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Math 540, Fall 2007
THEORY OF MODULES
Lecture Notes
Dr. Fahd Alshammari
Department of Mathematics
King Saud University
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1. Introduction to Modules
Dr. Alshammari
1. Introduction to Modules
1.1. Modules
NOTE. Throughout this course a ring means a ring with 1, a ring homomorphism f : R ® S is a map
that satisfies:
(1) f (a + b ) = f (a ) + f (b )
(2) f (ab ) = f (a ) f (b )
(3) f (1 ) = 1 .
Definition 1. Let R be a ring. A left R-module M is an abelian group, written additively, with a
multiplication operation of R on M, i.e. a map R ´ M ® M : (r , m ) a rm , such that
(M1)
(M2)
(M3)
(M4)
r (m 1 + m 2 ) = rm 1 + rm 2
(r1 + r2 )m = r1m + r2m
(r1r2 )m = r1 (r2m )
1m = m
For all r , r1, r2 Î R and m , m 1, m 2 Î M .
A right R-module is defined where the multiplication operation of R on M is on the right,
M ´ R ® M : (m , r ) a mr and the corresponding axioms are
(M1) (m 1 + m 2 )r = m 1r + m 2r
(M2) m (r1 + r2 ) = mr1 + mr2
(M3)
m (r1r2 ) = mr1r2
(M4)
m1 = m
All results proved for left R-modules have analogues ones for right R-modules. If R is commutative, then
any left R-module M can be viewed as a right R-module by defining mr := rm for r Î R , m Î M . If R is
not commutative then M3' dose not follow in general. Nonetheless we can always view a left R-module as
right module over an isomorphic ring.
Given a left module M over any ring R, let R op = R as abelian groups but with multiplication r ×s in
R op defined by r ×s := sr , where sr is the old multiplication in R. The left action of R on M induces a
right action of R op on M by defining mr := rm which makes M a right Rop-module. Furthermore, if R is
commutative then R op = R and in general we have R op @ R .
Since we mainly deal with left R-modules, unless otherwise specified, by an R-module we mean a left Rmodule. Let M be an R-module. The following basic properties follow from the above axioms.
1.
r 0M = 0M
2.
0R m = 0M
3.
(- r )m = r (- m ) = - (rm )
Where 0R Î R , 0M Î M are the additive identities. From now on we use 0 for both 0R and 0M as it is
clear from context. We will also denote the trivial module {0 } by 0.
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Example 2.
1. Let A be an (additive) abelian group. Then A is a ¢ -module where na for n Î ¢ , a Î A is given
by 0a = 0 and

for n > 0 , then na = a14444442
+ a + 444444
K + a3
n - times

2.
3.
4.
5.
6.
7.
for n < 0  - n > 0 , then na = - (- na ) .
E.g. ¤ , and, ¡ , £ , n ¢ = {, 0 ± n , ± 2n , ± 3n , K }, and ¢ n = {0,1, 2, K , n - 1 }.
If R is a ring, then R is an R-module the operation being ring multiplication.
If I is a left ideal of R then I is a left R-module (If x Î I and r Î R , then rx belongs to I. Similarly,
can see if I is a right ideal, then I is a right R-module.
If V is a vector space over a field F, then V is an F-module.
Let R be a ring. Let R n be the set of all n-tuples over R, then R n is an R-module with the usual
definitions of additions an scalar multiplication e.g. r (r1, r2, K , rn ) = (rr1, rr2, K , rrn ) .
Let R be a ring and M n [R ] the set of all n ´ n matrices over R. The usual definitions of matrix
addition and scalar multiplication make M n [R ] an R-module.
In the last example M n [R ] is also a ring. Recalling the column space R n which admits a
multiplication on the left by elements of the ring M n [R ] making R n a left M n [R ]-module.
1.2. Submodules & Quotient Modules
Definition 3. Let M be an R-module. A submodule of M is a nonempty subset N of M which is an Rmodule. Equivalently, N is a submodule of M iff N is subgroup of M with rn Î N for all r Î R and
n Î N .
Example 4. 0, M are, trivial, submodules of any given module M.
Example 5. For R as an R-module, a submodule of R is by definition just an ideal of R. E.g.
 Over a field F, the only submodules of F are 0 and F.
 Over a PID R, any submodule of R has the form Rx, where x Î R .
Example 6. Since a -module is simply an abelian group A, then the submodules of A are exactly its
subgroups. E.g.
 The ¢ -submodules of 3¢ are of the form (3k )¢ for k Î ¢ .
 4 = {0, 1, 2, 3 } has three submodules over ¢ : {0 } Í {0, 2 } Í ¢ 4 .
Proposition 7. Let M be an R-module and let N 1, N 2 be submodules of M. Then the following are also
submodules
a. N 1 Ç N 2 ,
b. N 1 + N 2 = {n 1 + n 2 : n 1 Î N 1, n 2 Î N 2 }.
Exercise 8. Let N 1, N 2 be R-submodules of M and let a, b be ideals of R. Define
aM = {a1x 1 + a2x 2 + K + an x n | ai Î a, x i Î M }
a. Show that aM is a submodule of M.
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b. Show that a(bM ) = (ab)M , (a + b)M = aM + bM , and a(N 1 + N 2 ) = aN 1 + aN 2 .
1.2.1. Quotient Modules
Let M be an R-module. If N is a submodule of M, then as abelian groups the quotient M N =
{m + N : m Î M } is an abelian group. We make M N into an R-module as follows. For a coset
m + N and r Î R , let r (m + N ) := rm + N . First we check the action is well-defined i.e. if
m 1 + N = m 2 + N , then rm 1 + N = rm 2 + N . But m 1 + N = m 2 + N  m 1 - m 2 Î N . Since N
is a submodule, then r (m 1 - m 2 ) = rm 1 - rm 2 Î N  rm 1 + N = rm 2 + N . The module axioms are
easily verified, Exercise9. The module M N is called the quotient module of M by N.
Exercise 9. Verify the module axioms for M N over R.
1.3. Homomorphisms & exact sequences
Definition 10. Let M and M ¢be R-modules. A map f : M ® N is called a homomorphism of R-modules,
or an R-homomorphism, if f is a group homomorphism with f (rm ) = rf (m ) for all r Î R , m Î M . A
homomorphism of R-modules is called an R-isomorphism if it is 1-1 and onto.
Example 11.
1. The zero map 0 : M ® M ¢ : 0(m ) = 0 is always an R-homomorphism.
2. For any abelian groups A and B as ¢ -modules any group homomorphism f : A ® B is a ¢ homomorphism: Let n Î ¢ and a Î A . To show f (n ×a ) = n ×f (a ), note that
 For n = 0, f (0 ×a ) = f (0 ) = 0 and 0 ×f (a ) = 0
 For n > 0, then
f (na ) = f (144444442
a + a + 4444444
K + a 3) = 144444444444
f (a ) + f (a4)2 +44444444444
K + f (a43) = nf (a )
n - t imes

n -t imes
For n < 0, then - n > 0 and
f (na ) = f (- (- n )a ) = - f
- n )a ) = - (- n ) f (a ) = nf (a )
((
3. Let M be an R-module. Then each r Î R defines a map r : M ® M : m a rm called the
multiplication by r map. The fact that it is an R-homomorphism follows from the module axioms.
Exercise 12. Let M and M ¢be R-modules. We denote by Hom R (M , M ¢) the set of all Rhomomorphisms from M to M ¢. Show that
a. Hom R (M , M ¢) is an abelian group where f + g is defined by ( f + g )(m ) = f (m ) + g(m ) .
b. If R is commutative, then Hom R (M , M ¢) is an R-module if we define rf Î Hom R (M , N ) by
(rf )(m ) = rf (m ) .
If f : M ® M ¢ is a homomorphism of R-modules, then as a group homomorphism f induces
 ker f = {m Î M : f (m ) = 0 } which is a subgroup of M. Recall that f is 1-1 iff
ker f = 0
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 Im f = {f (m ) : m Î M } a subgroup of M ¢.
In fact,
Lemma13.
ker f and Im f are submodules.
Proof. If m Î ker f , r Î R , then f (rm ) = rf (m ) = r 0 = 0  rm Î ker f . Similarly, if f (m ) Î Im f ,
r Î R , then rf (m ) = f (rm ) Î Im f .

Exercise 14. Let j : M ® M ¢ be a homomorphism of R-modules. Show that
a. For any submodule N of M, j (N ) is a submodule of M ¢.
b. For any submodule N ¢ of M ¢, j - 1(N ¢) is a submodule of M.
Example 15. Let N be an R-submodule of M. The canonical group homomorphism p : M ® M N is
actually an R-homomorphism, since p (rm ) = rm + N = r (m + N ) = r p (m ) .  is surjective with
ker p = N . p : M ® M N has a universal property in the category of homomorphisms of M whose
kernel contains N.
1.3.1. Universal Property of Quotients
Let N be a submodule of M. For any R-homomorphism f : M ® M ¢ with N Í ker f , then there is a
unique R-homomorphism f : M N ® M ¢ such that f o p = f . This equality is expressed by saying that
the following diagram of maps commutes:
M
f
M
!f
M N
Theorem 16. Furthermore, (i) f is onto iff f is onto. (ii) f is 1-1 iff kerf = N .
Proof. To see this, note that for a map f to satisfy f o p = f means f (m + N ) = f (m ) , but this defines
a homomorphism. For it is well-defined,
m 1 + N = m 2 + N  m 1 - m 2 Î N Í ker f  f (m 1 - m 2 ) = 0  f (m 1 ) = f (m 2 ) .
It follows that f is an R-homomorphism, since f is an R-homomorphism. Since f (m + N ) = f (m ) , then
Im f = Im f which proves (i). The fact f (m + N ) = f (m ) also implies that
ker f = {m + N : f (m ) = 0 } = {m + N : m Î ker f }.
So if ker f = N , then ker f = {0 + N } and f is 1-1. Conversely, if ker f = {0 + N }, then m Î N
for all m Î ker f which means ker f Í N , but N Í ker f . Therefore, N = ker f .
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1.3.2. Isomorphism Theorems
The following results are proved by applying the corresponding result from groups.
Theorem 17. [1st Iso Thm] An R-homomorphism f : M ® M ¢ with kernel N induces an
isomorphism M N @ Im f .
Proof. Apply the universal property of quotients to produce f : M N ® M ¢ which is 1-1 and is onto the
image of f.

Theorem 18. [2nd Iso Thm] Let R be a ring and M be a module over R. If N 1, N 2 are any two Rsubmodules of M, then (N 1 + N 2 ) N 2 @ N 1 (N 1 Ç N 2 ) .
Proof. Define f : N 1 ® M N 2 by f (n 1 ) = n 1 + N 2 . Then f is an R-homomorphism with
ker f = N 1 Ç N 2 and Im f = {n 1 + N 2 : n 1 Î N 1 } = (N 1 + N 2 ) N 2 . The 1st Iso. Thm gives an
isomorphism (N 1 + N 2 ) N 2 @ N 1 (N 1 Ç N 2 ) .

Theorem 19. [3rd Iso Thm] If K Í L Í M are R-submodules of M, then M L @ M K L K .
Proof. Define f : M K ® M L by f (m + K ) = m + L . Since, ker f = {m + K : m Î L } = L K
and Im f = M L , then by the 1st Iso. Thm we have an isomorphism M L @ M K L K .

Theorem 20. [Correspondence Thm] Let N is a submodule of an R-module M. There is a one to one
correspondence map L a L N (with inverse given by Q a p - 1(Q ) from submodules of M containing N
and submodules of M N .
- 1
Q Í M N
p - 1 (Q )
¬ ¾p ¾¾
of M
{Submodules
cont aining N }
«
of
{t heSubmodules
quot ient M N }
L Í M containing N
¾ ¾p ¾
®
p (L ) = L N
Proof. Since there is a one to one correspondence between subgroups of M containing N and subgroups of
M N , one needs to only check that submodules correspond to submodules.
Definition 21. An R-module M is called cyclic if it is generated by a single element, that if there is
x Î M such that
M = R x = {rx : r Î R }
Example 22.
1. R as an R- module is cyclic since any r Î R is generated by 1.
2. A vector space V over a field F is cyclic iff V is one dimensional.
3. An abelian group A is cyclic -module iff A is a cyclic abelian group, i.e. A @¢ n for some n.
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In fact this is a special case of the following.
Proposition 23. Every cyclic R-module M has the form M @ R J for some ideal J of R.
Proof. In this case, the map f : R ® M : f (r ) = rm which is clearly a homomorphism is in fact
surjective. Hence, we have an isomorphism M @ R ker f .
If R is a PID, then every ideal of R is principal and we have
Corollary 24. If M is a cyclic module over a PID R, then M @ R (a ) for some a Î R .
1.4. Exact Sequences
By an exact sequence of R-modules we mean a finite (or infinite) sequence
f 1
f
L ¾ ¾® M i - 1 ¾ ¾i ¾
® M i ¾ ¾i +¾¾
® M i + 1 ¾ ¾® L
such that ker fi + 1 = Im fi for all i. We can express that a map f : M ¢ ® M is injective by saying that
the sequence
f
0 ¾ ¾® M ¢¾ ¾¾
®M
is exact. Similarly, the exactness of the following sequence
g
M ¾ ¾¾
® M ¢¢¾ ¾® 0
is equivalent to saying that g is onto. Now a sequence
f
g
0 ¾ ¾® M ¢¾ ¾¾
® M ¾ ¾¾
® M ¢¢¾ ¾® 0
is exact iff f is 1-1, ker g = Im f , and g is onto.
Example 25. Associated to any submodule N of M the exact sequence
i
p
0 ¾ ¾® N ¾ ¾ ® M ¾ ¾ ¾
® M N ¾ ¾® 0
where the maps i and p are respectively the inclusion and projection maps.
Example 26. Given a homomorphism f : M ® M ¢¢, one can easily verify that f induces an exact
sequence
i¢
p¢
0 ¾ ¾® ker f ¾ ¾ ¾
® M ¾ ¾ ¾® M ¢¢ Im f ¾ ¾® 0 .
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