module_theory (13).pdf

A study on module theory, Msc project 2023

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

2. 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

3. CERTIFICATE
This is to certify that the dissertation entitledA Study on Module Theory
is a bonafide record of the work carried out bySreelekshmi M Sunder 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

4. 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 toDr. 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 toMrs. 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

5. 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.

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

7. 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

8. 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 alla, b∈Gwe havea∗bis inG. For
example, the addition and usual multiplication on the set of all integersZare
binary operations onZ.
A group is an ordered pair (G,∗) whereGis a set and∗is a binary operation
onGsatisfying the following axioms:
1. (a∗b) =a∗(b∗c), for alla, b, c∈G, that is,∗is associative.
2. there is an elemente∈G, called an identity of G, such that for alla∈G,
we havea∗e=e∗a=a,
3. for eacha∈Gthere is an elementa
−1
ofG, called an inverse ofa, such that
a∗a
−1
=a
−1
∗a=e.
The group (G,∗) is said to be an abelian or commutative group ifa∗b=b∗a, for
alla, b∈G.A subsetHofGis said to be a subgroup if it is non empty andH
itself is a group under the operation induced fromGand we denote it byH≤G.
For example, the set of all real numbersRunder addition is a commutative group
whileRis not a group under usual multiplication. Also the set of alln×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 fromRto
2

9. Ris a commutative group under point wise addition and is a group under function
composition.
A ringRis a set with two binary operations, addition (denoted bya+b) and
multiplication (denoted byab), such that for alla, 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∈Rsuch that
a+ 0 =afor alla∈R.
4. there is element−a∈Rsuch thata+ (−a) = 0.
5.a(bc) = (ab)c.
6.a(b+c) =ab+acand (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,ais a unit ifa
−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. Letn∈N, the set of alln×nmatrices over
Ris a ring with respect to usual addition and multiplication of matrices.
A subsetSof a ringRis a subring ofRifSitself is a ring under the operations
ofR. For example, the subsets 0 andRare subrings of any ringRand are called
trivial subrings ofR. The subsetZ(R) ={a∈R:ax+xa,for everyx∈R}
is a subring ofR, called the center ofR. Any subring ofZ(R) is called the
central subring ofR. The subsetsZ⊂Q⊂Rare all subrings ofC. A field is a
3

10. commutative ring with unity in which every non zero element is a unit. The rings
Q,R,Care examples of fields.
LetRbe a ring, a subsetIofRis called a left ideal ofRif,Iis a subgroup of
(R,+), that is,a, b∈I=⇒a−b∈IandIis closed for arbitrary multiplication
on the left by elements inR, that is,a∈Iandx∈R=⇒ax∈I.A subsetIof
Ris called a right ideal ofRifa, b∈I=⇒a−b∈Ianda∈I, x∈R=⇒ax∈I.
For example, in any ringRthe subsets{0}andRare both ideals. IfRis a field
these are the only ideals. LetFbe a field, the set of all polynomials overF,
denoted byF[x] ={
P
n
i=1
aix
i
:ai∈F}is a ring under usual polynomial addition
and multiplication, called polynomial ring overF.
LetRandSbe rings, a ring homomorphism is a mapϕ:R−→Ssatisfying
1.ϕ(a+b) =ϕ(a+b), for alla, b∈R.
2.ϕ(ab) =ϕ(a) +ϕ(b),for alla, b∈R
The kernel of the ring homomorphismϕ, denoted bykerϕis the set of all elements
ofRthat maps to 0 inS. That iskerϕ={x∈R:ϕ(x) = 0s}. An isomorphism
is a ring homomorphism which is both injective and surjective. For example,
consider two ringsZand 2Z. These are isomorphic as groups, since the function
Z−→2Zwhich sendsn−→2n, is a group homomorphism which is one to one and
onto. Howeverϕis not an isomorphism of rings (infact they are not isomorphic
as rings). Indeed,ϕ(1×1) =ϕ(1) = 2 whileϕ(1)ϕ(1) = 2×2 = 4̸= 2.Thus
ϕ(1×1)̸=ϕ(1)ϕ(1).LetR[x] denote the ring of polynomials with real coefficients.
The mappingf(x)−→f(1) is a ring homomorphism fromR[X] ontoR.
A vector spaceVover the fieldFis a set with two binary operations - addition
and scalar multiplication onVsatisfying the following:
1.Vis an abelian group under addition.
2. 1v=vfor allv∈V.
3.a(u+v) =au+av, (a+b)u=au+buand
4.a(b.v) = (a.b)vfor alla, b∈F,v∈V.
4

11. A subsetUof a vector spaceVis called a subspace ofVifUitself is a vector space
(using the same addition and scalar multiplication onV). A list (v1, v2. . . vm) of
vectors inVis called linearly independent if the only choices ofa1, a2, . . . am∈F
that makesa1v1+a2v2+. . . amvm= 0 isa1=a2=. . . am= 0.
A list of vectors inVis called linearly dependent if it is not linearly indepen-
dent. A basis ofVis a list of vectors inVthat is linearly independent and spans
V. A list (v1, v2, . . . vn) of vectors inVis a basis ofVif and only if everyv∈Vcan
be written uniquely in the formv=a1v1+a2v2+. . . anvn, wherea1, a2, . . . an∈F.
For example, the spaceV=F[x] of polynomials in the variablexwith coefficients
from the fieldFis in particular a vector space overF. The elements 1, x, x
2
, . . .
are linearly independent by definition, that is, a polynomial is 0 if and only if all
its coefficients are 0. Since these elements spansV, by definition, they are basis
forV.
LetVandWbe vector spaces over a fieldF, a linear mapTfromVinto
Wis a function which satisfies:T(u+v) =T(u) +T(v) andT(av) =aT(v),for
alla∈Fandu, v∈V. The set of all linear maps fromVintoWis denoted
byL(V, W) and we can see thatL(V, W) is a vector space over the fieldFunder
usual addition and scalar multipication.
5

12. Chapter 2
Modules
2.1Definitions and examples
In this section we give the definition, examples and basic results on modules.
Definition 2.1.1.LetRbe a ring (not necessarily commutative nor with 1). A
leftR-module or a left module overRis a setMtogether with
1. a binary operation+onMunder whichMis an abelian group,and
2. an action ofRonM(that is,a mapR×M→M) denoted byrm, for all
r∈Rand for allm∈Mwhich satisfies
(a)(r+s)m=rm+sm, for allr, s∈R, m∈M
(b)(rs)m=r(sm), for allr, s∈R, m∈M,and
(c)r(m+n) =rm+rn, for allr∈R, m, n∈M.
If the ring R has a1we impose the additional axiom:
(d)1m=m, for allm∈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 ringRis
commutative andMis a leftR-module we can makeMinto a rightR-module by
definingmr=rmform∈Mandr∈R. IfRis not commutative, axiom 2(b) in
general will not hold with this definition (so not every leftR-module is also a right
6

13. 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
havingrm= 0 for allr∈Randm∈M).
WhenRis a fieldFthe axioms for anR-module are precisely the same as
those for a vector space overF, so that modules over a fieldFand vector spaces
overFare the same.
Before giving other examples ofR-modules we record the obvious definition of
submodules.
Definition 2.1.2.LetRbe a ring and letMbe anR-module. AnR-submodule
ofMis a subgroupNofMwhich is closed under the action of ring elements, that
is,rn∈N, for allr∈R,n∈N.
Submodules ofMare therefore just subsets ofMwhich are themselves modules
under the restricted operations. In particular, ifR=Fis a field, submodules are
the same as subspaces. EveryR-moduleMhas the two submodulesMand 0 (the
latter is called the trivial submodule).
Example 2.1.3.LetRbe any ring. ThenM=Ris a leftR-module, where the
action of a ring element on a module element is just the usual multiplication in the
ringR(similarly,Ris a right module over itself). In particular, every field can be
considered as a (1-dimensional) vector space over itself. WhenRis considered as
a left module over itself in this fashion, the submodules ofRare precisely the left
ideals ofR(and ifRis considered as a rightR-module over itself, its submodules
are the right ideals). Thus ifRis not commutative it has a left and right module
structure over itself and these structures may be different.
Example 2.1.4.LetR=Z, letAbe any abelian group (finite or infinite) and
write the operation ofAas +. MakeAintoZ-module as follows : for anyn∈Z
7

14. anda∈Adefine
na=













a+a+· · ·+a(n times) ifn >0
0 if n= 0
−a−a− · · · −a(-n times) ifn <0
(here 0 is the identity of the additive groupA). This definition of an action of the
integers onAmakesAinto aZ-module, and the module axioms show that this
is the only possible action ofZonAmaking it a (unital)Z-module. Thus every
abelian group is aZ-module. Conversely, ifMis anyZ-module, a fortioriMis
an abelian group, soZ-modules are the same as abelian groups. Furthermore, it
is immediate from the definition thatZ-submodules are the same as subgroups.
Example 2.1.5.LetLbe a left ideal ofR. Then,
1.Lis an additive abelian group induced by that ofR, and
2.rm∈L, for allm∈Landr∈R.
Moreover, (a), (b) in module definition are clearly satisfied for allm, n∈Land
r, s∈R. Hence every idealLofRhas the structure of leftR-module in which
1. the addition is the one induced by addition inR, and
2. the scalar multiplicationr·mof an elementm∈Lby an elementr∈Ris
the productrmofrandmin the ringR
Similarly, right idealIofRhas the structure of a rightR-module in which ad-
dition and scalar multiplication are induced respectively from the addition and
multiplication in the ringR.
In particular, takingL=R,I=R, we see that the ringR, has in the above
manner, the structure of left as well as rightR-module.
Example 2.1.6.IfRis any ring, then,R
n
, the set of all n-tuples with components
inRis anR-module, with usual definitions of addition and scalar multiplication.
8

15. Example 2.1.7.LetFbe a field, letxbe an indeterminate and letRbe the
polynomial ringF[x]. LetVbe a vector space overFand letTbe a linear
transformation fromVtoV. We have already seen thatVis anF-module; the
linear mapTwill enable us to makeVinto anF[x]-module.
First, for the nonnegative integer n, define
T0=I
.
.
.
T
n
=T◦T◦. . .◦T(n times)
whereIis the identity map fromVtoVand◦denotes function composition
(which makes sense because the domain and codomain ofTare the same). Also,
for any two linear transformationsA, BfromVtoVand elementsα, β∈F, let
αA+βBdefined by
(αA+βB)(v) =α(A(v)) +β(B(v))
(that is addition and scalar multiplication of linear transformations are defined
pointwise). ThenαA+βBis easily seen to be a linear transformation from
VtoV, so that linearcombinations of linear transformations are again linear
transformations.
We now define the action of any polynomial inxonV. Letp(x) be the polynomial
p(x) =anx
n
+an−1x
n−1
+....+a1+a0, wherea0, a1, .., an∈F. For eachv∈V
define an action of the ring elementp(x) on the module elementvby
p(x)v= (anT
n
+an−1Tn−1+. . .+a1T+a0)(v)
=anT
n
(v) +an−1Tn−1(v) +. . .+a1T(v) +a0v
(that isp(x) acts by substituting the linear transformationTforxinp(x) and
applying the resulting linear transformation tov). Put another way,xacts onV
as the linear transformationTand we extend this to an action of all ofF[x] onV
in a natural way. It is easy to check that this definition of an action ofF[x] onV
satisfies all the module axioms and makesVinto anF[x]-module.
9

16. Remark.We can easily verify that the polynomial ringR[x]over a ringRis an
R-module.
Proposition 2.1.8.LetRbe a ring and letMbe anR-module. A subsetNof
Mis a submodule ofMif and only if
1. N̸=ϕand
2.x+ry∈Nfor allr∈Rand for allx, y∈N
Proof.IfNis a submodule, then 0∈NsoN=I= 0. AlsoNis closed under
addition and is sent to itself under the action of elements ofR. Conversely, suppose
(1) and (2) hold. Letr=−1 and apply the subgroup criterion (in additive form)
to see thatNis a subgroup ofM. In particular, 0∈N. Now letx= 0 and apply
hypothesis (2) to see thatNis sent to itself under the action ofR. This completes
the proof.
Proposition 2.1.9.LetMbe anR-module. Then for allr∈Rand 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+ 0Rmfrom which 0Rm=0MbecauseMis a group. Next,
from 0M+ 0M= 0Mfollowsr0M=r(0M+ 0M) =r0M+r0M, whencer0M= 0M.
Thus (1) and (2) are proved. Again 0R=r+ (−r) and therefore, by (1), 0M=
0Rm= (r+ (−r))m=rm+ (−r)mwhich yields (−r)m=−(rm). A similar
argument shows thatr(−x) =−(rx). Finally, using (3), we obtain (−r)(−x) =
−((−r)x)=−(−rx)) =rx. This completes the proof.
10

17. Proposition 2.1.10.For an abelian groupM, letEndZ(M)be the ring of all
(additive) endomorphisms ofM. LetRbe any ring. Then we have the following.
1.Mis a leftR-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
`
ψ:R→EndZ(M)that is
`
ψpreserves addition but reverses the multiplica-
tion.
3.MisR- unitary⇐⇒ψ(1R)=idM(resp.
`
ψ(1R)=idM)
Proof.LetMbe a leftR-module with scalar multiplication,
R×M→M, (a, x)7→ax. Now defineψ:R→EndZ(M) bya7→ψ(a), where
ψ(a) :M→Mis defined byψ(a)(x) =ax, for all a∈Randx∈M.
Claim(i):ψis a homomorphism of rings.
For leta, b∈Randx∈M. We have
ψ(a+b) = (a+b)x =ψ(a)(x) +ψ(b)(x) =[ψ(a) +ψ(b)](x).
Thusψ(a+b) =ψ(a) +ψ(b). Similarly, we have forx∈M,
ψ(ab)(x) = (ab)(x) =a(bx) =a(ψ(b)(x))
=ψ(a)(ψ(b)(x)) = (ψ(a)oψ(b))(x).
Thusψ(ab) =ψ(a)ψ(b) and henceψis a homomorphism of rings.
Conversely, suppose thatψ: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, leta, b∈Randx, y∈M.Sinceψ(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

18. and (ab)(x) =ψ(ab)(x) = (ψ(a)oψ(b))(x) =ψ(a)(ψ(b)(x)) =ψ(a)(bx) =a(bx).
ThusMis anR-module. Proof of (2) is similar to (1).
(3) SupposeMisR-unitary andψ:R→EndZ(M) is the corresponding homo-
morphism of rings . We haveψ(1) :M→M, x7→ψ(1)(x) = 1·x=x.
Henceψ(1R) =idM. Conversely, suppose thatψ(1R) =idMwhereψ: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.Mis a leftR-module implies thatMis a rightR
0
-module
whereR
0
is the ring opposite toR.
Proof.We have a homomorphism of ringsψ:R→EndZ(M).Compose this
with the identity mapid:R
0
→Rwhich is an anti-isomorphism, to get an anti-
homomorphismR
0
→EndZ(M) which meansMis a rightR
0
-module.
Conversely, suppose that we have an anti-homomorphism of rings
`
ψ:R
0
→
EndZ(M).Compose this with the identity mapid:R→R
0
which is an anti-
isomorphism, to get a homomorphismR→EndZ(M). ThereforeMis a left
R-module.
Definition 2.1.12.LetRbe a commutative ring with identity. AnR-algebra is
a ringAwith identity together with a ring homomorphismf:R→A mapping1R
to1Asuch that the subringf(R)ofAis contained in the center ofA.
IfAis anR-algebra then it is easy to check thatAhas a natural left and right
(unital)R-module structure defined byr·a=a·r=f(r)awheref(r)ais just
the multiplication in the ringAand this is the same asaf(r)since by assumption
f(r)lies in the center ofA. In general it is possible for anR-algebraAto 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.IfAandBare twoR-algebras, anR-algebra homomorphism
(or isomorphism) is a ring homomorphism (isomorphism, respectively)ϕ:A→B
mapping1Ato1Bsuch thatϕ(r·a) =r·ϕ(a)for allr∈Randa∈A.
12

19. Example 2.1.14.LetRbe a commutative ring with 1.
1. Any ring with identity is aZ-algebra
2. For any ringAwith identity, ifRis a subring of the center ofAcontaining
the identity ofAthenAis anR-algebra. In particular, a commutative ring
Acontaining 1 is anR-algebra for any subringRofAcontaining 1.
2.2Quotient modules and module homomorphisms
This section contains the basics of quotient modules and module homomorphisms.
Definition 2.2.1.LetRbe a ring and letMandNbeR-modules.
1. A mapϕ:M→Nis anR-module homomorphism if it respects theR-
module structures ofMandN, that is,
(a)ϕ(x+y) =ϕ(x) +ϕ(y),for allx, y∈Mand
(b)ϕ(rx) =rϕ(x),for allr∈R, x∈M.
2. AnR-module homomorphism is an isomorphism (ofR-modules) if it is both
injective and surjective. The modulesMandNare said to be isomorphic,
denotedM
∼
=N, if there is someR-module isomorphismϕ:M→N.
3. Ifϕ:M→Nis anR-module homomorphism, letkerϕ={m∈M|ϕ(m) =
0}(the kernel ofϕ) and letϕ(M) ={n∈N|n=ϕ(m)for some m∈M}
(the image ofϕ, as usual).
4. LetMandNbeR-modules and defineHomR(M, N)to be the set of all
R-module homomorphisms fromMintoN.
Example 2.2.2.IfRis a ring andM=Ris a module over itself, thenR-module
homomorphisms (even fromRto itself) need not be ring homomorphisms and ring
homomorphisms need not beR-module homomorphisms. For example, when
R=ZtheZ-module homomorphismx7→2xis not a ring homomorphism (1 does
not map to 1). WhenR=F[x] the ring homomorphismϕ:f(x)7→f(x
2
) is not
13

20. anF[x]-module homomorphism (if it were, we would havex
2
=ϕ(x) =ϕ(x·1) =
xϕ(1) =x).
Example 2.2.3.LetRbe a ring, letn∈Z
+
and letM=R
n
, for each
i∈ {1, ..., n}the projection mapπi:R
n
→Rbyπi(x1, ..., xn) =xiis a surjective
R-module homomorphism with kernel equal to the submodule of n-tuples which
have a zero in positioni.
Example 2.2.4.IfRis a field,R-module homomorphisms are called linear trans-
formations.
Example 2.2.5.For the ringR=Zthe 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 thatZ-module homomorphisms are the same as abelian group
homomorphisms.
Proposition 2.2.6.LetM, NandLbeR-modules
1. A mapϕ:M→Nis anR-module homomorphism if and only if
ϕ(rx+y) =rϕ(x) +ϕ(y)for allx, y∈Mand allr∈R.
2. Letϕ, ψbe elements ofHomR(M, N). Defineϕ+ψby , for allm∈M,
(ϕ+ψ)(m) =ϕ(m) +ψ(m).
Thenϕ+ψ∈HomR(M, N)and with this operationHomR(M, N)is an
abelian group. IfRis a commutative ring then forr∈Rdefinerϕby for
allm∈M,
(rϕ)(m) =r(ϕ(m))
Thenrϕ∈HomR(M, N)and with this action of the commutative ringR
the abelian groupHomR(M, N)is anR-module.
3. Ifϕ∈HomR(L, M)andψ∈HomR(M, N)thenψ◦ϕ∈HomR(L, N).
14

21. 4. With addition as above and multiplication defined as function composition,
HomR(M, M)is a ring with 1. WhenRis commutativeHomR(M, M)is
anR-algebra.
Proof.1. Certainlyϕ(rx+y) =ϕ(x)+ϕ(y) ifϕis anR-module homomorphism.
Conversely, ifϕ(rx+y) =ϕ(x) +ϕ(y), take r = 1 to see thatϕis additive
and take y = 0 to see thatϕcommutes with the action ofRonM(that is,
is homogeneous).
2. It is straightforward to check that all the abelian group andR-module axioms
hold with these definitions. We note that the commutativity ofRis used
to show thatrϕsatisfies the second axiom of anR-module homomorphism,
namely,
(r1ϕ)(r2m) =r1ϕ(r2m) (by the definition ofr1ϕ)
=r1r2(ϕ(m) (Sinceϕis a homomorphism)
=r2r1ϕ(m) (SinceRcommutative)
=r2r1ϕ(m) ( by the definition ofr1ϕ)
Verification of the axioms relies ultimately on the hypothesis thatNis an
R-module. The domainMcould in fact be any set - it does not have to be
anR-module nor an abelian group.
3. Letϕandψbe as given and letr∈R, x, y,∈L.Then
(ψ◦ϕ)(rx+y) =ψ(ϕ(rx+y)
=ψ(rϕ(x+ϕ(y)) (by (1) applied toϕ)
=rψ(ϕ(x)) +ψ(ϕ(y)) (by(1) applied toψ)
=r(ψ◦ϕ)(x) + (ψ◦ϕ)(y)
so, by (1),ψ◦ϕis an R-module homomorphism.
4. Note that since the domain and codomain of the elements ofHomR(M, M)
are the same, function composition is defined. By (3), it is a binary op-
eration onHomR(M, M). As usual, function composition is associative.
15

22. The remaining ring axioms are straightforward to check. The identity func-
tion,I, (as usual,I(x) =x, for allx∈M) is seen to be the multiplica-
tive identity ofHomR(M, M). IfRis commutative, then (2) shows that
the ringHomR(M, M) is a leftR-module and definingϕr=rϕfor all
ϕ∈HomR(M, M) andr∈RmakesHomR(M, M) into anR-algebra.
Definition 2.2.7.The ringHomR(M, M)is called the endomorphism ring of M
and will often be denoted byEndR(M), or justEnd(M)when the ringRis clear
from the context. Elements ofEnd(M)are called endomorphisms.
WhenRis commutative there is a natural map fromRintoEnd(M) given
byr7→rI, where the latter endomorphism ofMis just multiplication byron
M. The image ofRis contained in the center ofEnd(M) so ifRhas an identity,
End(M) is anR-algebra. The ring homomorphism fromRtoEndR(M) may not
be injective since for somerwe may haverm= 0 for allm∈M
(example,R=Z,M=Z/2Z,andr= 2). WhenRis a field, however, this map
is injective (in general, no unit is in the kernel of this map) and the copy ofRin
EndR(M) is called the (subring of) scalar transformations. Next we prove that
every submoduleNof anR-moduleMis ”normal” in the sense that we can always
form the quotient moduleM/N, and the natural projectionπ:M→M/Nis anR-
module homomorphism with kernelN. 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 ofRis
compatible with these group quotients and homomorphisms.
Proposition 2.2.8.LetRbe a ring, letMbe anR-module and letNbe a
submodule ofM. The (additive, abelian) quotient groupM/Ncan be made into
16

23. anR-module by defining an action of elements ofRby
r(x+N) = (rx) +N,for allr∈R,x+N∈M/N.
The natural projection mapπ:M→M/Ndefined byπ(x) = x + N is anR-
module homomorphism with kernelN.
Proof.SinceMis an abelian group under + the quotient groupM/Nis defined
and is an abelian group. To see that the action of the ring elementron the coset
x+Nis well defined, supposex+N=y+N,that isx−y∈N.SinceNis a (left)
R-submodule,r(x−y)∈N. Thusrx−ry∈Nandrx+N=ry+N,as desired.
Now since the operations inM/Nare ”compatible” with those ofM, the axioms for
anR-module are easily checked in the same way as was done for quotient groups.
For example, axiom 2(b) holds as follows: for allr1, r2∈Randx+N∈M/N,by
definition of the action of ring elements on elements ofM/N
(r1r2)(x+N) = (r1r2x) +N
=r1(r2x+N)
=r1(r2(x+N))
The other axioms are similarly checked. The natural projection mapπdescribed
above is, in particular, the natural projection of the abelian groupMonto the
abelian groupM/Nhence is a group homomorphism with kernelN. 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 showπis a module
homomorphism, that isπ(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

24. Definition 2.2.9.LetA, Bbe submodules of theR-moduleM. The sum ofA
andBis the set
A+B={a+b;a∈A, b∈B}.
The sum of two submodulesAandBis a submodule and is the smallest submodule
which contains bothAandB.
Theorem 2.2.10. 1. (The First Isomorphism Theorem for Modules) LetM, N
beR-modules and letϕ:M→Nbe an R-module homomorphism. Then
kerϕis a submodule ofMandM/kerϕ
∼
=ϕ(M)
2. (The Second Isomorphism Theorem) LetA, Bbe submodules of theR-module
M. Then(A+B)/B
∼
=A/(A∩B).
3. (The Third Isomorphism Theorem) LetMbe anR-module, and letAand
Bbe submodules ofMwithA⊆B. Then(M/A)/(B/A)
∼
=M/B.
Proof.1. By definition ofkerϕ,kerϕ={m∈M;ϕ(m) = 0}.Sinceϕ(0) =
0 , 0∈kerϕ,kerϕis a non empty subset ofM. Now, letm, nbe any two
arbitrary elements ofkerϕ. Then we haveϕ(m) = 0 andϕ(n) = 0.
ϕ(m−n) =ϕ(m)−ϕ(n) = 0.This impliesm, n∈kerϕ,for every m,n∈
kerϕ. Hencekerϕis an additive subgroup ofM. Again forr∈Randm∈
M, we haveϕ(rm) =rϕ(m) =r.0 = 0. This impliesrm∈kerϕ, for allr∈
R.Therefore from (2.1) and (2.2), we havekerϕis anR-submodule ofM.
Sincekerϕis a submodule ofM,M/kerϕis anR-module defined byr(x+
kerϕ) =rx+kerϕfor allr∈R, x+kerϕ∈M/kerϕ. We need to show
thatM/kerϕ
∼
=ϕ(M). For consider a map,ψ: M / kerϕ→ϕ(M). We
will prove thatψis anR- module homomorphism that is both injective and
18

25. surjective. To proveψis 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 proveψis onto, consider for anyϕ(m)∈ϕ(M) we can findm+kerϕ∈
M/kerϕsuch thatψ(m + kerϕ) =ϕ(m).
To proveψis anR-module homomorphism, letm+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. Letf:M→Nbe homomorphism ofR-modules, letAbe a submodule
ofMandBbe a submodule ofN. Thenf(A) andf
−1
(B) are submoodules
ofNandMrespectively. In particular,Imfis a submodule ofNwhile
kerfis a submodule ofM. Finallyf
−1
(f(A)) =A+kerf=A+B
whencef(A+B) =f(A). Butf(A+B) = (A+B)/Band therefore
f(A) + (A+B)/B. Thus by restrictingftoAthere is produced an epi-
morphismϕ:A→(A+B)/B. First isomorphism theorem for modules
shows that we have an isomorphismA/kerϕ
∼
=(A+B)/B. However, the
19

26. elements ofAwhich are mapped byϕinto zero are just the elements ofA
which belong toB. Accordinglykerϕ=A∩Band the proof is complete.
3. SinceAandBare submodules ofMwithA⊆B, we haveB/Ais a
submodule ofM/A. Defineϕ:M/A→M/Bbyϕ(m+A) =m+B, for all
m∈M.
To proveϕis well defined,
Letm1+A,m2+A∈M/A. Supposem1+A=m2+A
=⇒m1−m2∈A
=⇒m1−m2∈B
=⇒m1+B=m2+B
=⇒ϕ(m1+A) =ϕ(m2+A).
To proveϕis anR-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)
andϕ(r(m+A)) =ϕ(rm+A)
=rm+B
=r(m+B)
=rϕ(m+A).
To proveϕis onto, for anym+B∈M/Bwe can findm+A∈M/Asuch
thatϕ(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

27. Therefore, (M/A)/(B/A)
∼
=M/B.
2.3Generation of modules, direct sums, and free modules
LetRbe 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.LetMbe anR-module and letN1, ..., Nnbe submodules ofM.
1. The sum ofN1, ..., Nnis the set of all finite sums of elements from the sets
Ni:{a1+a2+ +an:ai∈Nifor alli}. Denote this sum byN1+. . .+Nn.
2. For any subsetAofMletRA={r1al+r2a2+ +rmam:r1, ..., rm∈
R, ai, ..., am∈A, m∈Z
+
}(where by conventionRA={0}ifA=ϕ). lf
Ais the finite set{a1, a2, ..., an}we shall writeRa1+Ra2+. . .+Ranfor
RA. CallRAthe submodule ofMgenerated byA. IfNis a submodule of
M(possiblyN=M) andN=RA, for some subsetAofM, we callAa
set of generators or generating set forN, and we sayNis generated byA.
3. A submoduleNofM(possiblyN=M) is finitely generated if there is some
finite subsetAofMsuch thatN=RA, that is, ifNis generated by some
finite subset.
4. A submoduleNofM(possiblyN=M) is cyclic if there exists an element
a∈Msuch thatN=Ra, that is, ifNis generated by one element:
N=Ra={ra:r∈R}
Note that these definitions do not require that the ringRcontain a 1, however
this condition ensures thatAis contained inRA. It is easy to see using the
submodule criterion that for any subsetAofM,RAis indeed a submodule ofM
and is the smallest submodule ofMwhich containsA(that is, any submodule of
21

28. Mwhich containsAalso containsRA). In particular, for submodulesN1, ..., Nn
ofM,N1+. . .+Nnis just the submodule generated by the setN1∪. . .∪Nnand
is the smallest submodule ofMcontainingNi, for alli. IfN1, ..., Nnare generated
by setsA1, . . . , Anrespectively, thenN1+. . .+Nnis generated byA1∪. . .∪An.
Note that cyclic modules are, a fortiori, finitely generated.
A submoduleNof anR-moduleMmay have many different generating sets (for
instance the setNitself always generatesN). IfNis finitely generated, then there
is a smallest nonnegative integer d such thatNis generated bydelements (and
no fewer). Any generating set consisting ofdelements will be called a minimal set
of generators forN(it is not unique in general). IfNis not finitely generated, it
need not have a minimal generating set.
The process of generating submodules of anR-moduleMby taking subsetsAofM
and forming all finite ”R-linear combinations” of elements ofAwill 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 ofA). 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 byA, for a subsetAofG. The situation
forR-modules is similar to that of abelian groups.
Example 2.3.2.LetR=Zand letMbe anyR-module, that is, any abelian
group. If a∈M, thenZis just the cyclic subgroup ofMgenerated bya:< a >.
More generally,Mis generated as aZ-module by a setAif and only ifMis
generated as a group byA(that is, the action of ring elements in this instance
produces no elements that cannot already be obtained fromAby addition and
subtraction). The definition of finitely generated forZ-modules is identical to that
for abelian groups.
Example 2.3.3.LetRbe a ring with 1 and letMbe the (left)R-moduleRitself.
Note thatRis a finitely generated, in fact cyclic,R-module becauseR=R1 (that
is we can takeA={1}). Recall that the submodules ofRare precisely the left
ideals ofR, so sayingIis a cyclicR-submodule of the leftR-moduleRis the
22

29. same as sayingIis a principal ideal ofR. (Also, sayingIis a finitely generated
R-submodule ofRis the same as sayingIis a finitely generated ideal. When
Ris a commutative ring we often writeARoraRfor the submodule (ideal)
generated byAorarespectively, as we have been doing forZwhen we wrote nZ.
In this situationAR=RAandaR=Ra(elementwise as well). Thus a Principal
Ideal Domain is a (commutative) integral domainRwith identity in which every
R-submodule ofRis cyclic.
Definition 2.3.4.LetM1, . . . , Mkbe a collection ofR-modules. The collection of
k-tuples(m1, m2, . . . , mk)wheremi∈M; with addition and action ofRdefined
componentwise is called the direct product ofM1, M2, . . . , Mkdenoted
M1⊕. . .⊕Mk.
Proposition 2.3.5.LeN1, N2, . . . , Nkbe submodules of theR-moduleM. Then
the following are equivalent:
1. The mapπ:N1×N2×. . .×Nk→N1+N2+. . .+Nkdefined by
π(a1, a2, . . . , ak) =a1+a2+. . .+ak
is an isomorphism ofR-modules :N1+N2+. . .+Nk
∼
=N1×N2×. . .×Nk.
2.Nj∩(N1+. . .+Nj−1+Nj+1+. . .+Nk) = 0for all j∈ {1,2, . . . , k}.
3. Everyx∈N1+. . .+Nkcan be written uniquely in the forma1+a2+. . .+ak
witha∈Ni.
Proof.To prove (1.) implies (2.), suppose for somejthat (2.) fails to hold and
letaj∈(N1+...+Nj−l+Nj+l+ +Nk)∩Nj,withai̸= 0 . Then
aj=a1+. . .+aj−1+aj+1+...+akfor someai∈Ni, and (a1, . . . , aj−l,−aj, aj+l, . . . ak)
would be a nonzero element ofkerπ, a contradiction.
Assume now that (2.) holds. If for some module elementsai, bi∈Niwe have
a1+a2+. . .+ak=b1+b2+. . .+bk
then for eachjwe have
aj−bj= (b1−a1)+. . .+(bj−1−aj−t1)+(bj+1−aj+1)+. . .+(bk−ak) . The left
23

30. hand side is inNjand the right side belongs toN1+. . .+Nj−1+Nj+1+ +Nk.
Thusaj−bj∈Ni∩(N1+. . .+Nj−1+Nj+1+. . .+Nk) = 0.This showsaj=bj
for allj, and so (2.) implies (3.). Finally, to see that (3.) implies (1.) observe
first that the mapπis clearly a surjectiveR-modu1e homomorphism. Then (3.)
simply impliesπis injective, hence is an isomorphism, completing the proof.
If anR-moduleM=N1+N2+. . .+Nkis the sum of submodulesN1, N2, . . . , Nk
ofMsatisfying the equivalent conditions of the proposition above, thenMis said
to be the (internal) direct sum ofN1, N2, ..., NkwrittenM=N1⊕N2⊕. . .⊕Nk
By the proposition, this is equivalent to the assertion that every elementmofM
can be written uniquely as a sum of elementsm=n1+n2+. . .+nkwithni∈Ni.
(Note that part (1) of the proposition is the statement that the internal direct sum
ofN1, N2, . . . , Nkis isomorphic to their external direct sum, which is the reason
we identify them and use the same notation for both.)
Definition 2.3.6.AnR-moduleMis called a free module ifMhas a basisB,
that is linearly independent subsetBofMsuch thatMis spanned byBoverR,
that is every elementx∈Mcan be written uniquely asx=
P
b∈B
λb·b,
λb∈R, λb= 0except for finitely manyb’s, that isxis finite linear combination
of elements inB, the scalars being unique forx.
Example 2.3.7.R
n
=R×. . .×R(n times), is a freeR-module ifRhas 1. The
setB={(1,0, . . . ,0),(0,1, . . . ,0), . . . ,(0,0, . . . ,1)}is anR-basis forR
n
, called
the standard basis ofR
n
.
Example 2.3.8.For, supposeMandNare freeR-modules with basesAand
Brespectively. NowM⊕N=M×Nis freeR-module because (A×{0})
∪({0} ×B) is anR-basis forM×N. More generally, for any family of freeR-
modules,{Mi:i∈I}, with basisAi, M=⊕i∈IMiis a free module with a basis
A=∪i∈IAi
Example 2.3.9.Any finite abelian group is not free as a module overZ. In fact,
any abelian groupMwhich has a non-trivial element of finite order cannot be free
as a module overZ. For, supposeMis free. SayBis a basis forMoverZ. Let 0
24

31. ̸=x∈Mbe such thatnx= 0 for somen∈N,mx̸= 0 form < nandn⩾2 Now
we havex=n1b1+n2b2+. . .+nrbrfor someb1, b2, . . . , br∈Bandn1, n2, . . . , nr∈
Z. Hence 0 =nx=n{n1b1+n2b2+. . .+nrbr}=nn1b1+nn2b2+. . . nnrbr
=⇒nn1= 0, nn2= 0, . . . , nnr= 0 (by linear independence ofB) =⇒n1= 0,
n2= 0, . . . , nr= 0 (sincen̸= 0 ), that is,x= 0, a contradiction.
Theorem 2.3.10.For any setAthere is a freeR-moduleF(A)on the setA
andF(A)satisfies the following universal property: ifMis anyR-module and
φ:A−→Mis any map of sets, then there is a uniqueR-module homomorphism
ϕ:F(A)−→Msuch thatϕ(a) =φ(A), for alla∈A, that is, the following
diagram (2.1) commutes.
Figure 2.1:
WhenAis the finite set{a1, a2, . . . , an},F(A)=Ra1⊕Ra2⊕. . .⊕Ran
∼
=R
n
.
Proof.LetF(A) ={0}ifA=∅. IfAis nonempty letF(A) be the collection of
all set functionsf:A−→Rsuch thatf(a) = 0 for all but finitely many a∈A.
MakeF(A) into anR-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 alla∈A, r∈Randf, g∈F(A).
It is an easy matter to check that all theR-module axioms hold. IdentifyAas
a subset ofF(A) by a7→fa, wherefais the function which is 1 at a and zero
elsewhere. We can, in this way, think ofF(A) as all finiteR-linear combinations of
elements ofAby identifying each function f with the sumr1a1+r2a2+. . .+rnan,
25

32. whereftakes on the valueriataiand is zero at all other elements ofA. Moreover,
each element ofF(A) has a unique expression as such a formal sum. To establish
the universal property ofF(A) supposeφ:A−→Mis a map of the setAinto
theR-moduleM. Defineϕ:F(A)−→Mby
ϕ:
n
X
i=1
riai7→
n
X
i=1
riφ(ai)
By the uniqueness of the expression for the elements ofF(A) as linear combinations
of theaiwe see easily thatϕis a well definedR-module homomorphism. By
definition, the restriction ofϕtoAequalsφ. Finally, sinceF(A) is generated by
A, once we know the values of anR-module homomorphism onAits values on
every element ofF(A) are uniquely determined, soϕis the unique extension ofφ
to all ofF(A).
WhenAis the finite seta1, a2, . . . , anwe have thatF(A) =Ra1⊕Ra2⊕. . .⊕Ran.
SinceR
∼
=Raifor alli(under the mapr7→rai) we have that the direct sum is
isomorphic toR
n
.
Result.
1. IfF1andF2are free modules on the same setA, there is a unique isomor-
phism betweenF1andF2which is the identity map onA.
2. IfFis any freeR-module with basisA, thenF
∼
=F(A). In particular,F
enjoys the same universal property with respect toAasF(A) does in above
theorem.
26

33. 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,Rstands for a ring with 1 (commuta-
tive or not) and all modules considered are assumed to be unitary modules.
3.1Artinian modules
Theorem 3.1.1.The following are equivalent for anR-moduleM.
1. Descending chain condition (d.c.c) hold for submodules ofM, that is any
descending chainM1⊇M2⊇. . .⊇Mn⊇. . .of submodules ofMis
stationary in the sense thatMr=Mr+1=. . .for somer. (We write this
Mr=Mr+1,for everyr≫0).
2. Minimum condition for submodules holds forM, in the sense that any non-
empty family of submodules ofMhas a minimal element.
Proof.(1) =⇒(2): LetF={Mi, i∈I}be a non-empty family of submodules
ofM. Pick any indexi1∈Iand look atMi1. IfMi1is minimal inF, we are
through. Otherwise, there is ani2∈Isuch thatMi1⊃Mi2,Mi1̸=Mi2. If this
Mi2is minimal inF, 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

34. descending chain of submodules ofM, namely,M1⊃M2⊃. . .⊃Mn⊃. . .
contradicting (1).
(2) =⇒(1): LetM1⊇M2⊇. . .⊇Mn⊇. . .be a descending chain of sub-
modules ofM. Consider the non-empty familyF={Mi:i∈N}of sub-
modules ofM. This must have a minimal element, sayMrfor somer. Now
we haveMs⊆Mr,for everys≥rwhich implies by minimality ofMrthat
Ms=Mr,for everys≥r
Definition 3.1.2.Artinian module: A moduleMis called Artinian if d.c.c
(or equivalently, the minimum condition) holds forM.
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 ofM.
Example 3.1.3.A module which has only finitely many submodules is Artinian.
In particular, finite abelian groups are Artinian as modules overZ.
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,Zhas a
nonstationary descending chain of subgroups, namely,
Z= (1)⊃(2)⊃(4)⊃. . .(2
n
)⊃. . .⊃. . .
Theorem 3.1.6.1. Submodules and quotient modules of Artinian modules are
Artinian.
2. If a moduleMis such that it has a submoduleNwith bothNandM/Nare
Artinian, thenMis Artinian.
Proof.1. LetMbe Artinian andNa submodule ofM. Any family of sub-
modules ofNis also one inMand hence the result follows. On the other
hand, any descending chain of submodules ofM/Ncorresponds to one inM
(wherein each member containsN) and hence the result.
28

35. 2. LetM1⊇M2⊇. . .⊇Mn⊇. . .be a descending chain inM. Intersecting
withNgives the descending chain inN, namely,N∩M1⊇N∩M2⊇. . .⊇
N∩Mn⊇. . .which must be stationary, sayN∩Mr=N∩Mr+l=. . .for
somer. On the other hand, we have the descending chain inM/N, namely,
(N+M1)/N⊇(N+M2)/N⊇. . .⊇(N+Mn)⊇. . .which must be also
stationary, say (N+Ms)/N= (N+Ms+1)/N=. . .for somes.
Next we prove thatMn=Mn+1,for everyn≥(r+s). This is an immediate
consequence of the following four facts:
1.Mn⊇Mn+1,for everyn∈N,
2.N∩Mn=N∩Mn+1,for everyn≥r,
3. (N+Mn)/N= (N+Mn+1)/N, for everyn≥sand
4. (N+Mn)/N
∼
=Mn/(N∩Mn),for everyn∈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, letM1, . . . , Mnbe Artinian submodules of a moduleM. LetN=
P
n
i=1
Mi. To proveNis Artinian, proceed by induction onn. Ifn= 1, there is
nothing to prove. Letn≥2 and assume, by induction, thatN
′
=
P
n−1
i=1
Miis
Artinian. Now look at
N/Mn= (N
′
+Mn)/Mn
∼
=N
′
/(N
′
∩Mn)
which is Artinian being a quotient of the Artinian moduleN
′
. Thus bothMnand
N/Mnare Artinian and henceNis Artinian, as required. The case of a direct
sum is an immediate consequence because ifM=⊕
n
i=1Mi, thenMis a finite sum
of the Artinian submodulesMiand hence Artinian.
29

36. 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 planeR
2
is a sum of all the lines
passing through the origin and is a direct sum of any two of them.)
3.2Noetherian Modules
Theorem 3.2.1.The following are equivalent for anR-moduleM.
1. Ascending chain condition (a.c.c) holds for submodule ofM, that is any
ascending chainM1⊆M2⊆. . .⊆Mn⊆. . .⊆. . .of submodules ofMis
stationary in the sense thatMr=Mr+1=. . .for somer. (We write this
Mr=Mr+1,for everyr≫0).
2. Maximum condition holds forMin the sense that any non-empty family of
submodules ofMhas a maximal element.
3. Finiteness condition holds forMin the sense that every submodule ofMis
finitely generated (that is, spanned).
Proof.(1)=⇒(2) : LetF={Mi, i∈I}be a non-empty family of submodules
ofM. Pick any indexi1∈Iand look atMi1. IfMi1, is maximal inF, we are
through. Otherwise, there is ani2∈Isuch thatMi1⊂Mi2, Mi1̸=Mi2. If this
Mi2is maximal inF, 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 ofM, namely,M1⊂M2⊂. . .⊂Mn⊂. . .⊂. . .
contradicting (1).
(2) =⇒(3): LetNbe a submodule ofM. Consider the familyFof all finitely
generated submodules ofN. This family is non-empty since the submodule (0)
is a member. This family has a maximal member, sayN0={x1, . . . , xr}. If
N0̸=N, pick anx∈N, x is not inN0. NowN1=N0+{x}={x1, . . . , xr}is a
30

37. finitely generated submodule ofNand henceN1∈ F. But then this contradicts
the maximality ofN0inFsinceN0⊂N1, N0̸=N1and soN0=Nis finitely
generated.
(3) =⇒(1) LetM1⊆M2⊆. . .⊆Mn⊆. . .⊆. . .be an ascending chain of
submodules ofM. Consider the submoduleN=∪
∞
i=1MiofMwhich must be
finitely generated, sayN={x1, . . . , xn}·It follows thatxi∈Mr,for everyi,1≤
i≤nfor somer(≫0). Now we haveN⊆Ms⊆N,for everys≥rand so
N=Mr=Mr+1=. . .
Definition 3.2.2.Noetherian modules :A moduleMis called Noetherian if
a.c.c (or equivalently, the maximum condition or the finiteness condition) holds
forM.
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)
submodulesNofM, N̸=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 overZ.
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

38. 2. If a moduleMis such that it has a submoduleNwith bothNandM/Nare
Noetherian, thenMis Noetherian.
Proof.1. LetMbe Noetherian andNa submodule ofM. Any family of
submodules ofNis also one inMand hence the result follows. On the
other hand, any ascending chain of submodules ofM/Ncorresponds to one
inM(wherein each member containsN) and hence the result.
2. LetM1⊆M2⊆. . .⊆Mn⊆. . .⊆. . .be an ascending chain inM.Intersecting
withNgives the ascending chain inN, 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 inM/N, namely, (N+M1)/N⊆(N+M2)/N⊆. . .⊆
(N+Mn)/N⊆. . .⊆. . .which must be also stationary, say, (N+Ms)/N=
(N+Ms+1) =. . .for somes.
Now we prove thatMn=Mn+1,for every n≥(r+s). This is an immediate
consequence of the following four facts,
1.Mn⊆Mn+1,for everyn∈N,
2.N∩Mn=N∩Mn+1, for everyn≥r,
3. (N+Mn)/N= (N+Mn+1)/N, for everyn≥sand
4. (N+Mn)/N
∼
=Mn/(N∩Mn),for everyn∈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

39. Proof.For, letM1, . . . , Mnbe Noetherian submodules of a moduleM. LetN=
P
n
i=1
Mi. To proveNis Noetherian, proceed by induction onn. Ifn= 1, there
is nothing to prove. Letn≥2 and assume, by induction, thatN
′
=
P
n−1
i=1
Miis
Noetherian. Now look at
N/Mn= (N
′
+Mn)/Mn
∼
=N
′
/(N
′
∩Mn)
which is Noetherian being a quotient of the Noetherian moduleN
′
. Thus both
MnandN/Mnare Noetherian and henceNis Noetherian, as required. The case
of a direct sum is an immediate consequence because ifM=⊕
n
i=1Mi, thenMis
a finite sum of the Noetherian submodulesMiand 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 groupµp
∗of all complex (p
n
)
th
roots of unity for a fixed prime number
p and alln∈N. For each positive integer n, letµp
ndenote the cyclic group of all
complex (p
n
)
th
roots of unity so that we haveµp⊂. . .⊂µp
n⊂. . .⊂and hence
µp
∗=
n
[
n=1
µp
n
Furthermore, we notice the following special features in this group.
33

40. 1. It is infinite and non-cyclic.
2. Every proper subgroup is finite and is equal toµp
nfor somen.
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 anyx∈µ
p
n+1xnot inµp
ngeneratesµ
p
n+1·Now
we give the required counter-examples.
Example 3.2.9.The groupµp
∗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.LetR=Z[X1, X2, . . . Xn, . . .] be the polynomial ring in in-
finetly many variable. We know thatR, as a module over itself, is generated by 1
butRis 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 groupZis 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

41. Conclusion
We see that vector spaces are special types of modules which arise when the
underlying ring is a field. IfRis a ring, the definition of anR-moduleMis
analogous to the definition of a group action whereRplays the role of the group
andMthe role of the set. Also we see that how the structure of the ringRis
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

42. 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