivalent:
1. The map π : N1 × N2 × . . . × Nk → N1 + N2 + . . . + Nk defined by
π(a1, a2, . . . , ak) = a1 + a2 + . . . + ak
is an isomorphism of R-modules : N1 + N2 + . . . + Nk
∼
= N1 × N2 × . . . × Nk.
2. Nj ∩ (N1 + . . . + Nj−1 + Nj+1 + . . . + Nk) = 0 for all j ∈ {1, 2, . . . , k}.
3. Every x ∈ N1 +. . .+Nk can be written uniquely in the form a1 +a2 +. . .+ak
with a ∈ Ni.
Proof. To prove (1.) implies (2.), suppose for some j that (2.) fails to hold and
let aj ∈ (N1 + ... + Nj−l + Nj+l + + Nk) ∩ Nj, with ai ̸= 0 . Then
aj = a1+. . .+aj−1+aj+1+...+ak for some ai ∈ Ni , and (a1, . . . , aj−l, −aj, aj+l, . . . ak)
would be a nonzero element of kerπ, a contradiction.
Assume now that (2.) holds. If for some module elements ai, bi ∈ Ni we have
a1 + a2 + . . . + ak = b1 + b2 + . . . + bk
then for each j we have
aj −bj = (b1 −a1)+. . .+(bj−1 −aj − t1)+(bj+1 −aj+1)+. . .+(bk −ak) . The left
23

hand side is in Nj and the right side belongs to N1 + . . . + Nj−1 + Nj+1 + + Nk.
Thus aj − bj ∈ Ni ∩ (N1 + . . . + Nj−1 + Nj+1 + . . . + Nk) = 0. This shows aj = bj
for all j, and so (2.) implies (3.). Finally, to see that (3.) implies (1.) observe
first that the map π is clearly a surjective R-modu1e homomorphism. Then (3.)
simply implies π is injective, hence is an isomorphism, completing the proof.
If an R-module M = N1+N2+. . .+Nk is the sum of submodules N1, N2, . . . , Nk
of M satisfying the equivalent conditions of the proposition above, then M is said
to be the (internal) direct sum of N1, N2, ..., Nk written M = N1 ⊕ N2 ⊕ . . . ⊕ Nk
By the proposition, this is equivalent to the assertion that every element m of M
can be written uniquely as a sum of elements m = n1 +n2 +. . .+nk with ni ∈ Ni.
(Note that part (1) of the proposition is the statement that the internal direct sum
of N1, N2, . . . , Nk is isomorphic to their external direct sum, which is the reason
we identify them and use the same notation for both.)
Definition 2.3.6. An R-module M is called a free module if M has a basis B,
that is linearly independent subset B of M such that M is spanned by B over R,
that is every element x ∈ M can be written uniquely as x =
P
b∈B λb · b,
λb ∈ R, λb = 0 except for finitely many b’s, that is x is finite linear combination
of elements in B, the scalars being unique for x.
Example 2.3.7. Rn
= R × . . . × R (n times), is a free R-module if R has 1. The
set B = {(1, 0, . . . , 0), (0, 1, . . . , 0), . . . , (0, 0, . . . , 1)} is an R-basis for Rn
, called
the standard basis of Rn
.
Example 2.3.8. For, suppose M and N are free R-modules with bases A and
B respectively. Now M ⊕ N = M × N is free R-module because (A ×{0})
∪({0} × B) is an R-basis for M × N. More generally, for any family of free R-
modules, {Mi : i ∈ I}, with basis Ai, M = ⊕i∈IMi is a free module with a basis
A = ∪i∈IAi
Example 2.3.9. Any finite abelian group is not free as a module over Z. In fact,
any abelian group M which has a non-trivial element of finite order cannot be free
as a module over Z. For, suppose M is free. Say B is a basis for M over Z. Let 0
24

̸= x ∈ M be such that nx = 0 for some n ∈ N, mx ̸= 0 for m < n and n ⩾ 2 Now
we have x = n1b1 +n2b2 +. . .+nrbr for some b1, b2, . . . , br ∈ B and n1, n2, . . . , nr ∈
Z. Hence 0 = nx = n {n1b1 + n2b2 + . . . + nrbr} = nn1b1 + nn2b2 + . . . nnrbr
=⇒ nn1 = 0, nn2 = 0, . . . , nnr = 0 (by linear independence of B) =⇒ n1 = 0,
n2 = 0, . . . , nr = 0 (since n ̸= 0 ), that is, x = 0, a contradiction.
Theorem 2.3.10. For any set A there is a free R-module F(A) on the set A
and F(A) satisfies the following universal property: if M is any R-module and
φ : A −→ Mis any map of sets, then there is a unique R-module homomorphism
ϕ : F(A) −→ M such that ϕ(a) = φ(A), for all a ∈ A, that is, the following
diagram (2.1) commutes.
Figure 2.1:
When A is the finite set {a1, a2, . . . , an}, F(A) = Ra1 ⊕Ra2 ⊕. . .⊕Ran
∼
= Rn
.
Proof. Let F(A) = {0} if A = ∅. If A is nonempty let F(A) be the collection of
all set functions f : A −→ R such that f(a) = 0 for all but finitely many a ∈ A.
Make F(A) into an R-module by pointwise addition of functions and pointwise
multiplication of a ring element times a function, that is,
(f + g)(a) = f(a) + g(a)
and (rf)(a) = r(f(a)). for all a ∈ A, r ∈ R and f, g ∈ F(A).
It is an easy matter to check that all the R-module axioms hold. Identify A as
a subset of F(A) by a 7→ fa, where fa is the function which is 1 at a and zero
elsewhere. We can, in this way, think of F(A) as all finite R-linear combinations of
elements of A by identifying each function f with the sum r1a1 + r2a2 + . . . + rnan,
25

where f takes on the value ri at ai and is zero at all other elements of A. Moreover,
each element of F(A) has a unique expression as such a formal sum. To establish
the universal property of F(A) suppose φ : A −→ Mis a map of the set A into
the R-module M. Define ϕ : F(A) −→ M by
ϕ :
n
X
i=1
riai 7→
n
X
i=1
riφ(ai)
By the uniqueness of the expression for the elements of F(A) as linear combinations
of the ai we see easily that ϕ is a well defined R-module homomorphism. By
definition, the restriction of ϕ to A equals φ. Finally, since F(A) is generated by
A, once we know the values of an R-module homomorphism on A its values on
every element of F(A) are uniquely determined, so ϕ is the unique extension of φ
to all of F(A).
When A is the finite set a1, a2, . . . , an we have that F(A) =Ra1 ⊕Ra2 ⊕. . .⊕Ran.
Since R ∼
= Rai for all i (under the map r 7→ rai) we have that the direct sum is
isomorphic to Rn
.
Result.
1. If F1 and F2 are free modules on the same set A, there is a unique isomor-
phism between F1 and F2 which is the identity map on A.
2. If F is any free R-module with basis A, then F ∼
= F(A). In particular, F
enjoys the same universal property with respect to A as F(A) does in above
theorem.
26

Chapter 3
Modules with chain conditions
In this concluding chapter, we shall study the basic properties of an important
class of modules and rings, ( ”Artinian” and ”Noetherian”), which have some very
special properties. Unless otherwise stated, R stands for a ring with 1 (commuta-
tive or not) and all modules considered are assumed to be unitary modules.
3.1 Artinian modules
Theorem 3.1.1. The following are equivalent for an R-module M.
1. Descending chain condition (d.c.c) hold for submodules of M, that is any
descending chain M1 ⊇ M2 ⊇ . . . ⊇ Mn ⊇ . . . of submodules of M is
stationary in the sense that Mr = Mr+1 = . . . for some r. (We write this
Mr = Mr+1, for every r ≫ 0).
2. Minimum condition for submodules holds for M, in the sense that any non-
empty family of submodules of M has a minimal element.
Proof. (1) =⇒ (2): Let F = {Mi, i ∈ I} be a non-empty family of submodules
of M. Pick any index i1 ∈ I and look at Mi1 . If Mi1 is minimal in F, we are
through. Otherwise, there is an i2 ∈ I such that Mi1 ⊃ Mi2 ,Mi1 ̸= Mi2 . If this
Mi2 is minimal in F, we are through again. Proceeding thus, if we do not find
a minimal element at any finite stage, we would end up with a non-stationary
27

descending chain of submodules of M, namely, M1 ⊃ M2 ⊃ . . . ⊃ Mn ⊃ . . .
contradicting (1).
(2) =⇒ (1): Let M1 ⊇ M2 ⊇ . . . ⊇ Mn ⊇ . . . be a descending chain of sub-
modules of M. Consider the non-empty family F = {Mi : i ∈ N} of sub-
modules of M. This must have a minimal element, say Mr for some r. Now
we have Ms ⊆ Mr, for every s ≥ r which implies by minimality of Mr that
Ms = Mr, for every s ≥ r
Definition 3.1.2. Artinian module: A module M is called Artinian if d.c.c
(or equivalently, the minimum condition) holds for M.
Remark. Minimal submodules exist in a non-zero Artinian module because a min-
imal submodule is simply a minimal element in the family of all non-zero submod-
ules of M.
Example 3.1.3. A module which has only finitely many submodules is Artinian.
In particular, finite abelian groups are Artinian as modules over Z.
Example 3.1.4. Finite dimensional vector spaces are Artinian (for reasons of
dimension) whereas infinite dimensional ones are not Artinian.
Example 3.1.5. Infinite cyclic groups are not Artinian. For instance, Z has a
nonstationary descending chain of subgroups, namely,
Z = (1) ⊃ (2) ⊃ (4) ⊃ . . . (2n
) ⊃ . . . ⊃ . . .
Theorem 3.1.6. 1. Submodules and quotient modules of Artinian modules are
Artinian.
2. If a module M is such that it has a submodule N with both N and M/N are
Artinian, then M is Artinian.
Proof. 1. Let M be Artinian and N a submodule of M. Any family of sub-
modules of N is also one in M and hence the result follows. On the other
hand, any descending chain of submodules of M/N corresponds to one in M
(wherein each member contains N) and hence the result.
28

2. Let M1 ⊇ M2 ⊇ . . . ⊇ Mn ⊇ . . . be a descending chain in M. Intersecting
with N gives the descending chain in N, namely, N ∩ M1 ⊇ N ∩ M2 ⊇ . . . ⊇
N ∩ Mn ⊇ . . . which must be stationary, say N ∩ Mr = N ∩ Mr+l = . . . for
some r. On the other hand, we have the descending chain in M/N, namely,
(N + M1)/N ⊇ (N + M2)/N ⊇ . . . ⊇ (N + Mn) ⊇ . . . which must be also
stationary, say (N + Ms)/N = (N + Ms+1)/N = . . . for some s.
Next we prove that Mn = Mn+1, for every n ≥ (r + s). This is an immediate
consequence of the following four facts:
1. Mn ⊇ Mn+1, for every n ∈ N,
2. N ∩ Mn = N ∩ Mn+1 ,for every n ≥ r,
3. (N + Mn)/N = (N + Mn+1)/N, for every n ≥ s and
4. (N + Mn)/N ∼
= Mn/(N ∩ Mn), for every n ∈ N.
Putting these we get that
Mn/(N ∩ Mn) = (N + Mn)/N = (N + Mn+1)/N = Mn+1/(N ∩ Mn+1) which
implies the claim and hence the result.
Corollory 3.1.7. Every non-zero submodule of an Artinian module contains a
minimal submodule.
Corollory 3.1.8. Sums and direct sums of finitely many Artinian modules are
Artinian.
Proof. For, let M1, . . . , Mn be Artinian submodules of a module M. Let N =
Pn
i=1 Mi. To prove N is Artinian, proceed by induction on n. If n = 1, there is
nothing to prove. Let n ≥ 2 and assume, by induction, that N′
=
Pn−1
i=1 Mi is
Artinian. Now look at
N/Mn = (N′
+ Mn)/Mn
∼
= N′
/(N′
∩ Mn)
which is Artinian being a quotient of the Artinian module N′
. Thus both Mn and
N/Mn are Artinian and hence N is Artinian, as required. The case of a direct
sum is an immediate consequence because if M = ⊕n
i=1Mi, then M is a finite sum
of the Artinian submodules Mi and hence Artinian.
29

1. Direct sum of an infinite family of non-zero Artinian modules is not Artinian
(because it contains non-stationary descending chains).
2. However, a sum of an infinite family of distinct Artinian modules could be
Artinian. (For example, the Euclidean plane R2
is a sum of all the lines
passing through the origin and is a direct sum of any two of them.)
3.2 Noetherian Modules
Theorem 3.2.1. The following are equivalent for an R-module M.
1. Ascending chain condition (a.c.c) holds for submodule of M, that is any
ascending chain M1 ⊆ M2 ⊆ . . . ⊆ Mn ⊆ . . . ⊆ . . . of submodules of M is
stationary in the sense that Mr = Mr+1 = . . . for some r. (We write this
Mr = Mr+1, for every r ≫ 0).
2. Maximum condition holds for M in the sense that any non-empty family of
submodules of M has a maximal element.
3. Finiteness condition holds for M in the sense that every submodule of M is
finitely generated (that is, spanned).
Proof. (1)=⇒(2) : Let F = {Mi, i ∈ I} be a non-empty family of submodules
of M. Pick any index i1 ∈ I and look at Mi1 . If Mi1 , is maximal in F, we are
through. Otherwise, there is an i2 ∈ I such that Mi1 ⊂ Mi2, Mi1 ̸= Mi2 . If this
Mi2 is maximal in F, we are through again. Proceeding thus, if we do not find
a maximal element at any finite stage, we would end up with a non-stationary
ascending chain of submodules of M, namely, M1 ⊂ M2 ⊂ . . . ⊂ Mn ⊂ . . . ⊂ . . .
contradicting (1).
(2) =⇒(3): Let N be a submodule of M. Consider the family F of all finitely
generated submodules of N. This family is non-empty since the submodule (0)
is a member. This family has a maximal member, say N0 = {x1, . . . , xr}. If
N0 ̸= N, pick an x ∈ N, x is not in N0. Now N1 = N0 + {x} = {x1, . . . , xr} is a
30

finitely generated submodule of N and hence N1 ∈ F. But then this contradicts
the maximality of N0 in F since N0 ⊂ N1, N0 ̸= N1 and so N0 = N is finitely
generated.
(3) =⇒ (1) Let M1 ⊆ M2 ⊆ . . . ⊆ Mn ⊆ . . . ⊆ . . . be an ascending chain of
submodules of M. Consider the submodule N = ∪∞
i=1Mi of M which must be
finitely generated, say N = {x1, . . . , xn}· It follows that xi ∈ Mr, for every i, 1 ≤
i ≤ n for some r(≫ 0). Now we have N ⊆ Ms ⊆ N, for every s ≥ r and so
N = Mr = Mr+1 = . . .
Definition 3.2.2. Noetherian modules : A module M is called Noetherian if
a.c.c (or equivalently, the maximum condition or the finiteness condition) holds
for M.
1. Maximal submodules exist in a non-zero Noetherian module (because a max-
imal submodule is simply a maximal element in the family of all (proper)
submodules N of M, N ̸= M)
2. However, maximal submodules exist in any finitely generated nonzero mod-
ule, even if the module is not Noetherian. (This is a simple consequence
of Zorn’s lemma applied to the family of all proper submodules of such a
module.)
Example 3.2.3. A module which has only finitely many submodules is Noethe-
rian. In particular, finite abelian groups are Noetherian as modules over Z.
Example 3.2.4. Finite dimensional vector spaces are Noetherian (for dimension
reasons) whereas infinite dimensional ones are not Noetherian.
Example 3.2.5. Unlike the Artinian case, infinite cyclic groups are Noetherian
because every subgroup of a cyclic group is again cyclic.
Theorem 3.2.6. 1. Submodules and quotient modules of Noetherian modules
are Noetherian.
31

2. If a module M is such that it has a submodule N with both N and M/N are
Noetherian, then M is Noetherian.
Proof. 1. Let M be Noetherian and N a submodule of M. Any family of
submodules of N is also one in M and hence the result follows. On the
other hand, any ascending chain of submodules of M/N corresponds to one
in M (wherein each member contains N) and hence the result.
2. Let M1 ⊆ M2 ⊆ . . . ⊆ Mn ⊆ . . . ⊆ . . . be an ascending chain in M.Intersecting
with N gives the ascending chain in N, namely, N ∩ M1 ⊆ N ∩ M2 ⊆ . . . ⊆
N ∩ Mn ⊆ . . . ⊆ . . . which must be stationary, say
N ∩ Mr = N ∩ Mr+1 = . . . = . . . for some r. On the other hand, we have
the ascending chain in M/N, namely, (N + M1)/N ⊆ (N + M2)/N ⊆ . . . ⊆
(N + Mn)/N ⊆ . . . ⊆ . . . which must be also stationary, say, (N + Ms)/N =
(N + Ms+1) = . . . for some s.
Now we prove that Mn = Mn+1, for every n ≥ (r+s). This is an immediate
consequence of the following four facts,
1. Mn ⊆ Mn+1, for every n ∈ N,
2. N ∩ Mn = N ∩ Mn+1, for every n ≥ r,
3. (N + Mn)/N = (N + Mn+1)/N, for every n ≥ s and
4. (N + Mn)/N ∼
= Mn/(N ∩ Mn), for every n ∈ N.
Putting these together we get that
Mn/(N ∩ Mn) = (N + Mn)/N = (N + Mn+1)/N = Mn+1/(N ∩ Mn+1) which
implies the claim and hence the result.
Corollory 3.2.7. Every non-zero submodule of a Noetherian module is contained
in a maximal submodule.
Corollory 3.2.8. Sums and direct sums of finitely many Noetherian modules are
Noetherian.
32

Proof. For, let M1, . . . , Mn be Noetherian submodules of a module M. Let N =
Pn
i=1 Mi. To prove N is Noetherian, proceed by induction on n. If n = 1, there
is nothing to prove. Let n ≥ 2 and assume, by induction, that N′
=
Pn−1
i=1 Mi is
Noetherian. Now look at
N/Mn = (N′
+ Mn)/Mn
∼
= N′
/(N′
∩ Mn)
which is Noetherian being a quotient of the Noetherian module N′
. Thus both
Mn and N/Mn are Noetherian and hence N is Noetherian, as required. The case
of a direct sum is an immediate consequence because if M = ⊕n
i=1Mi, then M is
a finite sum of the Noetherian submodules Mi and hence Noetherian.
Direct sum of an infinite family of non-zero Noetherian modules is not Noetherian
(because it contains non-stationary ascending chains).
Note:
1. An Artinian module need not be finitely generated.
2. Maximal submodules need not exist in an Artinian module.
3. An Artinian module need not be Noetherian.
4. A finitely generated module need not be Noetherian.
5. Minimal submodules need not exist in a Noetherian module.
6. A Noetherian module need not be Artinian.
7. There are modules which are neither Artinian nor Noetherian.
Before we give counter-examples to justify the statements above, first we consider
the abelian group µp∗ of all complex (pn
)th
roots of unity for a fixed prime number
p and all n ∈ N. For each positive integer n, let µpn denote the cyclic group of all
complex (pn
)th
roots of unity so that we have µp ⊂ . . . ⊂ µpn ⊂ . . . ⊂ and hence
µp∗ =
n
[
n=1
µpn
Furthermore, we notice the following special features in this group.
33

1. It is infinite and non-cyclic.
2. Every proper subgroup is finite and is equal to µpn for some n.
3. Every finitely generated subgroup is proper and hence finite.
4. In particular, µp∗ is not finitely generated. These properties can be easily
verified using the fact that any x ∈ µpn+1 x not in µpn generates µpn+1 · Now
we give the required counter-examples.
Example 3.2.9. The group µp∗ is Artinian but not Noetherian, not finitely gen-
erated and does not have maximal subgroups. This justifies the statements (1),(2)
and (3).
Example 3.2.10. Let R = Z[X1, X2, . . . Xn, . . .] be the polynomial ring in in-
finetly many variable. We know that R, as a module over itself, is generated by 1
but R is not Noetherian because it has a non-stationary ascending chain of ideals,
namely, (X1) ⊂ (X1, X2) ⊂ . . . ⊂ (X1, . . . , Xn) ⊂ . . . This serves the purpose for
statement (4).
Example 3.2.11. The infinte cyclic group Z is Noetherian but not Artinian and
it has no minimal subgroups. This justifies statements (5) and (6).
Example 3.2.12. Direct sum of any infinite family of non-zero modules, in par-
ticular, an infinite dimensional vector space, is neither Artinian nor Noetherian.
34

Conclusion
We see that vector spaces are special types of modules which arise when the
underlying ring is a field. If R is a ring, the definition of an R-module M is
analogous to the definition of a group action where R plays the role of the group
and M the role of the set. Also we see that how the structure of the ring R is
reflected by the structure of its modules and vice versa in the same way that the
structure of the collection of normal subgroups of a group was reflected by its
permutation representations.
35

BIBLIOGRAPHY
[1] David S. Dummit and Richard M. Foote, Abstract Algebra, Third Edition,
Wiley India Pvt. Ltd., 2004.
[2] C Musili, Introduction to Rings and Modules, Narosa Publishing House, New
Delhi, 1994.
[3] John B. Fraleigh, A First Course in Abstract Algebra, Seventh Edition, Pear-
son Education, Inc., 2003.
[4] I.N. Herstein, Topics in Algebra, Second Edition, John Wiley and Sons,
United States of America., 2002.
[5] Michael Artin, Algebra, Second Edition, Prentice-Hall of India, New Delhi,
2006.
36

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